For the reaction N_2O_{4_{(g)}} rightleftharp | vedclass

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(A) The relationship between $K_p$ and $K_c$ is given by the equation: $K_p = K_c(RT)^{\Delta n_g}$.For the reaction $N_2O_{4_{(g)}} \rightleftharpoon

Vedclass · Accepted Answer

$5.73 	imes 10^{-3} \ M$

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(A) The relationship between $K_p$ and $K_c$ is given by the equation: $K_p = K_c(RT)^{\Delta n_g}$.For the reaction $N_2O_{4_{(g)}} ightleftharpoons 2NO_{2_{(g)}}$,the change in the number of moles of gaseous products and reactants is $\Delta n_g = 2 - 1 = 1$.Given $K_p = 0.14 \ atm$,$R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$,and $T = 298 \ K$.Substituting these values into the equation: $0.14 = K_c 	imes (0.082 	imes 298)^1$.$K_c = \frac{0.14}{0.082 	imes 298}$.$K_c = \frac{0.14}{24.436} \approx 5.73 	imes 10^{-3} \ M$.

For the reaction $N_2O_{4_{(g)}} \rightleftharpoons 2NO_{2_{(g)}}$ at $298 \ K$,the equilibrium constant $K_p$ is $0.14 \ atm$. Calculate the value of $K_c$. $(R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1})$

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