Integrated Rate Law for Zero and First-Order Reactions: Can you determine how rapidly a pressure cooker cooks the food? How fast does an ice cream melt in your hand? Or what controls the rate at which diamond converts to graphite? The answer to all these questions depends on the rate at which a particular reaction proceeds and the various factors that affect the rate of a reaction. Let’s learn about the rate law, order, molecularity and the rate law for zero and first-order reactions.

Rate Law

The rate law is a mathematical expression in which the rate of the reaction is given in terms of the molar concentration of reactants with each term raised to some power, which may or may not be equal to the stoichiometric coefficient of the reacting species in a balanced chemical equation.

Let us consider a general reaction,

\({\rm{aA}}\,{\rm{ + bB}}\, \to {\rm{cC}}\,{\rm{ + }}\,{\rm{dD}}\)

The rate law for the above reaction is as follows,

\({\rm{Rate}}\,\alpha \,{[{\rm{A}}]^{\rm{x}}}{[{\rm{B}}]^{\rm{y}}}\)

Where \({\rm{a,b,c,d}}\) represents the stoichiometric coefficients of the reactants and the products, respectively, \({\text{x}}\) and \({\text{y}}\) may or may not be equal to the stoichiometric coefficients of the reactants \({\rm{(a}}\,{\rm{and}}\,{\rm{b)}}\).

The above reaction can also be written as:

\({\rm{Rate}}\,{\rm{ = }}\,{\rm{k[A}}{{\rm{]}}^{\rm{x}}}{{\rm{[B]}}^{\rm{y}}}\)

To express the rate at a particular time, we determine the instantaneous rate ( i.e. when \({\rm{\Delta t}}\, \to {\rm{0}}\)). Hence, for an infinitesimally small dt instantaneous rate is given by:

\( – \,\frac{{{\rm{d}}[{\rm{R}}]}}{{{\rm{dt}}}} = \,{\rm{k[A}}{{\rm{]}}^{\rm{x}}}{{\rm{[B]}}^{\rm{y}}}\)

As the reactants are getting consumed in the reaction, it is indicated by a negative sign.

Order of Reaction

The sum of the powers of the concentrations of the reactants in the rate law expression gives the order of a reaction. The rate law expression which is given by:

\({\rm{Rate}}\,\,{\rm{ = }}\,{\rm{k[A}}{{\rm{]}}^{\rm{x}}}{{\rm{[B]}}^{\rm{y}}}\)

\({\rm{x}}\,{\rm{and}}\,{\rm{y}}\) are the powers or the concentration coefficients in the general chemical equation. Hence, the order of a reaction \({\rm{(n)}}\) is given by:

\({\rm{n}}\,{\rm{ = }}\,{\rm{x}}\,{\rm{ + }}\,{\rm{y}}\)

The order of a reaction can take values \(0,\,1,2,\,3\) and even a fraction. A zero-order reaction is independent of the concentration of reactants.

Units of Rate Constants

From rate law expression, we know that:

\({\rm{Rate}}\,{\rm{ = }}\,{\rm{k[A}}{{\rm{]}}^{\rm{x}}}{{\rm{[B]}}^{\rm{y}}}\)

\({\rm{k}}\,{\rm{ = }}\,\,\frac{{{\rm{Rate}}}}{{{{{\rm{[A]}}}^{\rm{x}}}{{{\rm{[B]}}}^{\rm{y}}}}}\)

If \({\rm{[A]}}\,\,{\rm{ = }}\,\,{\rm{[B]}}\) then, \({{\rm{[A]}}^{{\rm{x + y}}}}\) where \({\rm{x}}\,{\rm{ + }}\,{\rm{y}}\,{\rm{ = }}\,{\rm{n}}\)

\({\rm{Rate}}\,\,{\rm{ = }}\,\,{\rm{k[A}}{{\rm{]}}^{\rm{n}}}\)

Considering the SI unit of concentration as mol \({{\text{L}}^{ – 1}}\)and time, \({\text{s}}\), the units of \({\text{k}}\) for different reaction order are listed below:

