A mixture of \(SO_2\) and \(O_2\) at atmosphere in the ratio of \(2:1\) is passed through a catalyst at \(117^0C\) for attainment of equilibrium. The exit gas is found to contain \(87\%\) \(SO_3\) by volume. Calculate \(K_p\) for the reaction.
\(SO_2(g) + \frac{1}{2}O_2 ⇌ SO_3(g)\)
Answer & explanation
Correct answer: option 3
The correct answer is option 3. \(48.24\, \ atm^{-1/2}\).
The given reaction is
\(SO_2(g) + \frac{1}{2}O_2 ⇌ SO_3(g)\)
The volume of \(SO_2\) and \(O_2\) at equilibrium \(= 100 - 87 = 13\, \ mL\)
Volume of \(SO_2\) \(= \frac{2}{3} \times 13 = 8.67\, \ mL\)
Volume of oxygen\(= \frac{1}{3} \times 13 = 4.33 mL \)
\(p_{SO_3} = \frac{87}{100} \times 1 = 0.87 atm\)
\(p_{SO_2} = \frac{8.67}{100} \times 1 = 0.0867 atm\)
\(p_{O_2} = \frac{4.33}{100} \times 1 = 0.0433atm\)
\(K_p = \frac{p_{SO_3}}{p_{SO_2}\times (p_{O_2})^{1/2}}\)
or, \(K_p = \frac{0.87}{0.0867\times (0.0433)^{1/2}}\)
or, \(K_p = \frac{0.87}{0.0867 \times 0.208}\)
or, \(K_p = 48.24\, \ atm^{-1/2}\)