Medium
The following are the rate constants of two different reactions. What is the overall order of reaction for each?
$(a)$ $2.418 \times 10^{-5} \ hr^{-1}$
$(b)$ $7.1 \times 10^{-4} \ atm \ s^{-1}$
$(a)$ $2.418 \times 10^{-5} \ hr^{-1}$
$(b)$ $7.1 \times 10^{-4} \ atm \ s^{-1}$
(A) The order of a reaction can be determined from the units of the rate constant $(k)$.
For a reaction of order $n$,the units of the rate constant are $(concentration)^{1-n} \ time^{-1}$.
$(a)$ The unit is $hr^{-1}$,which corresponds to $time^{-1}$. This implies $1-n = 0$,so $n = 1$. Thus,it is a $1^{st}$ order reaction.
$(b)$ The unit is $atm \ s^{-1}$,which corresponds to $(pressure)^1 \ time^{-1}$. This implies $1-n = 1$,so $n = 0$. Thus,it is a $0^{th}$ order reaction.
For a reaction of order $n$,the units of the rate constant are $(concentration)^{1-n} \ time^{-1}$.
$(a)$ The unit is $hr^{-1}$,which corresponds to $time^{-1}$. This implies $1-n = 0$,so $n = 1$. Thus,it is a $1^{st}$ order reaction.
$(b)$ The unit is $atm \ s^{-1}$,which corresponds to $(pressure)^1 \ time^{-1}$. This implies $1-n = 1$,so $n = 0$. Thus,it is a $0^{th}$ order reaction.
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$2 A + B \rightarrow C + D$
Experiment $[A] / mol \, L^{-1}$ $[B] / mol \, L^{-1}$ Initial rate of formation of $D / mol \, L^{-1} \, min^{-1}$ $I$ $0.1$ $0.1$ $6.0 \times 10^{-3}$ $II$ $0.3$ $0.2$ $7.2 \times 10^{-2}$ $III$ $0.3$ $0.4$ $2.88 \times 10^{-1}$ $IV$ $0.4$ $0.1$ $2.40 \times 10^{-2}$
Determine the rate law and the rate constant for the reaction.
$2 A + B \rightarrow C + D$
| Experiment | $[A] / mol \, L^{-1}$ | $[B] / mol \, L^{-1}$ | Initial rate of formation of $D / mol \, L^{-1} \, min^{-1}$ |
|---|---|---|---|
| $I$ | $0.1$ | $0.1$ | $6.0 \times 10^{-3}$ |
| $II$ | $0.3$ | $0.2$ | $7.2 \times 10^{-2}$ |
| $III$ | $0.3$ | $0.4$ | $2.88 \times 10^{-1}$ |
| $IV$ | $0.4$ | $0.1$ | $2.40 \times 10^{-2}$ |
Determine the rate law and the rate constant for the reaction.
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View Solution$A$ reaction between two different reactants cannot be a .....
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