Chemical Reaction Rate and Equilibrium

Question 1

Question 1: 1. The rate of a chemical reaction is affected by external conditions such as reaction temperature, ...

1. The rate of a chemical reaction is affected by external conditions such as reaction temperature, concentration of reactants, pressure, catalyst, etc. When barbecuing, fanning red-hot charcoal with a fan makes the fire hotter, for the following reasons

A. A. Lower pressure slows down the reaction
B. B. Lower temperature, faster reaction
C. C. Increase the concentration of $\mathrm { CO } _ { 2 }$ to speed up the reaction.
D. D. Increase the concentration of $\mathrm { O } _ { 2 }$ to speed up the reaction.

Answer

Answer: D

Solution: Carbon and oxygen reaction, the surface area of carbon and the concentration of oxygen is the main factors affecting the rate of reaction, fanning the red-hot charcoal with a fan can increase the concentration of oxygen, the rate of reaction is accelerated, the fire will be more vigorous, the answer choice D.

Question 2

Question 2: 2. The following chemical facts are not consistent with the idea that "quantitative change leads to ...

2. The following chemical facts are not consistent with the idea that "quantitative change leads to qualitative change".

A. A. Pass $\mathrm { CO } _ { 2 }$ into the clarified lime water, the solution first becomes turbid and then clarified
B. B. Rapidly stretching a syringe containing a mixture of ${ } ^ { \mathrm { NO } _ { 2 } }$ and ${ } ^ { \mathrm { N } _ { 2 } \mathrm { O } _ { 4 } }$ gas, the color of the gas becomes lighter and then darker.
C. C. Hydrochloric acid is added dropwise to ${ } ^ { \mathrm { Fe } ( \mathrm { OH } ) _ { 3 } }$ colloid, which settles and then dissolves.
D. D. Increase ${ } ^ { \mathrm { HNO } _ { 3 } }$ concentration, iron passivation from violent dissolution

Answer

Answer: B

Solution: A. $\mathrm { CO } _ { 2 }$ is added to lime water, and $\mathrm { CaCO } _ { 3 }$ precipitate is formed with $\mathrm { Ca } ( \mathrm { OH } ) _ { 2 }$ first, and appears turbid, and soluble $\mathrm { Ca } \left( \mathrm { HCO } _ { 3 } \right) _ { 2 }$ is generated after $\mathrm { Ca } \left( \mathrm { HCO } _ { 3 } \right) _ { 2 }$ is added. The solution is clarified. The change in quantity leads to a change in the product, which is consistent with the principle of "quantitative change leads to qualitative change", and A does not fit the question; B. When the syringe is stretched to reduce the pressure, the equilibrium between $\mathrm { NO } _ { 2 }$ and $\mathrm { N } _ { 2 } \mathrm { O } _ { 4 }$ shifts, resulting in a change in color, but no new substances are produced, which is only an adjustment of the concentration of the same equilibrium system, and it is a quantitative change that causes a quantitative change; C. A small amount of hydrochloric acid causes $\mathrm { Fe } ( \mathrm { OH } ) _ { 3 }$ colloid to polymerize and precipitate, and too much hydrochloric acid dissolves the precipitate, which is a quantitative change that leads to a different phenomenon, and C does not meet the meaning of the question; D. The concentration of $\mathrm { HNO } _ { 3 }$ is increased so that the iron is passivated instead of reacting violently, and the quantitative change in concentration triggers a qualitative change in the type of reaction, which is not in accordance with the meaning of the question.

Question 3

Question 3: 3. The reaction $\mathrm { H } _ { 2 } ( \mathrm {~g} ) + \mathrm { I } _ { 2 } ( \mathrm {~g} ) \ri...

3. The reaction $\mathrm { H } _ { 2 } ( \mathrm {~g} ) + \mathrm { I } _ { 2 } ( \mathrm {~g} ) \rightleftharpoons 2 \mathrm { HI } ( \mathrm { g } )$ occurs in a closed container of constant volume at a certain temperature. When the concentration of $\mathrm { H } _ { 2 } , \mathrm { I } _ { 2 } , \mathrm { HI }$ no longer changes, the following statement is correct ( )

A. A. $\mathrm { H } _ { 2 }$ and $\mathrm { I } _ { 2 }$ are all converted to HI
B. B. The reaction has reached chemical equilibrium
C. C. The concentrations of $\mathrm { H } _ { 2 } , \mathrm { I } _ { 2 } , \mathrm { HI }$ must be equal.
D. D. Forward and reverse reaction rates are equal and equal to zero

Answer

Answer: B

Solution: A. Since $\mathrm { H } _ { 2 } ( \mathrm {~g} ) + \mathrm { I } _ { 2 } ( \mathrm {~g} ) \rightleftharpoons 2 \mathrm { HI } ( \mathrm { g } )$ is a reversible reaction, $\mathrm { H } _ { 2 }$ and $\mathrm { I } _ { 2 }$ are all converted to HI; B. When the concentration of $\mathrm { H } _ { 2 } , \mathrm { I } _ { 2 } , \mathrm { HI }$ no longer changes, the reaction has reached chemical equilibrium, B is correct; C. The reaction reaches chemical equilibrium when the concentrations of the components no longer change and the concentrations of $\mathrm { H } _ { 2 } , \mathrm { I } _ { 2 } , \mathrm { HI }$ are not necessarily equal, C is wrong; D. When chemical equilibrium is reached, the rates of forward and reverse reactions are equal, but neither is equal to zero, D is wrong; Therefore, the answer is B.

Question 4

Question 4: 5. A closed container is filled with 4 mol A and 2 mol B at constant temperature and pressure, and t...

5. A closed container is filled with 4 mol A and 2 mol B at constant temperature and pressure, and the following reaction occurs: $2 \mathrm {~A} ( \mathrm {~g} ) + \mathrm { B } ( \mathrm { g } )$ $\approx 2 \mathrm { C } ( \mathrm { g } ) \Delta \mathrm { H } < 0$. After 2 min, the reaction reaches equilibrium and 1.6 mol of C is produced. The following analysis is correct

A. A. If the volume of the container at the start of the reaction is 2 L, then $\mathrm { V } _ { \mathrm { c } } = 0.4 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 } \cdot \mathrm {~min} ^ { - 1 }$
B. B. If reacted under constant pressure adiabatic conditions, after equilibrium $\mathrm { n } _ { \mathrm { c } } < 1.6 \mathrm {~mol}$
C. C. If, after 2 min, an equal amount of $A , C , B$ is added to the container, the conversion rate remains unchanged.
D. D. If the reaction is carried out at constant temperature and constant volume, the amount of heat released will increase by

Answer

Answer: B

Solution: 试题分析:A,由于该反应时体积减少的反应,在恒压条件下,体积减少,故 c的反应速率 $> 1.6 \div 2 \div 2 = 0.4 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 } \cdot \mathrm {~min} ^ { - 1 }$ 故不选 A;B,由于反应是放热的,若在绝热的条件下,体系温度升高,平衡逆向移向, c的物质的量减低,故选 B ; C ,再投入等量的AC ,相当于再次通入 C, and then put the same amount of AC, is equivalent to re-admission of the ratio of the amount of substances is greater than $2 : 1$ of A and B, so the conversion rate of B increases, so do not choose C; D, if the reaction is carried out at constant temperature and constant capacity, the reaction is not utilized in the positive direction, the heat will be reduced, so do not choose D. Points: chemical equilibrium shift influencing factors, equivalent equilibrium.

Question 5

Question 5: 7. Chemistry is an important foundation for materials science, life science, environmental science, ...

7. Chemistry is an important foundation for materials science, life science, environmental science, energy science and information science. The following processes do not involve chemical changes

A. A. Coal denitrification and desulfurization
B. B. wind power
C. C. Clay-based sintered ceramics
D. D. Synthesis of ammonia and ammonium salts from $\mathrm { N } _ { 2 }$

Answer

Answer: B

Question 6

Question 6: 8. The following practices in daily life are not related to the regulation of reaction rates

8. The following practices in daily life are not related to the regulation of reaction rates

A. A. Food Vacuum Packaging
B. B. Freezing food in the refrigerator
C. C. Filling the bearings with lubricant
D. D. Using pulverized coal instead of briquettes makes for a brighter fire.

Answer

Answer: C

Solution: A. Vacuum packaging of food is to prevent food from being oxidized and spoiled and to reduce the rate of chemical reaction in food, so A is correct; B. Lowering the temperature reduces the reaction rate, so B is correct; C.The lubricant is only added to reduce friction, so D is wrong; D. Increase the contact area can improve the rate of chemical reaction, so D is correct; so the question C. To master the influencing factors of chemical reaction rate, we should judge according to the actual examples. ## B [Knowledge Points]The effect of pressure on the movement of chemical equilibrium, the factors affecting the chemical equilibrium constant and its application, the factors affecting chemical equilibrium A. As can be seen from the figure, at the temperature corresponding to point E, the concentration of M in the 2 L container is about $0.7 \mathrm {~mol} / \mathrm { L } , 6 \mathrm {~L}$ and the concentration of M in the container is about $0.4 \mathrm {~mol} / \mathrm { L }$, and if the container is pressurized from 6 L to 2 L, the concentration of M does not increase by a factor of 3, so the equilibrium moves in the opposite direction, so $\mathrm { a } < \mathrm { m } + \mathrm { n } , \mathrm { A }$ is wrong. FORMULA_2]] is incorrect; B. From the diagram, it can be seen that when the volume of the container is constant, the temperature increases, the concentration of M increases, and the equilibrium is shifted in the positive direction, and the positive direction of the reaction is a heat-absorbing reaction, $\mathrm { Q } > 0$, B is correct; C. From the image, we can see that as the temperature increases, the concentration of M increases, the equilibrium moves forward, so the equilibrium constant increases, and the temperature at point F is higher than that at point E, so the equilibrium constant at point E is smaller than the equilibrium constant at point F. C. Wrong; [The temperature at point F is higher than that at point E, so the equilibrium constant at point E is smaller than the equilibrium constant at point F. C is wrong;

Question 7

Question 7: 9. Under certain conditions, a certain amount of $X$ and $Y$ react in a constant volume closed conta...