Order	Unit of rate constant
\({\rm{0}}\)	\(\frac{{{\rm{mol}}\,{{\rm{L}}^{{\rm{ – 1}}}}}}{{\rm{s}}}{\rm{ \times }}\frac{{\rm{l}}}{{{\rm{[mol}}\,{{\rm{L}}^{{\rm{ – 1}}}}{\rm{]}}\,}}{\rm{ = }}\,{\rm{mol}}\,{{\rm{L}}^{{\rm{ – 1}}}}{{\rm{s}}^{{\rm{ – 1}}}}\)
\({\rm{1}}\)	\(\frac{{{\rm{mol}}\,{{\rm{L}}^{{\rm{ – 1}}}}}}{{\rm{s}}}{\rm{ \times }}\frac{{\rm{l}}}{{{\rm{[mol}}\,{{\rm{L}}^{{\rm{ – 1}}}}{\rm{]}}\,}}{\rm{ = }}\,{{\rm{s}}^{{\rm{ – 1}}}}\)
\({\rm{2}}\)	\(\frac{{{\rm{mol}}\,{{\rm{L}}^{{\rm{ – 1}}}}}}{{\rm{s}}}{\rm{ \times }}\frac{{\rm{l}}}{{{\rm{[mol}}\,{{\rm{L}}^{{\rm{ – 1}}}}{\rm{]}}{\,^2}}}{\rm{ = }}\,{\rm{mo}}{{\rm{l}}^{ – 1}}{\rm{L}}\,{{\rm{s}}^{{\rm{ – 1}}}}\)

Molecularity of a Reaction

The number of reacting species that participate in an elementary reaction to bring about a chemical reaction is called the molecularity of a reaction. The reacting species consists of atoms, ions or molecules that must collide simultaneously to bring about the reaction. The reaction can be:

Molecularity	Number of reacting species	Examples
Unimolecular	\({\rm{1}}\)	Decomposition of ammonium nitrate \({\rm{N}}{{\rm{H}}_{\rm{4}}}{\rm{N}}{{\rm{O}}_{\rm{2}}} \to {{\rm{N}}_2} + 2{{\rm{H}}_{\rm{2}}}{\rm{O}}\)
Bimolecular	\({\rm{2}}\)	dissociation of hydrogen iodide \({\rm{2HI}}\, \to {{\rm{H}}_2} + {{\rm{I}}_2}\)
Trimolecular or termolecular	\({\rm{3}}\)	Formation of nitrogen dioxide \(2{\rm{NO}}{\mkern 1mu} {\mkern 1mu}  + {{\rm{O}}_2} \to 2{\rm{N}}{{\rm{O}}_{\rm{2}}}\)

However, reactions involving more than three molecules in the stoichiometric equation must occur in more than one step. In this case, the number of reacting species in the rate-determining step accounts for the molecularity of the reaction.

If two reacting species combine in the rate-determining step, then the reaction is bimolecular. If a single species makes up the transition state in the rate-determining step, the reaction would be called unimolecular. The relatively unlikely case of three independent species coming together in the transition state would be called termolecular.

For example: The catalytic decomposition of hydrogen peroxide by iodide ion in an alkaline medium.

\(2{{\rm{H}}_{\rm{2}}}{{\rm{O}}_2}\frac{{{{\rm{I}}^ – }}}{{{\rm{Alkaline}}\,{\rm{medium}}}}2{{\rm{H}}_{\rm{2}}}{\rm{O}}\,{\rm{ + }}\,{{\rm{O}}_2}\)

Experimental evidence suggests that this reaction takes place in two steps:

\((1)\,{{\rm{H}}_{\rm{2}}}{{\rm{O}}_2} + {\rm{I}} \to {{\rm{H}}_{\rm{2}}}{\rm{O}}\,{\rm{ + }}\,{\rm{I}}{{\rm{O}}^ – }\) [slow step-rate determining step]

\((2)\,{{\rm{H}}_{\rm{2}}}{{\rm{O}}_2} + {\rm{IO}} \to {{\rm{H}}_{\rm{2}}}{\rm{O}}\,{\rm{ + }}\,{{\rm{I}}^ – }{\rm{ + }}{{\rm{O}}_2}\)

Hence, the reaction rate is given by the first step, which is the rate-determining step of the reaction.

\({\rm{Rate}}\, = \, – \,\frac{{{\rm{d}}{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}}}{{{\rm{dt}}}} = {\rm{k[}}{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}{\rm{][}}{{\rm{I}}^{\rm{ – }}}{\rm{]}}\)

This reaction is bimolecular and first order with respect to both \({{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}\,{\rm{and}}\,\,{{\rm{I}}^ – }\).

