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. \(SO_2(g) + \frac{1}{2}O_2 ⇌ SO_3(g)\) |
\(24.48\, \ atm^{-1/2}\) \(84.42\, \ atm^{-1/2}\) \(48.24\, \ atm^{-1/2}\) \(82.44\, \ atm^{-1/2}\) |
\(48.24\, \ atm^{-1/2}\) |
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, \(K_p = 48.24\, \ atm^{-1/2}\) |