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(C) The equilibrium constant $K_c$ for the reaction $2 \ NO_{(g)} \rightleftharpoons N_{2_{(g)}} + O_{2_{(g)}}$ is given by:$K_c = \frac{[N_2][O_2]}{[

Vedclass · Accepted Answer

$0.717$

Vedclass · Answer

(C) The equilibrium constant $K_c$ for the reaction $2 \ NO_{(g)} ightleftharpoons N_{2_{(g)}} + O_{2_{(g)}}$ is given by:$K_c = \frac{[N_2][O_2]}{[NO]^2} = \frac{(3.0 	imes 10^{-3})(4.2 	imes 10^{-3})}{(2.8 	imes 10^{-3})^2} = \frac{12.6 	imes 10^{-6}}{7.84 	imes 10^{-6}} \approx 1.607$For the reaction starting with $0.1 \ M$ of $NO_{(g)}$:$2 \ NO_{(g)} ightleftharpoons N_{2_{(g)}} + O_{2_{(g)}}$Initial: $0.1 \quad 0 \quad 0$At equilibrium: $0.1(1 - \alpha) \quad 0.05 \alpha \quad 0.05 \alpha$$K_c = \frac{(0.05 \alpha)(0.05 \alpha)}{(0.1(1 - \alpha))^2} = \frac{0.0025 \alpha^2}{0.01(1 - \alpha)^2} = 0.25 \left( \frac{\alpha}{1 - \alpha} ight)^2$$1.607 = 0.25 \left( \frac{\alpha}{1 - \alpha} ight)^2$$\left( \frac{\alpha}{1 - \alpha} ight)^2 = \frac{1.607}{0.25} = 6.428$$\frac{\alpha}{1 - \alpha} = \sqrt{6.428} \approx 2.535$$\alpha = 2.535 - 2.535 \alpha$$3.535 \alpha = 2.535$$\alpha = \frac{2.535}{3.535} \approx 0.717$

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