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(C) The rate law is given by $r = k[A][B]^{2}$.When the volume of the vessel is reduced to half $(V_{2} = V_{1}/2)$,the concentration of the gaseous r

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$8$

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(C) The rate law is given by $r = k[A][B]^{2}$.When the volume of the vessel is reduced to half $(V_{2} = V_{1}/2)$,the concentration of the gaseous reactants doubles because concentration is inversely proportional to volume $(C = n/V)$.Thus,the new concentrations are $[A]' = 2[A]$ and $[B]' = 2[B]$.The new rate $r'$ is calculated as:$r' = k[A]'([B]')^{2} = k(2[A])(2[B])^{2}$.$r' = k 	imes 2[A] 	imes 4[B]^{2} = 8 	imes k[A][B]^{2}$.Since $r = k[A][B]^{2}$,we have $r' = 8r$.Therefore,the reaction rate increases by $8$ times.

The rate of a gaseous reaction is given by the expression $k[A][B]^{2}$. If the volume of the vessel is reduced to one-half of the initial volume,the reaction rate as compared to the original rate is:

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