At 527 , ^oC,the reaction given below has K_c | vedclass

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(C) For the reaction $NH_{3(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)}$,$K_c = 4$.The target reaction is $N_{2(g)} + 3H_{2(g)}

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$\left( \frac{1}{4 	imes 800R} ight)^2$

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(C) For the reaction $NH_{3(g)} ightleftharpoons \frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)}$,$K_c = 4$.The target reaction is $N_{2(g)} + 3H_{2(g)} ightleftharpoons 2NH_{3(g)}$.This is the reverse of the first reaction multiplied by $2$. Therefore,the equilibrium constant for the target reaction is $K_c' = (1/K_c)^2 = (1/4)^2 = 1/16$.Using the relation $K_p = K_c(RT)^{\Delta n}$,where $\Delta n = 2 - (1 + 3) = -2$ and $T = 527 + 273 = 800 \, K$:$K_p = (1/16) 	imes (R 	imes 800)^{-2} = (1/4)^2 	imes (800R)^{-2} = \left( \frac{1}{4 	imes 800R} ight)^2$.

At $527 \, ^oC$,the reaction given below has $K_c = 4$:
$NH_{3(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)}$
What is the $K_P$ for the following reaction?
$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$

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At $527 \, ^oC$,the reaction given below has $K_c = 4$:$NH_{3(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)}$What is the $K_P$ for the following reaction?$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$