Question

$5$ moles of $S{O}_2$ and $5$ moles of $O_2$ are allowed to react. At equilibrium, it was found that $60\%$ of $S{O}_2$ is used up. If the pressure of the equilibrium mixture is one atmosphere, the partial pressure of $O_2$ is:

• A. $0.52$ atm
• B. $0.21$ atm
• C. $0.41$ atm
• D. $0.82$ atm

Answer 1

Answer: (C)

$0.41$ atm

$S{O}_2+\frac{1}{2}{O}_2\rightleftharpoons S{O}_3$

$5$ $5$ $0$ Initial

$2$ $3.5$ $3$ Eqm

$60$% $S{O}_2$ reacted

$\therefore \left[S{O}_2\right]=5-5\times 0.6$

$=2$ mole

$\therefore \left[O_2\right]=5-5\times 0.3$

$=3.5$ mole

$\therefore \left[S{O}_3\right]=5\times 0.6$

$=3$ mole

Now total moles at equilibrium

$=2+3.5+3$

$=8.5$ mole

Hence mole fraction of $O_2$

$=\frac{3.5}{8.5}$

Total pressure at eqm $=1atm$

Hence $P_{O_2}=\left(\frac{3.5}{8.5}\right)\times 1=0.41atm$