For a reaction 2 A + B rightarrow P, when th | vedclass

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(B) The rate law is given by $r = k[A]^{\alpha}[B]^{\beta}$.When the concentration of $B$ is doubled, the rate does not change, which implies that the

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$L \ mol^{-1} \ s^{-1}$

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(B) The rate law is given by $r = k[A]^{\alpha}[B]^{\beta}$.When the concentration of $B$ is doubled, the rate does not change, which implies that the reaction is of zero order with respect to $B$. Therefore, $\beta = 0$.When the concentrations of both $A$ and $B$ are doubled, the rate increases by a factor of $4$:$\frac{r_2}{r_1} = \frac{k[2A]^{\alpha}[2B]^{\beta}}{k[A]^{\alpha}[B]^{\beta}} = 4$$2^{\alpha} \cdot 2^{\beta} = 4$Since $\beta = 0$, we have $2^{\alpha} = 4$, which gives $\alpha = 2$.The overall order of the reaction is $\alpha + \beta = 2 + 0 = 2$.The unit of the rate constant for a second-order reaction is $L \ mol^{-1} \ s^{-1}$.

For a reaction $2 \ A + B \rightarrow P$, when the concentration of $B$ alone is doubled, the rate does not change, and when the concentrations of both $A$ and $B$ are doubled, the rate increases by a factor of $4$. The unit of the rate constant is,

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