Question

For the reaction, $2NO+2H_2\to N_2+2H_{2}O$, the following kinetic data were obtained:

Expt. No.	$\left[H_2\right]$ $\left(mol{L}^{-1}\right)$	$\left[NO\right]$ $\left(mol{L}^{-1}\right)$	Rate of reaction $\left(mol{L}^{-1}{s}^{-1}\right)$
$1.$	$0.12$	$0.12$	$0.25\times {10}^{-4}$
$2.$	$0.12$	$0.24$	$1.0\times {10}^{-4}$
$3.$	$0.24$	$0.24$	$2.0\times {10}^{-4}$

Determine the rate law and the rate constant.

Answer 1

Rate $=K{\left[A\right]}^m{\left[B\right]}^n$

$=K{\left[H_2\right]}^m{\left[NO\right]}^n$

${\left(\frac{0.12}{0.12}\right)}^m{\left(\frac{0.12}{0.24}\right)}^n=\frac{0.25\times {10}^{-4}}{1.0\times {10}^{-4}}$

${\left(0.5\right)}^n=0.25$

$n=2$

${\left(\frac{01.12}{0.24}\right)}^m{\left(\frac{0.24}{0.24}\right)}^n=\frac{1\times {10}^{-4}}{2\times {10}^{-4}}$

${\left(0.5\right)}^m=\left(0.5\right)$

$m=1$

Rate $=K\left[H_2\right]{\left[NO\right]}^2$

$K=\frac{0.25\times {10}^{-4}}{\left[0.12\right]{\left[0.12\right]}^2}$

$K=0.0144{mol}^{-2}{l}^{+2}{S}^{-1}$