Medium
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 $(M \cdot s^{-1})$ |
| $1$ | $0.1$ | $0.1$ | $1.6 \times 10^{-4}$ |
| $2$ | $0.1$ | $0.2$ | $3.2 \times 10^{-4}$ |
| $3$ | $0.2$ | $0.1$ | $3.2 \times 10^{-4}$ |
- A$1$
- B$2$
- C$0$
- D$3$
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Similar Questions
For the reaction $A + 2B \longrightarrow C$, the reaction rate is doubled if the concentration of $A$ is doubled. The rate is increased by four times when concentrations of both $A$ and $B$ are increased by four times. The order of the reaction is
$A_2 + B_2 \to 2AB$; $R.O.R = k[A_2]^a[B_2]^b$
Initial $[A_2]$ Initial $[B_2]$ $R.O.R. (r) \ M s^{-1}$ $0.2$ $0.2$ $0.04$ $0.1$ $0.4$ $0.04$ $0.2$ $0.4$ $0.08$
Order of reaction with respect to $A_2$ and $B_2$ are respectively:
| Initial $[A_2]$ | Initial $[B_2]$ | $R.O.R. (r) \ M s^{-1}$ |
|---|---|---|
| $0.2$ | $0.2$ | $0.04$ |
| $0.1$ | $0.4$ | $0.04$ |
| $0.2$ | $0.4$ | $0.08$ |
Order of reaction with respect to $A_2$ and $B_2$ are respectively:
Medium
View SolutionHow can you determine the rate law of the following reaction?
$2 \, NO \, (g) + O_2 \, (g) \to 2 \, NO_2 \, (g)$
$2 \, NO \, (g) + O_2 \, (g) \to 2 \, NO_2 \, (g)$
Easy
View SolutionWhich of the following is an elementary reaction?
For the reaction $2A + B \rightarrow A_2B$,if the concentration of reactant $A$ is doubled and the concentration of reactant $B$ is halved,the rate of the reaction will:
Easy
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