Question

The data below are for the reaction of $NO$ and ${Cl}_2$ to form $NOCl$ at $295$ $K$.

Expt. No.	$\left[{Cl}_2\right]$ $\left(mol{L}^{-1}\right)$	$\left[NO\right]$ $\left(mol{L}^{-1}\right)$	Initial rate $\left(mol{L}^{-1}{s}^{-1}\right)$
$1.$	$0.05$	$0.05$	$1.0\times {10}^{-3}$
$2.$	$0.15$	$0.05$	$3.0\times {10}^{-3}$
$3.$	$0.05$	$0.15$	$9.0\times {10}^{-3}$

(a) What is the order with respect to $NO$ and ${Cl}_2$ in the reaction?
(b) Write the rate expression.
(c) Calculate the rate constant.
(d) Determine the reaction rate when concentration of ${Cl}_2$ and $NO$ are $0.2$ $M$ and $0.4$ $M$ respectively.

Answer 1

(a) Comparing trial 1 to trial 2. $\left[C{l}_2\right]$ triples & [NO] remains same and rate increase by 3 times. This means that it is first order with respect to $\left[C{l}_2\right]$

Comparing trial 1 to trial 3. $\left[NO\right]$ triples & $\left[C{l}_2\right]$ remains same and rate increase by 9 times, This means that it is second order with respect to $\left[NO\right]$.

Order of reaction $=1+2=3$

(b) Rate expression, Rate $=k\left[C{l}_2\right][NO{]}^2$

(c) Putting value from trial 1,

$k=\frac{Rate}{\left[C{l}_2\right][NO{]}^2}=\frac{1.0\times {10}^{-3}mol{L}^{-1}{s}^{-1}}{\left(0.05mol{L}^{-1}\right)(0.05mol{L}^{-1}{)}^2}$

$=0.4L^{2}mo{l}^{-2}{s}^{-1}$

(d) Rate $=k\left[C{l}_2\right][NO{]}^2=(0.4L^{2}mo{l}^{-2}{s}^{-1}\left)\right(0.2mol{L}^{-1})\times (0.4mol{L}^{-1}{)}^2$

$=0.0128mol{L}^{-1}{s}^{-1}$