InterviewSolution
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InterviewSolution
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The reaction , `CO(g) + 3 H_(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 is` 3*90`. |
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Answer» ` K_(c) = ([ CH_(4)][H_(2)O])/([CO][H_(2)]^(3))` ` :. 3* 90 = ([CH_(4)][H_(2)O])/([CO][H_(2)]^(3)) " " ("Molar conc = No. of moles because volume of flask = 1 L")` `[CH_(4)] = 0*0585 M = 5*85 xx 10^(-2) M` |
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