In a reversible reaction,K_c K_p and Delta | vedclass

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(A) For a reversible reaction,the relationship between $K_p$ and $K_c$ is $K_p = K_c(RT)^{\Delta n}$.Given $K_c > K_p$,it implies that $(RT)^{\Delta n

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Decrease in pressure and temperature

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(A) For a reversible reaction,the relationship between $K_p$ and $K_c$ is $K_p = K_c(RT)^{\Delta n}$.Given $K_c > K_p$,it implies that $(RT)^{\Delta n} $A$ negative $\Delta n$ indicates that the number of moles of gaseous products is less than the number of moles of gaseous reactants.According to Le Chatelier's principle,an increase in pressure shifts the equilibrium towards the side with fewer moles,favoring the product.Conversely,a decrease in pressure will shift the equilibrium towards the reactant side,resulting in less product formation.Given $\Delta H = +40 \, kcal$,the reaction is endothermic.According to Le Chatelier's principle,an increase in temperature favors the endothermic direction (product side).Conversely,a decrease in temperature will shift the equilibrium towards the reactant side,resulting in less product formation.Therefore,a decrease in both pressure and temperature will result in less product formation.

In a reversible reaction,$K_c > K_p$ and $\Delta H = +40 \, kcal$. The formation of product will be less at:

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