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 :
$$ \frac{R _3}{R _2}=\frac{1.6 \times 10^{-2}}{4 \times 10^{-3}}=\left(\frac{1.0 \times 10^{-3}}{5 \times 10^{-4}}\right)^{m} $$
$\Rightarrow$
$$ m=2 \text {, order w.r.t. } A $$
Now comparing the data of experiment number 1 and 2 :
$$ \begin{aligned} & \frac{R _2}{R _1}=\frac{4 \times 10^{-3}}{5 \times 10^{-4}}=\left(\frac{5 \times 10^{-4}}{2.5 \times 10^{-4}}\right)^{2}\left(\frac{6.0 \times 10^{-5}}{3.0 \times 10^{-5}}\right)^{n} \ & \Rightarrow \quad 8=(2)^{2}(2)^{n} \Rightarrow n=1, \text { order w.r.t. } B . \end{aligned} $$
(i) Order with respect to $A=2$, order with respect to $B=1$.
(ii) At $300 K, R=k[A]^{2}[B]$
$$ \begin{aligned} \Rightarrow \quad k & =\frac{R}{[A]^{2}[B]}=\frac{5.0 \times 10^{-4}}{\left(2.5 \times 10^{-4}\right)^{2}\left(3.0 \times 10^{-5}\right)} \ & =2.66 \times 10^{8} s^{-1} L^{2} mol^{-2} \end{aligned} $$
(iii) From first experiment :
Rate $(320 K)=k(320 K)\left(2.5 \times 10^{-4}\right)^{2}\left(3.0 \times 10^{-5}\right)$
$$ \begin{gathered} \Rightarrow \quad k(320 K)=\frac{2 \times 10^{-3}}{\left(2.5 \times 10^{-4}\right)^{2}\left(3.0 \times 10^{-5}\right)} \ =1.066 \times 10^{9} s^{-1} L^{2} mol^{-2} . \ \Rightarrow \quad \ln {\frac{k(320 K)}{k(300 K)} }=\frac{E _a}{R}\left(\frac{T _2-T _1}{T _1 T _2}\right) \ \Rightarrow \quad \ln \left(\frac{1.066 \times 10^{9}}{2.66 \times 10^{8}}\right)=\frac{E _a}{8.314}\left(\frac{20}{300 \times 320}\right) \ \Rightarrow \quad E _a=55.42 kJ mol^{-1} \end{gathered} $$
$$ \text { Now } \quad \ln k=\ln A-\frac{E _a}{R T} $$
At $300 K: \ln \left(2.66 \times 10^{8}\right)=\ln A-\frac{55.42 \times 10^{3}}{8.314 \times 300}$
Solving : $\quad \ln A=41.62 \Rightarrow A=1.2 \times 10^{18}$
