Question

The reaction in which the yield of the product will increase in pressure is
A.${{H}_{2}}(g)+{{I}_{2}}(g)\rightleftharpoons 2HI(g)$
B.${{H}_{2}}O(g)+CO(g)\rightleftharpoons C{{O}_{2}}(g)+{{H}_{2}}(g)$
C.${{H}_{2}}O(g)+C(s)\rightleftharpoons CO(g)+{{H}_{2}}(g)$
D.$CO(g)+3{{H}_{2}}(g)\rightleftharpoons C{{H}_{4}}(g)+{{H}_{2}}O(g)$

Answer 1

Hint: When in a chemical system pressure increases then the system equilibrium will shift towards a direction with a lesser number of moles so that the effect of change can be neutralized. So, to solve this problem we will have to find the $\Delta {{n}_{g}}$ difference in the number of moles in gaseous products and reactants.

Formula Used:$\Delta {{n}_{g}}={{n}_{p}}(g)-{{n}_{r}}(g)$
${{n}_{p}}(g)=$number of moles present is gaseous product molecules
${{n}_{r}}(g)=$number of moles present in gaseous reactant molecules

Complete answer:According to Le-Chatelier’s principle increasing pressure will cause the equilibrium shift towards a direction with fewer moles of gas to neutralize the change and hence increase the pressure.
Here we have four reactions at equilibrium, let us determine $\Delta {{n}_{g}}$the value of each given reaction.

In (A) ${{H}_{2}}(g)+{{I}_{2}}(g)\rightleftharpoons 2HI(g)$
${{n}_{p}}(g)=2$ , ${{n}_{r}}(g)=(1+1)=2$
$\therefore \Delta {{n}_{g}}=(2-2)=0$
As the difference between the number of moles is zero, hence no effect of pressure at equilibrium.

In (B) ${{H}_{2}}O(g)+CO(g)\rightleftharpoons C{{O}_{2}}(g)+{{H}_{2}}(g)$
${{n}_{p}}(g)=(1+1)=2$,${{n}_{r}}(g)=(1+1)=2$
$\therefore \Delta {{n}_{g}}=(2-2)=0$
Here zero value $\Delta {{n}_{g}}$indicates that no effect in pressure at equilibrium in (B)

In (C) ${{H}_{2}}O(g)+C(s)\rightleftharpoons CO(g)+{{H}_{2}}(g)$
${{n}_{p}}(g)=(1+1)=2$,${{n}_{r}}(g)=1$as carbon is solid and has no effect on the equilibrium.
$\therefore \Delta {{n}_{g}}=(2-1)=1$
Here$\Delta {{n}_{g}}=1$ indicates the number of moles in the gaseous product is greater than the number of moles in gaseous reactants molecules. As no. of moles increases, the volume of the system also increases, and thereby pressure decreases. So, at high pressure, this equilibrium is not favorable.

In (D) $CO(g)+3{{H}_{2}}(g)\rightleftharpoons C{{H}_{4}}(g)+{{H}_{2}}O(g)$
${{n}_{p}}(g)=(1+1)=2$;${{n}_{r}}(g)=(1+3)=4$
$\therefore \Delta {{n}_{g}}=(2-4)=-2$
Hence $\Delta {{n}_{g}}=-2$indicates that the number of moles in gaseous products is lesser than the number of moles in gaseous reactants molecules. As no. of moles decreases, the volume of the system also decreases, and thereby pressure the increases. So, at high pressure this equilibrium is favorable.

Thus, option (D) is correct.

Note: Pure solids and liquids are omitted from the equilibrium constant expression as they do not change their effective concentration throughout the reactions to reach equilibrium. That’s why pure solids and liquids do not affect equilibrium.