Order of a Reaction: Half-Life Period, Differences, Methods & More

The Order of Reaction indicates how changing the concentration of the reactant will affect the reactions speed. Its determination carries essential applications in chemistry. It explains the relationship between the rate of chemical reactions and the species concentration in it.

The composition of the mixture of all the species in the reaction can be understood once the rate equation is obtained. In this article, let’s explore everything about the order of a reaction in detail. Continue reading to know more.

What is Order of a Reaction?

The rate of reaction is determined by the concentration of each reactant. The dependence of diverse reactions on the concentration of reactants leads to a classification of the reaction in terms of order. For example,

(i) If the reaction rate (as determined experimentally) is directly proportional to the first power to the concentration of a reactant \({\rm{A}}\) the reaction is said to be of the first order with respect to \({\rm{A}}\)

(ii) If the reaction rate is proportional to the second, third or half power of the concentration of \({\rm{A}}\), it is of the second, third or half order with respect to \({\rm{A}}\). In a similar manner, the order of the reaction can be determined in respect of the other reactants, viz. \({\rm{B,}}\,{\rm{C}}\), etc.

The total or overall order of a reaction is the sum of its orders in respect of the individual reactants. Though theoretically, reactions of high order are also possible but it is doubtful if reactions higher than third order exist.

In short, if

(i) \({\rm{r}}\,{\rm{ = }}\,{\rm{k[A]}}\) the reaction is of the first order
(ii) \({\rm{r}}\,{\rm{ = }}\,{\rm{k[A}}{{\rm{]}}^2}\) the reaction is of second order
(iii) \({\rm{r}}\,\,{\rm{ = }}\,\,{\rm{k[A][B]}}\) the reaction is of second order the reaction is of first order with respect to \({\rm{A}}\) the reaction is of first order with respect to \({\rm{B}}\).
(iv). \({\rm{r}}\,\,{\rm{ = }}\,\,{\rm{k[A}}{{\rm{]}}^3}\) the reaction is of third order
(v). \({\rm{r}}\,\,{\rm{ = }}\,\,{\rm{k[A}}{{\rm{]}}^2}[{\rm{B}}]\) the reaction is of first order with respect to \({\rm{B}}\) second-order with respect to \({\rm{A}}\) and third-order overall
(vi). \({\rm{r}}\,\,{\rm{ = }}\,\,{\rm{k[A]}}{[{\rm{B}}]^2}\) the reaction is of first order with respect to \({\rm{A}}\) second-order with respect to \({\rm{B}}\) and overall third order
(vii). \({\rm{r}}\,\,{\rm{ = }}\,\,{\rm{k[A]}}[{\rm{B}}][{\rm{C}}]\) the reaction is of first order with respect to \({\rm{A,}}\,{\rm{B}}\) and \({\rm{C}}\) but the total order of the reaction is three.

Thus, the order of a reaction can be defined as the sum of the power of the concentration terms.

Although most reactions have a whole number order, certain reactions having fractional order (example., \(1/2,\,3/2,\,\), etc.) are also known. For example, the ortho- para-hydrogen conversion reaction order is 3/2 since its rate is expressed as below.

\(\frac{{{\rm{dx}}}}{{{\rm{dt}}}} = {\rm{k}}{[{{\rm{H}}_2}]^{3/2}}\)

Similarly, the order of dissociation of acetaldehyde in the gas phase is \(3/2\).

\(\frac{{{\rm{dx}}}}{{{\rm{dt}}}} = {\rm{k}}{[{\rm{C}}{{\rm{H}}_3}{\rm{CHO}}]^{3/2}}\)

In some heterogeneous or surface reactions, the rate of reaction is independent of the concentration of the reactant; such reactions are said to be of zero order.

It is important to note that the order of a reaction should be determined experimentally; it cannot be deduced from the chemical equation. Thus, nitrous oxide decomposes as below according to \(2{{\rm{N}}_2}{\rm{O}} \to {\rm{2}}{{\rm{N}}_2} + {{\rm{O}}_2}\) which the reaction should be of second-order; but actually, the reaction is found to be of the first-order, i.e.,

Rate \( = \,{\rm{K}}\,{\rm{[}}{{\rm{N}}_2}{\rm{O]}}\)

This is explained based on the fact that the reaction takes place in the following two separate steps.

Now, since the second step is very rapid, the reaction rate is governed by the first step, which is the first order. There are several such reactions, and actually, most of the chemical reactions are complex and involve a large number of separate steps.

