Question

In which of the following ${K_P}$ is less than ${K_C}$?
(A) ${{N}_{2}}{{O}_{4}}(g)\rightleftharpoons 2N{{O}_{2}}(g)$
(B) $2HI(g)\rightleftharpoons {{H}_{2}}(g)+{{I}_{2}}(g)$
(C) $2S{{O}_{2}}(g)+{{O}_{2}}(g)\rightleftharpoons 2S{{O}_{3}}(g)$
(D) $PC{{l}_{5}}(g)\text{ }\rightleftharpoons PC{{l}_{3}}(g)+C{{l}_{2}}(g)$

Answer 1

Hint: For an ideal gaseous mixture, ${K_P}$ and ${K_C}$ are the equilibrium constants which represent the equilibrium state of a reaction using two different quantities. ${K_P}$ is the equilibrium constant defined when the equilibrium conditions are given in the form of partial pressure while ${K_C}$ is the equilibrium constant defined when the equilibrium conditions are given in the form of molarity.

Complete Step by Step Solution:
The relationship between ${K_P}$ and ${K_C}$ can be mathematically defined as follows:
${K_P} = {K_C}{(RT)^{\Delta {n_g}}}$
Where, R is the universal gas constant, T is the temperature and $\Delta {n_g}$ is the difference between gaseous moles in products and reactants.

For a given reaction:
If, $\Delta {n_g} > 0$ i.e., the number of gaseous products is more than that of gaseous reactants which implies that ${K_P} > {K_C}$.
If, $\Delta {n_g} < 0$ i.e., the number of gaseous products are less than that of gaseous reactants which implies that ${K_P} < {K_C}$.
If, $\Delta {n_g} = 0$ i.e., the number of gaseous products are equal to gaseous reactants which implies that ${K_P} = {K_C}$.
Thus, we can calculate the value of $\Delta {n_g}$ and the relationship between ${K_P}$ and ${K_C}$ for the given reactions as follows:

Reaction given in option (A):
${{N}_{2}}{{O}_{4}}(g)\rightleftharpoons 2N{{O}_{2}}(g)$
$\Delta {n_g} = 2 - 1 = 1$
$ \Rightarrow {K_P} > {K_C}$

Reaction given in option (B):
$2HI(g)\rightleftharpoons {{H}_{2}}(g)+{{I}_{2}}(g)$
$\Delta {n_g} = 2 - 2 = 0$
$ \Rightarrow {K_P} = {K_C}$

Reaction given in option (C):
$2S{{O}_{2}}(g)+{{O}_{2}}(g)\rightleftharpoons 2S{{O}_{3}}(g)$
$\Delta {n_g} = 2 - 3 = - 1$
$ \Rightarrow {K_P} < {K_C}$

Reaction given in option (D):
$PC{{l}_{5}}(g)\text{ }\rightleftharpoons PC{{l}_{3}}(g)+C{{l}_{2}}(g)$
$\Delta {n_g} = 2 - 1 = 1$
$ \Rightarrow {K_P} > {K_C}$

Therefore, option (C) is the correct answer.

Note: It is important to note that we can also determine the change in volume within the reaction using ${K_P}$ and ${K_C}$ relation. If ${K_P} > {K_C}$, an increase in volume is observed, if ${K_P} < {K_C}$ then a decrease in volume is observed while for ${K_P} = {K_C}$, no change in volume is observed.