Question

The reaction, $2N_2{O}_5\to 4N{O}_2+O_2$, was studied and the following data were collected:

Expt. No.	$\left[N_2{O}_5\right]$ $\left(mol{L}^{-1}\right)$	Rate of disappearance of $N_2{O}_5$ $\left(mol{L}^{-1}{min}^{-1}\right)$
$1.$	$1.13\times {10}^{-2}$	$34\times {10}^{-5}$
$2.$	$0.84\times {10}^{-2}$	$25\times {10}^{-5}$
$3.$	$0.62\times {10}^{-2}$	$18\times {10}^{-5}$

Determine (i) order the reaction, (ii) the rate law, and (iii) rate constant for the reaction.

Answer 1

Let the rate law for the reaction be
Rate $=k{\left[N_2{O}_5\right]}^x$
$34\times {10}^{-5}=k{[1.13\times {10}^{-2}]}^x$
$25\times {10}^{-5}=k{[0.84\times {10}^{-2}]}^x$
or $\frac{34\times {10}^{-5}}{25\times {10}^{-5}}=\frac{{[1.13\times {10}^{-2}]}^x}{{[0.84\times {10}^{-2}]}^x}$
$1.36={\left[1.345\right]}^x$
$x=1$
Order of the reaction is 1. Rate law $=k\left[N_2{O}_5\right]$
$k=\frac{Rate}{\left[N_2{O}_5\right]}=\frac{34\times {10}^{-5}}{1.13\times {10}^{-2}}=30.08\times {10}^{-3}{min}^{-1}$