9. Under certain conditions, a certain amount of $X$ and $Y$ react in a constant volume closed container: $a X ( g ) + b Y ( s ) \rightleftharpoons m M ( g ) + \mathrm { nN } ( \mathrm { g } ) \Delta \mathrm { H } = \mathrm { Q } \mathrm { kJ } \cdot \mathrm { mol } ^ { - 1 }$, and when equilibrium is reached, the concentration of M is related to the temperature and the volume of the container as shown in the figure. The following judgment must be correct ![](/images/questions/chem-reaction-rate/image-001.jpg)

A. A. $\mathrm { a } > \mathrm { m } + \mathrm { n }$
B. B. $\mathrm { Q } > 0$
C. C. The equilibrium constant at point E is greater than the equilibrium constant at point F.
D. D. After equilibrium is reached, increasing the amount of Y will increase the conversion of X

Answer

Answer: B

Question 8

Question 8: 10. It is reported that Chinese scientists have developed a catalyst with graphene as a carrier, whi...

10. It is reported that Chinese scientists have developed a catalyst with graphene as a carrier, which directly converts $25 ^ { \circ } \mathrm { C }$ into oxygen-containing organic matter with $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ under $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$, and the main principle is as shown in the figure: ![](/images/questions/chem-reaction-rate/image-002.jpg) The following statement is incorrect

A. A. $\mathrm { OO } ^ { \mathrm { O } }$ represents $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ in the diagram.
B. B. The total reaction equation for steps i and ii is $\mathrm { CH } _ { 4 } + \mathrm { H } _ { 2 } \mathrm { O } _ { 2 } \xrightarrow { \text { catalyst } } \mathrm { CH } _ { 3 } \mathrm { OH } + \mathrm { H } _ { 2 } \mathrm { O }$
C. C. As can be seen from the figure, the $\mathrm { H } _ { 2 } \mathrm { O }$ generated in step iv, in which all the H atoms come from $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$, is the same as the $\mathrm { H } _ { 2 } \mathrm { O }$ generated in step iii.
D. D. Based on the above rationale, it is hypothesized that step vi generates HCOOH and $\mathrm { H } _ { 2 } \mathrm { O }$

Answer

Answer: C

Solution: A. According to the question, $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ and $\mathrm { CH } _ { 4 }$ are converted to oxygenated organic matter, and 000 is the scale model of $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$, so A is correct; B. According to steps i and ii in the diagram, the products water and $\mathrm { CH } _ { 3 } \mathrm { OH }$ are obtained, and the equation $\mathrm { CH } _ { 4 } + \mathrm { H } _ { 2 } \mathrm { O } _ { 2 } \xrightarrow { \text { 催化楼 } } \mathrm { CH } _ { 3 } \mathrm { OH } + \mathrm { H } _ { 2 } \mathrm { O } , \mathrm { B }$ is correct; C. According to the diagram, $\mathrm { H } _ { 2 } \mathrm { O }$ generated in step iv, where H comes from $\mathrm { CH } _ { 3 } \mathrm { OH }$ and $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ generated in step ii, C is incorrect; D. Mimicking step ii, $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ is decomposed into 2 $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$, and $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ is decomposed into 2 $- \mathrm { OH } , 1$. [INLINE_FORMULA_9]] that combine with H atoms to produce $\mathrm { H } _ { 2 } \mathrm { O } , 1$ that combine with - CHO produced by v to give HCOOH, so step vi produces HCOOH and $\mathrm { H } _ { 2 } \mathrm { O }$, D is correct;

Question 9

Question 9: 11. the same volume, the same pH of a monobasic strong acid solution (1) and a monobasic weak acid s...

11. the same volume, the same pH of a monobasic strong acid solution (1) and a monobasic weak acid solution (2) were reacted with a sufficient amount of zinc powder, the following on the volume of hydrogen (V) with the time (t) change in the schematic diagram correct is

A. A. ![](/images/questions/chem-reaction-rate/image-001.jpg)
B. B. ![](/images/questions/chem-reaction-rate/image-002.jpg)
C. C. ![](/images/questions/chem-reaction-rate/image-003.jpg)
D. D. ![](/images/questions/chem-reaction-rate/image-004.jpg)

Answer

Answer: C

Solution: In the same volume, same pH of a one-membered strong acid solution (1) and a one-membered weak acid solution (2), the weak acid solution (2) is partially ionized. So the $n \left( \mathrm { H } ^ { + } \right)$ of the weak acid solution (2) is greater than that of the monobasic strong acid solution (1), then the weak acid solution (2) reacts to get more hydrogen than the monobasic strong acid solution (1), and as the reaction proceeds, the $\mathrm { c } \left( \mathrm { H } ^ { + } \right)$ decreases but the weak acid solution (2) will continue to ionize so that after a period of time after the reaction the weak acid solution (2) has a $\mathrm { c } \left( \mathrm { H } ^ { + } \right)$ of $c \left( \mathrm { H } ^ { + } \right)$ will be greater than the $c \left( \mathrm { H } ^ { + } \right)$ of the monobasic strong acid solution (1). Then the weak acid solution (2) reaches the reaction endpoint first. Therefore, choose C. [点睛]This question requires attention to analyzing the ionization of the weak acid solution (2) and the change in the concentration of hydrogen ions in order to accurately select the answer. 12 The D Knowledge]Thermochemical equations, radical reactions and reaction history A. From the diagram, it can be seen that the transition state TS2 has higher energy than TS1, then the transition state TS1 is more stable, A is correct; B. From the diagram, we can see that the process $\mathrm { P } \rightarrow \mathrm { TS } 2$ is the reaction with higher activation energy in the two-step reaction, and it is a slow reaction, that is, it is the decisive speed step for compound M to produce compound N. B is correct; C. M is a catalyst, and the structure of M will not change before and after the whole reaction. Comparing the structural characteristics of M and N, it can be seen that M can only be generated when the $\mathrm { Fe } - \mathrm { O }$ bond in N is broken and H atoms are reconnected, so there must be a $\mathrm { Fe } - \mathrm { O }$ bond in the hydrogenation of compound M catalyzed by $\mathrm { Fe } - \mathrm { O }$. FORMULA_7]] bond must be broken during the hydrogenation of compound M. C is correct; D. The diagram shows the reaction of a $\mathrm { CO } _ { 2 }$ molecule, and according to the diagram, we can know that the energy change of the process $\Delta \mathrm { E } = - 11.63 \mathrm { eV }$ is $\Delta \mathrm { E } = - 11.63 \mathrm { eV }$, so the energy change when $1 \mathrm { molCO } _ { 2 }$ is involved in the reaction is - , so the reaction The enthalpy change $\Delta \mathrm { H } = - 1120 \mathrm {~kJ} / \mathrm { mol }$ of the reaction is $\Delta \mathrm { H } = - 1120 \mathrm {~kJ} / \mathrm { mol }$, and D is incorrect;

Question 10

Question 10: 12. Scientists in China study the mechanism of hydrogenation of the compound $\mathrm { M } ( \mathr...

12. Scientists in China study the mechanism of hydrogenation of the compound $\mathrm { M } ( \mathrm { s } )$ by catalyzing ${ } ^ { \mathrm { CO } _ { 2 } }$. The mechanism and relative energy curve of the process of generating compound $\mathrm { N } ( \mathrm { s } )$ from compound $\mathrm { M } ( \mathrm { s } )$ are shown in the figure (known as $1 \mathrm { eV } = 1.6 \times 10 ^ { - 19 } \mathrm {~J}$).TS1 and TS2 are transition states. The following statement is false. ![](/images/questions/chem-reaction-rate/image-003.jpg)

A. A. Transition state TS1 is more stable than TS2.
B. B. Procedure $P \rightarrow T S 2$ is the deceleration step for compound $M$ to produce compound $N$.
C. C. Compound M catalyzes ${ } ^ { \mathrm { CO } _ { 2 } }$ hydrogenation, which must involve the breaking of $\mathrm { Fe } - \mathrm { O }$ bonds.
D. D. The thermochemical equation for this process is: $M ( \mathrm {~s} ) + \mathrm { CO } _ { 2 } ( \mathrm {~g} ) = \mathrm { N } ( \mathrm { s } ) \quad \Delta \mathrm { H } = - 11.63 \mathrm {~kJ} \| \mathrm { mol } ^ { - 1 }$

Answer

Answer: D

Question 11

Question 11: 13. 500 mL of aqueous hydrogen peroxide decomposes in the presence of a small amount of $\mathrm { I...

13. 500 mL of aqueous hydrogen peroxide decomposes in the presence of a small amount of $\mathrm { I } ^ { - }$: $2 \mathrm { H } _ { 2 } \mathrm { O } _ { 2 } = 2 \mathrm { H } _ { 2 } \mathrm { O } + \mathrm { O } _ { 2 } \uparrow$. The amount of $\mathrm { O } _ { 2 }$ released is converted to the $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ concentration (c) as shown in the table below at room temperature and pressure: | $\mathrm { t } / \mathrm { min }$ | 0 | 20 | 40 | 60 | 80 | | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | $\mathrm { c } / \left( \mathrm { mol } \cdot \mathrm { L } ^ { - 1 } \right)$ | 0.80 | 0.40 | 0.20 | 0.10 | 0.050 | The following statements are correct

A. A. At 20 minutes into the reaction, the volume of $\mathrm { O } _ { 2 }$ was measured to be 448 mL
B. B. $40 \sim 60 \mathrm {~min}$, consuming $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ at an average rate of $0.2 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 } \cdot \mathrm {~min} ^ { - 1 }$
C. C. Instantaneous rate at 30 min is greater than instantaneous rate at 50 min
D. D. At 100 min, $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ is completely decomposed.

Answer

Answer: C

Solution: A. The gas condition is unknown and the molar volume of the gas cannot be determined, so the volume of the gas cannot be calculated; B. The average rate at which $0 \sim 60 \mathrm {~min}$ , consumes $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ is $\mathrm { v } = \frac { \Delta \mathrm { c } } { \Delta \mathrm { t } } = \frac { ( 0.20 - 0.10 ) \mathrm { mol } / \mathrm { L } } { ( 60 - 40 ) \mathrm { min } } = 0.005 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 } \cdot \mathrm {~mol} ^ { - 1 }$, $\mathrm { v } = \frac { \Delta \mathrm { c } } { \Delta \mathrm { t } } = \frac { ( 0.20 - 0.10 ) \mathrm { mol } / \mathrm { L } } { ( 60 - 40 ) \mathrm { min } } = 0.005 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 } \cdot \mathrm {~mol} ^ { - 1 }$ B is incorrect ; C. As the reaction proceeds, the concentration of reactants decreases and the reaction rate decreases, so the instantaneous rate at 30 min is greater than the instantaneous rate at 50 min; D. $60 \sim 80 \mathrm {~min}$, the reaction consumes $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ at a concentration of $0.050 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 }$, and the remaining $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ at a concentration of $0.050 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 }$, and as the reaction proceeds, the concentration of reactants decreases. As the reaction progresses, the concentration of reactants decreases and the rate of reaction decreases, so it is unlikely that the remaining $0.050 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 } \mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ will be completely decomposed by the 100th minute; Answer choice C.

