Question

Consider the following reaction at equilibrium:

C(s) + H2O (g) <--> CO (g) + H2 (g)

Which of the following conditions will increase the partial pressure of CO?
A) decreasing the partial pressure of H2O (g)
B) removing H2O (g) from the system
C) decreasing the volume of the reaction vessel
D) decreasing the pressure in the reaction vessel
E) increasing the amount of carbon in the system

Answer 1

Answer on Question #76717 – Chemistry – Other

Task:

Consider the following reaction at equilibrium:

$$\mathrm{C}(s) + \mathrm{H}_2\mathrm{O}(g) \rightleftharpoons \mathrm{CO}(g) + \mathrm{H}_2(g)$$

Which of the following conditions will increase the partial pressure of CO?

A) decreasing the partial pressure of $\mathrm{H}_2\mathrm{O}(g)$;

B) removing $\mathrm{H}_2\mathrm{O}(g)$ from the system;

C) decreasing the volume of the reaction vessel;

D) decreasing the pressure in the reaction vessel;

E) increasing the amount of carbon in the system.

Solution:

A) Wrong. By Le Chateliers principle equilibrium will adjust to oppose the change, produce more $\mathrm{H}_2\mathrm{O}$, reducing CO.

B) Wrong. Same explanation as for A.

C) Wrong. Decreasing volume equivalent to increasing pressure, by Le Chatelier this is opposed by system decreasing its volume, ie equilibrium shifts to left, more $\mathrm{H}_2\mathrm{O}$ and less CO (and $\mathrm{H}_2$).

D) Correct. By Le Chatelier equilibrium shifts to right, to produce a larger volume of gas so opposing the pressure decrease.

E) Wrong. The C in the solid state (has a constant and small vapour pressure) is irrelevant to the equilibrium in the gaseous phase.

Answer: D) decreasing the pressure in the reaction vessel.