Chemical and Ionic Equilibrium 1 Question 47
49. The equilibrium constant $K_{p}$ of the reaction,
$$ 2 \mathrm{SO}{2}(g)+\mathrm{O}{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) $$
is $900 \mathrm{~atm}$ at $800 \mathrm{~K}$. A mixture containing $\mathrm{SO}{3}$ and $\mathrm{O}{2}$ having initial pressure of 1 and $2 \mathrm{~atm}$ respectively is heated at constant volume to equilibrate. Calculate the partial pressure of each gas at $800 \mathrm{~K}$.
$(1989,3 \mathrm{M})$
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Solution:
$$ \begin{array}{lccc} & 2 \mathrm{SO}{2}(g)+\mathrm{O}{2}(g) & \rightleftharpoons \mathrm{SO}{3}(g) \ \text { Initial } p{i}: & 0 & 2 & 1 \ \text { Equilibrium } p_{i}: 2 p & 2+p & & 1-2 p \end{array} $$
$$ K_{p}=900=\frac{(1-2 p)^{2}}{(2+p)(2 p)^{2}} \quad[\text { Ignoring } p \text { in comparison to 2] } $$
$$ p=\frac{1}{87} \text { atm } $$
Partial pressure of $\mathrm{SO}_{2}=2 p=\frac{2}{87}$ atm
$$ \text { Partial pressure of } \mathrm{O}_{2}=2+p=2+\frac{1}{87}=\frac{175}{87} \mathrm{~atm} $$
Partial pressure of $\mathrm{SO}_{3}=1-2 p=1-2\left(\frac{1}{87}\right)=\frac{85}{87} \mathrm{~atm}$