Difference between Molecularity and Order

Molecularity	Order
The number of ions or molecules that participate in the rate-determining step to bring about a chemical reaction is molecularity.	The order of the reaction is given by the sum of powers to which the reactant concentrations are raised in the rate law equation.
The molecularity of a reaction is always a whole number. It can have values from \(1\) to \(3\). It cannot be zero or a non-integer.	It can either be a whole number or a fraction. The order of a reaction can be \({\rm{0,}}\,{\rm{1,}}\,{\rm{2,}}\,{\rm{3}}\) or even a fraction.
The molecularity of the reaction is determined by looking at the reaction mechanism.	The experimental methods determine the order of the reaction.
It is obtained by the rate-determining step, which is the slowest step.	It is obtained by adding the powers to which the reactant concentrations are raised in the rate law equation.
Molecularity is applicable only for elementary reactions.	The order applies to elementary as well as complex reactions.

For complex reactions, the rate law is given by the slowest step, the rate-determining step. The order is given by adding the powers of the reactant in the rate law equation of the slowest step. The molecularity of the slowest step is the same as the order of the overall reaction.

Integrated Rate Laws

Zero-Order Reaction

In a zero-order reaction, the rate of the reaction is independent of the concentration of the reactants. Thus, it means the sum of the powers of the reaction concentrations in the rate law equation is zero.

Let us consider a reaction, \({\rm{R}}\, \to {\rm{P}}\). Therefore, the rate law of this reaction is,

\({\rm{Rate}}\,\alpha \,{[{\rm{R}}]^0}\)

\(\therefore \,{\rm{Rate}}\,{\rm{ = }}\, – \,\frac{{{\rm{d[R]}}}}{{{\rm{dt}}}} = \,{\rm{k}}\)

\(\therefore  – {\rm{d}}\,{\rm{[R]}}\,{\rm{ = }}\,{\rm{kdt}}………….{\rm{Equation(1)}}\)

Integrating both sides of Eqn\({\rm{(1)}}\), we get:

\({\rm{[R]}}\,{\rm{ = }}\,\, – \,{\rm{kt}}\, + \,{\rm{I}}\,……………{\rm{Equation (2)}}\)

Here, I is the constant of integration. At \({\rm{t}}\,\,{\rm{ = }}\,\,{\rm{0}}\), the concentration of the reactant \({\rm{[R]}}\,\, = \,{{\rm{[R]}}_0}\) where \({{\rm{[R]}}_0}\) is the initial concentration of the reactant. Substituting this value in Eqn\({\rm{(2)}}\), we get-

\({{\rm{[R]}}_0}\, = \,{\rm{k}} \times 0 + {\rm{I}}\)

\({{\rm{[R]}}_0}\, = \,{\rm{I}}\)

Substituting this value of I in the Eqn\({\rm{(2)}}\), we get:

\(\left[ R \right] = – kt + {\left[ R \right]_0}\,\,\,\,…..{\rm{Equation}}\left( 3 \right)\)

Comparing Eqn\({\rm{(3)}}\) with the equation of a straight line \({\rm{y}}\,{\rm{ = }}\,{\rm{mx}}\,{\rm{ + c}}\), we get \({\rm{y}}\,{\rm{ = }}\,{\rm{[R],}}\,{\rm{m}}\,{\rm{ = }}\, – {\rm{k}}\) and \({\rm{x}}\,{\rm{ = }}\,{\rm{t}}\).

Plotting \({\rm{[R]}}\) against \({\rm{t}}\), we get a straight line with slope \({\rm{ = }}\, – {\rm{k}}\) and intercept \({\rm{ = }}\,\,{{\rm{[R]}}_0}\)

Therefore, on simplifying Eqn\((3)\), we get:

\({\rm{k}}\, = \,\frac{{{{[{\rm{R}}]}_0} – [{\rm{R}}]}}{{\rm{t}}}……………….{\rm{Equation}}(4)\)

Example of Zero Order Reaction

The decomposition of gaseous ammonia in the presence of a platinum catalyst at high pressure is a zero-order reaction.

\({\rm{Rate}}\, = \,{\rm{k}}{[{\rm{N}}{{\rm{H}}_3}]^0} = \,{\rm{k}}\)

As the reaction proceeds, the metal surface gets saturated with gas molecules due to high pressure. So, a further change in reaction conditions cannot alter the amount of ammonia on the surface of the catalyst, making the reaction rate independent of its concentration.

First Order Reaction

In a first-order reaction, the sum of the powers of concentrations of reactants in the rate law is equal to \(1\).

