InterviewSolution
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InterviewSolution
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The reaction, `CO(g)+3H_(2)(g) hArr CH_(4)(g)+H_(2)O(g)` is at equilibrium at `1300 K` in a `1 L` flask. It also contains `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 the given temperature us `3.90`. |
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Answer» Correct Answer - `5.85 xx 10^(-3)` Let the concentration of methane at equilibrium be x. `CO_(g) + 3H_(2(g)) harr CH_(4(g)) + H_(2)O_((g))` At equalibrium `(0.3)/(1)= 0.3M (0.1)/(1) = 0.1M , x , (0.02)/(1) = 0.02M` It is given that `K_(c) = 3.90` Therefore, `([CH_(4(g))][H_(2)O_((g))])/([CO_((g))][H_(2(g))]) K_(c)` `rArr (x xx 0.02)/(0.3 xx (0.1)^(3)) = 3.90` `rArr x = (3.90 xx 0.3 xx (0.1)^(3))/(0.02)` `= (0.00117)/(0.02)` `= 0.0585 M` `= 5.85 xx 10^(-2) M` Hence, the concentration of `CH_(4)` at equilibrium is `5.85 xx 10^(-2) M`. |
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