Experimental Determination of Rate Law, Rate Constant and Order of Reaction: Ever imagined how much time a bud takes to bloom into a flower? What is the time taken by fossil fuels to turn into coal? A bud can blossom into a flower within a day; however, fossils take millions of years to become coal. The study of the kinetics of these processes requires accurate time measurement. Some reactions will be quick, while others will be slow – the reaction rate, which is determined by a rate law, is the speed of the reaction. We’ll learn about reaction rates, rate laws, the rate constant, and reaction order in this post.

Many factors influence the rate of a chemical reaction, including the nature (of reactivity) of reactants, surface area, temperature, concentration, and catalysts. Rate laws can be represented as a rate law equation for each individual chemical reaction to explain how the concentrations of reactants affect the rate of the reaction. It’s vital to remember that rate law can only be determined by experimentation! Let’s understand how the rate of the reaction is determined.

Determining the Rate Law from Experimental Data

A chain of experiments is executed with numerous preliminary concentrations of reactants to determine the rate law experimentally. The initial rate law is then measured for each of the reactions.

Let us consider the following results that have been obtained during the kinetic studies of the reaction: \(2{\text{A}} + {\text{B}} \to {\text{C}} + {\text{D}}\)

Experiment	\({\text{A}}\left[ {{\text{mol}}\,{{\text{L}}^{ – 1}}} \right]\)	\({\text{B}}\left[ {{\text{mol}}\,{{\text{L}}^{ – 1}}} \right]\)	Initial Formation of \({\text{D}}\left[ {{\text{mol}}\,{{\text{L}}^{ – 1}}{{\min }^{ – 1}}} \right]\)
\(1\)	\(0.1\)	\(0.1\)	\(6.0 \times {10^{ – 3}}\)
\(2\)	\(0.3\)	\(0.2\)	\(7.2 \times {10^{ – 2}}\)
\(3\)	\(0.3\)	\(0.4\)	\(2.88 \times {10^{ – 1}}\)
\(4\)	\(0.4\)	\(0.1\)	\(2.40 \times {10^{ – 2}}\)

