The reaction, \( CO_{(g)} \; + \; 3H_{2(g)} \; ⇌ \; CH_{4(g)} \; + \; H_{2}O_{(g)}\) at 1300K is at equilibrium in a 1L container. It has 0.30 mol of CO, 0.10 mol of \( H_{2}\) and 0.02 mol of \( H_{2}O\) and an unknown amount of \( CH_{4}\) in the flask. Determine the concentration of \( CH_{4}\) in the mixture. The equilibrium constant, \( K_{c}\) for the reaction at given temperature is 3.90.

Asked by Pragya Singh | 1 year ago | 89

Let the concentration of \( CH_{4}\) at equilibrium be y.

\( CO_{(g)} \; + \; 3H_{2(g)} \; ⇌ \; CH_{4(g)} \; + \; H_{2}O_{(g)}\)

At equilibrium,

For \( CO – \dfrac{0.3}{1} \; = \; 0.3M\)

For \( H_{2} – \dfrac{0.1}{1} \; = \; 0.1M\)

For \( H_{2}O – \dfrac{0.02}{1} \; = 0.02M K_{c} = 3.90\)

Therefore,

\( \dfrac{[CH_{4(g)}][H_{2}O_{(g)}]}{[CO_{(g)}][H_{2(g)}]^{3}} \;\)

\( =Kc \dfrac{y \times 0.02}{0.3 \times (0.1)^{3}} =3.9 y \; \)

\( = \; \dfrac{3.9 \times 0.3 \times (0.1)^{3}}{0.02}y\)

\( =\dfrac{0.00117}{0.02}y=0.0585M\)

\( y=5.85×10^{ −2} M\)

Therefore, the concentration of \( CH_{4}\) at equilibrium is \( 5.85 \times 10^{-2}M\)

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