For the reaction 2NO_{2(g)} rightleftharpoons | vedclass

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(D) For the reaction $2NO_{2(g)} \rightleftharpoons N_2O_{4(g)}$,$K_p = \frac{P_{N_2O_4}}{P_{NO_2}^2} = 2 \ atm^{-1}$.Let $P_{NO_2} = P_1$ and $P_{N_2

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$5.19$

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(D) For the reaction $2NO_{2(g)} \rightleftharpoons N_2O_{4(g)}$,$K_p = \frac{P_{N_2O_4}}{P_{NO_2}^2} = 2 \ atm^{-1}$.Let $P_{NO_2} = P_1$ and $P_{N_2O_4} = P_2$. Given $P_1 + P_2 = 10$ and $\frac{P_2}{P_1^2} = 2$.Substituting $P_2 = 10 - P_1$ into the $K_p$ expression: $\frac{10 - P_1}{P_1^2} = 2 \implies 2P_1^2 + P_1 - 10 = 0$.Solving the quadratic equation: $P_1 = \frac{-1 \pm \sqrt{1 + 80}}{4} = \frac{-1 \pm 9}{4}$. Taking positive value,$P_1 = 2 \ atm$,so $P_2 = 8 \ atm$.When volume is doubled,pressure of each gas is halved: $P'_{NO_2} = 1 \ atm$ and $P'_{N_2O_4} = 4 \ atm$.Let $x$ be the decrease in pressure of $N_2O_4$ to reach new equilibrium: $2NO_2 \rightleftharpoons N_2O_4$.New partial pressures: $P_{NO_2} = 1 + 2x$,$P_{N_2O_4} = 4 - x$.$K_p = \frac{4 - x}{(1 + 2x)^2} = 2 \implies 4 - x = 2(1 + 4x + 4x^2) = 2 + 8x + 8x^2$.$8x^2 + 9x - 2 = 0$. Solving for $x$: $x = \frac{-9 \pm \sqrt{81 + 64}}{16} = \frac{-9 + 12.04}{16} \approx 0.19$.Total pressure $P_T = (1 + 2x) + (4 - x) = 5 + x = 5 + 0.19 = 5.19 \ atm$.

For the reaction $2NO_{2(g)} \rightleftharpoons N_2O_{4(g)}$ at $300 \ K$,the value of $K_p$ is $2 \ atm^{-1}$. The total pressure at equilibrium is $10 \ atm$. If the volume of the container becomes two times its original volume,what will be its equilibrium pressure at $300 \ K$ (in $atm$)?

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