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(C) The rate of a reaction is determined by its rate-determining step. Given the rate-determining step is $Y + \frac{1}{2}Z 	o Q$,the rate law is exp

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become $1.414$ times

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(C) The rate of a reaction is determined by its rate-determining step. Given the rate-determining step is $Y + \frac{1}{2}Z 	o Q$,the rate law is expressed as: $Rate = k[Y][Z]^{1/2}$. If the concentration of $Z$ is doubled,the new rate $(r')$ becomes: $r' = k[Y][2Z]^{1/2} = k[Y] 	imes \sqrt{2} 	imes [Z]^{1/2}$. $r' = \sqrt{2} 	imes Rate = 1.414 	imes Rate$. Therefore,the rate of reaction will become $1.414$ times the original rate.

For a chemical reaction $Y + 2Z \to$ Product,the rate-controlling step is $Y + \frac{1}{2}Z \to Q$. If the concentration of $Z$ is doubled,the rate of reaction will:

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