N_2O_{(g)} to N_{2(g)} + frac{1}{2}O_{2(g)}In | vedclass

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(A) Let the initial pressure of $N_2O$ be $P_0$. The reaction is $N_2O_{(g)} 	o N_{2(g)} + \frac{1}{2}O_{2(g)}$.At $t=0$,pressure is $P_0, 0, 0$. Tot

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$K = \frac{1}{t} \ln \left( \frac{P_{\infty}}{3P_{\infty} - 3P_t} \right)$

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(A) Let the initial pressure of $N_2O$ be $P_0$. The reaction is $N_2O_{(g)} 	o N_{2(g)} + \frac{1}{2}O_{2(g)}$.At $t=0$,pressure is $P_0, 0, 0$. Total pressure $P_0 = P_{\infty} / 1.5 = \frac{2}{3}P_{\infty}$.At time $t$,pressure is $(P_0 - x), x, 0.5x$. Total pressure $P_t = P_0 - x + x + 0.5x = P_0 + 0.5x$.So,$0.5x = P_t - P_0$,which means $x = 2(P_t - P_0)$.The pressure of $N_2O$ at time $t$ is $P_{N_2O} = P_0 - x = P_0 - 2(P_t - P_0) = 3P_0 - 2P_t$.Substituting $P_0 = \frac{2}{3}P_{\infty}$,we get $P_{N_2O} = 3(\frac{2}{3}P_{\infty}) - 2P_t = 2P_{\infty} - 2P_t$.For a first-order reaction,$K = \frac{1}{t} \ln \left( \frac{P_0}{P_{N_2O}} ight) = \frac{1}{t} \ln \left( \frac{2P_{\infty}/3}{2P_{\infty} - 2P_t} ight) = \frac{1}{t} \ln \left( \frac{2P_{\infty}}{3(2P_{\infty} - 2P_t)} ight) = \frac{1}{t} \ln \left( \frac{P_{\infty}}{3P_{\infty} - 3P_t} ight)$.

$N_2O_{(g)} \to N_{2(g)} + \frac{1}{2}O_{2(g)}$
In a closed vessel,the reaction follows first-order kinetics. If starting with pure $N_2O_{(g)}$,the total pressure after time $t$ is $P_t$ and after a very long time is $P_{\infty}$,then which of the following expressions is correct?

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$N_2O_{(g)} \to N_{2(g)} + \frac{1}{2}O_{2(g)}$In a closed vessel,the reaction follows first-order kinetics. If starting with pure $N_2O_{(g)}$,the total pressure after time $t$ is $P_t$ and after a very long time is $P_{\infty}$,then which of the following expressions is correct?