Let us consider the above reaction \({\text{R}} \to {\text{P}}\) again. Therefore, the rate law of this reaction is,

\({\rm{Rate}}\,{\rm{\alpha  [R]}}\)

\({\rm{Rate}}\, = \, – \,\frac{{{\rm{d}}[{\rm{R}}]}}{{{\rm{dt}}}} = {\rm{k}}[{\rm{R}}]\)

\( – \,\frac{{{\rm{d}}[{\rm{R}}]}}{{[{\rm{R}}]}} = {\rm{kdt}}\)

Integrating the above equation, we get:

\({\rm{ln[R]}}\,{\rm{ = }}\,{\rm{ – kt}}\,{\rm{ + }}\,{\rm{I}}……….{\rm{Equation(5)}}\)

Where I is the constant of integration.

When \({\rm{t}}\, = \,0\,[{\rm{R}}]\, = \,{[{\rm{R}}]_0}\) where, \({[{\rm{R}}]_0}\) is the initial concentration of the reactant. Hence, \({\rm{Eqn}}.(5)\) can be written as-

\(\ln {[{\rm{R}}]_0}\, = \, – \,{\rm{k}} \times 0 + {\rm{I}}\)

\({\rm{ln[R}}{{\rm{]}}_0}\, = \,{\rm{I}}\)

Substituting the value of I in Eqn. (5), we get:

\({\rm{ln[R]}}\,{\rm{ = }}\,\, – {\rm{kt}}\,{\rm{ + }}\,{\rm{ln[R}}{{\rm{]}}_{\rm{0}}}……………{\rm{Eqn(6)}}\)

Comparing \({\rm{Eqn(6)}}\) with the equation of a straight line \({\rm{y}}\,{\rm{ = }}\,{\rm{mx}}\,{\rm{ + c}}\) we get

\({\rm{y}}\,\,{\rm{ = }}\,\,{\rm{ln[R],}}\,{\rm{m}}\,{\rm{ = }}\, – {\rm{k}}\,and\,{\rm{x}}\,\,{\rm{ = }}\,{\rm{t}}\)

Plotting \({\rm{ln[R]}}\) against t, we get a straight line with slope \({\rm{ = }}\,\, – {\rm{k}}\) and intercept \({\rm{ = }}\,{\rm{ln}}\,{{\rm{[R]}}_0}\)

Eqn. (6) can also be written as:

\(\ln \,\frac{{[{\rm{R}}]}}{{{{[{\rm{R}}]}_0}}} = \, – {\rm{kt}}\)

Taking antilog on both the sides of the above equation, we get:

\({\rm{[R]}}\,\,{\rm{ = }}\,\,{{\rm{[R]}}_0}{{\rm{e}}^{{\rm{ – kt}}}}\,\)

Also, from Eqn. (6), we have:

\({\rm{k = }}\,\frac{{\rm{l}}}{{\rm{t}}}{\rm{ln}}\frac{{{\rm{[R]}}}}{{{{{\rm{[R]}}}_{\rm{0}}}}}…………..{\rm{Eqn}}.(7)\)

The above equation can also be written as:

\({\rm{k}} = {\mkern 1mu} \frac{{2.303}}{{\rm{t}}}\log {\mkern 1mu} \frac{{[{\rm{R}}]}}{{{{[{\rm{R}}]}_0}}}\)

\(\log \,\frac{{{{[{\rm{R}}]}_o}}}{{[{\rm{R}}]}} = \,\frac{{{\rm{kt}}}}{{2.303}}\)

Plotting a graph between \(\log \,\frac{{{{[{\rm{R}}]}_0}}}{{[{\rm{R}}]}}\,{\rm{vs}}\,{\rm{t}}\) the slope \({\rm{ = }}\,\frac{{\rm{k}}}{{2.303}}\)

At \({\rm{t}}\,{\rm{ = }}\,{{\rm{t}}_{\rm{1}}}\)

\(\ln {[{\rm{R}}]_1}\, = \, – {\rm{k}}{{\rm{t}}_1}\, + \,\ln \,{[{\rm{R}}]_0}………..{\rm{Equation}}.(8)\)

\(\ln {[R]_2}\, = \, – \,{\rm{k}}{{\rm{t}}_2}\, + \,\ln \,{[{\rm{R}}]_0}…………….{\rm{Equation}}.(9)\)

where, \({[{\rm{R}}]_1}\) and \({[{\rm{R}}]_2}\) are the concentrations of the reactants at the time \({{\rm{t}}_1}\) and \({{\rm{t}}_2}\) respectively.

