Question

What is the unit of ${{\text{K}}_{\text{p}}}$ for the reaction?
(A) atm
(B) ${\text{at}}{{\text{m}}^{{\text{ - 2}}}}$
(C) ${\text{at}}{{\text{m}}^{\text{2}}}$
(D) ${\text{at}}{{\text{m}}^{{\text{ - 1}}}}$

Answer 1

To solve this question we must know about the equilibrium constant for a chemical reaction. The equilibrium constant for a chemical reaction can be defined as a relation between the corresponding product and the given reactants when the chemical equation reaches an equilibrium.

Complete step by step solution:
The equilibrium constant in terms of concentration ${{\text{K}}_{\text{C}}}$ is defined as the ratio of the concentration of products to the concentration of the reactants.
The equilibrium constant in terms of partial pressure ${{\text{K}}_{\text{p}}}$ is defined as the ratio of the partial pressure of products to the partial pressure of the reactants.
For the reaction,

$${K_C} = \dfrac{{{{\left[ C \right]}^c}{{\left[ D \right]}^d}}}{{{{\left[ A \right]}^a}{{\left[ B \right]}^b}}} \\\ {K_p} = \dfrac{{{{\left[ {pC} \right]}^c}{{\left[ {pD} \right]}^d}}}{{{{\left[ {pA} \right]}^a}{{\left[ {pB} \right]}^b}}} \\\$$

So for the reaction given in question ${{\text{K}}_{\text{p}}}$ can be written as,

$${K_p} = \dfrac{{{{\left[ {pC{H_4}} \right]}^1}{{\left[ {p{H_2}S} \right]}^2}}}{{{{\left[ {pC{S_2}} \right]}^1}{{\left[ {{H_2}} \right]}^4}}} \\\ {K_p} = \dfrac{{{{\left[ {atm} \right]}^1}{{\left[ {atm} \right]}^2}}}{{{{\left[ {atm} \right]}^1}{{\left[ {atm} \right]}^4}}} = {\left[ {atm} \right]^{ - 2}} \\\$$

So the unit of ${{\text{K}}_{\text{p}}}$ is ${\left[ {{\text{atm}}} \right]^{{\text{ - 2}}}}$.

Hence the correct option is (B).

Note: Using the ideal gas equation i.e. pV = nRT, ${{\text{K}}_{\text{p}}}$ can be written in terms of ${{\text{K}}_{\text{C}}}$.
${K_p} = {K_C}{(RT)^{(c + d) - (a + b)}}$, where ${{\Delta n = (c + d) - (a + b)}}$, the number of moles of product subtracted the number of moles reactants in gaseous phase in the balanced chemical equation.