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In a reversible reaction K_c> K_p and $∆ H$ = + 40 K.Cal. The product will be obtained in less amount on

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Decreasing pressure & increasing temperature

Increasing pressure & decreasing temperature

Increasing both pressure & temperature

Decreasing both pressure & temperature

answer is B.

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Detailed Solution

Given K_C > K_P. This indicates Δn = negative n_P(g) < n_R(g)

Thus forward reaction is favoured at high pressure

$\large \because \Delta H = +40K cals$ (endothermic), High temperature favours forward reaction.

Product obtained is less when

Temperature is low

Pressure is low

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