3H\(_{2(g)}\)+ N\(_{2(g)}\)⇔ 2NH\(_{3(g)}\) ; H= -ve...
3H\(_{2(g)}\)+ N\(_{2(g)}\)⇔ 2NH\(_{3(g)}\) ; H= -ve
In the reaction above, lowering of temperature will
have no effect on the equilibrium position
increase the rate of both the forward and reverse reactions equally
favour the forward reaction
favour the reverse reaction
Correct answer is C
N\(_{2(g)}\) + 3H\(_{2(g)}\) ⇔ 2NH\(_{3(g)}\) In the Haber process, the forward reaction is exothermic and the backward reaction is endothermic. If the temperature is decreased, the yield from the exothermic direction is increased i.e. the forward reaction is increased.
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