Furthermore, the order of the reaction is sometimes altered by the conditions. For example, consider the hydrolysis of an ester with an excess of water (say \(100\) moles of water to one mole of ester). The amount of the various constituents at the start and finish will be as follows:

\({\rm{C}}{{\rm{H}}_3}{\rm{COO}}{{\rm{C}}_2}{{\rm{H}}_5}\,\, + \,\,{{\rm{H}}_2}{\rm{O}} \to {\rm{C}}{{\rm{H}}_3}{\rm{COOH}}\,\,{\rm{ + }}\,\,{{\rm{C}}_2}{{\rm{H}}_5}{\rm{OH}}\)

Now since the concentration of water molecules nearly remains constant during the reaction, the term \([{{\rm{H}}_2}{\rm{O}}]\) in the following. Rate \( = {\rm{K}}[{\rm{C}}{{\rm{H}}_3}{\rm{COO}}{{\rm{C}}_2}{{\rm{H}}_5}][{{\rm{H}}_2}{\rm{O]}}\) equation can very safely be taken as constant; and thus, the equation takes the following form with the result, the reaction is

Rate \( = {\rm{k}}[{\rm{C}}{{\rm{H}}_3}{\rm{COO}}{{\rm{C}}_2}{{\rm{H}}_5}]\)

of the first order although, it involves two molecules. Thus, we can now define the order of reaction as the total number of molecules whose concentration changes during the reaction.

Half-Life Period of a First-Order Reaction

The half-life period of a first-order reaction \(\left( {{{\rm{t}}_{\frac{{\rm{1}}}{{\rm{z}}}}}} \right)\) : We know that, for 1st order reaction, rate constant, \({\rm{k}}\,{\rm{ = }}\,\frac{{2.303}}{{\rm{t}}}\log \frac{{{{\rm{C}}_0}}}{{\rm{C}}}\) can also be used to calculate the time taken for any given proportion of the reactant to disappear. In general half-life period of a reaction is the time required to convert the concentration \(({{\rm{C}}_0})\) of reactants to half \(({{\rm{C}}_0}/2)\). Thus, by substituting the value \({{\rm{C}}_0}/2\) in place of \({\rm{C}}\) in equation \({\rm{k}}\, = \,\frac{{2.303}}{{\rm{t}}}\log \frac{{{{\rm{C}}_0}}}{{\rm{C}}}\), the half-life of the reaction can be ascertained.

\({{\rm{t}}_{\frac{1}{{\rm{2}}}}} = \frac{{2.303}}{{\rm{k}}}\log \frac{{{{\rm{C}}_0}}}{{{{\rm{C}}_0}/2}}\)

\( = \frac{{2.303}}{{\rm{k}}}\log 2\)

\({{\rm{t}}_{\frac{{\rm{1}}}{{\rm{2}}}}}{\rm{ = }}\frac{{{\rm{0}}{\rm{.693}}}}{{\rm{k}}}\)

Note that the half-life period of a first-order reaction is independent of the initial concentration, a characteristic of the first-order reactions.

The quantity of the reactant present after two half-times will be \(1/2 \times 1/2\,{{\rm{C}}_0},\) i.e., \({(1/2)^2}\,{{\rm{C}}_0};\) similarly, after n half-times, the quantity of the reactant left, i.e., \({{\rm{C}}_{\rm{n}}}\) may be given by

\({{\rm{C}}_{\rm{n}}}\, = \,{[1/2]^{\rm{n}}} \times {{\rm{C}}_0}\)

Decomposition of \({{\rm{H}}_2}{{\rm{O}}_2}\) Follows a First-Order Reaction

The decomposition of hydrogen peroxide in an aqueous solution (catalysed by the presence of finely divided platinum) takes place according to the equation:

\({{\rm{H}}_2}{{\rm{O}}_2} \to {{\rm{H}}_2}{\rm{O}}\,{\rm{ + }}\,\frac{1}{2}{{\rm{O}}_2}\)

The kinetics of this reaction may be studied either by the same method as done earlier (that is, by collecting the oxygen gas produced and noting its volume at different intervals of time) or by making use of the fact that \({{\rm{H}}_2}{{\rm{O}}_2}\) solution can be titrated against \({\rm{KMn}}{{\rm{O}}_4}\) solution. Thus, by withdrawing equal amounts of the solution (usually \(5{\rm{cc}}\) at regular intervals of time and titrating against the same \({\rm{KMn}}{{\rm{O}}_4}\) solution, the amount of \({{\rm{H}}_2}{{\rm{O}}_2}\) present can be found every time. It is obvious that the same volume of the reaction solution was withdrawn.

