Question

Expt. No.	$\left[A\right]$ $\left(mol{L}^{-1}\right)$	$\left[B\right]$ $\left(mol{L}^{-1}\right)$	Initial rate $\left(mol{L}^{-1}{s}^{-1}\right)$
			300K	320K
(1)	$2.5\times {10}^{-4}$	$3.0\times {10}^{-5}$	$5.0\times {10}^{-4}$	$2.0\times {10}^{-3}$
(2)	$5.0\times {10}^{-4}$	$6.0\times {10}^{-5}$	$4.0\times {10}^{-3}$	$...$
(3)	$1.0\times {10}^{-3}$	$6.0\times {10}^{-5}$	$1.6\times {10}^{-2}$	$...$

From the data given above for the reaction between $A$ and $B$, calculate the following:
(i) The order of the reaction with respect to $A$ and with respect to $B$,
(ii) The rate constant at $300$ $K$
(iii) The energy of activation and
(iv) The pre-exponential factor.

Answer 1

(i) Let the rate law be:
Rate $=k{\left[A\right]}^x{\left[B\right]}^y$
From expt. (1), $5.0\times {10}^{-4}=k{[2.5\times {10}^{-4}]}^x{[3.0\times {10}^{-5}]}^y$ ...(i)
From expt. (2), $4.0\times {10}^{-3}=k{[5.0\times {10}^{-4}]}^x{[6.0\times {10}^{-5}]}^y$ ...(ii)
Dividing equation (ii) by equation (i), $\frac{4.0\times {10}^{-3}}{5.0\times {10}^{-4}}=2^x\cdot 2^y=8$
From expt. (3), $1.6\times {10}^{-2}=k{[1.0\times {10}^{-3}]}^x{[6.0\times {10}^{-5}]}^y$ ...(iii)
Dividing equation (iii) by equation (ii), $\frac{1.6\times {10}^{-2}}{4.0\times {10}^{-3}}=2^x=4$
or $x=2$ and $y=1$
Hence, order with respect to $A$ is $2$nd and order with respect to $B$ is $1$st.
(ii) Rat $=k\cdot {\left[A\right]}^2\left[B\right]$
From expt. (1), $5\times {10}^{-4}=k{[2.5\times {10}^{-4}]}^2[3.0\times {10}^{-5}]$
or $k=\frac{5\times {10}^{-4}}{{[2.5\times {10}^{-4}]}^2[3.0\times {10}^{-5}]}=2.67\times {10}^8{L}^2{mol}^{-2}{s}^{-1}$
(iii) From Arrhenius equation,
${\log }_{10}\frac{2.0\times {10}^{-3}}{5.0\times {10}^{-4}}=\frac{E_a}{2.303\times 8.314}\times \frac{20}{300\times 320}$
$E_a=\frac{2.303\times 8.314\times 300\times 320}{20}\times {\log }_{10}4$
$=55.333$ $kJ$ ${mol}^{-1}$
(iv) Applying ${\log }_{10}k={\log }_{10}A-\frac{E_a}{2.303RT}$
${\log }_{10}\frac{A}{k}=\frac{55333}{2.303\times 8.314\times 300}=9.633$
or $\frac{A}{k}=4.29\times {10}^9$
or $A=4.29\times {10}^9\times 2.67\times {10}^8$
$=1.145\times {10}^{18}$