Subtracting Eqn \((9)\) from Eqn \((8)\) we get-

\(\ln {[{\rm{R}}]_1}\, – \,\ln {[{\rm{R}}]_2} = \,{\rm{k}}{{\rm{t}}_1} – ( – {\rm{k}}{{\rm{t}}_2})\)

\(\frac{{\ln \,{{[{\rm{R]}}}_1}\,}}{{\ln \,{{[{\rm{R]}}}_2}\,}} = {\rm{k}}({{\rm{t}}_2} – {{\rm{t}}_1})\)

\({\rm{k = }}\frac{{\rm{1}}}{{\left( {{{\rm{t}}_{\rm{2}}}{\rm{ – }}{{\rm{t}}_{\rm{1}}}} \right)}}{\rm{ln}}\frac{{{{\left[ {\rm{R}} \right]}_{\rm{1}}}}}{{{{\left[ {\rm{R}} \right]}_{\rm{2}}}}}\)

Example of First Order Reaction

The hydrogenation of ethene is an example of a first-order reaction.

\({{\rm{C}}_2}{{\rm{H}}_4}\, + \,{{\rm{H}}_2} \to {{\rm{C}}_2}{{\rm{H}}_6}\)

Therefore, the rate of reaction for the above reaction is \({\rm{k}}[{{\rm{C}}_2}{{\rm{H}}_4}]\)

Summary

In a chemical reaction, the rate law gives the relationship between reaction rate and the concentrations of reactants. There are two types of rate laws – Differential rate law and Integrated rate law. The major difference between the integrated rate law and differential rate law is that the integrated rate law expresses the reaction rate as a function of the initial concentration of one or more reactants after a specific time, whereas the differential rate law expresses the reaction as a function of the change in concentration of one or more reactants during a particular period of time. For elementary reactions, the reaction order is determined experimentally through the rate law, and the number of reacting species gives the molecularity of the chemical reaction. However, in complex reactions (order > 3), order and molecularity are determined through the rate law of the slowest step, the rate-determining step. This article taught us the rate law, types of rate laws, order and molecularity of an elementary and complex reaction. We also learned to determine the integrated rate law of zero and first-order reaction.  

Frequently Asked Questions

Q.1. What is the integrated rate law formula of zero and first-order reactions?
Ans: The integrated rate law formula of zero and first-order reactions are-
Zero-order :\([{\rm{R}}]\, = \, – {\rm{kt}}\, + \,{[{\rm{R}}]_0}\)
First order: \(\ln [{\rm{R}}]\, = \, – {\rm{kt}}\,{\rm{ + }}\,{\rm{ln}}\,{{\rm{[R]}}_0}\,or\,\log \,\frac{{{{[{\rm{R}}]}_0}}}{{[{\rm{R}}]}} = \frac{{{\rm{kt}}}}{{2.303}}\)
Here, \(\,{{\rm{[R]}}_0}\) and \({[{\rm{R}}]}\) are the initial and final concentrations of the reactants, respectively.

Q.2. What is the difference between differential rate law and integrated rate law?
Ans: The key difference between the integrated rate law and differential rate law is that the integrated rate law expresses the reaction rate as a function of the initial concentration of one or more reactants after a specific time, whereas the differential rate law expresses the reaction rate as a function of the change in concentration of one or more reactants during a particular period of time.

Q.3. What is \({\rm{K}}\) in the integrated rate law?
Ans: \({\rm{K}}\) in the integrated rate law represents the rate constant of the reaction.

Q.4. What is the second-order integrated rate law?
Ans: The integrated rate law for the second-order reaction \({\rm{R}}\, \to \) products is: \(\frac{1}{{[{\rm{R}}]}} = \,{\rm{kt}}\,{\rm{ + }}\,\frac{1}{{{{[{\rm{R}}]}_0}}}\) Comparing the above equation with a straight line, \({\rm{y}}\, = \,{\rm{mx}}\,{\rm{ + c}}\) , a plot of the inverse of \({\rm{[R]}}\) as a function of time yields a straight line. The rate constant for the reaction can be determined from the slope of the line, which is equal to \({\rm{k}}\).

Q.5. How do you prove a reaction is first-order?
Ans: If a linear graph is obtained when the natural logarithm of a reactant concentration is plotted against time, the reaction is first order. The slope of such a linear graph is negative.

Q.6. Can the order of reaction be zero?
Ans: The reverse Haber process, which is the decomposition of ammonia into its constituents, is an example of a zero-order reaction. This is because its rate is independent of the concentration of ammonia. These are the reactions in which the rate of the reaction is independent of the concentration of the reactants.

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