For the reaction SnO_{2(s)} + 2H_{2(g)} right | vedclass

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Question

(A) For an ideal gas,mole $\%$ is equivalent to volume $\%$.Given that the mixture contains $45 \% \, H_2$ by volume,the mole fraction of $H_2$ is $0.

Vedclass · Accepted Answer

$1.49$

Vedclass · Answer

(A) For an ideal gas,mole $\%$ is equivalent to volume $\%$.Given that the mixture contains $45 \% \, H_2$ by volume,the mole fraction of $H_2$ is $0.45$.Therefore,the mole fraction of $H_2O$ is $1 - 0.45 = 0.55$.The equilibrium constant $K_p$ for the reaction is given by the expression $K_p = \frac{(P_{H_2O})^2}{(P_{H_2})^2}$.Since $P_i = x_i 	imes P_{total}$,the $P_{total}$ terms cancel out in the ratio:$K_p = \frac{(x_{H_2O})^2}{(x_{H_2})^2} = \frac{(0.55)^2}{(0.45)^2} = \frac{0.3025}{0.2025} \approx 1.49$.

For the reaction $SnO_{2(s)} + 2H_{2(g)} \rightleftharpoons 2H_2O_{(g)} + Sn_{(l)}$,calculate $K_p$ at $900 \, K$,where the equilibrium steam-hydrogen mixture was $45 \% \, H_2$ by volume.

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