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(A) The reaction is: $SO_2(g) + NO_2(g) \rightleftharpoons SO_3(g) + NO(g)$Initial concentrations (moles in $1 \ L$): $[SO_2] = 1 \ M, [NO_2] = 1 \ M,

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$1.6, 0.4$

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(A) The reaction is: $SO_2(g) + NO_2(g) \rightleftharpoons SO_3(g) + NO(g)$Initial concentrations (moles in $1 \ L$): $[SO_2] = 1 \ M, [NO_2] = 1 \ M, [SO_3] = 1 \ M, [NO] = 1 \ M$Let $x$ be the change in concentration at equilibrium:Equilibrium concentrations: $[SO_2] = (1-x) \ M, [NO_2] = (1-x) \ M, [SO_3] = (1+x) \ M, [NO] = (1+x) \ M$$K_C = \frac{[SO_3][NO]}{[SO_2][NO_2]} = \frac{(1+x)(1+x)}{(1-x)(1-x)} = \frac{(1+x)^2}{(1-x)^2} = 16$Taking the square root on both sides: $\frac{1+x}{1-x} = 4$$1+x = 4 - 4x \implies 5x = 3 \implies x = 0.6$Equilibrium concentrations:$[NO] = 1 + x = 1 + 0.6 = 1.6 \ M$$[NO_2] = 1 - x = 1 - 0.6 = 0.4 \ M$Thus,the equilibrium concentrations of $NO$ and $NO_2$ are $1.6 \ M$ and $0.4 \ M$ respectively.

At a certain temperature,the equilibrium constant $(K_C)$ is $16$ for the reaction:
$SO_2(g) + NO_2(g) \rightleftharpoons SO_3(g) + NO(g)$
If we take one mole of each of the four gases in a $1 \ L$ container,what would be the equilibrium concentration of $NO$ and $NO_2$ respectively?

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At a certain temperature,the equilibrium constant $(K_C)$ is $16$ for the reaction:$SO_2(g) + NO_2(g) \rightleftharpoons SO_3(g) + NO(g)$If we take one mole of each of the four gases in a $1 \ L$ container,what would be the equilibrium concentration of $NO$ and $NO_2$ respectively?