Question

From the following data for the reaction between $A$ and $B$

[A] (mol\L)	[B] (mol\L)	Initial rate $(mol\,L^{-1}\,n^{-1})m$
		$300 \,K$	$320 \,K$
$2.5 \times 10^{-4}$	$3.0 \times 10^{-5}$	$5.0 \times 10^{-4}$	$2.0 \times 10^{-3}$
$5.0 \times 10^{-4}$	$6.0 \times 10^{-5}$	$4.0 \times 10^{-3}$	-
$1.0 \times 10^{-3}$	$6.0 \times 10^{-5}$	$1.6 \times 10^{-2}$	-

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

Answer 1

Comparing the data of experiment number $2$ and $3$ :
$\frac{R_3}{R_2}=\frac{1.6\times10^{-2}}{4\times10^{-3}}=\bigg(\frac{1.0\times10^{-3}}{5\times10^{-4}}\bigg)^m$
$\Rightarrow m=2$, order w.r.t. $A$
Now comparing the data of experiment number $1$ and $2$ :
$\frac{R_2}{R_1}=\frac{4\times10^{-3}}{5\times10^{-4}}=\bigg(\frac{5\times10^{-4}}{2.5\times10^{-4}}\bigg)^2\bigg(\frac{6.0\times10^{-5}}{3.0\times10^{-5}}\bigg)^n$
$\Rightarrow 8=(2)^2(2)^n\Rightarrow n=1,$ order w.r.t. $B$
(i) Order with respect to $A = 2$, order with respect to $B = 1$
(ii) At $300\,K, R=k[A]^2[B]$
$\Rightarrow k=\frac{R}{[A]^2[B]}=\frac{5.0\times10^{-4}}{(2.5\times10^{-4})^2(3.0\times10^{-5})}$
$=2.66\times10^8 \,s^{-1}\, L^2 mol^{-2}$
(iii) From first experiment:
$R(320\, K)=k(320 \,K)(2.5\times10^{-4})^2(3.0\times10^{-5})$
$\Rightarrow k(320\, K)=\frac{2\times10^{-3}}{(2.5\times10^{-4})^2(3.0\times10^{-5})}$
$=1.066\times10^9 \,s^{-1}\,L^2 \,mol^{-2}$
$\Rightarrow \ln\biggl\{\frac{k (320 \,K)}{k (300\, K)}\biggl\}=\frac{E_a}{R}\bigg(\frac{T_2-T_1}{T_1T_2}\bigg)$
$\Rightarrow \ln\bigg(\frac{1.066\times10^9}{2.66\times10^8}\bigg)=\frac{E_a}{8.314}\bigg(\frac{20}{300\times320}\bigg)$
$\Rightarrow E_a=55.42\, kJ \, mol^{-1}$
Now $\ln k=\ln A-\frac{E_a}{RT}$
At $300 \,K: \ln(2.66\times10^8)=\ln A-\frac{55.42\times10^3}{8.314\times300}$
Solving: $\ln A =41.62$
$\Rightarrow A=1.2\times10^{18}$