Substituting these values in the first-order equation, we get

\({\rm{k}}\,\, = \,\,\frac{{2.303}}{{\rm{t}}}\log \frac{{\rm{a}}}{{({\rm{a}} – {\rm{x}})}}\)

As tested by this equation, the decomposition of hydrogen peroxide is found to be of the first order.

Difference Between Order and Molecularity of a Reaction

The difference between order and molecularity of reaction are tabulated below:

Methods of Determining the Order of a Reaction

The rate constant cannot be calculated until the order of the reaction is known. The latter can be determined by using one of the following methods.

Integration Method or Trial Method

This method consists of carrying out the reaction with known quantities of reactant and determining the amount of reactant consumed after different intervals of time and then substituting the data in the equation for first and second-order reactions derived earlier.

(i) Rate equations for first order reaction:

\({\rm{k}}\,\, = \,\,\frac{{2.303}}{{\rm{t}}}\log \frac{{{{\rm{C}}_0}}}{{\rm{C}}}\)

\({\rm{k}} = \frac{{2.303}}{{\rm{t}}}\log \frac{{\rm{a}}}{{{\rm{(a}}\,{\rm{ – }}\,{\rm{x)}}}}\)

(ii) Rate equation for second-order reactions

\({\rm{k}}\,\, = \,\,\frac{1}{{\rm{t}}}\left[ {\frac{1}{{\rm{C}}} – \frac{1}{{{{\rm{C}}_0}}}} \right]\) or

\({\rm{k}}\,\, = \,\,\frac{1}{{\rm{t}}}\left[ {\frac{1}{{({\rm{a}} – {\rm{x}})}} – \frac{1}{{\rm{a}}}} \right]\)

The equation, which gives the constant value of k, the velocity constant, indicates the appropriate order of the reaction.

Alternatively, the validity of the various rate equations can be tested by the graphical method. Since the initial concentration \({{\rm{C}}_0}\) or a is constant for each complete experiment, the equation for first and second-order reduce to:

\({\rm{k}} = \frac{{2.303}}{{\rm{t}}}\log ({\rm{a}} – {\rm{x}}) + \) constant, for first-order reactions \({\rm{k}} = \frac{1}{{\rm{t}}}.\frac{1}{{({\rm{a}} – {\rm{x}})}} + \) constant, for second-order reactions

(i) Thus, the first-order reaction will give a straight line when the \(\log ({\rm{a}} – {\rm{x}})\) is plotted versus \({\rm{t}}\).

(ii) While the second-order reaction will give a straight line when \(\frac{1}{{({\rm{a}} – {\rm{x}})}}\) is plotted versus \({\rm{t}}\).

However, the integration method suffers because it cannot determine the fractional order of their reactions.

Half-Life Method (Fractional Change Method)

As discussed, the earlier half-life method period of a reaction (that is, the time required to convert the original concentration of the reactants to half) of different orders follows the following relationship.

For first-order reaction \({{\rm{t}}_{\frac{1}{{\rm{2}}}}} = \frac{{0.693}}{{\rm{k}}}\)

For second order reaction \({{\rm{t}}_{\frac{1}{{\rm{2}}}}} = \frac{1}{{{\rm{ka}}}}\)

Thus, the half-life (or any definite fraction) period of a reaction is

(i) Independent of the initial concentration for a first-order reaction
(ii) Inversely proportional to the initial concentration in case of a second-order reaction.

Thus, in general, \({{\rm{t}}_{\rm{f}}}{\rm{\alpha }}{\left( {\frac{1}{a}} \right)^{{\rm{n}} – 1}}\)

Where n is the order of the reaction, \({{\rm{t}}_{\rm{f}}}\) is the time required to complete a definite fraction (say half) of the reactant, and a is the initial concentration of the reactant.

Suppose separate experiments are carried out starting with different concentrations of the reactant. In that case, the order can be determined by finding the time taken to complete any definite fraction (not necessarily half) of the reactants. Let \({{\rm{t}}_{\rm{1}}}\) and \({{\rm{t}}_{\rm{2}}}\) are the time taken to complete the same fraction of the change with initial concentration \({{\rm{a}}_1}\) and \({{\rm{a}}_2}\) Then according to general relationship \({{\rm{t}}_{\rm{f}}}\,{\rm{\alpha }}\,{\left( {\frac{1}{{\rm{a}}}} \right)^{{\rm{n}} – 1}}\)

\({{\rm{t}}_{1/2}}{\rm{\alpha }}\,\frac{1}{{{{\rm{a}}_1}^{{\rm{n}} – 1}}}\)

And \({\rm{t}}{{\rm{‘}}_{1/2}}{\rm{\alpha }}\,\frac{1}{{{{\rm{a}}_2}^{{\rm{n}} – 1}}}\)

