Medium
For the reaction $2A + B \rightarrow \text{Product}$,the rate constant $(K)$ is $2.5 \times 10^{-5} \text{ L mol}^{-1} \text{ s}^{-1}$ after $15 \text{ s}$,$2.60 \times 10^{-5} \text{ L mol}^{-1} \text{ s}^{-1}$ after $30 \text{ s}$,and $2.55 \times 10^{-5} \text{ L mol}^{-1} \text{ s}^{-1}$ after $50 \text{ s}$. What is the order of the reaction?
- A$2$
- B$3$
- C$0$
- D$1$
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Similar Questions
For the reaction $A_2 + B_2 \to 2AB$,the experimental data is given below. Determine the order of the reaction.
Experiment No. $[A_2] \text{ (M)}$ $[B_2] \text{ (M)}$ Rate $(M \cdot s^{-1})$ $1$ $0.1$ $0.1$ $1.6 \times 10^{-4}$ $2$ $0.1$ $0.2$ $3.2 \times 10^{-4}$ $3$ $0.2$ $0.1$ $3.2 \times 10^{-4}$
| Experiment No. | $[A_2] \text{ (M)}$ | $[B_2] \text{ (M)}$ | Rate $(M \cdot s^{-1})$ |
| $1$ | $0.1$ | $0.1$ | $1.6 \times 10^{-4}$ |
| $2$ | $0.1$ | $0.2$ | $3.2 \times 10^{-4}$ |
| $3$ | $0.2$ | $0.1$ | $3.2 \times 10^{-4}$ |
Medium
View SolutionWhat is the order of reaction for $A + B \to C$?
$Observation$ $[A] \ (mol \ L^{-1})$ $[B] \ (mol \ L^{-1})$ $Rate \ (mol \ L^{-1} \ sec^{-1})$ $1$ $0.1$ $0.1$ $2 \times 10^{-3}$ $2$ $0.2$ $0.1$ $4 \times 10^{-3}$ $3$ $0.1$ $0.2$ $8 \times 10^{-3}$
| $Observation$ | $[A] \ (mol \ L^{-1})$ | $[B] \ (mol \ L^{-1})$ | $Rate \ (mol \ L^{-1} \ sec^{-1})$ |
| $1$ | $0.1$ | $0.1$ | $2 \times 10^{-3}$ |
| $2$ | $0.2$ | $0.1$ | $4 \times 10^{-3}$ |
| $3$ | $0.1$ | $0.2$ | $8 \times 10^{-3}$ |
Medium
View SolutionWhich one of the following statements is correct for the reaction?
$CH_3COOC_2H_5(aq) + NaOH(aq) \longrightarrow CH_3COONa(aq) + C_2H_5OH(aq)$
$CH_3COOC_2H_5(aq) + NaOH(aq) \longrightarrow CH_3COONa(aq) + C_2H_5OH(aq)$
For the reaction $A + B \rightarrow \text{product}$,the rate of reaction is $3.6 \times 10^{-2} \ mol \ dm^{-3} \ sec^{-1}$. When $[A] = 0.2 \ mol \ dm^{-3}$ and $[B] = 0.1 \ mol \ dm^{-3}$,calculate the rate constant of the reaction if the reaction is first order in $A$ and second order in $B$.
The conversion of $A \to B$ follows second order kinetics. Doubling the concentration of $A$ will increase the rate of formation of $B$ by a factor of:
Medium
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