$2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})$ In an equilibrium mixture, the partial pressures are $\mathrm{P}_{\mathrm{SO}_{3}}=43 \mathrm{kPa} ; \mathrm{P}_{\mathrm{O}_{2}}=530 \mathrm{~Pa}$ and $\mathrm{P}_{\mathrm{SO}_{2}}=45 \mathrm{kPa}$. The equilibrium constant $\mathrm{K}_{\mathrm{p}}=$ \_\_\_\_ $\times 10^{-2}$ (Nearest integer)

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Equilibrium

2 SO_{2} (g) + O_{2} (g) ⇌ 2 SO_{3} (g)

In an equilibrium mixture, the partial pressures are

P_{SO_{3}} = 43 kPa; P_{O_{2}} = 530 Pa

and

P_{SO_{2}} = 45 kPa

. The equilibrium constant

K_{p} =

____

\times 1 0^{- 2}

(Nearest integer)

Answer: 172

All the pressures must be in one unit, so

530 Pa = 0.53 kPa

. For

2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)

the equilibrium constant in terms of pressures is

K_{p} = \frac{p _{S O_{3}}^{2}}{p _{S O_{2}}^{2} p _{O_{2}}} = \frac{( 43 ) ^{2}}{( 45 ) ^{2} \times 0.53} = \frac{1849}{1073.25} = 1.723 kP a^{- 1}

To the nearest integer the equilibrium constant is

172 \times 1 0^{- 2}

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Integer answer