Question

In which of the following equilibria, ${{K}_{P}}\ne {{K}_{c}}$?
(a) $N{{O}_{2}}(g)+S{{O}_{2}}(g)\rightleftharpoons NO(g)+S{{O}_{3}}(g)$
(b) $2HI(g)\rightleftharpoons {{H}_{2}}(g)+{{I}_{2}}(g)$
(c) $2NO(g)\rightleftharpoons {{N}_{2}}(g)+{{O}_{2}}(g)$
(d) $2C(s)+{{O}_{2}}(g)\rightleftharpoons 2CO(g)$

Answer 1

Hint:. At equilibrium the rate of forward reaction is always equal to rate of backward reaction and both ${{K}_{P}}$ and ${{K}_{c}}$ are the equilibrium constant but when the reactants and products are in gaseous phase, then state of equilibrium is expressed in terms of partial pressures and equilibrium constant is expressed as ${{K}_{P}}$.${{K}_{P}}$ and ${{K}_{c}}$ are related as; \[{{K}_{P}}={{K}_{c}}{{(RT)}^{\Delta n}}\] and \[{{K}_{P}}={{K}_{c}}\] if the $\Delta n$=0, then RT is also 0. Now you can easily find the equilibria in which ${{K}_{P}}\ne {{K}_{c}}$.

Complete step by step answer:
First of all let’s discuss what is ${{K}_{P}}$ and ${{K}_{c}}$. By the term ${{K}_{c}}$, we mean the equilibrium constant. When a reaction is at equilibrium , the rate of the forward reaction and the rate of backward reaction is always equal to each other through the reactions even at the equilibrium and the equilibrium constant tells us about the relationship between reactants and the products.
Consider general reaction at equilibrium as:
\[aA +b B\rightleftharpoons cC+ d D\]
Then, the equilibrium constant of the reaction is as:
${{K}_{c}}={{{C}^{c}}{{D}^{d}}}{{{A}^{a}}{{B}^{b}}}$

Here, C and D are the products , A and B are the reactants and a, b , c and d are the number of moles of reactants and products respectively and${{K}_{c}}$ is the equilibrium constant.
But when the reactants and the products both are in the gaseous phase , then the equilibrium constant is expressed in form of p[atrial pressures and is represented as ${{K}_{P}}$.
Consider the general reaction in gaseous phase at equilibrium as:
 \[a{{A}_{(g)}}+b{{B}_{(g)}}\rightleftharpoons c{{C}_{(g)}}+d{{D}_{(g)}}\]

Then the equilibrium constant is:
${{K}_{P}}={P_{C}^{c}P_{D}^{d}}{P_{A}^{a}P_{B}^{b}}$
Here,${{P}_{A}}$, ${{P}_{A}}$,${{P}_{A}}$ and ${{P}_{A}}$ are the partials pressure of the gases A, B,C and D and ${{K}_{P}}$ is the equilibrium constant.
${{K}_{P}}$ and ${{K}_{c}}$ are related to each other as:
\[{{K}_{P}}={{K}_{c}}{{(RT)}^{\Delta n}}\]
Here R is the gas constant and T is the temperature and $\Delta n $ is the no of moles of gaseous products- no of moles gaseous reactants.
And \[{{K}_{P}}={{K}_{c}}\] if the $\Delta n$=0, then RT is also 0.

Now considering the statement:
(a) $N{{O}_{2}}(g)+S{{O}_{2}}(g)\rightleftharpoons NO(g)+S{{O}_{3}}(g)$
\[{{K}_{P}}={{K}_{c}}{{(RT)}^{\Delta n}}\]
Here, $\Delta n$= 2-2
 =0
So, \[{{K}_{P}}={{K}_{c}}\]

(b) $2HI(g)\rightleftharpoons {{H}_{2}}(g)+{{I}_{2}}(g)$
\[{{K}_{P}}={{K}_{c}}{{(RT)}^{\Delta n}}\]
Here, $\Delta n$= 2-2
 =0
So, \[{{K}_{P}}={{K}_{c}}\]

(c) $2NO(g)\rightleftharpoons {{N}_{2}}(g)+{{O}_{2}}(g)$
\[{{K}_{P}}={{K}_{c}}{{(RT)}^{\Delta n}}\]
Here, $\Delta n$= 2-2
 =0
So, \[{{K}_{P}}={{K}_{c}}\]

(d) $2C(s)+{{O}_{2}}(g)\rightleftharpoons 2CO(g)$
\[{{K}_{P}}={{K}_{c}}{{(RT)}^{\Delta n}}\]
Here, $\Delta n$= 2-1
 =1
So, ${{K}_{P}}\ne {{K}_{c}}$.
So, the correct answer is “Option D”.

Note: If equilibrium constant (${{K}_{c}}$) is greater than one, then the equilibrium shifts towards the products and if the equilibrium constant is smaller than 1 , then the equilibrium constant shifts towards the reactants.