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In a homogenous reaction $A \longrightarrow B+C+D$ the initial pressure was $P_0$ and after time $t$ it was $P$. Expression for rate constant $k$ in terms of $P_0$, $P$ and $t$ will be
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The correct answer is:
$k=\frac{2.303}{t} \log \frac{2 P_0}{3 P_0-P}$
$A \longrightarrow B+C+D$
$\begin{array}{lllll}\text { Initial } & a & 0 & 0 & 0 \\ \text { After time } t a-x & x & x & x\end{array}$
From (i),
$P_0+2 x=P \quad$ or $\quad x=\frac{P-P_0}{2}$
From rate equation
$k=\frac{2.303}{t} \log \frac{a}{a-x}$
$=\frac{2.303}{t} \log \frac{P_0}{P_0-\left(\frac{P-P_0}{2}\right)}=\frac{2.303}{t} \log \frac{2 P_0}{3 P_0-P}$
$\begin{array}{lllll}\text { Initial } & a & 0 & 0 & 0 \\ \text { After time } t a-x & x & x & x\end{array}$
From (i),
$P_0+2 x=P \quad$ or $\quad x=\frac{P-P_0}{2}$
From rate equation
$k=\frac{2.303}{t} \log \frac{a}{a-x}$
$=\frac{2.303}{t} \log \frac{P_0}{P_0-\left(\frac{P-P_0}{2}\right)}=\frac{2.303}{t} \log \frac{2 P_0}{3 P_0-P}$
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