2SO_2}(g)+O

Question:
$2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})$

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

$\mathrm{P}_{\mathrm{SO}_{2}}=45 \quad \mathrm{kPa}$. The equilibrium constant

$\mathrm{K}_{\mathrm{P}}=$ _________ $\times 10^{-2} \cdot$ (Nearest integer)

Solution:

$2 \mathrm{SO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}=2 \mathrm{SO}_{3(\mathrm{~g})}$

$\mathrm{K}_{\mathrm{p}}=\frac{\left(\mathrm{pSO}_{3(\mathrm{~g})}\right)^{2}}{\mathrm{pSO} 2(\mathrm{~g})} \times \mathrm{pO}_{2(\mathrm{~g})}$

$=172.28 \times 10^{-5} \mathrm{~Pa}^{-1}$

= 172.28 atm

$=17228 \times 10^{-2} \mathrm{~atm}$

Ans is 17228

2SO_2}(g)+O_2(g)

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