Suppose the order of the reaction with respect to \({\text{A}}\) is \({\text{x}}\) and with respect to \({\text{B}}\) is \({\text{y}}{\text{.}}\) Therefore, the rate of the reaction is given by, \({\text{Rate}} = {\text{k}}{\left[ {\text{A}} \right]^{\text{x}}}{\left[ {\text{B}} \right]^{\text{y}}}\)
According to the question,
\(6.0 \times {10^{ – 3}} = {\text{k}}{\left[ {0.1} \right]^{\text{x}}}{\left[ {0.1} \right]^{\text{y}}} \ldots \ldots \ldots \ldots \ldots \ldots \left( i \right)\)
\(7.2 \times {10^{ – 2}} = {\text{k}}{\left[ {0.3} \right]^{\text{x}}}{\left[ {0.2} \right]^{\text{y}}} \ldots \ldots \ldots \ldots \ldots \ldots \left( {ii} \right)\)
\(2.88 \times {10^{ – 1}} = {\text{k}}{\left[ {0.3} \right]^{\text{x}}}{\left[ {0.4} \right]^{\text{y}}} \ldots \ldots \ldots \ldots \ldots \ldots \left( {iii} \right)\)
\(2.40 \times {10^{ – 2}} = {\text{k}}{\left[ {0.4} \right]^{\text{x}}}{\left[ {0.1} \right]^{\text{y}}} \ldots \ldots \ldots \ldots \ldots \ldots \left( {iv} \right)\)
Dividing eqn (iv) by eqn (i), we get-
\(\frac{{2.40 \times {{10}^{ – 2}}}}{{6.0 \times {{10}^{ – 3}}}} = \frac{{{\text{k}}{{\left[ {0.4} \right]}^{\text{x}}}{{\left[ {0.1} \right]}^{\text{y}}}}}{{{\text{k}}{{\left[ {0.1} \right]}^{\text{x}}}{{\left[ {0.1} \right]}^{\text{y}}}}}\)
\(4 = \frac{{{{\left[ {0.4} \right]}^{\text{x}}}}}{{{{\left[ {0.1} \right]}^{\text{x}}}}}\)
\(4 = {\left[ {\frac{{0.4}}{{0.1}}} \right]^{\text{x}}}\)
\({4^1} = {4^{\text{x}}}\)
\({\text{x = 1}}\)
Dividing eqn (iii) by eqn (ii), we get-
\(\frac{{2.88 \times {{10}^{ – 1}}}}{{7.2 \times {{10}^{ – 2}}}} = \frac{{{\text{k}}{{\left[ {0.3} \right]}^{\text{x}}}{{\left[ {0.4} \right]}^{\text{y}}}}}{{{\text{k}}{{\left[ {0.3} \right]}^{\text{x}}}{\text{k}}{{\left[ {0.2} \right]}^{\text{y}}}}}\)
\(4 = \frac{{{{[0.4]}^{^{\text{y}}}}}}{{{{[0.2]}^{^{\text{y}}}}}}\)
\({2^2} = {2^{\text{y}}}\)
\({\text{y}} = 2\)
Therefore, the rate law is-
\({\text{Rate}} = {\text{k}}{\left[ {\text{A}} \right]^1}{\left[ {\text{B}} \right]^2}\)
The order of the reaction is \({\text{x}} + {\text{y}} = 1 + 2 = 3\)
From experiment (I), we get-
\({\text{k}} = \frac{{6.0 \times {{10}^{ – 3}}\;{\text{mol }}{{\text{L}}^{ – 1}}\;{\text{mi}}{{\text{n}}^{ – 1}}}}{{\left( {0.1\;{\text{mol }}{{\text{L}}^{ – 1}}} \right){{\left( {0.1\;{\text{mol }}{{\text{L}}^{ – 1}}} \right)}^2}}} = 6.0\,{{\text{L}}^2}\;{\text{mo}}{{\text{l}}^{ – 2}}\;{\text{mi}}{{\text{n}}^{ – 1}}\)
From experiment (II), we get-
\({\text{k}} = \frac{{7.2 \times {{10}^{ – 2}}\;{\text{mol }}{{\text{L}}^{ – 1}}\;{\text{mi}}{{\text{n}}^{ – 1}}}}{{\left( {0.3\;{\text{mol }}{{\text{L}}^{ – 1}}} \right){{\left( {0.2\;{\text{mol }}{{\text{L}}^{ – 1}}} \right)}^2}}} = 6.0\,{{\text{L}}^2}\;{\text{mo}}{{\text{l}}^{ – 2}}\;{\text{mi}}{{\text{n}}^{ – 1}}\)
From experiment (III), we get-
\({\text{k}} = \frac{{2.88 \times {{10}^{ – 1}}\;{\text{mol }}{{\text{L}}^{ – 1}}\;{\text{mi}}{{\text{n}}^{ – 1}}}}{{\left( {0.3\;{\text{mol }}{{\text{L}}^{ – 1}}} \right){{\left( {0.4\;{\text{mol }}{{\text{L}}^{ – 1}}} \right)}^2}}} = 6.0\,{{\text{L}}^2}\;{\text{mo}}{{\text{l}}^{ – 2}}\;{\text{mi}}{{\text{n}}^{ – 1}}\)
From experiment (IV), we get-
\({\text{k}} = \frac{{2.40 \times {{10}^{ – 2}}\;{\text{mol }}{{\text{L}}^{ – 1}}\;{\text{mi}}{{\text{n}}^{ – 1}}}}{{\left( {0.4\;{\text{mol }}{{\text{L}}^{ – 1}}} \right){{\left( {0.1\;{\text{mol }}{{\text{L}}^{ – 1}}} \right)}^2}}} = 6.0\,{{\text{L}}^2}\;{\text{mo}}{{\text{l}}^{ – 2}}\;{\text{mi}}{{\text{n}}^{ – 1}}\)
Therefore, the rate constant is \({\text{k}} = 6.0\,{{\text{L}}^2}\;{\text{mo}}{{\text{l}}^{ – 2}}\;{\text{mi}}{{\text{n}}^{ – 1}}\) and the reaction is of the third order.
Let us consider the reaction between nitrogen monoxide gas and hydrogen gas to form nitrogen gas and water vapour.
\(2{\text{NO}}\left( {\text{g}} \right) + 2{{\text{H}}_2}\left( {\text{g}} \right) \to {{\text{N}}_2}\left( {\text{g}} \right) + 2{{\text{H}}_2}{\text{O}}\left( {\text{g}} \right)\)
For this reaction at \({1280^ \circ }{\text{C,}}\) the following data was collected.

