Chemical Kinetics - Result Question 69
68. From the following data for the reaction between $A$ and $B$
| $[\boldsymbol{A}],(mol / L)$ | $[\boldsymbol{B}],(mol / L)$ | Initial rate $\left(mol L^{-1} s^{-1}\right)$ at | |
|---|---|---|---|
| $\mathbf{3 0 0} \mathbf{~ K}$ | $\mathbf{3 2 0} \mathbf{~ K}$ | ||
| $2.5 \times 10^{-4}$ | $3.0 \times 10^{-5}$ | $5.0 \times 10^{-4}$ | $2.0 \times 10^{-3}$ |
| $5.0 \times 10^{-4}$ | $6.0 \times 10^{-5}$ | $4.0 \times 10^{-3}$ | - |
| $1.0 \times 10^{-3}$ | $6.0 \times 10^{-5}$ | $1.6 \times 10^{-2}$ | - |
Calculate
(i) the order of the reaction with respect to $A$ and with respect to $B$.
(ii) the rate constant at $300 $ $K$.
(iii) the pre-exponential factor.
(1994, 5M)
Show Answer
Solution:
- Comparing the data of experiment number 2 and 3 :
$ \dfrac{R _3}{R _2}=\dfrac{1.6 \times 10^{-2}}{4 \times 10^{-3}}=\left(\dfrac{1.0 \times 10^{-3}}{5 \times 10^{-4}}\right)^{m} $
$\Rightarrow \quad$ $ m=2 \text {, order w.r.t. } A $
Now comparing the data of experiment number 1 and 2 :
$ \begin{aligned} & \dfrac{R _2}{R _1}=\dfrac{4 \times 10^{-3}}{5 \times 10^{-4}}=\left(\dfrac{5 \times 10^{-4}}{2.5 \times 10^{-4}}\right)^{2}\left(\dfrac{6.0 \times 10^{-5}}{3.0 \times 10^{-5}}\right)^{n} \\ & \Rightarrow \quad 8=(2)^{2}(2)^{n} \Rightarrow n=1, \text { order w.r.t. } B . \end{aligned} $
(i) Order with respect to $A=2$, order with respect to $B=1$.
(ii) At $300 K, R=k[A]^{2}[B]$
$ \begin{aligned} \Rightarrow \quad k & =\dfrac{R}{[A]^{2}[B]}=\dfrac{5.0 \times 10^{-4}}{\left(2.5 \times 10^{-4}\right)^{2}\left(3.0 \times 10^{-5}\right)} \\ & =2.66 \times 10^{8} s^{-1} L^{2} mol^{-2} \end{aligned} $
(iii) From first experiment :
Rate $(320 K)=k(320 K)\left(2.5 \times 10^{-4}\right)^{2}\left(3.0 \times 10^{-5}\right)$
$\Rightarrow \quad k(320 K)=\dfrac{2 \times 10^{-3}}{\left(2.5 \times 10^{-4}\right)^{2}\left(3.0 \times 10^{-5}\right)} $
$=\quad 1.066 \times 10^{9}$ $ s^{-1}$ $ L^{2}$ $ mol^{-2} . $
$\Rightarrow \quad \ln \{\dfrac{k(320 K)}{k(300 K)} \}=\dfrac{E _a}{R}\left(\dfrac{T _2-T _1}{T _1 T _2}\right) $
$\Rightarrow \quad \ln \left(\dfrac{1.066 \times 10^{9}}{2.66 \times 10^{8}}\right)=\dfrac{E _a}{8.314}\left(\dfrac{20}{300 \times 320}\right) $
$\Rightarrow \quad E _a=55.42$ $ kJ$ $ mol^{-1}$
$ \text { Now } \quad \ln k=\ln A-\dfrac{E _a}{R T} $
At $300 K: \ln \left(2.66 \times 10^{8}\right)=\ln A-\dfrac{55.42 \times 10^{3}}{8.314 \times 300}$
Solving : $\quad \ln A=41.62 \Rightarrow A=1.2 \times 10^{18}$
BETA