\(\frac{{{{\rm{t}}_{1/2}}}}{{{\rm{t}}{‘_{1/2}}}} = \frac{{{{\rm{a}}_2}^{{\rm{n}} – 1}}}{{{{\rm{a}}_1}^{{\rm{n}} – 1}}} = {\left( {\frac{{{{\rm{a}}_2}}}{{{{\rm{a}}_1}}}} \right)^{{\rm{n}} – 1}}\)

Taking logarithms, \(\log \frac{{{{\rm{t}}_{1/2}}}}{{{\rm{t}}{‘_{1/2}}}} = ({\rm{n}} – 1)\log \frac{{{{\rm{a}}_2}}}{{{{\rm{a}}_1}}}\)

\({\rm{n}} = 1 + \frac{{\log \frac{{{{\rm{t}}_{1/2}}}}{{{\rm{t}}{‘_{1/2}}}}}}{{\log \frac{{{{\rm{a}}_2}}}{{{{\rm{a}}_1}}}}}\)

\({\rm{n}} = 1 + \frac{{\log {{\rm{t}}_{1/2}} – \log \,{\rm{t}}{‘_{1/2}}}}{{\log \,{{\rm{a}}_2} – \,\log \,{{\rm{a}}_1}}}\)

Where \({\rm{n}}\) is the order of reaction and thus can be calculated by knowing the values of \({{\rm{a}}_1},\,{{\rm{a}}_2},\,\log {{\rm{t}}_{1/2}},\,{\rm{t}}{‘_{1/2}}\)

Vant Hoff’s Differential Method

From our earlier discussion, we know that the reaction rate varies as the nth power of the concentration of the reactant, where n is the order of the reaction. Thus, in general.

\({\rm{r}}{\mkern 1mu}  = {\mkern 1mu} \, – {\mkern 1mu} \frac{{{\rm{dC}}}}{{{\rm{dt}}}}{\mkern 1mu}  = {\mkern 1mu} {\rm{k}}{{\rm{C}}^{\rm{n}}}\)

For two different concentrations \({{\rm{C}}_1}\) and \({{\rm{C}}_2}\) the rates \({{\rm{r}}_1}\) and \({{\rm{r}}_2}\) will be given by

\({{\rm{r}}_1} = \, – \,\frac{{{\rm{d}}{{\rm{C}}_1}}}{{{\rm{dt}}}} = {\rm{k}}{{\rm{C}}_1}^{\rm{n}}\)

\({{\rm{r}}_2} = \, – \,\frac{{{\rm{d}}{{\rm{C}}_2}}}{{{\rm{dt}}}} = {\rm{k}}{{\rm{C}}_2}^{\rm{n}}\)

On taking logarithms, we get

\(\log \,{{\rm{r}}_1}\, = \,\log {\rm{k}}\,{\rm{ + }}\,{\rm{nlog}}\,{{\rm{C}}_1}\)

and \(\log \,{{\rm{r}}_2}\, = \,\log {\rm{k}}\,{\rm{ + }}\,{\rm{nlog}}\,{{\rm{C}}_2}\)

On subtracting equation \(\log \,{{\rm{r}}_1}\, = \,\log {\rm{k}}\,{\rm{ + }}\,{\rm{nlog}}\,{{\rm{C}}_1}\) from equation \(\log \,{{\rm{r}}_2}\, = \,\log {\rm{k}}\,{\rm{ + }}\,{\rm{nlog}}\,{{\rm{C}}_2}\)

we get

\(\log \,{{\rm{r}}_2}\, – \,\log \,{{\rm{r}}_1}\, = \,{\rm{n}}\,\log \,{{\rm{C}}_2}\, – \,{\rm{n}}\,\log \,{{\rm{C}}_1}\)

\(\log {{\rm{r}}_2}\, – \,\log {{\rm{r}}_1}\,{\rm{ = }}\,\,{\rm{n}}\left( {\log {\mkern 1mu} {{\rm{C}}_2} – \log {\mkern 1mu} {{\rm{C}}_1}} \right)\)

\({\rm{n}}\, = \,\frac{{\log \,{{\rm{r}}_2}\, – \,\log \,{{\rm{r}}_1}}}{{\log \,{{\rm{C}}_2} – \log \,{{\rm{C}}_1}}}\)

Thus, in short, the method involves the determination of the concentration of a reactant at different intervals of time.

