At 450 , K,K_{p} = 2.0 times 10^{10} , bar^{- | vedclass

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(A) For the given reaction,$\Delta n = 2 - 3 = -1$.Given: $T = 450 \, K$,$R = 0.0831 \, L \, bar \, K^{-1} \, mol^{-1}$,$K_{p} = 2.0 	imes 10^{10} \,

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$7.48 	imes 10^{11} \, M^{-1}$

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(A) For the given reaction,$\Delta n = 2 - 3 = -1$.Given: $T = 450 \, K$,$R = 0.0831 \, L \, bar \, K^{-1} \, mol^{-1}$,$K_{p} = 2.0 	imes 10^{10} \, bar^{-1}$.The relationship between $K_{p}$ and $K_{c}$ is $K_{p} = K_{c}(RT)^{\Delta n}$.Substituting the values: $2.0 	imes 10^{10} = K_{c} (0.0831 	imes 450)^{-1}$.$K_{c} = 2.0 	imes 10^{10} 	imes (0.0831 	imes 450) = 2.0 	imes 10^{10} 	imes 37.395$.$K_{c} = 74.79 	imes 10^{10} = 7.48 	imes 10^{11} \, M^{-1}$.

At $450 \, K$,$K_{p} = 2.0 \times 10^{10} \, bar^{-1}$ for the given reaction at equilibrium.
$2 \, SO_{2(g)} + O_{2(g)} \longleftrightarrow 2 \, SO_{3(g)}$
What is $K_{c}$ at this temperature?

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At $450 \, K$,$K_{p} = 2.0 \times 10^{10} \, bar^{-1}$ for the given reaction at equilibrium.$2 \, SO_{2(g)} + O_{2(g)} \longleftrightarrow 2 \, SO_{3(g)}$What is $K_{c}$ at this temperature?