Question 12

Question 12: 14. Methanol vapor catalytic reforming is an important method to obtain $\mathrm { H } _ { 2 }$, and...

14. Methanol vapor catalytic reforming is an important method to obtain $\mathrm { H } _ { 2 }$, and the energy change of reaction $\mathrm { CH } _ { 3 } \mathrm { OH } \left( \mathrm { g } + \mathrm { H } _ { 2 } \mathrm { O } ( \mathrm { g } ) \rightleftharpoons \mathrm { CO } _ { 2 } ( \mathrm {~g} ) + 3 \mathrm { H } _ { 2 } ( \mathrm {~g} ) \right.$ is shown in Fig. 1. When $830 ^ { \circ } \mathrm { C }$, a certain amount of CaO is injected into the reaction system at the same time to do a comparative experiment, and the result is shown in Fig. 2. ![](/images/questions/chem-reaction-rate/image-004.jpg) Figure 1 ![](/images/questions/chem-reaction-rate/image-005.jpg) Figure 2

A. A. The $\Delta \mathrm { H } = ( \mathrm { a } - \mathrm { b } - \mathrm { c } ) \mathrm { kJ } \cdot \mathrm { mol } ^ { - 1 }$ for this response
B. B. Selection of an appropriate catalyst to reduce $\Delta \mathrm { H } _ { 3 }$ can significantly increase the rate of catalytic reforming reaction.
C. C. CaO increases the rate of catalytic reforming reaction
D. D. The larger the surface area of CaO, the better the absorption $\mathrm { CO } _ { 2 }$.

Answer

Answer: D

Solution: A. According to the information in the figure and Gais' law, the enthalpy change of the reaction is only related to the starting and ending states, so the $\Delta \mathrm { H } = ( \mathrm { a } - \mathrm { b } + \mathrm { c } ) \mathrm { kJ } \cdot \mathrm { mol } ^ { - 1 }$ of the reaction is $\Delta \mathrm { H } = ( \mathrm { a } - \mathrm { b } + \mathrm { c } ) \mathrm { kJ } \cdot \mathrm { mol } ^ { - 1 }$, so A is wrong; B. The activation energy of the first step is large, so the reaction rate of the reaction is mainly determined by the first step, so choosing an appropriate catalyst to reduce $\Delta \mathrm { H } _ { 3 }$ cannot significantly increase the rate of catalytic reforming reaction, so B is wrong; C. CaO is not a catalyst for the reaction, so it cannot increase the rate of the catalytic reforming reaction, so C is wrong; D. According to the information in Fig. 2, the larger the surface area of CaO, the better the absorption of $\mathrm { CO } _ { 2 }$, the lower the concentration of carbon dioxide, the equilibrium shifts positively and the percentage of hydrogen increases, so D is correct. Therefore, D is correct.

Question 13

Question 13: 15. Other conditions remain unchanged only to increase the temperature, must be increased by

15. Other conditions remain unchanged only to increase the temperature, must be increased by

A. A. Solubility of strong bases in water
B. B. Thermal effects of heat-absorbing reactions
C. C. Chemical equilibrium constant K
D. D. Degree of salt hydrolysis

Answer

Answer: D

Solution: A. The solubility of strong bases in water is determined by the nature of the substance, and the solubility of calcium hydroxide decreases as the temperature rises, A is wrong ; B. The thermal effect of a heat-absorbing reaction is determined by the energy contained in the substance and has nothing to do with the ambient temperature; C. The chemical equilibrium constant K is affected by the enthalpy change of the reaction, when the positive reaction absorbs heat, increase the temperature, the reaction moves positively, the chemical equilibrium constant increases, C error; D. Hydrolysis of salts is a heat-absorbing reaction, and as the temperature rises, the equilibrium moves in the positive direction and the degree of hydrolysis increases, D is correct. The answer is D.

Question 14

Question 14: 16. At a certain temperature, in a 2 L container, $\mathrm { X } , \mathrm { Y } , \mathrm { Z }$ th...

16. At a certain temperature, in a 2 L container, $\mathrm { X } , \mathrm { Y } , \mathrm { Z }$ the curves of the quantities of three gases are shown in the figure. The following statement is false. ![](/images/questions/chem-reaction-rate/image-006.jpg)

A. A. The reaction equation is $X + 3 Y \rightleftharpoons 2 Z$
B. B. Under these conditions, the conversion of X was maximized at 2 min.
C. C. Chemical reaction rate at 2 min $V ( \mathrm { Z } ) = 0.02 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 } \cdot \min ^ { - 1 }$
D. D. After 2 min, although the concentration of X no longer changes, X is still reacting.

Answer

Answer: C

Solution: A. According to the diagram of $\mathrm { X } , \mathrm { Y } , \mathrm { Z }$, the changes in the amount of substances are $0.1 \mathrm {~mol} , 0.3 \mathrm {~mol} , 0.2 \mathrm {~mol}$, and according to the ratio of the amount of change of substances is equal to the ratio of the stoichiometric numbers, we can tell that $\mathrm { X } , \mathrm { Y }$ is a reactant, Z is a product and the ratio of its stoichiometric numbers is $1 : 3 : 2$, A is correct. INLINE_FORMULA_3]], A is correct; B. The reaction reaches equilibrium at 2 min when the amount of each substance does not change anymore and the reaction reaches its maximum limit and the conversion rate of X reaches its maximum, B is correct; C. The rate of a chemical reaction represents the average rate over a period of time rather than the instantaneous rate, C is wrong; D. The reaction is reversible, and the reaction is still in progress when it reaches equilibrium, D is correct;

Question 15

Question 15: 17. Under certain conditions, the following cannot be explained by Le Chatelier's principle

17. Under certain conditions, the following cannot be explained by Le Chatelier's principle

A. A. Ammonia liquefaction separation in ammonia synthesis can improve the utilization of raw materials
B. B. $\mathrm { H } _ { 2 } , \mathrm { I } _ { 2 } , \mathrm { HI }$ Gas mixture darkens under pressure
C. C. Laboratory methods of draining saturated NaCl solutions are commonly used to collect $\mathrm { Cl } _ { 2 }$
D. D. Freshly made chlorine water, drop silver nitrate solution, the color of the solution becomes lighter

Answer

Answer: B

Solution: A. When ammonia is synthesized, ammonia is liquefied and separated to reduce the concentration of products, and the chemical equilibrium of ammonia synthesis is shifted in the positive direction to improve the utilization of raw materials, which is related to the shift of chemical equilibrium, so A is not selected; B. $\mathrm { H } _ { 2 } , \mathrm { I } _ { 2 } , \mathrm { HI }$ The chemical equilibrium reached by the mixture of gases is $\mathrm { H } _ { 2 } + \mathrm { I } _ { 2 } \rightleftharpoons 2 \mathrm { HI }$, the volume of the gas before and after the reaction remains unchanged, after mixing and pressurization, the equilibrium does not move, the color becomes darker because of the volume of the system decreases the concentration of the substance increases, and the equilibrium has nothing to do with the, therefore, the B choice; C. $\mathrm { Cl } _ { 2 } + \mathrm { H } _ { 2 } \mathrm { O } \rightleftharpoons \mathrm { H } ^ { + } + \mathrm { Cl } ^ { - } + \mathrm { HClO }$, there is a large amount of $\mathrm { Cl } ^ { - }$ in the saturated NaCl solution, which inhibits the dissolution of $\mathrm { Cl } _ { 2 }$, so C is not chosen; D. $\mathrm { Cl } _ { 2 } + \mathrm { H } _ { 2 } \mathrm { O } \rightleftharpoons \mathrm { H } ^ { + } + \mathrm { Cl } ^ { - } + \mathrm { HClO }$ exists in chlorinated water, and when silver nitrate solution is added, $\mathrm { Ag } +$ and $\mathrm { Cl } ^ { - }$ produce a white precipitate, and $\mathrm { Cl } ^ { - }$ is dissolved in $\mathrm { Cl } _ { 2 } + \mathrm { H } _ { 2 } \mathrm { O } \rightleftharpoons \mathrm { H } ^ { + } + \mathrm { Cl } ^ { - } + \mathrm { HClO }$. The solution becomes lighter in color, which cannot be explained by Le Chatelier's principle, so D is not chosen.

Question 16

Question 16: 18. For the reaction $2 \mathrm { SO } _ { 2 } ( \mathrm {~g} ) + \mathrm { O } _ { 2 } ( \mathrm {~...

18. For the reaction $2 \mathrm { SO } _ { 2 } ( \mathrm {~g} ) + \mathrm { O } _ { 2 } ( \mathrm {~g} )$ in the mouth of the Jihua Department also tons $2 \mathrm { SO } _ { 3 }$ is a step in the industrial production of sulfuric acid, the following statements are correct

A. A. Warming up the temperature slows down the reaction rate
B. B. Increasing the pressure slows down the reaction rate
C. C. Using the right catalyst speeds up the rate of the reaction
D. D. Upon reaching chemical equilibrium, $\mathrm { SO } _ { 2 }$ can be $100 \%$ converted to $\mathrm { SO } _ { 3 }$.

Answer

Answer: C

Solution: A. Increasing the temperature speeds up the rate of reaction, A error ; B. Increasing the pressure can speed up the reaction rate, B error; C. Using the right catalyst can lower the activation energy and speed up the reaction rate; D. $2 \mathrm { SO } _ { 2 } ( \mathrm {~g} ) + \mathrm { O } _ { 2 } ( \mathrm {~g} )$ micronized particles $2 \mathrm { SO } _ { 3 }$ is a reversible reaction, so $\mathrm { SO } _ { 2 }$ and $\mathrm { O } _ { 2 }$ are unlikely to be converted to $100 \%$. INLINE_FORMULA_5]], D is wrong;

Question 17

Question 17: 19. The following experimental purpose corresponds to the correct experimental method is | Select <...

