At 600 K,2 mol of NO are mixed with 1 mol | vedclass

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(C) The reaction is $2 \ NO_{(g)} + O_{2(g)} \rightleftarrows 2 \ NO_{2(g)}$.Initial moles: $NO = 2, O_2 = 1, NO_2 = 0$.At equilibrium,moles of $O_2 =

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$2$

Vedclass · Answer

(C) The reaction is $2 \ NO_{(g)} + O_{2(g)} ightleftarrows 2 \ NO_{2(g)}$.Initial moles: $NO = 2, O_2 = 1, NO_2 = 0$.At equilibrium,moles of $O_2 = 1 - x = 0.6$,so $x = 0.4$.Equilibrium moles: $NO = 2 - 2(0.4) = 1.2$,$O_2 = 0.6$,$NO_2 = 2(0.4) = 0.8$.Total moles at equilibrium = $1.2 + 0.6 + 0.8 = 2.6$.Mole fractions: $X_{NO} = \frac{1.2}{2.6}$,$X_{O_2} = \frac{0.6}{2.6}$,$X_{NO_2} = \frac{0.8}{2.6}$.Partial pressures $(P_{total} = 1 \ atm)$: $P_{NO} = \frac{1.2}{2.6} \ atm$,$P_{O_2} = \frac{0.6}{2.6} \ atm$,$P_{NO_2} = \frac{0.8}{2.6} \ atm$.$K_p = \frac{(P_{NO_2})^2}{(P_{NO})^2 	imes P_{O_2}} = \frac{(\frac{0.8}{2.6})^2}{(\frac{1.2}{2.6})^2 	imes (\frac{0.6}{2.6})} = \frac{0.8^2 	imes 2.6}{1.2^2 	imes 0.6} = \frac{0.64 	imes 2.6}{1.44 	imes 0.6} = \frac{1.664}{0.864} \approx 1.926$.The nearest integer is $2$.

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