The reaction N_2O_{4(g)} rightleftharpoons 2N | vedclass

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(A) The reaction is $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$.Initial moles: $N_2O_4 = 0.8 \ mol$,$NO_2 = 0 \ mol$.At equilibrium,let the amount of

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$0.06$

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(A) The reaction is $N_2O_{4(g)} ightleftharpoons 2NO_{2(g)}$.Initial moles: $N_2O_4 = 0.8 \ mol$,$NO_2 = 0 \ mol$.At equilibrium,let the amount of $N_2O_4$ reacted be $x \ mol$.Then,$[N_2O_4] = (0.8 - x) \ M$ and $[NO_2] = 2x \ M$ (since volume is $1 \ L$).$K_c = \frac{[NO_2]^2}{[N_2O_4]} = \frac{(2x)^2}{0.8 - x} = 0.00466$.$4x^2 = 0.00466(0.8 - x) \implies 4x^2 + 0.00466x - 0.003728 = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $x \approx 0.03$.Therefore,$[NO_2] = 2x = 2 	imes 0.03 = 0.06 \ M$.

The reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$ is started by taking $0.8 \ mol$ of $N_2O_4$ in a $1 \ L$ flask. If the equilibrium constant at $298 \ K$ is $0.00466 \ M$,the equilibrium concentration of $NO_2$ will be ........... $M$.

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