# Question 13.6: Determining the Equilibrium Constant Kp Methane (CH4) reacts......

Determining the Equilibrium Constant $K_p$

Methane ($CH_4$) reacts with hydrogen sulfide to yield $H_2$ and carbon disulfide, a solvent used in manufacturing rayon and cellophane:

$\mathrm{CH}_4(g)+2 \mathrm{H}_2 \mathrm{~S}(g) \rightleftharpoons \mathrm{CS}_2(g)+4 \mathrm{H}_2(g)$

What is the value of $K_p$ at 1000 K if the partial pressures in an equilibrium mixture at 1000 K are 0.20 atm of $CH_4$, 0.25 atm of $H_2S$, 0.52 atm of $CS_2$, and 0.10 atm of $H_2$?

STRATEGY

Write the equilibrium equation by setting $K_p$ equal to the equilibrium-constant expression using partial pressures. Put the partial pressures of products in the numerator and the partial pressures of reactants in the denominator, with the pressure of each substance raised to the power of its coefficient in the balanced chemical equation. Then substitute the partial pressures into the equilibrium equation and solve for $K_p$.

Step-by-Step
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$K_{\mathrm{p}}=\frac{\left(P_{\mathrm{CS}_2}\right)\left(P_{\mathrm{H}_2}\right)^{4 \ \leftarrow Coefficient \ of \ H_2}}{\left(P_{\mathrm{CH}_4}\right)\left(P_{\mathrm{H}_2 \mathrm{~S}}\right)^{2 \ \leftarrow \ Coefficient of H_2S}} \\ K_{\mathrm{p}}=\frac{\left(P_{\mathrm{CS}_2}\right)\left(P_{\mathrm{H}_2}\right)^4}{\left(P_{\mathrm{CH}_4}\right)\left(P_{\mathrm{H}_2 \mathrm{~S}}\right)^2}=\frac{(0.52)(0.10)^4}{(0.20)(0.25)^2}=4.2 \times 10^{-3}$

Note that the partial pressures must be in units of atmospheres (not mm Hg) because the standard-state partial pressure for gases is 1 atm

Question: 13.16

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Question: 13.9

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