At 673 K,for the reaction N_{2(g)} + 3H_{2(g | vedclass

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(A) 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

$1.64 	imes 10^{-4}$

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(A) 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 $\Delta n_g = n_p - n_r = 2 - (1 + 3) = 2 - 4 = -2$.Given $K_c = 0.50$,$R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$,and $T = 673 \ K$.Substituting these values into the formula: $K_p = 0.50 	imes (0.082 	imes 673)^{-2}$.$K_p = 0.50 	imes (55.186)^{-2}$.$K_p = 0.50 	imes \frac{1}{(55.186)^2} = 0.50 	imes \frac{1}{3045.49} \approx 1.64 	imes 10^{-4}$.

At $673 \ K$,for the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,the equilibrium constant $K_c = 0.50$. Calculate $K_p$ at this temperature. (Given $R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$)

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