2. At a temperature of $\mathrm { T } ^ { \circ } \mathrm { C }$, a sufficient amount of activated carbon and $1 \mathrm { molNO } _ { 2 }$ are added to closed containers of unequal volume, and the following reaction occurs: $2 \mathrm { C } ( \mathrm { S } ) + 2 \mathrm { NO } _ { 2 } ( \mathrm {~g} ) = \mathrm { N } _ { 2 } ( \mathrm {~g} ) + 2 \mathrm { CO } _ { 2 } ( \mathrm {~g} )$. At the same reaction time, the conversion rate of $\mathrm { NO } _ { 2 }$ in each container is measured as a function of the volume of the container as shown in the graph ($\mathrm { V } _ { 1 } < \mathrm { V } _ { 2 } < \mathrm { V } _ { 3 }$ ). Known: The ideal gas equation of state is $\mathrm { pV } = \mathrm { nRT }$. The following statement is false ( )
Volume of the container / L
- A. A. At $T ^ { \circ } \mathrm { C }$, the equilibrium constant of the reaction at $a$ is $\frac { 4 } { 45 V _ { 1 } }$
- B. B. According to the ideal gas equation, the pressure inside the container: $\mathrm { Pa } : \mathrm { Pb } = 6 : 7$
- C. C. $T ^ { \circ } \mathrm { C }$ when point b is in chemical equilibrium, point c is not in chemical equilibrium
- D. D. Pressurize the container at point c to reduce the volume of the container, then at this point $\mathrm { v } _ { \text {正 } } > \mathrm { v }$ inverse
Answer: B
Solution: A. Based on the data at point a, the reaction reaches equilibrium at point a. The $\mathrm { NO } _ { 2 }$ conversion rate is $40 \%$ and the reaction triplet at point a is:
| | $2 \mathrm { C } ( \mathrm { s } ) + 2 \mathrm { NO } _ { 2 } ( \mathrm {~g} )$ | $= \mathrm { N } _ { 2 } ( \mathrm {~g} )$ | $+ 2 \mathrm { CO } _ { 2 } ( \mathrm {~g} )$ | |
| :--- | :---: | :---: | :---: | :---: | :---: |
| Starting amount $( \mathrm { mol } )$ | | 1 | 0 | 0 |
| change amount $( \mathrm { mol } )$ | | 0.4 | 0.2 | 0.4 |
| Equilibrium $( \mathrm { mol } )$ | | 0.6 | 0.2 | 0.4 |
At an equilibrium concentration of each substance of $\mathrm { c } \left( \mathrm { NO } _ { 2 } \right) = \frac { 0.6 } { \mathrm {~V} _ { 1 } } \mathrm { mo } / \mathrm { L } , \mathrm { c } \left( \mathrm { N } _ { 2 } \right) = \frac { 0.2 } { \mathrm {~V} _ { 1 } } \mathrm {~mol} / \mathrm { L } , \mathrm { c } \left( \mathrm { CO } _ { 2 } \right) = \frac { 0.4 } { \mathrm {~V} _ { 1 } } \mathrm {~mol} / \mathrm { L } , \mathrm { T } ^ { \circ } \mathrm { C }$, the reaction
The equilibrium constant of the reaction is $= \frac { \mathrm { c } \left( \mathrm { N } _ { 2 } \right) \cdot \mathrm { c } ^ { 2 } \left( \mathrm { CO } _ { 2 } \right) } { \mathrm { c } ^ { 2 } \left( \mathrm { NO } _ { 2 } \right) } = \frac { \frac { 0.2 } { \mathrm {~V} _ { 1 } } \times \left( \frac { 0.4 } { \mathrm {~V} _ { 1 } } \right) ^ { 2 } } { \left( \frac { 0.6 } { \mathrm {~V} _ { 1 } } \right) ^ { 2 } } = \frac { 4 } { 45 \mathrm {~V} _ { 1 } }$ at an equilibrium concentration of $\mathrm { c } \left( \mathrm { NO } _ { 2 } \right) = \frac { 0.6 } { \mathrm {~V} _ { 1 } } \mathrm { mo } / \mathrm { L } , \mathrm { c } \left( \mathrm { N } _ { 2 } \right) = \frac { 0.2 } { \mathrm {~V} _ { 1 } } \mathrm {~mol} / \mathrm { L } , \mathrm { c } \left( \mathrm { CO } _ { 2 } \right) = \frac { 0.4 } { \mathrm {~V} _ { 1 } } \mathrm {~mol} / \mathrm { L } , \mathrm { T } ^ { \circ } \mathrm { C }$;
