For a reaction,if K_p K_c,then the forward | vedclass

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(C) The relationship between $K_p$ and $K_c$ is given by the equation: $K_p = K_c(RT)^{\Delta n_g}$.Given $K_p > K_c$,it implies that $(RT)^{\Delta n_

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Low pressure

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(C) The relationship between $K_p$ and $K_c$ is given by the equation: $K_p = K_c(RT)^{\Delta n_g}$.Given $K_p > K_c$,it implies that $(RT)^{\Delta n_g} > 1$.Since $T > 15 \ K$ and $R$ is a positive constant,this condition is satisfied when $\Delta n_g > 0$.$\Delta n_g$ is the change in the number of moles of gaseous products and reactants,defined as $\Delta n_g = n_p - n_r$.When $\Delta n_g > 0$,the number of moles of gaseous products is greater than the number of moles of gaseous reactants.According to Le Chatelier's principle,for a reaction with $\Delta n_g > 0$,decreasing the pressure favors the forward reaction (the side with more moles of gas).Therefore,the forward reaction is favored by low pressure.

For a reaction,if $K_p > K_c$,then the forward reaction is favored by ..... $(T > 15 \ K)$

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