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The reaction CO(g) + 3H(2)(g) hArr CH(4...

The reaction `CO_(g) + 3H_(2)(g) hArr CH_(4)(g) + H_(2)O(g)`
is at equilibrium at 1300 K in a 1L flask. It also contain 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.

Text Solution

Verified by Experts

`K_(c )=([CH_(4)][H_(2)O])/([CO][H_(2)]^(3))`
`:. 3.90=([CH_(4)][0.02])/((0.30)(0.10)^(3))`
(Molar concentration =Number of moles because volume of flask `=1 L`)
or `[CH_(4)]=0.0585 M=5.89xx10^(-2) M`
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