From the following data, determine the rate law, the rate constant, and the order with respect to A for the reaction \({\bf{A}} \to {\bf{2C}}\).

Rate law: General rate of reaction = \({\bf{k(A}}{{\bf{)}}^{\bf{m}}}\)

\(\begin{align}Experiment\,1:3.80 \times {10^{ - 7}}mol{L^{ - 1}}{h^{ - 1}} &= k{(1.33 \times {10^{ - 2}})^m}\\Experiment\,2:1.52 \times {10^{ - 6}}mol{L^{ - 1}}{h^{ - 1}} &= k{(2.66 \times {10^{ - 2}})^m}\\Experiment\,3:3.42 \times {10^{ - 6}}mol{L^{ - 1}}{h^{ - 1}} &= k{(3.99 \times {10^{ - 2}})^m}\end{align}\)

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), we get

\(\begin{align}\frac{{3.80 \times {{10}^{ - 7}}mol{L^{ - 1}}{h^{ - 1}} = k{{(1.33 \times {{10}^{ - 2}})}^m}}}{{1.52 \times {{10}^{ - 6}}mol{L^{ - 1}}{h^{ - 1}} = k{{(2.66 \times {{10}^{ - 2}})}^m}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{1}{4} &= {(\frac{1}{2})^m}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,m &= 2\end{align}\)

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.

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

Hence, the rate of reaction for a given reaction is

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

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

\(k = \frac{{3.80 \times {{10}^{ - 7}}mol{L^{ - 1}}{h^{ - 1}}}}{{{{(1.33 \times {{10}^{ - 2}})}^2}}} = 2.14 \times {10^{ - 3}}mol{L^{ - 1}}{h^{ - 1}}\)

For experiment 2, rate constant

\(k = \frac{{1.52 \times {{10}^{ - 6}}mol{L^{ - 1}}{h^{ - 1}}}}{{{{(2.66 \times {{10}^{ - 2}})}^2}}} = 2.14 \times {10^{ - 3}}mol{L^{ - 1}}{h^{ - 1}}\)

For experiment 3, rate constant

\(k = \frac{{3.42 \times {{10}^{ - 6}}mol{L^{ - 1}}{h^{ - 1}}}}{{{{(3.99 \times {{10}^{ - 2}})}^2}}} = 2.14 \times {10^{ - 3}}mol{L^{ - 1}}{h^{ - 1}}\)