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The rate of a gaseous reaction is given by the expression $k {[A]}^{2} {[B]}^{3} .$ The volume of the reaction vessel is suddenly reduced to one half of the initial volume. The reaction rate relative to the original rate will be:

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Detailed Solution

$Rate = k [A]^{2} [B]^{3}$
When volume is halved ,the concentration will become double.
$∴ Rate = [2 A]^{2} [2 B]^{3}$

$Rate = 32 k [A]^{2} [B]^{3}$

$32 \times Original rate from . . . . (i)$

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