At T (K),K_{p} value for the reaction,2 AO_ | vedclass

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(D) The given reaction is: $2 \ AO_{2(g)} + O_{2(g)} ightleftharpoons 2 \ AO_{3(g)}$ with equilibrium constant $K_{p} = 4 	imes 10^{10}$.We need to

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$8 	imes 10^{15}$

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(D) The given reaction is: $2 \ AO_{2(g)} + O_{2(g)} ightleftharpoons 2 \ AO_{3(g)}$ with equilibrium constant $K_{p} = 4 	imes 10^{10}$.We need to find the equilibrium constant $K_{p}^{\prime}$ for the reaction: $3 \ AO_{2(g)} + \frac{3}{2} \ O_{2(g)} ightleftharpoons 3 \ AO_{3(g)}$.Notice that the second reaction is obtained by multiplying the first reaction by a factor of $n = \frac{3}{2}$.According to the properties of equilibrium constants,if a reaction is multiplied by a factor $n$,the new equilibrium constant $K_{p}^{\prime}$ is given by $K_{p}^{\prime} = (K_{p})^n$.Substituting the values: $K_{p}^{\prime} = (4 	imes 10^{10})^{3/2}$.$K_{p}^{\prime} = (\sqrt{4 	imes 10^{10}})^3 = (2 	imes 10^5)^3$.$K_{p}^{\prime} = 8 	imes 10^{15}$.

At $T \ (K)$,$K_{p}$ value for the reaction,$2 \ AO_{2(g)} + O_{2(g)} \rightleftharpoons 2 \ AO_{3(g)}$ is $4 \times 10^{10}$. What is the $K_{p}^{\prime}$ value for $3 \ AO_{2(g)} + \frac{3}{2} \ O_{2(g)} \rightleftharpoons 3 \ AO_{3(g)}$ at $T \ (K)$?

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