Answer: B
Solution: A. Industrial ammonia synthesis of raw gas may contain impurities (such as sulfide, etc.), these impurities will lead to poisoning of the catalyst inactivity, step (1) "purification" can remove impurities, to prevent poisoning of the catalyst, A is correct;
B. ammonia synthesis reaction for the gas volume reduction reaction, pressurization can accelerate the reaction rate and make the equilibrium positive shift to improve the ammonia yield, but the pressure is too high will significantly increase the cost of equipment and energy consumption, industrial production needs to be considered to choose the appropriate pressure (such as the title of the $2 \times 10 ^ { 7 } \mathrm {~Pa}$), is not the higher the better, B error;
C. The iron catalyst used in ammonia synthesis has the greatest activity at $400 \sim 500 ^ { \circ } \mathrm { C }$, and the main purpose of controlling this temperature in step (3) is to ensure the activity of the catalyst, C is correct;
D. Ammonia is easy to liquefy, while $\mathrm { N } _ { 2 } , \mathrm { H } _ { 2 }$ is difficult to liquefy, step (4) can be separated from ammonia liquefaction by rapid cooling, prompting the equilibrium to move forward, D is correct;
D is correct. Therefore, B is wrong.
$40 . \mathrm { C }$
[INLINE_FORMULA_3]] The writing of ionic equations, the preparation of common inorganic substances, the concept of redox reactions to determine, acids, bases, salts and oxides
[Analysis]From the process given in the question, reaction 1 is the reaction of white phosphorus with excess sodium hydroxide solution to form $\mathrm { NaH } _ { 2 } \mathrm { PO } _ { 2 }$ and phosphine, the chemical equation of the reaction is $\mathrm { P } _ { 4 } + 3 \mathrm { NaOH } + 3 \mathrm { H } _ { 2 } \mathrm { O } = 3 \mathrm { NaH } _ { 2 } \mathrm { PO } _ { 2 } + \mathrm { PH } _ { 3 } \uparrow$, and reaction 2 is the reaction of $\mathrm { NaH } _ { 2 } \mathrm { PO } _ { 2 }$ solution with dilute sulfuric acid to form $\mathrm { H } _ { 3 } \mathrm { PO } _ { 2 }$, and reaction 2 is the reaction of $\mathrm { H } _ { 3 } \mathrm { PO } _ { 2 }$ with dilute sulfuric acid to form $\mathrm { H } _ { 3 } \mathrm { PO } _ { 2 }$. FORMULA_7]] and sodium sulfate to form $2 \mathrm { NaH } _ { 2 } \mathrm { PO } _ { 2 } + \mathrm { H } _ { 2 } \mathrm { SO } _ { 4 } = 2 \mathrm { H } _ { 3 } \mathrm { PO } _ { 2 } + \mathrm { Na } _ { 2 } \mathrm { SO } _ { 4 }$, and reaction 3 is the decomposition of $\mathrm { H } _ { 3 } \mathrm { PO } _ { 2 }$ to form $\mathrm { H } _ { 3 } \mathrm { PO } _ { 4 }$ and $\mathrm { PH } _ { 3 }$, and the equation is $\mathrm { PH } _ { 3 }$. The chemical equation of the reaction is $2 \mathrm { H } _ { 3 } \mathrm { PO } _ { 2 } = \mathrm { H } _ { 3 } \mathrm { PO } _ { 4 } + \mathrm { PH } _ { 3 } \uparrow$, which is analyzed accordingly;
A. White phosphorus and phosphine can spontaneously combust in air, so the preparation of phosphine needs to be carried out under oxygen-free conditions, A is correct;
B. In reaction 1, ${ } ^ { \mathrm { P } _ { 4 } }$ produces $\mathrm { NaH } _ { 2 } \mathrm { PO } _ { 2 }$ (P is +1 valence) and $\mathrm { PH } _ { 3 }$ (P is -3 valence), and the P elemental valence is increased.
$( 0 \rightarrow + 1$, oxidized $)$ and reduced $( 0 \rightarrow - 3$, and reduced $)$, white phosphorus acts as an oxidizing agent and a reducing agent, B is correct;
C. In reaction 1, white phosphorus reacts with excess sodium hydroxide solution to form $\mathrm { NaH } _ { 2 } \mathrm { PO } _ { 2 }$ and phosphine.
Reacting with sodium hydroxide solution, $\mathrm { NaH } _ { 2 } \mathrm { PO } _ { 2 }$ cannot ionize hydrogen ions in the solution and belongs to the positive salt, C is wrong;
D. In reaction 2, $\mathrm { NaH } _ { 2 } \mathrm { PO } _ { 2 }$ (ionized $\mathrm { H } _ { 2 } \mathrm { PO } _ { 2 } ^ { - }$) reacts with dilute sulfuric acid (providing $\mathrm { H } ^ { + }$) to form $\mathrm { H } _ { 3 } \mathrm { PO } _ { 2 }$ (a weak monoprotic acid, $\mathrm { H } ^ { + }$).
difficult to ionize) with the ionic equation $\mathrm { H } _ { 2 } \mathrm { PO } _ { 2 } ^ { - } + \mathrm { H } ^ { + } = \mathrm { H } _ { 3 } \mathrm { PO } _ { 2 }$ , D is correct;