Question

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

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

Calculate
(i) the order of the reaction with respect to $A$ and with respect to $B$. (ii) the rate constant at $300K$.
(iii) the pre-exponential factor

Answer 1

Comparing the data of experiment number $2$ and $3$ :
$\frac{R_3}{R_2}=\frac{1.6\times 10^{-2}}{4\times 10^{-3}}=(\frac{1.0\times 10^{-3}}{5\times 10^{-4}}{)}^m$
$\Rightarrow m=2$, order w.r.t. $A$
Now comparing the data of experiment number $1$ and $2$ :
$\frac{R_2}{R_1}=\frac{4\times 10^{-3}}{5\times 10^{-4}}=\left(\frac{5\times 10^{-4}}{2.5\times 10^{-4}}{)}^2\right(\frac{6.0\times 10^{-5}}{3.0\times 10^{-5}}{)}^n$
$\Rightarrow 8=(2{)}^2(2{)}^n\Rightarrow n=1,$ order w.r.t. $B$
(i) Order with respect to $A=2$, order with respect to $B=1$
(ii) At $300K,R=k[A{]}^2[B]$
$\Rightarrow k=\frac{R}{[A{]}^2[B]}=\frac{5.0\times 10^{-4}}{(2.5\times 10^{-4}{)}^2(3.0\times 10^{-5})}$
$=2.66\times 10^8{s}^{-1}{L}^{2}mo{l}^{-2}$
(iii) From first experiment:
$R(320K)=k(320K)(2.5\times 10^{-4}{)}^2(3.0\times 10^{-5})$
$\Rightarrow k(320K)=\frac{2\times 10^{-3}}{(2.5\times 10^{-4}{)}^2(3.0\times 10^{-5})}$
$=1.066\times 10^9{s}^{-1}{L}^{2}mo{l}^{-2}$
$\Rightarrow \ln \left\{\frac{k(320K)}{k(300K)}\right\}=\frac{E_a}{R}\left(\frac{T_2-T_1}{T_1{T}_2}\right)$
$\Rightarrow \ln \left(\frac{1.066\times 10^9}{2.66\times 10^8}\right)=\frac{E_a}{8.314}\left(\frac{20}{300\times 320}\right)$
$\Rightarrow E_a=55.42kJmo{l}^{-1}$
Now $\ln k=\ln A-\frac{E_a}{RT}$
At $300K:\ln (2.66\times 10^8)=\ln A-\frac{55.42\times 10^3}{8.314\times 300}$
Solving: $\ln A=41.62$
$\Rightarrow A=1.2\times 10^{18}$