Experiment	\(\left[ {{\text{NO}}} \right]\)	\(\left[ {{{\text{H}}_2}} \right]\)	Initial Rate \(\left( {{\text{M}}/{\text{S}}} \right)\)
\(1\)	\(0.0050\)	\(0.0020\)	\(1.25{\text{ }} \times {10^{ – 5}}\)
\(2\)	\(0.010\)	\(0.0020\)	\(5.00{\text{ }} \times {10^{ – 5}}\)
\(3\)	\(0.010\)	\(0.0040\)	\(1.00{\text{ }} \times {10^{ – 4}}\)

To determine the order and rate of the reaction, the initial concentrations of each reactant must be altered while the other is held constant. Assume the reaction order is \({\text{x}}\) with respect to \({\text{NO}}\) and \({\text{y}}\) with respect to \({{\text{H}}_2}.\) As a result, the reaction rate is given by,
\({\text{Rate}} = {\text{k}}{\left[ {{\text{NO}}} \right]^{\text{x}}}{\left[ {{{\text{H}}_2}} \right]^{\text{y}}}\)
According to the question,
\(1.25 \times {10^{ – 5}} = {\text{k}}{\left[ {0.005} \right]^{\text{x}}}{\left[ {0.002} \right]^{\text{y}}} \ldots \ldots \ldots \ldots \ldots \ldots \left( i \right)\)
\(5.00 \times {10^{ – 5}} = {\text{k}}{\left[ {0.01} \right]^{\text{x}}}{\left[ {0.002} \right]^{\text{y}}} \ldots \ldots \ldots \ldots \ldots \ldots \left( {ii} \right)\)
\(1.00 \times {10^{ – 4}} = {\text{k}}{\left[ {0.01} \right]^{\text{x}}}{\left[ {0.004} \right]^{\text{y}}} \ldots \ldots \ldots \ldots \ldots \ldots \left( {iii} \right)\)
Comparing experiments \(1\) and \(2\):
While the concentration of \({{\text{H}}_2}\) was held constant, the concentration of \({\text{NO}}\) was doubled.
Dividing eqn (i) by eqn (ii), we get-
\(\frac{{1.25 \times {{10}^{ – 5}}}}{{5.00 \times {{10}^{ – 5}}}} = \frac{{{\text{k}}{{\left[ {0.005} \right]}^{\text{x}}}{{\left[ {0.002} \right]}^{\text{y}}}}}{{{\text{k}}{{\left[ {0.01} \right]}^{\text{x}}}{{\left[ {0.002} \right]}^{\text{y}}}}}\)
\(0.25 = \frac{{{{\left[ {0.005} \right]}^{\text{x}}}}}{{{{\left[ {0.01} \right]}^{\text{x}}}}}\)
\(0.25 = {\left[ {0.5} \right]^{\text{x}}}\)
\({\left[ {0.5} \right]^2} = {\left[ {0.5} \right]^2}\)
\({\text{x = 2}}\)
As a result, the reaction order with respect to \({\text{NO}}\) is \(2.\)
In other words, rate \(\alpha {\left[ {{\text{NO}}} \right]^2}.\)
Comparing experiments \(2\) and \(3\):
The concentration of \({{\text{H}}_2}\) was doubled while \({\text{NO}}\) concentration was held constant.
Dividing eqn (ii) by eqn (iii), we get-
\(\frac{{5.00 \times {{10}^{ – 5}}}}{{1.00 \times {{10}^{ – 4}}}} = \frac{{{\text{k}}{{\left[ {0.01} \right]}^{\text{x}}}{{\left[ {0.002} \right]}^{\text{y}}}}}{{{\text{k}}{{\left[ {0.01} \right]}^{\text{x}}}{{\left[ {0.004} \right]}^{\text{y}}}}}\)
\(0.5 = \frac{{{{\left[ {0.002} \right]}^{\text{y}}}}}{{{{\left[ {0.004} \right]}^{\text{y}}}}}\)
\(0.5 = {\left[ {0.5} \right]^{\text{y}}}\)
\({\left[ {0.5} \right]^1} = {\left[ {0.5} \right]^{\text{y}}}\)
\({\text{y = 1}}\)
Therefore, the order of the reaction with respect to \({{\text{H}}_2}\) is \(1,\) or rate \(\alpha {\left[ {{{\text{H}}_2}} \right]^1}.\)
Both of these results are subsequently incorporated into the overall rate law.
\({\text{Rate}} = {\text{k}}{\left[ {{\text{NO}}} \right]^2}{\left[ {{{\text{H}}_2}} \right]^1}\)