The values of \({{\rm{r}}_1}\) and \({{\rm{r}}_2}\) are evaluated by plotting concentration-time graphs corresponding to \({{\rm{C}}_1}\) and \({{\rm{C}}_2}\) concentration. Substitution of these values in equation \({\rm{n}}\, = \,\frac{{\log \,{{\rm{r}}_2} – \log \,{{\rm{r}}_1}}}{{\log \,{{\rm{C}}_2} – \,\log \,{{\rm{C}}_1}}}\) leads to the evaluation of the value of \({\rm{n}}\).

Ostwald’s Isolation Method

This method is applicable for reactions involving two or more reactants. In this method, the experiment is carried out several times, each time taking all the reactants, except one in excess turn-by-turn. The reactant which is not taken in excess is said to be isolated. The kinetics of the reaction then gives the order of the reaction to the isolated reactant. The orders found for the different reactants are then added up to get the overall order of the reaction.

Consider a reaction involving three reactants \({\rm{X,}}\,{\rm{Y}}\) and \({\rm{Z}}\). Further, the order of the reaction be \({{\rm{n}}_1}\) where \({\rm{X}}\) is isolated, \({{\rm{n}}_2}\) when \({\rm{Y}}\) is isolated and \({{\rm{n}}_3}\) when \({\rm{Z}}\) is isolated. Thus, the overall order of the reaction will be \({{\rm{n}}_1} + {{\rm{n}}_2} + {{\rm{n}}_3}\).

Summary

The order of the reaction is the power dependence of the rate on the concentration of reactants. Therefore, for a first-order reaction to occur, the dependence of the rate is certainly on the concentration of a single species. In this article, we learnt about the half-life period of a first-order reaction, decomposition of hydrogen peroxide, methods of determining the order of a reaction, differentiation between the order of a reaction and molecularity of a reaction.

Frequently Asked Questions

Q.1. How do you determine the order of a reaction example?
Ans: There is a relation between the reaction rate and the concentration of a reactant by the rate constant. Furthermore, its representation is by the letter K. Moreover, there is a constant change in the rate as the temperature changes.
However, the rate remains the same when change occurs but only in the concentration. The reaction takes place at constant pressure and temperature. For this, the rate equals the rate constant times the concentration of reactants to the power. Mostly, this power is of the order of each reactant.
Example: It is simple to calculate the iodine reaction rate. This is because after the reaction is complete, the solution in the reaction container becomes blue. Furthermore, the time it takes for the reaction to turn blue is proportional to the reaction rate.

Q.2. How do you determine the order of a reaction from experimental data?
Ans: The integrated rate law or the differential rate law can determine the reaction order from experimental data. Often, the exponents in the rate law are the positive integers: \(1\) and \(2\) or even \(0\). Thus, the reactions are zeroth, first, or second-order in each reactant.

Q.3. How do you know if it is a first-order reaction?
Ans: Plot the natural logarithm of a reactant concentration versus time and see if the graph is linear, which indicates that the reaction is a first-order reaction. The reaction must be a first-order reaction if the graph is linear with a negative slope.

Q.4. What are the examples of zero-order reactions?
Ans: Photochemical reactions: Photochemical reactions are those reactions that take place in the presence of light. The combination of hydrogen and chlorine in the presence of light to form hydrogen chloride is a typical example.
\({{\rm{H}}_2} + {\rm{C}}{{\rm{l}}_2} \to 2{\rm{HCl}}\)
Heterogeneous reactions: A heterogeneous reaction means a reaction in which the reactants, products and catalysts are present in different phases. Some of these reactions are of zero order. Two well-known examples are
(a) the decomposition of hydrogen iodide on the surface of gold.
(b) the decomposition of ammonia on the surface of tungsten, molybdenum or platinum at high temperatures.

Q.5. What is a second-order reaction? Give an example?
Ans: Second order reactions can be defined as chemical reactions wherein the sum of the exponents in the corresponding rate law of the chemical reaction is equal to two. The rate of such a reaction can be written either as \({\rm{r}}\, = \,{\rm{k}}{[{\rm{A}}]^2}\), as \({\rm{r}}\, = \,{\rm{k}}[{\rm{A}}]\,[{\rm{B}}]\).
Gaseous reactions of second order are very common. Some important examples are thermal dissociation of oxygen, ozone, chlorine monoxide, nitrous oxide, formaldehyde, acetaldehyde, hydrogen iodide, nitrogen oxide, etc. Similarly, the combination of hydrogen and iodine to form hydrogen iodide, polymerization of ethylene, hydrogenation of ethylene, etc., are also examples of second-order reactions.

Q.6. Why is the reaction of higher-order rare?
Ans: The reactions of higher order are rare because many body collisions have a very low probability. The chances of three or more molecules colliding simultaneously to give a product are very low.