The standard equilibrium constant,K_p at 298 | vedclass

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Question

(C) The relationship between $K_p$ and $K_c$ is given by the formula: $K_p = K_c(RT)^{\Delta n_g}$.For the reaction $N_{2(g)} + 3H_{2(g)} \rightleftha

Vedclass · Accepted Answer

$3.56 	imes 10^8$

Vedclass · Answer

(C) The relationship between $K_p$ and $K_c$ is given by the formula: $K_p = K_c(RT)^{\Delta n_g}$.For the reaction $N_{2(g)} + 3H_{2(g)} ightleftharpoons 2NH_{3(g)}$,the change in the number of moles of gaseous species is $\Delta n_g = 2 - (1 + 3) = -2$.Given $K_p = 5.8 	imes 10^5$,$T = 298 \, K$,and $R = 0.08314 \, L \, bar \, K^{-1} \, mol^{-1}$.Rearranging the formula to solve for $K_c$: $K_c = K_p(RT)^{-\Delta n_g} = K_p(RT)^2$.Substituting the values: $K_c = (5.8 	imes 10^5) 	imes (0.08314 	imes 298)^2$.$K_c = (5.8 	imes 10^5) 	imes (24.77572)^2$.$K_c = (5.8 	imes 10^5) 	imes 613.836$.$K_c \approx 3.56 	imes 10^8$.

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