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A mixture of 1.57 mol of N(2), 1.92 mol ...

A mixture of `1.57` mol of `N_(2), 1.92` mol of `H_(2)` and `8.13` mol of `NH_(3)` is introduced into a `20 L` reaction vessel at `500 K`. At this temperature, the equilibrium constant, `K_(c)` for the reaction `N_(2)(g) + 3H_(2)(g) hArr 2NH_(3)(g) ` is `1.7 xx 10^(2)`. Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?

Text Solution

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`[N_(2)] = 1.57/20, [H_(2)] = 1.92/20, [NH_(3)] = 8.17/20`
`K_(c) = [NH_(3)]/([N_(2)][H_(2)]^(3))`
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