19. The following experimental purpose corresponds to the correct experimental method is | Select <br> Item | Purpose of the experiment | Experimental method | | :--- | :--- | :--- | | A | Compare the strength of the metallic properties of Mg and Al | Add ammonia dropwise to $\mathrm { MgCl } _ { 2 }$ and $\mathrm { AlCl } _ { 3 }$ solutions to excess | B | Test spinach to determine the strength of the metallic properties of spinach. | B | Test for iron in spinach | Grind spinach and add dilute hydrochloric acid, filter and take the filtrate and add an excess of concentrated bromine water and then add phenol solution | | C | Investigate the effect of different catalysts on the rate of decomposition of $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ <br> | Add ${ } ^ { 2 \mathrm {~mL} 10 \% \mathrm { H } _ { 2 } \mathrm { O } _ { 2 } }$ solution to each of the three test tubes, and then add 2 drops of ${ } ^ { 1 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 } \mathrm { CuCl } _ { 2 } }$ solution and 2 drops of ${ } ^ { 1 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 } \mathrm { FeCl } _ { 3 } \text { 溶 } }$ solution to two of them respectively. FORMULA_5]] solution and 2 drops of ${ } ^ { 1 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 } \mathrm { FeCl } _ { 3 } \text { 溶 } }$ solution to each of the two test tubes. | D | Test for aldehyde groups in glucose | Prepare about 2 mL of silver-ammonia solution in a clean test tube, add the $1 \mathrm {~mL} 10 \%$ glucose <br> solution, and shake, then heat with an alcohol lamp | | :--- | :--- | :--- | :--- | .

A. A. A
B. B. B
C. C. C
D. D. D

Answer

Answer: C

Solution: A. When excess ammonia is added dropwise to $\mathrm { MgCl } _ { 2 }$ and $\mathrm { AlCl } _ { 3 }$ solutions, hydroxide precipitates insoluble in the excess ammonia are formed, and it is not possible to compare the metal properties of $M g$ and $A l$ by the formation or dissolution of the precipitates. metal properties. The correct method is to use a strong base (e.g. NaOH) to observe the dissolution of the precipitates; B. Excessive bromine will react with phenol to form a white precipitate, and the dark color of the bromine may cover the color of $F e ^ { 3 + }$ and phenol. The dark color of bromine may mask the color reaction between $F e ^ { 3 + }$ and phenol and interfere with the detection of iron. The correct method is to use $K S C N$ solution to test $F e ^ { 3 + }$, B is wrong; C. Add $\mathrm { CuCl } _ { 2 } , \mathrm { FeCl } _ { 3 }$ and no catalyst to the three $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ solutions, and observe the rate of bubble production to compare the catalytic effect of the different catalysts, the experimental design is in line with the method of controlling variables, C is correct; D. The silver mirror reaction needs water bath heating, direct heating of alcohol lamp may lead to the reaction is out of control or can not form silver mirror, D error; Therefore, the answer is C.

Question 18

Question 18: 20. The following statements about experimental apparatus and operations are incorrect. ![](/images/...

20. The following statements about experimental apparatus and operations are incorrect. ![](/images/questions/chem-reaction-rate/image-007.jpg) ![](/images/questions/chem-reaction-rate/image-008.jpg) ![](/images/questions/chem-reaction-rate/image-009.jpg)

A. A. Study of the effect of concentration on chemical equilibrium
B. B. Preparation and testing of ethylene
C. C. Titration of sodium hydroxide solution with hydrochloric acid solution
D. D. Removal of benzene from bromobenzene

Answer

Answer: B

Solution: A. As can be seen from the figure, only the concentration of KSCN is different in the two experiments, so the effect of concentration on chemical equilibrium can be studied, A is correct; B. From the device can be seen, ethylene with ethanol, concentrated sulfuric acid as a catalyst, concentrated sulfuric acid has dehydration and strong oxidation, so the ethylene mixed with carbon dioxide, sulfur dioxide and water vapor and other impurities, sulfur dioxide has the reductive properties, but also make the bromine water discoloration, so you can not use the bromine water to test the ethylene, B error; C. Titration of sodium hydroxide solution with hydrochloric acid solution, the left hand rotating piston, the right hand holding the conical flask, C is correct; D. Benzene in bromobenzene is soluble in each other, to remove the benzene in bromobenzene can be separated by distillation, D is correct; Answer choice B.

Question 19

Question 19: 21. Reversible reactions $\mathrm { mA } ( \mathrm { g } ) + \mathrm { nB } ($ ? $) \rightleftharpoo...

21. Reversible reactions $\mathrm { mA } ( \mathrm { g } ) + \mathrm { nB } ($ ? $) \rightleftharpoons \mathrm { pC } ( \mathrm { g } ) + \mathrm { qD } ($ ?) In $)$, A and C are colorless gases, and when equilibrium is reached, the following is true

A. A. If the equilibrium is shifted positively after changing the conditions, the percentage of D must increase
B. B. If the temperature is increased, the concentration of A increases, indicating that the positive reaction is heat-absorbing.
C. C. If the pressure is increased, the equilibrium does not shift, which means $m + n$ must be equal to $p + q$
D. D. If the color of the system deepens after the equilibrium shifts by increasing the amount of B, then B must be a gas.

Answer

Answer: D

Solution: A. If the equilibrium is shifted positively after changing the conditions, the percentage of D does not necessarily increase because the state of B is uncertain, so A is wrong; B. If the temperature is increased, the concentration of A increases, which means that the reverse reaction is a heat-absorbing reaction, so B is wrong; C. If the pressure is increased, the equilibrium does not move, and if $B$, $D$ are gases, then $m + n$ must be equal to $p + q$, if $B$ and D are not gases, then $B$ is not a gas. ]] and D are not gases, $m = p$, so C is wrong; D. The amount of solid and liquid does not affect the equilibrium, if the amount of $B$ is increased the color of the system is deepened after the equilibrium is shifted, which means $B$ must be a gas, so D is correct; Therefore, the answer is D.

Question 20

Question 20: 22. In order to compare the catalytic effect of $\mathrm { Fe } ^ { 3 + }$ and $\mathrm { Cu } ^ { 2...

22. In order to compare the catalytic effect of $\mathrm { Fe } ^ { 3 + }$ and $\mathrm { Cu } ^ { 2 + }$ on the decomposition reaction of $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$, the two students designed the experiments as shown in Figs. A and B respectively. The following description is not correct ![](/images/questions/chem-reaction-rate/image-010.jpg) A ![](/images/questions/chem-reaction-rate/image-011.jpg) B

A. A. The experiment shown in Figure A can be used to compare reaction rates by observing how quickly bubbles are produced
B. B. If the experimental reaction rate shown in Figure A is (1) > (2), then it means that $\mathrm { Fe } ^ { 3 + }$ must catalyze the decomposition of $\mathrm { Cu } ^ { 2 + }$ better than $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$.
C. C. The rate of the reaction is determined using the apparatus shown in Figure B. The time required for a given volume of gas to be produced can be determined
D. D. In order to check the airtightness of the device shown in Figure B, close the piston at A, pull the syringe piston out for a certain distance, and after a period of time, release the piston and observe whether the piston returns to its original position.

Answer

Answer: B

Solution: A. In the experiment shown in Figure A, the fast production of bubbles indicates that a large volume of $\mathrm { O } _ { 2 }$ was produced in the same amount of time, and the rate of reaction was fast, A is correct; B. The reaction rate (1) > (2) in the same environment indicates that the catalytic effect of $\mathrm { FeCl } _ { 3 }$ is better than that of $\mathrm { CuSO } _ { 4 }$, but since the anions and cations are different, it is not clear what the catalytic effect of $\mathrm { Cl } ^ { - }$ and $\mathrm { SO } _ { 4 } ^ { 2 - }$ is, so we can't be sure that $\mathrm { SO } _ { 4 } ^ { 2 - }$ has a better catalytic effect than $\mathrm { CuSO } _ { 4 }$. Therefore, it is not certain that $\mathrm { Fe } ^ { 3 + }$ catalyzes $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ decomposition better than $\mathrm { Cu } ^ { 2 + }$, and B is incorrect; C. Using the device in Figure B to determine the rate of reaction, you can determine the size of the volume of gas produced in the same time, you can also determine the length of time used to produce the same volume of gas, C is correct; D. Close the piston at A, pull the syringe piston out a certain distance, after a period of time to release the piston, if the piston back to the original position, indicating that the device is well sealed, if the piston can not return to the original position, it shows that the device leaks, D is correct;

Question 21

Question 21: 23. In recent years, scientists have studied a new method for the catalytic synthesis of ethyl aceta...

23. In recent years, scientists have studied a new method for the catalytic synthesis of ethyl acetate from ethanol: $2 \mathrm { C } _ { 2 } \mathrm { H } _ { 5 } \mathrm { OH } ( \mathrm { g } )$ ' like the Ministry of China $\mathrm { t } \mathrm { CH } _ { 3 } \mathrm { COOC } _ { 2 } \mathrm { H } _ { 5 } ( \mathrm {~g} ) + 2 \mathrm { H } _ { 2 } ( \mathrm {~g} )$, the reaction at atmospheric pressure, condensation collection, measured at room temperature in the liquid collector of the main products of the mass fraction of the figure shows. The following speculations about the method are not reasonable. ![](/images/questions/chem-reaction-rate/image-012.jpg)

A. A. The reaction temperature should not exceed $300 ^ { \circ } \mathrm { C }$.
B. B. The reaction is an elimination reaction
C. C. In the presence of a catalyst, acetaldehyde is an intermediate product in the reaction process
D. D. Improving catalyst activity and selectivity and reducing by-products such as ether and ethylene are key to the process

Answer

Answer: B

Solution: A. Through the image analysis: when the temperature is higher than $300 ^ { \circ } \mathrm { C }$, the mass fraction of ethyl acetate starts to decrease, so the reaction temperature should not be higher than $300 ^ { \circ } \mathrm { C }$, A is correct; B. According to the equation of the synthesis reaction, the reaction is a redox reaction, not an elimination reaction, B is wrong; C. From the diagram, we know that the yield of acetaldehyde is large at the beginning, and then decreases with the increase of the yield of ethyl acetate, so it should be an intermediate product, C is correct; D. ethanol reaction may generate ethylene or ether and other by-products, catalyst selectivity, if you choose a high-quality catalyst can improve the reaction rate, while reducing the occurrence of side-reaction process can improve the yield, D is correct;

Question 22

Question 22: 24. $\mathrm { CO } _ { 2 }$ One of the reaction courses for the production of methane by catalytic ...

