Question

For the equilibrium
$CO+H_{2}O\rightleftharpoons C{O}_2+H_2$
The relation between $K_P$ and $K_C$ at 25°C and at 100°C are:

• A. $K_P=K_C,K_P=K_C$
• B. $K_P=K_C(RT{)}^{-1},K_P=K_C$
• C. $K_P=K_C\left(RT\right),K_P=K_C\left(RT\right)$
• D. $K_P=K_C\left(RT\right),K_P=K_C$

Answer 1

The correct option is D $K_P=K_C\left(RT\right),K_P=K_C$

We know,
$K_p=K_c(RT{)}^{\Delta n_g}$
Where $\Delta n_g$ is change in number of moles of gaseous molecules.

The given reaction is,
$CO+H_{2}O\rightleftharpoons C{O}_2+H_2$

At $25{}^{o}C$, $H_{2}O$ will be in liquid form
So,
$K_p=K_c(RT{)}^{2-1}$
$\Rightarrow K_p=K_c\left(RT\right)$

At $100{}^{o}C$, $H_{2}O$ will be in gaseous form,
$K_p=K_c(RT{)}^{2-2}$
$\Rightarrow K_p=K_c(RT{)}^0=K_c$