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

##### Solution :-

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 $$5.85 \times 10^{-2}M$$

Answered by Abhisek | 1 year ago

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