For a particular reaction,the rate expression | vedclass

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(B) The rate law is given by $r = k[A][B]^{0.5}$.Concentration is defined as $C = n/V$,where $n$ is the number of moles and $V$ is the volume.If the v

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increase by $8$ times

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(B) The rate law is given by $r = k[A][B]^{0.5}$.Concentration is defined as $C = n/V$,where $n$ is the number of moles and $V$ is the volume.If the volume $V$ is reduced to $V/4$,the concentration becomes $C' = n/(V/4) = 4(n/V) = 4C$.Substituting the new concentrations $[A]' = 4[A]$ and $[B]' = 4[B]$ into the rate law:$r' = k[A]'([B]')^{0.5} = k(4[A])(4[B])^{0.5}$.$r' = k 	imes 4[A] 	imes 2[B]^{0.5} = 8 	imes k[A][B]^{0.5}$.$r' = 8r$.Therefore,the rate of reaction increases by $8$ times.

For a particular reaction,the rate expression is given as $r = k[A][B]^{0.5}$. If the volume of the vessel is reduced to one-fourth of the initial volume,the rate of reaction would

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