B. According to the data in B, the amount of gas in the container at point a is $1.2 \mathrm {~mol} ; \mathrm { b }$, and the reaction is trivial:
| | $2 \mathrm { C } ( \mathrm { s } )$ | $+ 2 \mathrm { NO } _ { 2 } ( \mathrm {~g} )$ | $=$ | $\mathrm { N } _ { 2 } ( \mathrm {~g} )$
| :--- | :---: | :---: | :---: | :---: | :---: |
| Starting amount $( \mathrm { mol } )$ | | 1 | 0 | $2 \mathrm { CO } _ { 2 } ( \mathrm {~g} )$ |
| change amount $( \mathrm { mol } )$ | | 0.8 | 0.4 | 0.8 |
| Equilibrium $( \mathrm { mol } )$ | | 0.2 | 0.4 | 0.8 |
According to the reaction triplet at point a, the triplet at point b and $\mathrm { pV } = \mathrm { nRT }$ in option A, $\mathrm { P } _ { \mathrm { a } } \mathrm { V } _ { 1 } = 1.2 \mathrm { RT } , ~ \mathrm { P } _ { \mathrm { b } } \mathrm { V } _ { 2 } = 1.4 \mathrm { RT }$ can be found, $\mathrm { P } _ { \mathrm { a } } \mathrm { V } _ { 1 } = 1.2 \mathrm { RT } , ~ \mathrm { P } _ { \mathrm { b } } \mathrm { V } _ { 2 } = 1.4 \mathrm { RT }$, and because of the $V _ { 2 } > V _ { 1 }$, the pressure inside the container: $\mathrm { Pa } : \mathrm { Pb } > 6 : 7$, so B is wrong;
C. The conversion rate of $\mathrm { NO } _ { 2 }$ is the highest at point b in the figure, then the equilibrium is reached at point b at the temperature of $\mathrm { T } ^ { \circ } \mathrm { C }$, and the volume of the corresponding container on the ab curve increases gradually, and the starting concentration of $\mathrm { NO } _ { 2 }$ decreases gradually, but the concentration is greater than that at point b, so it is wrong. As the volume of the corresponding container on the ab curve gradually increases, the starting concentration of $\mathrm { NO } _ { 2 }$ gradually decreases, but the concentration is greater than that at point b. The greater the concentration of $\mathrm { NO } _ { 2 }$, the greater the reaction rate, and the shorter the time to reach equilibrium, so the reactions on the ab curve reach equilibrium, and the positive direction of the reaction is the reaction with the increase of the gas volume. As the volume of the container increases, the starting concentration of $\mathrm { NO } _ { 2 }$ decreases, the reaction rate slows down, and the time to reach equilibrium is prolonged, so the reaction on the bc curve has not reached equilibrium, C is correct;
D. In the same reaction time, the container volume corresponding to point c is the largest and the pressure is the smallest, the chemical reaction rate is slower, so the conversion rate of nitrogen dioxide is smaller, pressurize the container at point c and reduce the container volume, then the chemical reaction rate is accelerated, and the reaction is carried out in the direction of the positive reaction, at this time $v$ (inverse) $< v$ (reverse). 30]] (positive), D is correct.