24. $\mathrm { CO } _ { 2 }$ One of the reaction courses for the production of methane by catalytic hydrogenation is shown in the figure. ![](/images/questions/chem-reaction-rate/image-013.jpg)

A. A. (1) is a catalyst activation process
B. B. The process involves the breaking and formation of polar bonds
C. C. $\mathrm { D } _ { 2 }$ instead of $\mathrm { H } _ { 2 }$ reaction, D detected only in water
D. D. $\mathrm { CO } _ { 2 }$ and $\mathrm { H } _ { 2 }$ are adsorbed on different parts of the catalyst.

Answer

Answer: C

Solution: A. From the initial substance, the catalyst is $\mathrm { Ni } / \mathrm { CeO } _ { 2 } - \mathrm { ZrO } _ { 2 }$, which is adsorbed on the CZ-carrier, and (1) yields $\mathrm { Ni } ^ { 0 }$, which is the main constituent of the catalyst, so (1) is the process of catalyst activation, and A is correct; B. The process (6) has the breaking of O-C bonds, which produces $\mathrm { CH } _ { 4 }$ There is the formation of $\mathrm { C } - \mathrm { H }$ bonds, which involves the breaking and formation of polar bonds. B is correct; C. From step (6), it can be seen that $\mathrm { H } _ { 2 }$ also participates in $\mathrm { CH } _ { 4 }$, and if $\mathrm { H } _ { 2 }$ is replaced by $\mathrm { D } _ { 2 }$, there is also a D in the resulting $\mathrm { CH } _ { 4 }$. C error; D. From (2), it can be seen that $\mathrm { CO } _ { 2 }$ is adsorbed at $\mathrm { Ce } ^ { x + }$, and from the following steps, it can be seen that $\mathrm { H } _ { 2 }$ is adsorbed on $\mathrm { Ni } ^ { 0 }$, and the adsorption sites are different. D is correct;

Question 23

Question 23: 25. The mechanism of liquid-phase catalytic oxidation of acetaldehyde by ethylene is shown in Fig: $...

25. The mechanism of liquid-phase catalytic oxidation of acetaldehyde by ethylene is shown in Fig: $\mathrm { HCl } ,$ ![](/images/questions/chem-reaction-rate/image-014.jpg) The following statement is incorrect

A. A. Generate $1 \mathrm { molCH } _ { 3 } \mathrm { CHO }$ transferring the number of electrons to $2 \mathrm {~N} _ { \mathrm { A } }$
B. B. $\mathrm { CuCl } _ { 2 }$ and $\mathrm { PdCl } _ { 2 }$ are catalysts, CuCl and Pd are intermediates.
C. C. Redox reactions occur in steps I, II, and III.
D. D. The reaction involves both polar and nonpolar bond breaking and polar and nonpolar bond formation

Answer

Answer: D

Solution: A. The valence of carbon in ethylene is -2, and in acetaldehyde is -1. 1 mol of ethylene is converted to acetaldehyde with a loss of 2 mol of electrons, so A is correct; B. From the diagram, it can be seen that the starting point of the reaction $\mathrm { PdCl } _ { 2 }$ is the reactant, generating Pd, and copper chloride is the reactant in the process II, reacting with Pd to generate CuCl, and it can be seen that $\mathrm { CuCl } _ { 2 }$ and $\mathrm { PdCl } _ { 2 }$ are the substances that exist at the beginning of the reaction, serving as catalysts, and CuCl and Pd are the products during the reaction. CuCl and Pd are the products in the process of reaction, and also the reactants in another reaction, as intermediate products, so B is correct; C. I, II, III step in the participation or generation of monomers, are valence changes, are redox reactions, so C is correct; D. From the figure, we can see that the final product of the process is acetaldehyde, which contains only polar and nonpolar bond breakage, no nonpolar bond formation, so D is wrong;

Question 24

Question 24: 26. The main reactions involved in the synthesis of $\mathrm { C } _ { 2 } \mathrm { H } _ { 5 } \ma...

26. The main reactions involved in the synthesis of $\mathrm { C } _ { 2 } \mathrm { H } _ { 5 } \mathrm { OH }$ from $\mathrm { CO } , { } ^ { H _ { 2 } }$ are as follows. (1) $2 \mathrm { CO } ( \mathrm { g } ) + 4 \mathrm { H } _ { 2 } ( \mathrm {~g} ) = \mathrm { C } _ { 2 } \mathrm { H } _ { 5 } \mathrm { OH } ( \mathrm { g } ) + \mathrm { H } _ { 2 } \mathrm { O } ( \mathrm { g } ) \quad \Delta \mathrm { H } _ { 1 }$ (2) $\mathrm { CO } ( \mathrm { g } ) + \mathrm { H } _ { 2 } \mathrm { O } ( \mathrm { g } ) = \mathrm { CO } _ { 2 } ( \mathrm {~g} ) + \mathrm { H } _ { 2 } ( \mathrm {~g} ) \quad \Delta H _ { 2 } = - 41.1 \mathrm {~kJ} \cdot \mathrm {~mol} ^ { - 1 }$ When other conditions such as reaction time are the same, the conversion of CO and the product selectivity at different temperatures when the feed gas is charged according to $n ( \mathrm { CO } ) : n \left( \mathrm { H } _ { 2 } \right) = 1 : 1$ are shown below. ![](/images/questions/chem-reaction-rate/image-015.jpg) Known : Selectivity of product A $= \frac { \text { 转化成 } \mathrm { A } \text { 所用的 } \mathrm { CO } \text { 量 } } { \text { 反应消耗的 } \mathrm { CO } \text { 量 } } \times 100 \%$ ## The following statements are incorrect

A. A. $\mathrm { CO } _ { 2 }$ Increased selectivity may be caused by the effect of temperature on reaction rates
B. B. It can be inferred that ${ } ^ { \Delta \mathrm { H } _ { 1 } > 0 }$ based on the fact that the conversion of CO increases when the temperature is increased.
C. C. More $\mathrm { C } _ { 2 } \mathrm { H } _ { 5 } \mathrm { OH }$ can be synthesized per unit of time compared to $230 ^ { \circ } \mathrm { C } , 270 ^ { \circ } \mathrm { C }$.
D. D. In addition to reaction (2), there are other side reactions in the system

Answer

Answer: B

Solution: A. Reaction (2) is exothermic, so the increase in selectivity may be due to the increase in temperature to reach the appropriate temperature for the catalyst, which increases the rate of the chemical reaction, A. Wrong; B. From the analysis, $\Delta \mathrm { H } _ { 1 } < 0$ , B is wrong; C. Compared with $230 ^ { \circ } \mathrm { C } , ~ 270 ^ { \circ } \mathrm { C }$ when the temperature is high, the chemical reaction rate increases, and more can be synthesized per unit time. INLINE_FORMULA_3]], C is correct; C. The temperature is higher compared to $230 ^ { \circ } \mathrm { C } , ~ 270 ^ { \circ } \mathrm { C }$. C is correct ; D. The selectivity of $\mathrm { C } _ { 2 } \mathrm { H } _ { 5 } \mathrm { OH }$ and $\mathrm { CO } _ { 2 }$ is decreasing while the conversion of CO is increasing, so there are other side reactions in the system besides reaction (2), D is correct;

Question 25

Question 25: 27. Scientists have used transition metal nitrides (TMNS) to catalyze the synthesis of ammonia at ro...

27. Scientists have used transition metal nitrides (TMNS) to catalyze the synthesis of ammonia at room temperature using the reaction mechanism shown in the figure. The following statements are correct. ![](/images/questions/chem-reaction-rate/image-016.jpg)

A. A. TMNS increases the equilibrium conversion of $\mathrm { N } _ { 2 }$ in ammonia synthesis.
B. B. Oxidation of N atoms on the surface of TMNS to ammonia
C. C. The synthesis reaction with ${ } ^ { 15 } \mathrm {~N} _ { 2 }$ resulted in a product with only ${ } ^ { 15 } \mathrm { NH } _ { 3 }$
D. D. Vacancies on the TMNS surface from ammonia desorption favor adsorption $\mathrm { N } _ { 2 }$

Answer

Answer: D

Solution: A. TMNS greatly reduces the activation energy of the ammonia reaction and speeds up the reaction rate, but it cannot affect the equilibrium shift and does not increase the conversion rate; B. The N atoms on the surface of TMNS combine with hydrogen atoms, the N elemental valence decreases, and it is reduced to ammonia; C. The synthesis of ammonia is a reversible reaction and there are side reactions. If ${ } ^ { 15 } \mathrm {~N} _ { 2 }$ is used to carry out the synthesis reaction, the product is not necessarily only ${ } ^ { 15 } \mathrm { NH } _ { 3 }$, but also other by-products containing ${ } ^ { 15 } \mathrm {~N}$, which is wrong in statement C. The synthesis of ammonia is a reversible reaction and there are side reactions; D. According to the picture, the vacancies on the surface of TMNS produced by ammonia desorption are favorable for the adsorption of $\mathrm { N } _ { 2 }$, and D is correct.

Question 26

Question 26: 28. The following statements are incorrect

28. The following statements are incorrect

A. A. Spontaneous reactions do not require any conditions for the reaction to take place
B. B. Increasing the temperature increases the rate of a chemical reaction, mainly because it increases the percentage of activated molecules in the reactant molecules
C. C. Whether or not a chemical reaction can actually take place should be considered in addition to the spontaneity of the reaction, but also the rate of the reaction D. The reaction of $\Delta \mathrm { H } < 0 , \Delta \mathrm {~S} > 0$ can take place spontaneously at any temperature
D. D. Whether or not a chemical reaction can actually take place should take into account the rate of the reaction in addition to the spontaneity of the reaction D. $\Delta \mathrm { H } < 0 , \Delta \mathrm {~S} > 0$

Answer

Answer: A

Solution: A. Spontaneous reactions also require certain conditions to occur; B.Elevating the temperature increases the percentage of activated molecules in the reactant molecules and increases the rate of reaction, B is correct; C. Whether a chemical reaction can actually take place, in addition to considering the spontaneity of the reaction, should also consider the reaction rate, only when $\Delta \mathrm { H } - \mathrm { T } \Delta \mathrm { S } < 0$ and the reaction rate is not infinite, we think that the reaction will take place, C correct; D. When the $\Delta H < 0 , \Delta S > 0$ of the reaction is $\Delta H - T \Delta S$, $\Delta H - T \Delta S$ must be less than 0. Therefore, the reaction can proceed spontaneously at any temperature, and D is correct; The answer is A.

