For the reaction A_{(g)} rightleftharpoons 2 | vedclass

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(C) The equilibrium constant $K_c$ is given by the ratio of the forward rate constant $K_f$ to the backward rate constant $K_b$: $K_c = \frac{K_f}{K_b

Vedclass · Accepted Answer

$0.033$

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(C) The equilibrium constant $K_c$ is given by the ratio of the forward rate constant $K_f$ to the backward rate constant $K_b$: $K_c = \frac{K_f}{K_b}$ Given that $K_b = 2500 \ K_f$,we have: $K_c = \frac{K_f}{2500 \ K_f} = \frac{1}{2500} = 4 	imes 10^{-4}$ For the reaction $A_{(g)} ightleftharpoons 2 B_{(g)}$,the change in the number of moles of gaseous species is $\Delta n_g = 2 - 1 = 1$. The relation between $K_p$ and $K_c$ is: $K_p = K_c(RT)^{\Delta n_g}$ Substituting the values: $K_p = \frac{1}{2500} 	imes (0.0831 	imes 1000)^1$ $K_p = \frac{83.1}{2500} = 0.03324 \approx 0.033$

For the reaction $A_{(g)} \rightleftharpoons 2 B_{(g)}$,the backward reaction rate constant is higher than the forward reaction rate constant by a factor of $2500$,at $1000 \ K$. [Given : $R = 0.0831 \ L \ atm \ mol^{-1} \ K^{-1}$] $K_p$ for the reaction at $1000 \ K$ is

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