Respuesta :
Answer:
To solve this problem, we can use the expression for the equilibrium constant (\(K_p\)) and the reaction quotient (\(Q_p\)).
The given reaction is: \[2H_2S(g) \rightleftharpoons 2H_2(g) + S_2(g)\]
The equilibrium constant (\(K_p\)) expression for this reaction is: \[K_p = \frac{{[H_2]^2 \cdot [S_2]}}{{[H_2S]^2}}\]
[tex]Given that \(K_p = 0.012\) and the equilibrium mixture contains \(0.250 \, \text{atm}\) of \(H_2(g)\), we need to find the initial pressure of \(H_2S\) (\([H_2S]_0\) or \(P_{H_2S_0}\)).[/tex]
[tex]The expression for \(Q_p\) is the same as \(K_p\) at equilibrium. At equilibrium, we have: \[Q_p = \frac{{[H_2]^2 \cdot [S_2]}}{{[H_2S]^2}}\][/tex]
[tex]Given that \([H_2]_{\text{equilibrium}} = 0.250 \, \text{atm}\), \([S_2]_{\text{equilibrium}}\) is not given, and \([H_2S]_{\text{equilibrium}}\) is also not given.[/tex]
[tex]However, since we are starting with pure \(H_2S\), initially, \([H_2S]_0 = P_{H_2S_0}\) (initial pressure of \(H_2S\)).[/tex]
[tex]Now, set up the expression for \(Q_p\) using the given information and solve for \(P_{H_2S_0}\):\[Q_p = \frac{{(0.250)^2 \cdot [S_2]_{\text{equilibrium}}}}{{P_{H_2S_0}^2}}\][/tex]
[tex]We know that \(Q_p = K_p\) at equilibrium, so set \(Q_p\) equal to \(K_p\) and solve for \(P_{H_2S_0}\):[/tex]
[tex]\[K_p = \frac{{(0.250)^2 \cdot [S_2]_{\text{equilibrium}}}}{{P_{H_2S_0}^2}}\]\[0.012 = \frac{{(0.250)^2 \cdot [S_2]_{\text{equilibrium}}}}{{P_{H_2S_0}^2}}\][/tex]
[tex]Now, solve for \(P_{H_2S_0}\):[/tex]
[tex]\[P_{H_2S_0}^2 = \frac{{(0.250)^2 \cdot [S_2]_{\text{equilibrium}}}}{{0.012}}\][/tex]
[tex]\[P_{H_2S_0} = \sqrt{\frac{{(0.250)^2 \cdot [S_2]_{\text{equilibrium}}}}{{0.012}}}\][/tex]
Without the value for [tex]\([S_2]_{\text{equilibrium}}\)[/tex], we cannot calculate the exact value for [tex]\(P_{H_2S_0}\)[/tex]. If you have additional information about the equilibrium concentration of[tex]\(S_2\)[/tex], you can substitute that into the equation to find [tex]\(P_{H_2S_0}\).[/tex]
Explanation:
