Find out the value of K_C for the following r | vedclass

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(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 $2NOCl_{(g)} \rightleftharpoons 2N

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$2 	imes 10^{11} \ mol \ L^{-1}$

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(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 $2NOCl_{(g)} ightleftharpoons 2NO_{(g)} + Cl_{2(g)}$,the change in the number of moles of gaseous products and reactants is $\Delta n_g = (2 + 1) - 2 = 1$.Given values are $K_P = 8 	imes 10^{12} \ atm$,$T = 500 \ K$,and $R = 0.08 \ L \ atm \ mol^{-1} \ K^{-1}$.Substituting these values into the formula:$8 	imes 10^{12} = K_C 	imes (0.08 	imes 500)^1$$8 	imes 10^{12} = K_C 	imes 40$$K_C = \frac{8 	imes 10^{12}}{40} = 0.2 	imes 10^{12} = 2 	imes 10^{11} \ mol \ L^{-1}$.Thus,the correct option is $C$.

Find out the value of $K_C$ for the following reaction from the value of $K_P$:
$2NOCl_{(g)} \rightleftharpoons 2NO_{(g)} + Cl_{2(g)}$
Given: $K_P = 8 \times 10^{12} \ atm$ at $500 \ K$,use $R = 0.08 \ L \ atm \ mol^{-1} \ K^{-1}$.

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Find out the value of $K_C$ for the following reaction from the value of $K_P$:$2NOCl_{(g)} \rightleftharpoons 2NO_{(g)} + Cl_{2(g)}$Given: $K_P = 8 \times 10^{12} \ atm$ at $500 \ K$,use $R = 0.08 \ L \ atm \ mol^{-1} \ K^{-1}$.