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Equilibrium constant, `K_(c)` for the reaction, `N_(2(g))+3H_(2(g))hArr2NH_(3(g))`, at `500 K` is `0.061 litre^(2) "mole"^(-2)`. At a particular time, the analysis shows that composition of the reaction mixture is `3.00 mol litre^(-1)N_(2)`, `2.00 mol litre^(-1)H_(2)`, and `0.500 mol litre^(-1)NH_(3)`. Is the reaction at equilibrium? If not, in which direction does the reaction tend to proceed to reach equilibrium? |
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Answer» `Q_(c)=([NH_(3)]^(2))/([N_(2)][H_(2)]^(3))=((0.5)^(2))/((3.0)(2.0)^(3))=0.0104` As `Q_(c ) ne K_(c )`, reaction is not in equilibrium. As `Q_(c ) lt K_(c )` reaction will proceed in the forward direction. |
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