Question 27

Question 27: 29. In the contact method $\mathrm { H } _ { 2 } \mathrm { SO } _ { 4 }$, the gas from the upper par...

29. In the contact method $\mathrm { H } _ { 2 } \mathrm { SO } _ { 4 }$, the gas from the upper part of the absorption tower is reintroduced into the contact chamber for the purpose of.

A. A. Increase in $\mathrm { SO } _ { 2 }$ utilization rate
B. B. Make full use of the reaction heat
C. C. Accelerating the rate of chemical reactions
D. D. Prevention of catalyst poisoning

Answer

Answer: A

Solution: The $2 \mathrm { SO } _ { 2 } + \mathrm { O } _ { 2 } \rightleftharpoons 2 \mathrm { SO } _ { 3 }$ reaction is reversible, and the gas exported from the upper part of the absorption tower contains sulfur dioxide and oxygen, which are passed into the contact chamber again for recycling, which can improve the $\mathrm { SO } _ { 2 }$ utilization rate, so choose A;

Question 28

Question 28: 30. Given that $\mathrm { N } _ { \mathrm { A } }$ represents the value of Avogadro's constant, the ...

30. Given that $\mathrm { N } _ { \mathrm { A } }$ represents the value of Avogadro's constant, the following statements are correct (1) The number of $\mathrm { H } ^ { + }$ in a $1 \mathrm {~L} \mathrm { pH } = 1$ solution of sulfuric acid is $0.2 \mathrm {~N} _ { \mathrm { A } }$. (2) The total number of carbon atoms contained in 2.24 L of a mixture of ethylene and propylene at standard conditions is $0.2 \mathrm {~N} _ { \mathrm { A } }$ (3) The total number of anions and cations in the $7.2 \mathrm {~g} \mathrm { CaO } _ { 2 }$ crystal is $0.3 \mathrm {~N} _ { \mathrm { A } }$ (4) The number of electrons transferred in the reaction $3 \mathrm { H } _ { 2 } ( \mathrm {~g} ) + \mathrm { N } _ { 2 } ( \mathrm {~g} ) \rightleftharpoons 2 \mathrm { NH } _ { 3 } ( \mathrm {~g} ) \Delta \mathrm { H } =$ " - 92 " $\mathrm { kJ } / \mathrm { mol }$ is 0.6 when the heat released is 9.2 kJ $\mathrm { N } _ { \mathrm { A } }$ (5) The number of extra-nuclear electrons in 1 mol of helium at room temperature and pressure is $4 \mathrm {~N} _ { \mathrm { A } }$. (6) At room temperature and pressure, the number of neutrons in 1 mol of methyl ( - ${ } ^ { 14 } \mathrm { CH } _ { 3 }$) is $8 \mathrm {~N} _ { \mathrm { A } }$.

A. A. (1) (2)
B. B. (4) (5)
C. C. (4) (6)
D. D. (3) (5)

Answer

Answer: C

Solution: 试题分析:(1) $1 \mathrm { LpH } = 1$ of sulfuric acid solution contains $\mathrm { H } ^ { + }$ more $0.1 \mathrm {~N} _ { \mathrm { A } }$ than $0.1 \mathrm {~N} _ { \mathrm { A } }$. (2) At standard conditions, 2.24 L of a mixture of ethylene and propylene contains a total number of carbon atoms greater than $0.2 \mathrm {~N} _ { \mathrm { A } }$ (3) $7.2 \mathrm {~g} \mathrm { CaO } _ { 2 \text { 晶体中阴离子 } \mathrm { Ca } ^ { 2 + } 0.1 \mathrm {~N} _ { \mathrm { A } } \text { 和阳离子 } \mathrm { O } _ { 2 } { } ^ { 2 - } 0.1 \mathrm {~N} _ { \mathrm { A } } \text { 总数为 } 0.2 \mathrm {~N} _ { \mathrm { A } } }$ (4) the number of electrons transferred in the reaction $3 \mathrm { H } _ { 2 } ( \mathrm {~g} ) + \mathrm { N } _ { 2 } ( \mathrm {~g} ) \rightleftharpoons 2 \mathrm { NH } _ { 3 } ( \mathrm {~g} ) \Delta \mathrm { H } =$ " - 92 " $\mathrm { kJ } / \mathrm { mol }$ releases heat 9.2 kJ $\mathrm { kJ } / \mathrm { mol }$ is 0.6 $\mathrm { N } _ { \mathrm { A } }$ (5) The number of extra-nuclear electrons in 1 mol of helium at room temperature and pressure is $2 \mathrm {~N} _ { \mathrm { A } }$ (helium is a monatomic molecule). (6) At room temperature and pressure, only $\left( - { } ^ { 14 } \mathrm { CH } _ { 3 } \right)$ contains 8 neutrons in methyl ${ } ^ { 14 } \mathrm { C }$, and the number of neutrons in 1 mol of methyl is $8 \mathrm {~N} _ { \mathrm { Ao } }$ Therefore, only (4) and (6) are correct. 考点:阿伏加德罗常数的判断 Involving the number of ions in solution, the number of atoms in a mixture of gases, $\mathrm { CaO } _ { 2 }$ the number of anions in crystals, the transfer of electrons in redox reactions, the structure of the helium molecule, and the calculation of the number of nuclei and neutrons.

Question 29

Question 29: 31. Automobile exhaust contains NO and other toxic gases, activated carbon treatment of exhaust: $\m...

31. Automobile exhaust contains NO and other toxic gases, activated carbon treatment of exhaust: $\mathrm { C } ( \mathrm { s } ) + 2 \mathrm { NO } ( \mathrm { g } ) \rightleftharpoons \mathrm { N } _ { 2 } ( \mathrm {~g} ) + \mathrm { CO } _ { 2 } ( \mathrm {~g} )$. Other conditions remain unchanged, only change one condition, can not make the rate of nitrogen production in the reaction increased is

A. A. Raise the temperature
B. B. Reduction of container volume
C. C. Constant capacity filled with Ne
D. D. Use of suitable catalysts

Answer

Answer: C

Solution: A. Raising the temperature increases the internal energy of the substance and accelerates the rate of molecular motion, resulting in a faster rate of reaction to produce nitrogen; B. When other conditions remain unchanged, reducing the volume of the container results in an increase in the pressure of the system, which accelerates the rate of the chemical reaction to produce nitrogen, and B does not meet the meaning of the question; C. When Ne is charged at constant volume, the concentration of the gas in the reaction system remains unchanged, and the rate of reaction also remains unchanged, so the rate of nitrogen production in the reaction cannot be increased, and C is consistent with the meaning of the question; D.Using a suitable catalyst can speed up the rate of chemical reaction greatly, so the rate of nitrogen production can be increased, D does not meet the meaning of the question; Therefore, the reasonable choice is C.

Question 30

Question 30: 32. The following measures are designed to accelerate the rate of a chemical reaction.

32. The following measures are designed to accelerate the rate of a chemical reaction.

A. A. Chlorine-containing disinfectants stored away from light
B. B. The use of brewer's yeast in grain brewing
C. C. Sulfur dioxide added to wine
D. D. Zinc inlay on the bottom of marine vessels

Answer

Answer: B

Solution: A. Chlorine disinfectants are stored away from light in order to prevent the decomposition of hypochlorous acid in the presence of light, which would render the chlorine disinfectant ineffective; B. Wine quartz is a catalyst that catalyzes the decomposition of glucose into alcohol, which can speed up the rate of grain brewing, B is consistent with the meaning of the question; C. The addition of sulfur dioxide to wine is to use the reducing properties of $\mathrm { SO } _ { 2 }$ to prolong the shelf life of wine, i.e., to slow down the rate of deterioration of wine, C is not the case; D. Since zinc is more active than iron, the bottom of the seafaring ship is inlaid with zinc in order to slow down the corrosion rate of the hull steel, D is not the case; Therefore, the answer is: B.

Question 31

Question 31: 33. Cuprous chloride (CuCl) is a white powder that is slightly soluble in water, insoluble in ethano...

33. Cuprous chloride (CuCl) is a white powder that is slightly soluble in water, insoluble in ethanol, and easily oxidized. The process of producing CuCl from chalcopyrite (the main component of which is $\mathrm { CuFeS } _ { 2 }$) is shown in the figure. The following statement is not correct. ![](/images/questions/chem-reaction-rate/image-017.jpg)

A. A. Adding water to $\left[ \mathrm { CuCl } _ { 3 } \right] ^ { 2 - }$ shifts the balance $\left[ \mathrm { CuCl } _ { 3 } \right] ^ { 2 - } ( \mathrm { aq } ) \rightleftharpoons \mathrm { CuCl } ( \mathrm { s } ) + 2 \mathrm { Cl } ^ { - } ( \mathrm { aq } )$ in the opposite direction.
B. B. The reaction that occurs when concentrated hydrochloric acid is added is $\operatorname { CuCl } ( \mathrm { s } ) + 2 \mathrm { Cl } ^ { - } ( \mathrm { aq } ) \rightleftharpoons \left[ \mathrm { CuCl } _ { 3 } \right] ^ { 2 - } ( \mathrm { aq } )$
C. C. Filtering, washing, drying, etc. must not be carried out in air.
D. D. Ethanol can be used to wash CuCl

Answer

Answer: A

Solution: A. Dilution shifts the equilibrium in the direction of increasing ion concentration. Adding water to $\left[ \mathrm { CuCl } _ { 3 } \right] ^ { 2 - }$ shifts the equilibrium $\left[ \mathrm { CuCl } _ { 3 } \right] ^ { 2 - } ( \mathrm { aq } ) \rightleftharpoons \mathrm { CuCl } ( \mathrm { s } ) + 2 \mathrm { Cl } ^ { - } ( \mathrm { aq } )$ in a positive direction, causing CuCl to be analyzed, which is incorrect; B. The addition of concentrated hydrochloric acid increases the concentration of $\mathrm { Cl } ^ { - }$, which moves the equilibrium $\left[ \mathrm { CuCl } _ { 3 } \right] ^ { 2 - } ( \mathrm { aq } ) \rightleftharpoons \mathrm { CuCl } ( \mathrm { s } ) + 2 \mathrm { Cl } ^ { - } ( \mathrm { aq } )$ in the opposite direction, i.e., the reaction $\mathrm { CuCl } ( \mathrm { s } ) + 2 \mathrm { Cl } ^ { - } ( \mathrm { aq } ) \rightleftharpoons \left[ \mathrm { CuCl } _ { 3 } \right] ^ { 2 - } ( \mathrm { aq } )$ occurs, and B is correct; C. From the information in the question, we know that CuCl is a kind of white powder that is easy to be oxidized, so the process of filtration, washing and drying should not be carried out in the air to prevent it from being oxidized, C is correct; D. Ethanol washing can reduce the loss due to dissolution and improve the yield, and vacuum drying can prevent CuCl from being oxidized and improve the purity, D is correct; Therefore, the answer is: A.

