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The rate equation of a gaseous reaction is given by $r=k[A][B]$. If the volume of the reaction vessel is suddenly reduced to $1 / 2$ of the initial volume, the reaction rate relating to the original rate will be
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The concentration of gaseous $\propto \frac{1}{\text { volume }}$
$\therefore$ If volume of vessel is reduced of initial volume to half then concentration increases to two-times.
$$
\begin{array}{llrl}
\therefore & & {\left[A^{\prime}\right]=2[A]} \\
& & {\left[B^{\prime}\right]=2[B]} \\
\text { Initially, } & r=k[A][B]
\end{array}
$$
After reduction of volume, $r_{1}=k_{2}\left[A^{\prime}\right] 2\left[B^{\prime}\right]$
$$
\begin{aligned}
&r_{1}=4 k\left[A^{\prime}\right]\left[B^{\prime}\right] \quad\left(k_{2}=k\right) \\
&r_{1}=4 r
\end{aligned}
$$
$\therefore$ Rate increases to 4 times of original rate.
$\therefore$ If volume of vessel is reduced of initial volume to half then concentration increases to two-times.
$$
\begin{array}{llrl}
\therefore & & {\left[A^{\prime}\right]=2[A]} \\
& & {\left[B^{\prime}\right]=2[B]} \\
\text { Initially, } & r=k[A][B]
\end{array}
$$
After reduction of volume, $r_{1}=k_{2}\left[A^{\prime}\right] 2\left[B^{\prime}\right]$
$$
\begin{aligned}
&r_{1}=4 k\left[A^{\prime}\right]\left[B^{\prime}\right] \quad\left(k_{2}=k\right) \\
&r_{1}=4 r
\end{aligned}
$$
$\therefore$ Rate increases to 4 times of original rate.
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