Question

Given, ${{C}_{(s)}}\,+\,{{H}_{2}}{{O}_{(g)}}\,\underset{{}}{\overset{{}}{\longleftrightarrow}}\,C{{O}_{(g)}}\,+\,{{H}_{2(g)}}$
The above equilibrium when subjected to pressure:
(A)- Remains unaffected
(B)- Proceeds in the backward direction
(C)- Proceeds in the forward direction
(D)- None of the above

Answer 1

To answer this question we will have to apply Le Chatelier’s Principle and then we will see the effect of pressure on this reaction. On increasing the pressure, the reaction proceeds in forward reaction when the moles of products are less than reactants.

Complete answer:
Let’s look at the solution of the given question:
First we will understand the Le Chatelier’s Principle.
According to this principle, “if any of the reaction condition is changed (T, P, concentration) changed at equilibrium then the reaction will proceed in the direction where the effect of this change is countered.”
Effect of pressure on the equilibrium:
On increasing pressure the equilibrium shifts to the side where the number of moles is less than the low volume side.
On decreasing pressure the equilibrium shifts to the side where the number of moles is more that is towards the high volume side.
Now, we have seen that the effect of pressure is visible when there is a difference in the number of moles. In the given reaction the total number of moles on the reactant side is greater than the total number of moles on the product side.
So, in increasing pressure the equilibrium will shift towards the backward direction.
Hence, the answer for the given question is option (B).

Note:
Students can get confused while calculating the number of moles. Only those moles are counted which are in gaseous phase. This means that we ignore the moles of pure solids and liquids. This is because the change in concentration of a pure solid or a pure liquid is very negligible.