For reaction : SO_2(~g)+\frac12 O_2(~g) leftharpoons SO_3(~g) K_p=2 \times 10^12 at 27^\circ C and 1 ~atm pressure. The K_c for the same reaction is \underline\hspace2cm \times 10^13 . (Nearest integer) (Given R=0.082 ~L ~atm ~K^-1 ~mol^-1 )

Chemical Equilibrium PYQ — JEE Main 2023 · 31 Jan · Shift 1 · Q24 with Solution

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Correct answer: 1

For the reaction $\mathrm{SO_2(g)}+\frac{1}{2}\mathrm{O_2(g)}\rightleftharpoons \mathrm{SO_3(g)}$ the relation between $K_p$ and $K_c$ is $K_p=K_c(RT)^{\Delta n}$ where $\Delta n = (\text{moles of gaseous products})-(\text{moles of gaseous reactants})$
Calculate $\Delta n$ : $\Delta n = 1-\left(1+\frac{1}{2}\right)=1-\frac{3}{2}=-\frac{1}{2}$
Substitute into the relation: $K_p=K_c(RT)^{-1/2}$ Hence, $K_c = K_p(RT)^{1/2}$
Given: $K_p=2\times 10^{12}, \quad R=0.082\,\text{L atm K}^{-1}\text{mol}^{-1}, \quad T=27^\circ C=300\,K$

First calculate $RT$ : $RT=0.082\times 300=24.6$

Then, $(RT)^{1/2}=\sqrt{24.6}\approx 4.96$
Now calculate $K_c$ : $K_c = 2\times 10^{12}\times 4.96 = 9.92\times 10^{12}$
Write in the asked form: $K_c \approx 0.992\times 10^{13}$

Nearest integer for the blank is: $1$

Therefore, $K_c = 1\times 10^{13}$ (approximately).