For a first order gas phase reaction: A_{(g)} | vedclass

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(B) The reaction is $A_{(g)} \rightarrow 2 B_{(g)} + C_{(g)}$. Let $P$ be the decrease in pressure of $A$ at time $t$. Initial pressure$P_0$$0$$0$Pres

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$\frac{2.303}{t} \log \left(\frac{2 P_0}{3 P_0 - P_t}\right)$

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(B) The reaction is $A_{(g)} \rightarrow 2 B_{(g)} + C_{(g)}$. Let $P$ be the decrease in pressure of $A$ at time $t$. Initial pressure$P_0$$0$$0$Pressure at time $t$$P_0 - P$$2P$$P$ Total pressure $P_t = (P_0 - P) + 2P + P = P_0 + 2P$. Therefore,$2P = P_t - P_0$,which gives $P = \frac{P_t - P_0}{2}$. The pressure of $A$ at time $t$ is $P_A = P_0 - P = P_0 - \frac{P_t - P_0}{2} = \frac{2P_0 - P_t + P_0}{2} = \frac{3P_0 - P_t}{2}$. The integrated rate equation for a first order reaction is $k = \frac{2.303}{t} \log \left( \frac{P_0}{P_A} \right)$. Substituting $P_A$: $k = \frac{2.303}{t} \log \left( \frac{P_0}{\frac{3P_0 - P_t}{2}} \right) = \frac{2.303}{t} \log \left( \frac{2P_0}{3P_0 - P_t} \right)$.

For a first order gas phase reaction:
$A_{(g)} \rightarrow 2 B_{(g)} + C_{(g)}$
Let $P_0$ be the initial pressure of $A$ and $P_t$ be the total pressure at time $t$. The integrated rate equation is:

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For a first order gas phase reaction: $A_{(g)} \rightarrow 2 B_{(g)} + C_{(g)}$ Let $P_0$ be the initial pressure of $A$ and $P_t$ be the total pressure at time $t$. The integrated rate equation is: