Question

In the reaction : $2S{O}_3\left(g\right)\leftrightharpoons 2S{O}_2\left(g\right)+O_2\left(g\right),S{O}_3\left(g\right)$ is 50% dissociated at ${27}^{\circ }C$ , when the equilibrium pressure is 0.5 atm. Hence, partial pressure of $S{O}_3\left(g\right)$ at equilibrium is:

• A. 0.5 atm
• B. 0.3 atm
• C. 0.2 atm
• D. 0.1 atm

Answer 1

The correct option is C 0.2 atm

$2S{O}_3\leftrightharpoons 2S{O}_2+O_2$

2 0 0 ----- initial

2-2x 2x x --------- final

Let initially $2$ moles of $S{O}_3$ are present.

Initial number of moles of $S{O}_2$ and $O_2$ are $0$ each. $2x$ moles of $S{O}_3$ dissociate to reach equilibrium

At equilibrium, total number of moles are $2+x$ and the number of moles of sulphur trioxide are $2-2x.50$% dissociation means out of $2$ moles of $S{O}_3,1$ mole dissociates. Hence $2x=1$ or $x=0.5$

The mole fraction of sulphur trioxide $X_{S{O}_3}=\frac{2-2x}{2+x}=\frac{2-1}{2+0.5}=\frac{1}{2.5}=0.4$

Partial pressure of $P_{S{O}_3}=0.5\times X_{S{O}_3}=0.2atm$

Option C is correct.