Question

Which of the following reactions proceed at low pressure
A.${{N}_{2}}+3{{H}_{2}}\rightleftharpoons 2N{{H}_{3}}$
B.${{H}_{2}}+{{I}_{2}}\rightleftharpoons 2HI$
C.$PC{{l}_{5}}\rightleftharpoons PC{{l}_{3}}+C{{l}_{2}}$
D.${{N}_{2}}+{{O}_{2}}\rightleftharpoons 2NO$

Answer 1

Hint: When a system is in equilibrium and undergoes a change in pressure, the overall system will shift to adjust to the change and reestablish a new equilibrium. When a reaction proceeds at low pressure, the equilibrium will shift towards the side of the reaction with higher moles of the constituents present in the system.

Complete answer:According to Le-chatelier’s principle, if any system is disturbed by pressure, concentration, or temperature, then the position of equilibrium shifts to neutralize the change to reestablish a new equilibrium.

We know the pressure is inversely proportional to the volume and the effects of volume change are opposite to the effects of change in pressure. When a reaction proceeds at low pressure the equilibrium shifts towards the side of the reaction with higher moles.
Here we have four reactions in equilibrium. Let us check these reactions one by one where the number of moles on the product side is greater than on the reactant side.

In the case of (A),${{N}_{2}}+3{{H}_{2}}\rightleftharpoons 2N{{H}_{3}}$ reactants are ${{N}_{2}}\And {{H}_{2}}$and the product is $N{{H}_{3}}$
Total number of moles at the reactant side$=(1+3)=4$
And total no. of moles present on the product side$=2$
Hence the no. of moles on the product side is less than on the reactant side and can’t proceed at low pressure.

In (B)${{H}_{2}}+{{I}_{2}}\to 2HI$
Total number of moles at the reactant (${{H}_{2}}\And {{I}_{2}}$) side$=(1+1)=2$
And total no. of moles on the product ($HI$) in the forward direction$=2$
Hence the no. of moles on the product side is equal to the reactant side and can’t proceed at low pressure.

In (C), $PC{{l}_{5}}\rightleftharpoons PC{{l}_{3}}+C{{l}_{2}}$
Total number of moles at the reactant ($PC{{l}_{5}}$) side$=1$
And total no. of moles on the product ($PC{{l}_{3}}\And C{{l}_{2}}$) in the forward side$=(1+1)=2$
Hence the no. of moles on the product side is greater than on the reactant side and can proceed at low pressure.

In the case of (D), ${{N}_{2}}+{{O}_{2}}\rightleftharpoons 2NO$
Total number of moles at the reactant (${{N}_{2}}\And {{O}_{2}}$) side$=(1+1)=2$
And total no. of moles on the product ($NO$) in the forward side$=2$
Hence the no. of moles on the product side is equal to the reactant side and can’t proceed at low pressure.



Thus, option (C) is correct.

Note: When the reaction proceeds under a higher pressure condition, Le-Chatelier's principle will shift the equilibrium toward lower moles of constituents. At high pressure, there is a decrease in space for the moving of molecules, and thereby the number of collisions between molecules increases.