Question 32

Question 32: 34. The following statements are incorrect

34. The following statements are incorrect

A. A. The use of catalysts can reduce the temperature of the reaction, and play an energy-saving effect
B. B. Most of the ceramic knives are processed with a nano-material "zirconium oxide", which has the advantages of high hardness, good thermal conductivity and corrosion resistance.
C. C. Graphene is a new material with a monolayer sheet-like structure composed of carbon atoms and thus has a huge surface area, which can be used to directly observe the adsorption and release processes of single atoms by penetrating electron microscopy
D. D. Promoting the renovation of facilities such as desulfurization, denitrification, and dust removal can reduce emissions $\mathrm { SO } _ { 2 } , \mathrm { NO } _ { 2 } , \mathrm { PM } 2.5$ substantially and make the sky reappear "APEC blue

Answer

Answer: B

Solution: Test Question Analysis: The use of catalysts can speed up or slow down the reaction rate, reduce the temperature of the reaction, play the effect of energy saving, A correct, ceramic knife are basically zirconium oxide-based, thermal conductivity is very low, B error. Graphene is a kind of carbon atoms to sp 2 hybridized orbitals composed of hexagonal honeycomb lattice planar film, only a carbon atom thickness of two-dimensional materials, C correct. Points: Chemistry and Life

Question 33

Question 33: 35. Scientists are investigating the use of catalytic technology to convert $\mathrm { NO } _ { \tex...

35. Scientists are investigating the use of catalytic technology to convert $\mathrm { NO } _ { \text {和 } } \mathrm { CO } _ { \text {转变为 } } \mathrm { CO } _ { 2 }$ and $\mathrm { N } _ { 2 }$ in exhaust gases to $2 \mathrm { NO } ( \mathrm { g } ) + 2 \mathrm { CO } ( \mathrm { g } ) - 4 \tau 2 \mathrm { CO } _ { 2 } ( \mathrm {~g} ) + \mathrm { N } _ { 2 } ( \mathrm {~g} ) \quad \Delta \mathrm { H } < 0$, and the equilibrium conversion rate of NO $( \alpha )$ with the variation of temperature and pressure is shown in the diagram. As shown in the figure, the following statements are correct ![](/images/questions/chem-reaction-rate/image-018.jpg)

A. A. X and Y represent pressure and temperature, respectively
B. B. $Y _ { 1 } > Y _ { 2 }$
C. C. Equilibrium constant: $\mathrm { K } _ { \mathrm { a } } < \mathrm { K } _ { \mathrm { b } } < \mathrm { K } _ { \mathrm { c } }$
D. D. The equilibrium conversion of NO is increased by lowering the temperature appropriately.

Answer

Answer: D

Solution: A. $2 \mathrm { NO } ( \mathrm { g } ) + 2 \mathrm { CO } ( \mathrm { g } )$ port $\downarrow 2 \mathrm { CO } _ { 2 } ( \mathrm {~g} ) + \mathrm { N } _ { 2 } ( \mathrm {~g} ) \quad \Delta \mathrm { H } < 0$ is an exothermic reaction with decreasing amount of gaseous substances, increasing the temperature, the equilibrium moves in the opposite direction, and the equilibrium rate of conversion of NO decreases, then X is the temperature and Y is the pressure, so A is wrong. B. At the same temperature, if the pressure is increased, the equilibrium conversion rate of NO increases, then $\mathrm { Y } _ { 2 } > \mathrm { Y } _ { 1 }$, so B is wrong; C. Increase the temperature, the equilibrium moves in the reverse direction, the equilibrium constant decreases, the temperature remains unchanged, the equilibrium constant remains unchanged, the equilibrium constant: $\mathrm { K } _ { \mathrm { a } } = \mathrm { K } _ { \mathrm { b } } < \mathrm { K } _ { \mathrm { c } }$, so C is wrong; D. $2 \mathrm { NO } ( \mathrm { g } ) + 2 \mathrm { CO } ( \mathrm { g } ) - 4$ $2 \mathrm { CO } _ { 2 } ( \mathrm {~g} ) + \mathrm { N } _ { 2 } ( \mathrm {~g} ) \quad \Delta \mathrm { H } < 0$ is an exothermic reaction in which the amount of gaseous material decreases, and the equilibrium is shifted positively by lowering the temperature, so the equilibrium conversion rate of NO increases, so D is correct;

Question 34

Question 34: 36. The following experiments correspond to the correct design of the experimental program. | | Exp...

36. The following experiments correspond to the correct design of the experimental program. | | Experiment | Experimental protocol | | :--- | :--- | :--- | | A | Prepare a $100 \mathrm {~mL} 1.0 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 }$ solution of $\mathrm { CuSO } _ { 4 }$ | Dissolve $25.0 \mathrm {~g} \mathrm { CuSO } _ { 4 } \cdot 5 \mathrm { H } _ { 2 } \mathrm { O }$ in water to make a 100 mL solution | | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | B | Compare the acidity of hypochlorite and acetic acid | Determine the pH of NaClO and $\mathrm { CH } _ { 3 } \mathrm { COONa }$ solutions at the same concentration using pH paper at room temperature. | C | Investigate whether the reaction between $\mathrm { Fe } ^ { 3 + }$ and $\mathrm { I } ^ { - }$ is a reversible reaction | Mix equal concentrations of KI solution and $\mathrm { FeCl } _ { 3 }$ solution, and after sufficient reaction, drop in KSCN solution, which turns red | D | Simulate Houstman's acidic strength of NaClO solution and $\mathrm { CH } _ { 3 } \mathrm { COONa }$ solution pH with pH paper at temperature. | D | simulate Hou's alkali production method to prepare $\mathrm { NaHCO } _ { 3 }$ solid | first to the saturated brine into a sufficient amount of $\mathrm { CO } _ { 2 }$, and then into the $\mathrm { NH } _ { 3 }$ after the precipitation of a solid, filtration, washing, drying |

A. A. A
B. B. B
C. C. C
D. D. D

Answer

Answer: A

Solution: A. Dissolve $25.0 \mathrm {~g} \mathrm { CuSO } _ { 4 } \cdot 5 \mathrm { H } _ { 2 } \mathrm { O }$ in water to make a 100 mL solution; the resulting solution has a concentration of substance of: $\frac { \frac { 25.0 \mathrm {~g} } { 250 \mathrm {~g} \cdot \mathrm {~mol} ^ { - 1 } } } { 0.1 \mathrm {~L} } = 1.0 \mathrm {~mol} / \mathrm { L } , \mathrm { A }$ is correct; B. Since HClO generated by hydrolysis of NaClO has strong oxidizing property, it can bleach the pH test paper, that is, we cannot use pH test paper to measure the pH value of NaClO solution, B is wrong; C. According to the reaction $2 \mathrm { Fe } ^ { 3 + } + 2 \mathrm { I } ^ { - } = 2 \mathrm { Fe } ^ { 2 + } + \mathrm { I } _ { 2 }$ can be known, did not tell the volume of the solution, will be equal to the concentration of KI solution and $\mathrm { FeCl } _ { 3 }$ solution mixing, is not sure which reactant is too much, if the $\mathrm { FeCl } _ { 3 }$ excess of fully reacted drops into the KSCN solution, the solution turns red, then can not be If $\mathrm { Fe } ^ { 3 + }$ is mixed with $\mathrm { I } ^ { - }$, the reaction is reversible. If KI is in excess, the reaction between $\mathrm { Fe } ^ { 3 + }$ and $\mathrm { I } ^ { - }$ can be investigated as a reversible reaction, C is wrong; D. Since the solubility of $\mathrm { CO } _ { 2 }$ in saturated brine is very small and that of $\mathrm { NH } _ { 3 }$ increases under alkaline conditions, the reaction between $\mathrm { NH } _ { 3 }$ and $\mathrm { NH } _ { 3 }$ is a reversible reaction. The solubility of $\mathrm { NH } _ { 3 }$ in saturated salt water is very large, and it is alkaline when dissolved in water. Therefore, in Hou's method of making alkali, we should first pass a sufficient amount of $\mathrm { NH } _ { 3 }$ into the saturated salt water, then pass $\mathrm { CO } _ { 2 }$ and precipitate the solid, then filter, wash, and dry, D is wrong; Therefore, the answer is: A.

Question 35

Question 35: 37. In the presence of a catalyst, $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ decomposes, and the...

37. In the presence of a catalyst, $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ decomposes, and the reaction energy varies with the course of the reaction as shown below: The following statement is correct. ![](/images/questions/chem-reaction-rate/image-019.jpg)

A. A. $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ Decomposition of the reaction of $\triangle \mathrm { H } > 0$
B. B. The addition of a catalyst reduces the thermal effect of the reaction
C. C. The addition of a catalyst increases the decomposition rate of $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$
D. D. Catalysts increase the $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ decomposition rate by altering the course of the reaction and lowering the activation energy of the reaction.

