For the reaction SO_{3(g)} rightleftharpoons | vedclass

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(D) Given reaction: $SO_{3(g)} ightleftharpoons SO_{2(g)} + 1/2 O_{2(g)}$ with $K_1 = 4.9 	imes 10^{-2}$.Target reaction: $2SO_{2(g)} + O_{2(g)}

Vedclass · Accepted Answer

$416$

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(D) Given reaction: $SO_{3(g)} ightleftharpoons SO_{2(g)} + 1/2 O_{2(g)}$ with $K_1 = 4.9 	imes 10^{-2}$.Target reaction: $2SO_{2(g)} + O_{2(g)} ightleftharpoons 2SO_{3(g)}$.To obtain the target reaction,we reverse the given reaction and multiply the coefficients by $2$.If a reaction is reversed,the equilibrium constant becomes $1/K$.If a reaction is multiplied by a factor $n$,the equilibrium constant becomes $K^n$.Therefore,$K_2 = (1/K_1)^2 = 1 / (4.9 	imes 10^{-2})^2$.$K_2 = 1 / (24.01 	imes 10^{-4}) \approx 1 / 0.002401 \approx 416.49$.Rounding to the nearest provided option,$K_c = 416$.

For the reaction $SO_{3(g)} \rightleftharpoons SO_{2(g)} + 1/2 O_{2(g)}$,the equilibrium constant $K_c = 4.9 \times 10^{-2}$. What is the value of $K_c$ for the reaction $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$?

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