Question

For which one of the following reactions ${K_p} = {K_c}$
A.${N_2} + 3{H_2} \rightleftarrows 2N{H_3}$
B.${N_2} + {O_2} \rightleftarrows 2NO$
C.$PC{l_5} \rightleftarrows PC{l_3} + C{l_2}$
D.$2S{O_3} \rightleftarrows 2S{O_2} + {O_2}$

Answer 1

We have to know that ${K_p}$ and ${K_c}$ are directly related to each other. We can write the relationship between ${K_p}$ and ${K_c}$ as ${K_p} = {K_c}{\left( {RT} \right)^{\Delta n}}$. We can say that partial pressures of gases in a closed system is ${K_p}$, an equilibrium constant that represents partial pressure of gases.

Complete answer:
We have to know that ${K_c}$ is an equilibrium constant that is used to express the molarity of reagents found in the solution. The c is ${K_c}$ indicates concentration.
We can write the relationship between ${K_p}$ and ${K_c}$ as ${K_p} = {K_c}{\left( {RT} \right)^{\Delta n}}$.
Here,
The equilibrium constant for partial pressure of gas is ${K_p}$.
The equilibrium constant for molarity of reagents is ${K_c}$.
The gas constant is R.
Temperature is T.
The change in quantity of gaseous moles in the reaction is n.
When the amount of gas molecules on the product side is equal to the amount of gas molecules in the reactant side, the value of ${K_p} = {K_c}$.
When the amount of gas molecules on the product side is more than the amount of gas molecules in the reactant side, the value of ${K_c}$ would be less than ${K_p}$.
When the amount of gas molecules on the product side is less than the amount of gas molecules in the reactant side, the value of ${K_c}$ would be higher than ${K_p}$.
For reaction A,
${N_2} + 3{H_2} \rightleftarrows 2N{H_3}$
The value of $\Delta {n_g}$ is calculated as,
$\Delta {n_g} = 2 - 4 = 2$
The value of $\Delta {n_g}$ is $- 2$. So, ${K_p} \ne {K_c}$. Option (A) is incorrect.
For reaction B,
${N_2} + {O_2} \rightleftarrows 2NO$
The value of $\Delta {n_g}$ is calculated as,
$\Delta {n_g} = 2 - 2 = 0$
The value of $\Delta {n_g}$ is $0$. So, ${K_p} = {K_c}$. Option (B) is correct.
For reaction C,
$PC{l_5} \rightleftarrows PC{l_3} + C{l_2}$
The value of $\Delta {n_g}$ is calculated as,
$\Delta {n_g} = 2 - 1 = 1$
The value of $\Delta {n_g}$ is $0$. So, ${K_p} \ne {K_c}$. Option (C) is incorrect.
For reaction D,
$2S{O_3} \rightleftarrows 2S{O_2} + {O_2}$
The value of $\Delta {n_g}$ is calculated as,
$\Delta {n_g} = 3 - 2 = 1$
The value of $\Delta {n_g}$ is $1$. So, ${K_p} \ne {K_c}$. Option (D) is incorrect.
So, we can conclude that for the reaction (B), the value of ${K_p} = {K_c}$.

Option (B) is correct.

Note:
We have to know that in a homogeneous mixture, all the mixtures are present in the same phase. We have to know that when we represent equilibrium for gases, we use ${K_p}$. When the number of moles of reactant would be equal to the number of moles of the product being the same, the ${K_p} = {K_c}$.