Medium
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:
- A$1/4$
- B$2$
- C$1/2$
- D$4$
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Similar Questions
If doubling the concentration of a reactant $A$ increases the rate $4$ times and tripling the concentration of $A$ increases the rate $9$ times,the rate is proportional to
MediumAIIMS 1991
View SolutionWhich of the following reactions is a termolecular reaction?
Medium
View SolutionThe following results have been obtained during the kinetic studies of the reaction: $2 \ NO + 2 \ H_2 \longrightarrow N_2 + 2 \ H_2O$
Expt $\frac{-d[NO]}{dt} \ (mol \ L^{-1} \ s^{-1})$ $[NO] \ (mol \ L^{-1})$ $[H_2] \ (mol \ L^{-1})$ $1$ $4.8 \times 10^{-5}$ $1 \times 10^{-2}$ $1 \times 10^{-3}$ $2$ $43.2 \times 10^{-5}$ $3 \times 10^{-2}$ $1 \times 10^{-3}$ $3$ $86.4 \times 10^{-5}$ $3 \times 10^{-2}$ $2 \times 10^{-3}$
| Expt | $\frac{-d[NO]}{dt} \ (mol \ L^{-1} \ s^{-1})$ | $[NO] \ (mol \ L^{-1})$ | $[H_2] \ (mol \ L^{-1})$ |
| $1$ | $4.8 \times 10^{-5}$ | $1 \times 10^{-2}$ | $1 \times 10^{-3}$ |
| $2$ | $43.2 \times 10^{-5}$ | $3 \times 10^{-2}$ | $1 \times 10^{-3}$ |
| $3$ | $86.4 \times 10^{-5}$ | $3 \times 10^{-2}$ | $2 \times 10^{-3}$ |
The order of the reaction occurring by the following mechanism should be:
$(i)$ $A_2 \to A + A$ (fast)
$(ii)$ $A + B_2 \to AB + B$ (slow)
$(iii)$ $A + B \to AB$ (fast)
$(i)$ $A_2 \to A + A$ (fast)
$(ii)$ $A + B_2 \to AB + B$ (slow)
$(iii)$ $A + B \to AB$ (fast)
Easy
View SolutionWhich is the unit of the rate constant for a second-order reaction?
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