Nitrosyl chloride, NOCl, decomposes to NO and \({\bf{C}}{{\bf{l}}_{\bf{2}}}\).

\({\bf{2NOCl(g)}} \to {\bf{2NO(g) + C}}{{\bf{l}}_{\bf{2}}}{\bf{(g)}}\)

Determine the rate law, the rate constant, and the overall order for this reaction from the following data:

The rate law for decomposition of nitrosyl chloride depends on the concentration of nitrosyl chloride.

rate of reaction = \({\bf{k(NOCl}}{{\bf{)}}^{\bf{m}}}\)

\(\begin{align}Experiment\,1:8 \times {10^{ - 10}}mol{L^{ - 1}}{h^{ - 1}} &= k{[0.10]^m}\,\,\,\,\,......(1)\\Experiment\,2:3.2 \times {10^{ - 9}}mol{L^{ - 1}}{h^{ - 1}} &= k{[0.20]^m}\,\,......(2)\\Experiment\,3:7.2 \times {10^{ - 9}}mol{L^{ - 1}}{h^{ - 1}} &= k{[0.30]^m}\,\,......(3)\end{align}\)

Value of m examining from experiments 1 and 2. It found the rate increased by a factor of 4 and concentration is increased by a factor of 2

Value of m examining from experiments 1 and 3. It found the rate increased by a factor of 9, and concentration is increased by a factor of 3.

To calculate the overall order of reaction, equation (2) is divided by (1), and we get

\(\begin{align}\frac{{8 \times {{10}^{ - 10}}mol{L^{ - 1}}{h^{ - 1}} = k{{(0.10)}^m}}}{{3.2 \times {{10}^{ - 9}}mol{L^{ - 1}}{h^{ - 1}} = k{{(0.20)}^m}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{1}{4} &= {\left( {\frac{1}{2}} \right)^m}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,m &= 2\end{align}\)

Since the value of m is 2 hence, the order of the reaction will be second.

Reactions in which reactants are identical and form a product can also be a second-order reaction like the decomposition of nitrosyl chloride.

A second-order reaction rate is proportional to the square of the concentration of a reactant.

Hence, the rate of reaction for nitrosyl chloride is

Rate of reaction =\({\bf{k(NOCl}}{{\bf{)}}^{\bf{2}}}\)

The rate constant is the proportionality constant in the equation that expresses the relationship between the rate of a chemical reaction and the concentration of reacting substances.

From the given table, the value of the rate constant can be calculated as;

for experiment 1, rate constant

\(\begin{align}k &= \frac{{8 \times {{10}^{ - 10}}mol{L^{ - 1}}{h^{ - 1}}}}{{{{(0.10)}^2}}}\\ &= 8 \times {10^{ - 8}}mol{L^{ - 1}}{h^{ - 1}}\end{align}\)

for experiment 2, rate constant

\(\begin{align}k &= \frac{{3.2 \times {{10}^{ - 9}}mol{L^{ - 1}}{h^{ - 1}}}}{{{{(0.20)}^2}}}\\ &= 8 \times {10^{ - 8}}mol{L^{ - 1}}{h^{ - 1}}\end{align}\)

for experiment 3, rate constant

\(\begin{align}k &= \frac{{7.2 \times {{10}^{ - 9}}mol{L^{ - 1}}{h^{ - 1}}}}{{{{(0.30)}^2}}}\\ &= 8 \times {10^{ - 8}}mol{L^{ - 1}}{h^{ - 1}}\end{align}\)