Medium
The half-life period of a $second$ order reaction is:
- AProportional to the initial concentration of reactants
- BIndependent of the initial concentration of reactants
- CInversely proportional to the initial concentration of reactants
- DInversely proportional to the square of the initial concentration of reactants
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Similar Questions
The rate of reaction,$A + B \rightarrow \text{product}$,is $7.2 \times 10^{-2} \ mol \ dm^{-3} \ s^{-1}$ at $[A] = 0.4 \ mol \ dm^{-3}$ and $[B] = 0.1 \ mol \ dm^{-3}$. The reaction is first order in $A$ and second order in $B$. Calculate the rate constant.
The rate of reaction $A + 2B \to 3C$ becomes $72$ times when the concentration of $A$ is tripled and the concentration of $B$ is doubled. The order of reaction with respect to $A$ and $B$ respectively is:
Medium
View SolutionIf 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 ......
Medium
View SolutionFor the reaction $2A + B \rightarrow \text{Product}$,the following mechanism is given. What is the order of the reaction?
$2A \rightleftharpoons A_2$ (fast)
$A_2 + B \rightarrow P$ (slow)
$2A \rightleftharpoons A_2$ (fast)
$A_2 + B \rightarrow P$ (slow)
Medium
View SolutionFor the reaction $A + B \to \text{products}$,what will be the order of reaction with respect to $A$ and $B$?
$Exp.$ $[A] \ (mol \ L^{-1})$ $[B] \ (mol \ L^{-1})$ Initial rate $(mol \ L^{-1} \ s^{-1})$ $1$ $2.5 \times 10^{-4}$ $3 \times 10^{-5}$ $5 \times 10^{-4}$ $2$ $5 \times 10^{-4}$ $6 \times 10^{-5}$ $4 \times 10^{-3}$ $3$ $1 \times 10^{-3}$ $6 \times 10^{-5}$ $1.6 \times 10^{-2}$
| $Exp.$ | $[A] \ (mol \ L^{-1})$ | $[B] \ (mol \ L^{-1})$ | Initial rate $(mol \ L^{-1} \ s^{-1})$ |
| $1$ | $2.5 \times 10^{-4}$ | $3 \times 10^{-5}$ | $5 \times 10^{-4}$ |
| $2$ | $5 \times 10^{-4}$ | $6 \times 10^{-5}$ | $4 \times 10^{-3}$ |
| $3$ | $1 \times 10^{-3}$ | $6 \times 10^{-5}$ | $1.6 \times 10^{-2}$ |
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