(D) The relationship between $K_p$ and $K_c$ is given by the formula: $K_p = K_c(RT)^{\Delta n}$.Here,$\Delta n = (2 + 1) - 2 = 1$.Given $K_p = 4.157

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(D) The relationship between $K_p$ and $K_c$ is given by the formula: $K_p = K_c(RT)^{\Delta n}$.Here,$\Delta n = (2 + 1) - 2 = 1$.Given $K_p = 4.157 \times 10^{-4} \ bar$,$T = 1000 \ K$,and $R = 0.083 \ L \ bar \ K^{-1} \ mol^{-1}$.Substituting the values: $4.157 \times 10^{-4} = K_c(0.083 \times 1000)^1$.$K_c = \frac{4.157 \times 10^{-4}}{83} = 0.05008 \times 10^{-4} = 5.008 \times 10^{-6} \ mol \ L^{-1}$.Thus,the correct value is approximately $5.0 \times 10^{-6}$.

Class 11 Chemistry 6-1.Equilibrium (Chemical Equilibrium)Kp and Kc Relationship

Medium

At $1000 \ K$,if the equilibrium constant $K_p$ for the reaction $2 \ NOCl_{(g)} \rightleftharpoons 2 \ NO_{(g)} + Cl_{2(g)}$ is $4.157 \times 10^{-4} \ bar$,the $K_c$ (in $mol \ L^{-1}$) is $(R = 0.083 \ L \ bar \ K^{-1} \ mol^{-1})$

AP EAMCET 2018 All AP EAMCET

A
$4.16 \times 10^{-7}$
B
$4.16 \times 10^{-4}$
C
$50 \times 10^{-4}$
D
$5.0 \times 10^{-6}$

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At $1000 \ K$,if the equilibrium constant $K_p$ for the reaction $2 \ NOCl_{(g)} \rightleftharpoons 2 \ NO_{(g)} + Cl_{2(g)}$ is $4.157 \times 10^{-4} \ bar$,the $K_c$ (in $mol \ L^{-1}$) is $(R = 0.083 \ L \ bar \ K^{-1} \ mol^{-1})$

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