Rate Constant

Once the rate law has been defined, the specific rate constant can be established by putting the data from any of the experiments into the rate law and solving for \({\text{k}}{\text{.}}\) From experiment \(\left( 1 \right),\) we get-
\({\text{k}} = \frac{{1.25 \times {{10}^{ – 5}}\;{\text{mol}}\;{{\text{s}}^{ – 1}}}}{{{{\left( {0.005\;{\text{mol}}} \right)}^2}\left( {0.002\;{\text{mol}}} \right)}} = 250\;\,{\text{mo}}{{\text{l}}^{ – 2}}\;{{\text{s}}^{ – 1}}\)

Order of the Reaction

The overall reaction is third order since the exponents add up to \(2 + 1 = 3.\) The reaction’s rate law has nothing to do with the balanced equation of the overall reaction. The coefficients of \({\text{NO}}\) and \({{\text{H}}_2}\) are both \(2,\) but the reaction order in relation to \({{\text{H}}_2}\) is just one. The units for the specific rate constant change depending on the reaction’s order. There are several reactions where the reaction rate is unaffected by the concentration of one of the reactants. With respect to that reactant, such a reaction is called zero-order.

Summary

The rate law of a particular reaction can be experimentally determined by conducting a series of experiments with various initial concentrations of reactants. Once the rate law is determined, the rate constant and order of the reaction can also be determined. Reactions can be first, second and zero-order with respect to the reactant. In this article, we learnt how the rate of the reaction could be determined by conducting experiments with a varying initial concentration of the reactants.

FAQs

Q.1. How do you determine the order of a reaction?
Ans: The overall order of the reaction is determined by adding up the individual orders. For example, if the reaction is first order with respect to the reactant \({\text{A}}\) and \({\text{B,}}\) the overall order is \(2.\) The overall order of such a reaction is of second order.

Q.2. How do you determine the rate of a reaction?
Ans: The rate of a reaction or reaction rate is calculated using the formula, rate \( = \Delta \left[ {\text{C}} \right]/\Delta {\text{t,}}\) where \(\Delta \left[ {\text{C}} \right]\) represents the change in the concentration of the product during time period \(\Delta {\text{t}}.\) Similarly, the reaction rate can be calculated by the disappearance of a reactant over time.

Q.3. What is the meaning of rate constant?
Ans: The rate constant is the proportionality constant in the rate law equation that expresses the link between the rate of a chemical reaction and the concentrations of the reacting species. It is also known as the specific rate constant.

Q.4. Why is it essential to determine the order of a reaction?
Ans: The overall order of reaction indicates how changing the concentration of the reactants changes the speed of the reaction. For higher orders of reaction, a slight change in the concentration of the reactants results in large changes in the rate of the reaction.

Q.5. What is the experimental rate law for the reaction?
Ans: The rate law can be calculated experimentally using the method of initial rates, which involves measuring the instantaneous reaction rate by mixing the reactants instantly. The process is repeated multiple times, changing the concentration of one component at a time.

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