Higher Order of Reaction, Chemistry tutorial

Zeroth Order Reaction:

It is simple to spot a Zeroth order reaction as the rate stays constant even although the reactant is being employed up. The rate law for a Zeroth order reaction is of the given form:

-d[A]/dt = k[A]o = k

As [A]o = 1

It will be noted that Zeroth order reaction is usually a heterogeneous reaction.

In the Zeroth reaction, the reaction rate is independent of the concentrations of the reactant.

Some illustrations of Zeroth order reactions are described below:

a) Decomposition of the ammonia on a hot platinum surface.

2NH3 (g) → (Pt) → 2N2 (g) + 3H2 (g)

b) Decomposition of the nitrous oxide on a hot platinum surface.

2N2O (g) → (Pt) → 2N2 (g) + O2 (g)

c) Decomposition of the hydrogen iodide on finely divided gold at 320 K

2HI (g) → (Au) → H2 (g) + I2 (g)

Integrated Rate Law for Zeroth Order Reaction:

Let us taken the Zeroth order reaction:

A → Products

Suppose the concentration of A at the starting be [A]o and its concentration at the time t be [A]t.

The integrated form of equation -d[A]/dt = k[A]o = k can be derived as:

[A]o[A]t - d[A] = ot kdt

- [A]o{[A]}[A]t = k (t - 0)

- [A]t + [A]o = kt

That is, [A]o - [A]t = kt

Or [A]t = [A]o - kt

On plotting [A]t, against t, a straight line is acquired for a Zeroth order rejection. The slope s is equivalent to - k

K = Slope

Half-Life of a Zeroth Order Reaction:

As explained for the first and second order reactions,

[A]t = [A]o/2 when t = t1/2

Therefore, Equation [A]t = [A]o - kt becomes,

[A]o/2 - [A]o = - kt1/2

Or, kt1/2 = [A]o/2

t1/2 = [A]o/2k

This means that the half-life of a Zeroth order reaction is directly proportional to the initial concentration of the reactant.

Third Order Reactions:

There are not many third order reactions; two illustrations are described below:

2NO (g) + O2 (g)   → 2NO2 (g)

2NO (g) + Br2 (g) → 2NOBr (g)

The procedures of arriving at differential and integrated rate laws for the third order reactions are identical to those of first and second order reactions.

Pseudo-First Order Reactions:

One of the modes of simplifying the study of reactions comprising more than one reactant is because of the study the pseudo-first order conditions. For illustration, considered the reaction: 

CH3Br (aq) + OH- (aq) → CH3OH (aq) + Br (aq)

Reactions rate = k [CH3Br] [OH-]

Whenever the concentrations of CH3Br and OH- are comparable, then the reaction is second order. Assume that the concentration of HO- is much bigger (say, 10 times or more) than the concentration of CH3Br. In such conditions, the concentration of OH- will not change much throughout the reaction and can be considered constant. Therefore, the reaction rate depends only on the concentration of CH3Br.

Reaction rate = k' [CH3Br]

Here k' = k [HO-]; k' is the pseudo-first order rate constant. The reaction can be treated as first order for computation purposes. Therefore, the integrated rate law is:

log [CH3Br]0/[CH3Br]t = k't/2.303

In common, the reactions similar to the above that are efficiently first order because of huge excess of one of the reactions are termed as pseudo-first order reactions.

Let us now discuss the given two reactions which are studied under pseudo-first order condition.

A) Acid Hydrolysis of Ester:

The hydrolysis of ethyl acetate in the presence of a mineral acid (state, HCl) can be symbolized by the given equation:

CH3COOC2H5 + H2O + H3O+ → CH3COOH + C2H5OH + H3O+

The reaction rate mainly depends on [ester], [water] and [H3O+]. Here H3O+ ion is a catalyst. As the concentration of the catalyst doesn't change throughout the reaction, and water is present in huge amount, the reaction becomes pseudo first order in ester.

Rate = k'[ester]

Here k' comprises concentration and H3O+. If the reaction is taken on a solvent other than water, the first order dependence on [water] as well could be seen.

The pseudo-first order rate constant is found out by titrating a definite volume of the reaction mixture having ester to HCl with standard alkali. