The equilibrium constants (K_C) of two reacti | vedclass

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(C) For the reactions:$1) \ H_2 + I_2 \rightleftharpoons 2 HI, K_{C_1} = 50$$2) \ N_2 + 3 H_2 \rightleftharpoons 2 NH_3, K_{C_2} = 1000$We need the eq

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$0.008$

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(C) For the reactions:$1) \ H_2 + I_2 ightleftharpoons 2 HI, K_{C_1} = 50$$2) \ N_2 + 3 H_2 ightleftharpoons 2 NH_3, K_{C_2} = 1000$We need the equilibrium constant for the reaction:$N_2 + 6 HI ightleftharpoons 2 NH_3 + 3 I_2$This reaction can be obtained by taking the second reaction and subtracting three times the first reaction:$(N_2 + 3 H_2 ightleftharpoons 2 NH_3) - 3 	imes (H_2 + I_2 ightleftharpoons 2 HI)$$= N_2 + 3 H_2 - 3 H_2 - 3 I_2 ightleftharpoons 2 NH_3 - 6 HI$Rearranging gives: $N_2 + 6 HI ightleftharpoons 2 NH_3 + 3 I_2$The equilibrium constant $K_{C_3}$ is given by:$K_{C_3} = \frac{K_{C_2}}{(K_{C_1})^3}$$K_{C_3} = \frac{1000}{(50)^3} = \frac{1000}{125000} = \frac{1}{125} = 0.008$

The equilibrium constants $(K_C)$ of two reactions $H_2 + I_2 \rightleftharpoons 2 HI$ and $N_2 + 3 H_2 \rightleftharpoons 2 NH_3$ are $50$ and $1000$,respectively. The equilibrium constant of the reaction $N_2 + 6 HI \rightleftharpoons 2 NH_3 + 3 I_2$ is closest to

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