For the reaction 2NOCl_{(g)} rightleftharpoon | vedclass

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(A) The relationship between $K_p$ and $K_c$ is given by the formula: $K_p = K_c(RT)^{\Delta n}$.For the reaction $2NOCl_{(g)} \rightleftharpoons 2NO_

Vedclass · Accepted Answer

$3.8 	imes 10^{-4}$

Vedclass · Answer

(A) The relationship between $K_p$ and $K_c$ is given by the formula: $K_p = K_c(RT)^{\Delta n}$.For the reaction $2NOCl_{(g)} ightleftharpoons 2NO_{(g)} + Cl_{2_{(g)}}$,the change in the number of moles of gaseous products and reactants is $\Delta n = (2 + 1) - 2 = 1$.Given $K_p = 0.033$,$T = 1060 \ K$,and $R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$.Substituting these values into the equation: $0.033 = K_c 	imes (0.082 	imes 1060)^1$.$K_c = \frac{0.033}{0.082 	imes 1060} = \frac{0.033}{86.92} \approx 3.796 	imes 10^{-4} \ mol \ L^{-1}$.Rounding to two significant figures,we get $K_c \approx 3.8 	imes 10^{-4} \ mol \ L^{-1}$.

For the reaction $2NOCl_{(g)} \rightleftharpoons 2NO_{(g)} + Cl_{2_{(g)}}$ at $1060 \ K$,the equilibrium constant $K_p$ is $0.033 \ atm$. Find the value of $K_c$. (Given $R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$)

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