Question

For which of the following reactions $K_P=K_C$

A) $PC{{l}_{3}}_{(g)}+C{{l}_{2}}_{(g)}\rightleftharpoons PC{{l}_{5}}_{(g)}$

B) ${{H}_{2(g)}}+C{{l}_{2(g)}}\rightleftharpoons 2HC{{l}_{(g)}}$

C) ${{N}_{2(g)}}+3{{H}_{2(g)}}\rightleftharpoons 2N{{H}_{3}}$

D) $CaC{{O}_{3(s)}}\rightleftharpoons Ca{{O}_{(S)}}+C{{O}_{2(g)}}$

Answer 1

${{K}_{P}}={{K}_{C}}(RT)\Delta n$, is the relation by which $K_P$ and $K_C$ are related to each other.

The partial pressures of the gases in a closed system are known as ${{K}_{P}}$. It is an equilibrium constant which denotes partial pressure of gases. Therefore, $P$ in its notation form represents partial pressure.

Complete step-by-step answer:

${{K}_{C}}$ is also an equilibrium constant on any chemical reaction. It expresses the molar concentration of reagents present in reaction. Therefore $C$ in its notation form stands for concentration.

${{K}_{C}}$ and ${{K}_{P}}$ both are equilibrium constants, however one tells about molar concentration that is and one tells about partial pressure in a closed system.

They both are related to each other by following formula:

${{K}_{P}}={{K}_{C}}(RT)\Delta n$

Where, R is known as gas constant, T stands for Temperature and n is the change in no. of gaseous moles in the reaction

$K_C$ will be equal to $K_P$, when number of gas molecules on the product side is the same as the number of gas molecules on the reactant side,

$K_C$ will be less than $K_P$ when number of gas molecules on the product side are greater than the number of gas molecules on the reactant side

$K_C$ will be more than $K_P$ when number of gas molecules on the product side are less than the number of gas molecules on the reactant side

$\Delta n$ is equal to the difference between the number of moles of gaseous products and the number of moles of gaseous reactants in a balanced equation.

For the reaction in option A:

$PC{{l}_{3}}_{(g)}+C{{l}_{2}}_{(g)}\rightleftharpoons PC{{l}_{5}}_{(g)}$

$\Delta n=1-\left( 1+1 \right)=\text{ }-1$

$\therefore K_P ​< K_C$​

For the reaction in option B:

${{H}_{2(g)}}+C{{l}_{2(g)}}\rightleftharpoons 2HC{{l}_{(g)}}$

$\Delta n=2-\left( 1+1 \right)=\text{ 0}$

$\therefore K_P=K_C​$

For the reaction in option C:

${{N}_{2(g)}}+3{{H}_{2(g)}}\rightleftharpoons 2N{{H}_{3}}$

$\Delta n=2-\left( 3+1 \right)=\text{ -2}$

$\therefore K_P​ < K_C$​

For the reaction in option D:

$CaC{{O}_{3(s)}}\rightleftharpoons Ca{{O}_{(S)}}+C{{O}_{2(g)}}$

$\Delta n=( 1+ 1 )- 1 = 1$

$\therefore K_P > K_C$

Therefore the correct answer is option B.

Note: An equilibrium in which everything present in the mixture is in the same phase is known as homogeneous mixture. So, everything must be a gas to use $K_P$.

For $K_P=K_C$, reagent in reaction should be in gaseous form and number of moles of reactant should be equal to number of moles of product.