For the equilibrium 2SO_{2(g)} + O_{2(g)} rig | vedclass

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(A) The equilibrium constant $K_p$ for the reaction $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$ is given by:$K_p = \frac{(P_{SO_3})^2}{(P_{S

Vedclass · Accepted Answer

$0.4$

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(A) The equilibrium constant $K_p$ for the reaction $2SO_{2(g)} + O_{2(g)} ightleftharpoons 2SO_{3(g)}$ is given by:$K_p = \frac{(P_{SO_3})^2}{(P_{SO_2})^2 	imes P_{O_2}}$Substituting the initial values:$K_p = \frac{(0.331)^2}{(0.662)^2 	imes 0.101} = \frac{(0.331)^2}{(2 	imes 0.331)^2 	imes 0.101} = \frac{1}{4 	imes 0.101} = \frac{1}{0.404} \approx 2.475$When the equilibrium partial pressures of $SO_2$ and $SO_3$ are equal $(P_{SO_2} = P_{SO_3})$,the expression becomes:$K_p = \frac{(P_{SO_3})^2}{(P_{SO_2})^2 	imes P_{O_2}'} = \frac{1}{P_{O_2}'}$Equating the two $K_p$ values:$\frac{1}{0.404} = \frac{1}{P_{O_2}'}$Therefore,$P_{O_2}' = 0.404 \ atm \approx 0.4 \ atm$.

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