Answer

Answer: D

Solution: 试题分析:根据图可知反应物能量高,生成物能量低,过氧化氢分解属于放热反应,选项 A错误;催化剂能改变反应的活化能量,不能改变反应的热效应,选项 B 错误;催化剂只能改变化学反应速率,不能該邊反应的转化率,选项 C 错;催化剂通过改变反应历程,降低反应的活化能来提高 [$\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ decomposition rate,选项 D正确. INLINE_FORMULA_0]] decomposition rate by changing the reaction course and reducing the activation energy of the reaction, option D is correct. Points: the effect of catalysts on chemical reactions

Question 36

Question 36: 38. The basic process of synthesizing phenylacetylene using the Sonogashira reaction mechanism is sh...

38. 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A. A. During step (4), Pd elements are reduced
B. B. $\mathrm { L } _ { \mathrm { n } } \mathrm { Pd }$ and ${ } _ { \mathrm { CuI } }$ are the catalysts for this reaction process, which can increase the
C. C. Atomic utilization of total reactions up to $100 \%$
D. D. If phenylacetylene is used instead of the same substance in the above process for the reaction with $\mathrm { CH } _ { 2 } = \mathrm { CHI }$, one can synthesize

Answer

Answer: A

Solution: A. In step (4), the valence of $\operatorname { Pd }$ changes from +2 to 0, the valence decreases and the element is reduced, so A is correct; B. The catalyst cannot shift the equilibrium, and the equilibrium conversion of iodobenzene remains unchanged, so B is wrong; C. From the figure, we can see that the reaction is acetylene and <img class="imgSvg" id = "mi1myhg9qtsksei9j4" src="data:image/svg+xml;base64, PHN2ZyBpZD0ic21pbGVzLW1pMW15aGc5cXRza3NlaTlqNCIgeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzIwMDAvc3ZnIiB2aWV3Qm94PSIwIDAgMTc4IDk2LjU1OTYyODcxMzAxMTk5IiBzdHlsZT0id2lkdGg6IDE3OC40OTk5OTk5OTk5NjE5N3B4OyBoZWlnaHQ6IDk2LjU1OTYyODcxMzAxMTk5cHg7IG92ZXJmbG93OiB2aXNpYmxlOyI +PGRlZnM+ PGxpbmVhckdyYWRpZW50IGlkPSJsaW5lLW1pMW15aGc5cXRza3NlaTlqNC0xIiBncmFkaWVudFVuaXRzPSJ1c2VyU3BhY2VPblVzZSIgeDE9IjEwNC45OTk5OTk5OTk5NzQ2MyIgeTE9IjQ4LjI3OTg0MjYzMTEyMDM0NiIgeDI9IjEzNi40OTk5OTk5OTk5NjE5NyIgeTI9IjQ4LjI3OTg3MDkwNTczNDcyNSI +PHN0b3Agc3RvcC1jb2xvcj0iY3VycmVudENvbG9yIiBvZmZzZXQ9IjIwJSI+. 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If we use phenylacetylene to replace the same kind of substance in the above process and $\mathrm { CH } _ { 2 } = \mathrm { CHI }$ to have the substitution reaction, we can synthesize <img class="imgSvg" id = "mi1myhgapb6agstzvf" src="data:image/svg+xml;base64, PHN2ZyBpZD0ic21pbGVzLW1pMW15aGdhcGI2YWdzdHp2ZiIgeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzIwMDAvc3ZnIiB2aWV3Qm94PSIwIDAgMjQ4IDEwNS4wMDAwNjEyMTYyOTYxMyIgc3R5bGU9IndpZHRoOiAyNDcuNjc4ODM2NjU4NDIzODVweDsgaGVpZ2h0OiAxMDUuMDAwMDYxMjE2Mjk2MTNweDsgb3ZlcmZsb3c6IHZpc2libGU7Ij48ZGVmcz48bGluZWFyR3JhZGllbnQgaWQ9ImxpbmUtbWkxbXloZ2FwYjZhZ3N0enZmLTEiIGdyYWRpZW50VW5pdHM9InVzZXJTcGFjZU9uVXNlIiB4MT0iMTc2Ljk4MTUyMTUyMjMzMzIyIiB5MT0iNTAuMDQ0OTM4ODIyNDc3Mzg0IiB4Mj0iMjA0LjI2MTMzOTQxMzE1OTg1IiB5Mj0iMzQuMjk0OTY5NDMwNjU1MiI +PHN0b3Agc3RvcC1jb2xvcj0iY3VycmVudENvbG9yIiBvZmZzZXQ9IjIwJSI+ PC9zdG9wPjxzdG9wIHN0b3AtY29sb3I9ImN1cnJlbnRDb2xvciIgb2Zmc2V0PSIxMDAlIj48L3N0b3A+ 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 +PC90ZXh0PjwvZz48L3N2Zz4="/> Therefore, D is wrong;

Question 37

Question 37: 39. It is known that when the acidity of three monobasic weak acids is $\mathrm { HX } > \mathrm { H...

39. It is known that when the acidity of three monobasic weak acids is $\mathrm { HX } > \mathrm { HY } > \mathrm { HZ } . 25 ^ { \circ } \mathrm { C }$, HCl gas is passed into a mixed solution of $\mathrm { KX } , \mathrm { KY } , \mathrm { KZ }$ of $0.1 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 }$ with the concentration of $0.1 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 }$ (the change in the volume of the solution is neglected), the distribution fraction of ${ } _ { \delta } \left[ \right.$ (acid radical ions) changes as $\delta ( \mathrm { HX } ) = \frac { \mathrm { c } ( \mathrm { HX } ) } { \mathrm { c } ( \mathrm { HX } ) + \mathrm { c } \left( \mathrm { X } ^ { - } \right) }$ changes. ] acid radical ion) distribution fraction ${ } _ { \delta } \left[ \right.$ as $\delta ( \mathrm { HX } ) = \frac { \mathrm { c } ( \mathrm { HX } ) } { \mathrm { c } ( \mathrm { HX } ) + \mathrm { c } \left( \mathrm { X } ^ { - } \right) }$ changes as shown in the figure. The equilibrium constant for the reaction is known greater than $10 ^ { 5 }$ is considered to be complete. The following statement is false. ![](/images/questions/chem-reaction-rate/image-021.jpg)

A. A. $25 ^ { \circ } \mathrm { C } _ { \text {时,} } \mathrm { K } _ { \mathrm { a } } ( \mathrm { HX } )$ of the order of $10 ^ { - 10 }$
B. B. The reaction $\mathrm { HX } + \mathrm { Z } ^ { - } \rightarrow \mathrm { HZ } + \mathrm { X } ^ { - }$ is able to proceed completely
C. C. KX, KY, KZ mixed solution in the presence of ${ } ^ { \mathrm { C } ( \mathrm { HX } ) + \mathrm { c } ( \mathrm { HY } ) + \mathrm { c } ( \mathrm { HZ } ) = \mathrm { c } \left( \mathrm { OH } ^ { - } \right) - \mathrm { c } \left( \mathrm { H } ^ { + } \right) }$
D. D. If the corresponding pH at $\mathrm { M } , \mathrm {~N}$ is equal, then the pH is 3.70.

Answer

Answer: A

Solution: A. According to the above analysis, ${ } ^ { 25 ^ { \circ } } \mathrm { C } _ { \text {时 } } , \mathrm { K } _ { \mathrm { a } } ( \mathrm { HX } ) _ { \text {的数量级为 } } 10 ^ { - 3 }$ , A is wrong; B. The reaction ${ } _ { \mathrm { HX } + \mathrm { Z } ^ { - } \rightarrow \mathrm { HZ } + \mathrm { X } ^ { - } }$ has an equilibrium constant of $\mathrm { K } = \frac { \mathrm { c } ( \mathrm { HZ } ) \cdot \mathrm { c } \left( \mathrm { X } ^ { - } \right) } { \mathrm { c } ( \mathrm { HX } ) \cdot \mathrm { c } \left( \mathrm { Z } ^ { - } \right) } = \frac { \mathrm { K } _ { \mathrm { a } } ( \mathrm { HX } ) } { \mathrm { K } _ { \mathrm { a } } ( \mathrm { HZ } ) } = \frac { 10 ^ { - 2.66 } } { 10 ^ { - 9.98 } } = 10 ^ { 7.32 } > 10 ^ { 5 }$, and the reaction can be carried out completely, and B is correct; C. In the mixed solution, according to the conservation of protons there is ${ } ^ { \mathrm { c } ( \mathrm { HX } ) + \mathrm { c } ( \mathrm { HY } ) + \mathrm { c } ( \mathrm { HZ } ) + \mathrm { c } \left( \mathrm { H } ^ { + } \right) = \mathrm { c } \left( \mathrm { OH } ^ { - } \right) \text {, option C is correct;} }$ D. $K _ { a } ( H X ) \cdot K _ { a } ( H Y ) = \frac { c \left( H ^ { + } \right) \cdot c \left( X ^ { - } \right) } { c ( H X ) } \frac { c \left( H ^ { + } \right) \cdot c \left( Y ^ { - } \right) } { c ( H Y ) } , M ^ { \text {点时 } } c ( H Y ) = c \left( X ^ { - } \right) , N ^ { \text {点时 } } c ( H X ) = c \left( Y ^ { - } \right)$, if $M , N$ corresponds to $p H$ at the same place, then [INLINE_FORMULA_7]] is equal, then A is the answer.

Question 38

Question 38: 40. Chemical potentials are important when dealing with gas-liquid transformations. That is, substan...

40. Chemical potentials are important when dealing with gas-liquid transformations. That is, substances are always transferred from the phase with the higher chemical potential to the phase with the lower potential. When the chemical potentials of substances in the gas-liquid are equal, the transformation process stops and the system reaches equilibrium. The chemical potential ($\mu$) versus pressure (p) of any one-component system in the gas (g) and liquid (l) states at constant temperature is correctly plotted as follows High School Chemistry Assignment, November 14, 2025

A. A. ![](/images/questions/chem-reaction-rate/image-005.jpg)
B. B. ![](/images/questions/chem-reaction-rate/image-006.jpg)
C. C. ![](/images/questions/chem-reaction-rate/image-007.jpg)
D. D. ![](/images/questions/chem-reaction-rate/image-008.jpg)

Answer

Answer: C

Solution: Pressure increases chemical potential increases, substances are always transferred from the phase with higher chemical potential to the phase with lower chemical potential, the chemical potential of the gaseous state is higher than that of the liquid state, so the answer is C

Chemical Reaction Rate and Equilibrium

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