Question

Expt. No.	Initial Concentration		Initial rate
	$\left[NO\right]$	$\left[C{l}_2\right]$
1.	0.010	0.010	$1.2\times {10}^{-4}$
2.	0.010	0.020	$2.4\times {10}^{-4}$
3.	0.020	0.020	$9.6\times {10}^{-4}$

For the reaction, $2NO+{Cl}_2⟶2NOCl$, at $300K$, following data are obtained:
Write rate law and order of the reaction? Also, calculate the specific rate constant.

Answer 1

Let the rate law for the reaction be
Rate $=k{\left[NO\right]}^x{\left[{Cl}_2\right]}^y$
From expt. (1), $1.2\times {10}^{-4}=k{\left[0.010\right]}^x{\left[0.010\right]}^y$ ...(i)
From expt. (2), $2.4\times {10}^{-4}=k{\left[0.010\right]}^x{\left[0.020\right]}^y$ ...(ii)
Dividing equation (ii) by equation (i),
$\frac{2.4\times {10}^{-4}}{1.2\times {10}^{-4}}=\frac{{\left[0.020\right]}^y}{{\left[0.010\right]}^y}$
or $2={\left(2\right)}^y$
$y=1$
From expt. (2), $2.4\times {10}^{-4}=k{\left[0.010\right]}^x{\left[0.020\right]}^y$
From expt. (3), $9.6\times {10}^{-4}=k{\left[0.020\right]}^x{\left[0.020\right]}^y$
Dividing equation (iii) by equation (ii),
$\frac{9.6\times {10}^{-4}}{2.4\times {10}^{-4}}=\frac{{\left[0.020\right]}^x}{{\left[0.010\right]}^x}$
or $4=2^x$
$x=2$
Order of reaction $=x+y=2+1=3$
Rate law for the reaction is
Rate $=k{\left[NO\right]}^2\left[{Cl}_2\right]$
Considering equation (i) again,
$1.2\times {10}^{-4}=k{\left[0.010\right]}^2\left[0.010\right]$
$k=\frac{1.2\times {10}^{-4}}{{\left[0.010\right]}^3}=1.2\times {10}^2{mol}^{-2}{L}^2{s}^{-1}$