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Question

# Variation of equilibrium constant K for the reaction 2A (g)+B (g)⇌C (g)+2D (g) is plotted against absolute temperature T in figure as: ln K vs 1/T;

A
The forward reaction is exothermic
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B
The forward reaction is endothermic
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C
The slope of line is equal to ΔH0R
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D
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Solution

## The correct options are A The forward reaction is exothermic C The slope of line is equal to −ΔH0R D Adding ′A′ favours forward reaction From the given graph, equilibrium constant increases with increase in 1T, i.e, with decrease in temperature. Thus the reaction is exothermic. We have the relation, ΔG∘=−RTlnK, where K is the equilibrium constant. Also, ΔG∘=ΔH∘−TΔS∘ Thus, −RTlnK=ΔH∘−TΔS∘ lnK=−ΔH∘RT+ΔS∘R From these relations we can conclude that the forward reaction is exothermic and the slope of the line is equal to −ΔH0R. Obviously, the addition of one of the reactant will favour the forward reaction as the reaction will proceed in a direction where 'A' is consumed.

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