(20 12? As we all know, CO2 can produce green fuel methanol. CO2 (g) +3H2 (g)? CH3OH(g)+H2O(g)△H =- 187.4 kj？ Health officer

(1) When catalyst I is used, the conversion of carbon dioxide is 0. 18, and the change of carbon dioxide concentration is 1.00mol? L-1× 0.18 = 0.18mol/l, then the average reaction rate of carbon dioxide in 10h is: v (CO2) = 0.18mol/l1. (L？ H)- 1, then v(H2)=3v(CO2)=0.054mol? (L？ h)- 1，

So the answer is: 0.054;

(2)A, before and after the reaction, the mass of the gas is constant, and the volume of the container is fixed, so the density of the mixed gas is always constant, so when the density of the gas in the container does not change, the reaction may not reach the equilibrium state, so A is correct;

B, argon is charged to increase the pressure, but the concentration of each component in the reaction system remains unchanged, and the chemical balance remains unchanged, so the CO2 conversion rate remains unchanged, so B is wrong;

C, the catalyst does not affect the chemical equilibrium, so under the above reaction conditions, the equilibrium conversion rate of CO2 is equal to that of catalyst I, so c is wrong;

D, under the above reaction conditions, the reaction rate of catalyst II is higher than that of catalyst I, indicating that the catalytic efficiency of catalyst II is higher than that of catalyst I, so d is correct;

So, choose AD;

(3)CO2 (g) +3H2 (g)? Methanol (g) +H2O (g)

Initial concentration (mol/L) 1.00 1.60? 0 0

Changing concentration (mol/L) 0.20? 0.60 0.20? 0.20

Equilibrium concentration (mol/L) 0.80? 1.0? 0.20? 0.20

k = c(CH3OH)×c(H2O)c(CO2)×C3(H2)= 0.20mol/L×0.20mol/L 0.80mol/L×( 1.0mol/L)3 = 0.050 L2？ Mol -2,

So the answer is: 0.050? L2？ mol-2；

(4) Because the reaction is exothermic, the chemical equilibrium moves in the opposite direction as the temperature rises, the reactant concentration decreases, the product concentration increases, and the chemical equilibrium constant increases, so the equilibrium constant k (400℃) < k (300℃).

So the answer is:

(5)CO2 (g) +3H2 (g)? CH3OH(g)+H2O(g)△H =- 187.4 kj？ mol- 1①

2h 2(g)+O2(g)= 2H2O(g)△H =-242.8 kj？ mol- 1②

According to Gass Law: ②×3-①×2, we can get: 2ch3oh (g)+3o2 (g) = 2co2 (g)+4h2o (g); △H=(-242.8kJ？ mol- 1)×3-(- 187.4kJ？ mol- 1)×2=-353.6kJ？ Mol-1

So the answer is: -353.6.