If the concentration of reactant B is doubled | vedclass

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(B) Let the rate law for the reaction be $Rate = k[A]^x[B]^y$. Initially,$Rate_1 = k[A]^x[B]^y$. When the concentration of $B$ is doubled,the new rate

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$-2$

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(B) Let the rate law for the reaction be $Rate = k[A]^x[B]^y$. Initially,$Rate_1 = k[A]^x[B]^y$. When the concentration of $B$ is doubled,the new rate is $Rate_2 = k[A]^x[2B]^y$. Given that $Rate_2 = (1/4)Rate_1$. Substituting the values: $k[A]^x[2B]^y = (1/4)k[A]^x[B]^y$. This simplifies to $2^y = 1/4$. Since $1/4 = 2^{-2}$,we have $2^y = 2^{-2}$. Therefore,$y = -2$. The order of the reaction with respect to reactant $B$ is $-2$.

If the concentration of reactant $B$ is doubled,the rate of the reaction between reactants $A$ and $B$ becomes $1/4$ of the initial rate. The order of the reaction with respect to reactant $B$ is ......

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