Medium
Calculate the overall order of a reaction which has the rate expression:
$(a)$ $\text{Rate} = k[A]^{1/2}[B]^{3/2}$
$(b)$ $\text{Rate} = k[A]^{3/2}[B]^{-1}$
$(a)$ $\text{Rate} = k[A]^{1/2}[B]^{3/2}$
$(b)$ $\text{Rate} = k[A]^{3/2}[B]^{-1}$
(N/A) The overall order of a reaction is the sum of the powers of the concentration terms in the rate law expression.
$(a)$ $\text{Rate} = k[A]^{1/2}[B]^{3/2}$
$\text{Order} = 1/2 + 3/2 = 4/2 = 2$
Thus,the reaction is of second order.
$(b)$ $\text{Rate} = k[A]^{3/2}[B]^{-1}$
$\text{Order} = 3/2 + (-1) = 3/2 - 2/2 = 1/2$
Thus,the reaction is of half order.
$(a)$ $\text{Rate} = k[A]^{1/2}[B]^{3/2}$
$\text{Order} = 1/2 + 3/2 = 4/2 = 2$
Thus,the reaction is of second order.
$(b)$ $\text{Rate} = k[A]^{3/2}[B]^{-1}$
$\text{Order} = 3/2 + (-1) = 3/2 - 2/2 = 1/2$
Thus,the reaction is of half order.
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Initial concentration of $A$ $(M)$ Initial concentration of $B$ $(M)$ Initial Rate $(M \cdot s^{-1})$ $1$ $10$ $100$ $1$ $1$ $1$ $10$ $1$ $10$
The order of the reaction with respect to $A$ and $B$ are:
| Initial concentration of $A$ $(M)$ | Initial concentration of $B$ $(M)$ | Initial Rate $(M \cdot s^{-1})$ |
| $1$ | $10$ | $100$ |
| $1$ | $1$ | $1$ |
| $10$ | $1$ | $10$ |
The order of the reaction with respect to $A$ and $B$ are:
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View SolutionThe results given in the below table were obtained during kinetic studies of the following reaction:
$2 A + B \longrightarrow C + D$
Experiment $[A] / mol \ L^{-1}$ $[B] / mol \ L^{-1}$ Initial rate / $mol \ L^{-1} \ min^{-1}$ $I$ $0.1$ $0.1$ $6.00 \times 10^{-3}$ $II$ $0.1$ $0.2$ $2.40 \times 10^{-2}$ $III$ $0.2$ $0.1$ $1.20 \times 10^{-2}$ $IV$ $X$ $0.2$ $7.20 \times 10^{-2}$ $V$ $0.3$ $Y$ $2.88 \times 10^{-1}$
$X$ and $Y$ in the given table are respectively :
$2 A + B \longrightarrow C + D$
| Experiment | $[A] / mol \ L^{-1}$ | $[B] / mol \ L^{-1}$ | Initial rate / $mol \ L^{-1} \ min^{-1}$ |
| $I$ | $0.1$ | $0.1$ | $6.00 \times 10^{-3}$ |
| $II$ | $0.1$ | $0.2$ | $2.40 \times 10^{-2}$ |
| $III$ | $0.2$ | $0.1$ | $1.20 \times 10^{-2}$ |
| $IV$ | $X$ | $0.2$ | $7.20 \times 10^{-2}$ |
| $V$ | $0.3$ | $Y$ | $2.88 \times 10^{-1}$ |
$X$ and $Y$ in the given table are respectively :
DifficultJEE MAIN 2020
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