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The rate of a gaseous reaction is given by the expression $k[A][B]^{2}$. If the volume of vessel is reduced to one half of the initial volume, the reaction rate as compared to original rate is
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8
Given, rate law expression $(r)=k[A][B]^{2}$
Final volume $\left(V_{2}\right)$ is reduced to half of initial volume $\left(V_{1}\right)$ i.e. $V_{2}=1 / 2 V_{1}$
$\therefore$ Final concentration $[A]$ and $[B]=[2 A]$ and $2[B]$
$\therefore$ Final rate $r^{\prime}=k[A]^{0}[B]^{2}=k[2 A][2 B]^{2}$
$\begin{aligned}
& r^{\prime}=k \times 2 \times 4[A][B]^{2} \Rightarrow r^{\prime}=8 k[A][B]^{2} \\
\therefore \quad & r^{\prime}=8 r
\end{aligned}$
i.e. rate increases by 8 times.
Final volume $\left(V_{2}\right)$ is reduced to half of initial volume $\left(V_{1}\right)$ i.e. $V_{2}=1 / 2 V_{1}$
$\therefore$ Final concentration $[A]$ and $[B]=[2 A]$ and $2[B]$
$\therefore$ Final rate $r^{\prime}=k[A]^{0}[B]^{2}=k[2 A][2 B]^{2}$
$\begin{aligned}
& r^{\prime}=k \times 2 \times 4[A][B]^{2} \Rightarrow r^{\prime}=8 k[A][B]^{2} \\
\therefore \quad & r^{\prime}=8 r
\end{aligned}$
i.e. rate increases by 8 times.
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