Question

For a given reversible reaction at a fixed temperature, equilibrium constant $K_p$ and $K_c$ are related by:

Answer 1

$K_C$ is defined by molar concentration.

$K_P$ is defined by partial pressure of gases.

For a given reaction,

$aA+bB\rightleftharpoons cC+dD$

Rate of forward reaction

$Rat{e}_f=K_f\left[A{]}^a\right[B{]}^b⟶\left(1\right)$

Rate of backward reaction,

$Rat{e}_b=K_b\left[C{]}^c\right[D{]}^d⟶\left(2\right)$

At equilibrium, $\left(1\right)=\left(2\right)$

$\therefore K_f\left[A{]}^a\right[B{]}^b=K_b\left[C{]}^c\right[D{]}^d$

$\therefore K_C=\frac{K_f}{K_b}=\frac{\left[C{]}^c\right[D{]}^d}{\left[A{]}^a\right[B{]}^b}⟶\left(I\right)$

Now, $K_p=\frac{\left[P_C{]}^c\right[P_D{]}^d}{\left[P_A^a\right]\left[P_B^b\right]}⟶\left(3\right)$

We know, $PV=nRT$

$P=\frac{n}{V}RT$ where $C=\frac{n}{V}$

$\therefore P_A=C_{A}RT,P_B=C_{B}RT$ $⟶\left(H\right)$

$P_B=C_{B}RT,P_D=C_{D}RT$

Using $\left(H\right)$ in equation $3$

$\Rightarrow K_P=\frac{\left[C_{C}RT{]}^c\right[C_{D}RT{]}^d}{\left[C_{A}RT{]}^a\right[C_{B}RT{]}^b}=K_C(RT{)}^{(c+d)-(a+b)}$

$\Rightarrow K_P=K_C(RT{)}^{\Delta ng}$ where $\Delta ng=c+d-a-b$