Difficult
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:
- Aremain the same
- Bbecome four times
- Cbecome $1.414$ times
- Dbecome double
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Similar Questions
The following are the rate constants of two different reactions. Determine the overall order of reaction for each case:
$(a)$ $6.66 \times 10^{-3} \, s^{-1}$
$(b)$ $4.5 \times 10^{-2} \, mol^{-1} \, L \, s^{-1}$
$(a)$ $6.66 \times 10^{-3} \, s^{-1}$
$(b)$ $4.5 \times 10^{-2} \, mol^{-1} \, L \, s^{-1}$
Medium
View SolutionThe rate constant value for a reaction is $1.75 \times 10^2 \ L^2 \ mol^{-2} \ sec^{-1}$. The half-life period $t_{1/2} \propto$ . . . . . . .
The mechanism of the reaction $A + 2B \to D$ is given by:
$2B \xrightarrow{k} B_2$ [Slow]
$B_2 + A \to D$ [Fast]
The rate law expression,order with respect to $A$,order with respect to $B$,and overall order are respectively:
$2B \xrightarrow{k} B_2$ [Slow]
$B_2 + A \to D$ [Fast]
The rate law expression,order with respect to $A$,order with respect to $B$,and overall order are respectively:
Medium
View SolutionIf the surface area of the reactants increases,then the order of the reaction:
Easy
View SolutionFrom the rate expression for the following reaction,determine its order of reaction and the dimensions of the rate constant.
$H_2 O_2+3 I^{-}+2 H^{+} \rightarrow 2 H_2 O+I_3^{-} \text { Rate }=k\left[H_2 O_2\right]\left[I^{-}\right]$
$H_2 O_2+3 I^{-}+2 H^{+} \rightarrow 2 H_2 O+I_3^{-} \text { Rate }=k\left[H_2 O_2\right]\left[I^{-}\right]$
Medium
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