Question

In which of the following equilibrium, ${{K}_{c}}$ and ${{K}_{p}}$ are not equal?
(A) $2{{C}_{(s)}}+{{O}_{2(g)}}\rightleftharpoons 2C{{O}_{(g)}}$
(B) $2N{{O}_{(g)}}\rightleftharpoons {{N}_{2(g)}}+{{O}_{2(g)}}$
(C) $S{{O}_{2(g)}}+N{{O}_{2(g)}}\rightleftharpoons S{{O}_{3(g)}}+N{{O}_{(g)}}$
(D) ${{H}_{2(g)}}+{{I}_{2(g)}}\rightleftharpoons 2H{{I}_{(g)}}$

Answer 1

Recollect what is ${{K}_{c}}$ and ${{K}_{p}}$. Find out the relationship between ${{K}_{c}}$ and ${{K}_{p}}$. Then accordingly substitute the values from the chemical reaction and find out the answer.

Complete answer:
-${{K}_{c}}$ and ${{K}_{p}}$ are both equilibrium constants. ${{K}_{c}}$ is measured in terms of molar concentration of reactants and products whereas, ${{K}_{p}}$ is measured in terms of partial pressures of reactants and products.
We know that, $~{{\text{K}}_{\text{c}}}\text{=}\dfrac{\left[ \text{products} \right]}{\left[ \text{reactants} \right]}$ and also, ${{\text{K}}_{\text{p}}}\text{=}\dfrac{\text{partial pressure of products}}{\text{partial pressure of reactants}}$
From ideal gas equation we have,
PV=nRT
$P=\dfrac{n}{V}RT$
Concentration, $C=\dfrac{n}{V}$. Substituting this in above equation, we get,
$P=CRT$
So, when we equate ${{K}_{c}}$ and ${{K}_{p}}$, we obtain the relation,
${{K}_{p}}={{K}_{c}}\times {{(RT)}^{\Delta n}}$ where $\Delta n$ is the change in number of moles of gaseous products to number of moles of gaseous reactants. $\Delta n$ should be zero in order to get ${{K}_{p}}={{K}_{c}}$.
Now, let’s calculate $\Delta n$ for each reaction.
For option (A),
$2{{C}_{(s)}}+{{O}_{2(g)}}\rightleftharpoons 2C{{O}_{(g)}}$
$\Delta n=2-3=-1$
So, ${{K}_{p}}\ne {{K}_{c}}$
For option (B),
$2N{{O}_{(g)}}\rightleftharpoons {{N}_{2(g)}}+{{O}_{2(g)}}$
$\Delta n=2-2=0$
So, ${{K}_{p}}={{K}_{c}}$
For option (C),
$S{{O}_{2(g)}}+N{{O}_{2(g)}}\rightleftharpoons S{{O}_{3(g)}}+N{{O}_{(g)}}$
$\Delta n=2-2=0$
So, ${{K}_{p}}={{K}_{c}}$
For option (D),
${{H}_{2(g)}}+{{I}_{2(g)}}\rightleftharpoons 2H{{I}_{(g)}}$
$\Delta n=2-2=0$
So, ${{K}_{p}}={{K}_{c}}$
So, we can conclude that, in reaction,
(A) $2{{C}_{(s)}}+{{O}_{2(g)}}\rightleftharpoons 2C{{O}_{(g)}}$
${{K}_{c}}$ and ${{K}_{p}}$ are not equal because $\Delta n\ne 0$

Therefore, option (A) is the correct answer.

Note:
Remember the equilibrium concept thoroughly. Make a note of what is partial pressure of a gas and how to derive the relation between both the equilibrium constants. Also, see in which type of problems ${{K}_{c}}$ is used and where ${{K}_{p}}$ is used.