In which of the following cases is K_p less t | vedclass

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(B) The relationship between $K_p$ and $K_c$ is given by the equation: $K_p = K_c(RT)^{\Delta n_g}$.For $K_p < K_c$,the value of $\Delta n_g$ must be

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$2SO_2 + O_2 \rightleftharpoons 2SO_3$

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(B) The relationship between $K_p$ and $K_c$ is given by the equation: $K_p = K_c(RT)^{\Delta n_g}$.For $K_p $\Delta n_g$ is calculated as (sum of stoichiometric coefficients of gaseous products) - (sum of stoichiometric coefficients of gaseous reactants).For option $A$: $\Delta n_g = 2 - (1 + 1) = 0$,so $K_p = K_c$.For option $B$: $\Delta n_g = 2 - (2 + 1) = -1$,so $K_p = K_c(RT)^{-1} = K_c/RT$,which means $K_p For option $C$: $\Delta n_g = 2 - (1 + 1) = 0$,so $K_p = K_c$.For option $D$: $\Delta n_g = (1 + 1) - 1 = 1$,so $K_p = K_c(RT)^1$,which means $K_p > K_c$.Therefore,the correct option is $B$.

In which of the following cases is $K_p$ less than $K_c$?

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