Question

S : $K_c$, $K_p$ and $K_x$ are the equilibrium constants of a reaction in terms of concentration, pressure and mole fraction respectively.
E : $K_c$ and $K_p$ do not depend on equilibrium pressure but $K_x$ depends upon equilibrium pressure if $\Delta n\ne 0$.

• A. S is correct but E is wrong
• B. S is wrong but E is correct
• C. Both S and E are correct and E is correct explanation of S
• D. Both S and E are correct but E is not correct explanation of S

Answer 1

The correct option is D Both S and E are correct but E is not correct explanation of S

Option (D) is correct.
$K_c$, $K_p$ and $K_x$ are the equilibrium constants of a reaction.
For gaseous reactions, the concentration terms are replaced by partial pressures because at a given temperature the partial pressure of gases is proportional to its concentration.
The reciprocal of $K_c$ is also a constant and it will be the equilibrium constant for the backward reaction.Equilibrium constant varies with temperature. But at a particular temperature, its value remains constant. Its value increases with temperature.It is independent of the initial concentration.
Consider a reaction : $aA\rightleftharpoons bB$
$K_c=\frac{{\left[B\right]}^b}{{\left[A\right]}^a}$; $K_p=\frac{{\left(P_B^{\prime }\right)}^b}{{\left(P_A^{\prime }\right)}^a}$
$\because$ $K_p=\frac{{\left(P_B^{\prime }\right)}^b}{{\left(P_A^{\prime }\right)}^a}=\frac{{(P_T\cdot X_B)}^b}{{(P_T\cdot X_A)}^a}=\frac{{\left(X_B\right)}^b}{{\left(X_A\right)}^a}\times {\left(P_T\right)}^{b-a}$
$K_p=K_x\times {\left(P_T\right)}^{b-a}$
or $K_x=K_p\times {\left(P_T\right)}^{a-b}=K_p{\left(P_T\right)}^{-\Delta n}$
Thus, $K_x$ equilibrium constant in terms of mole fraction depends upon $P$ if $\Delta n\ne 0$.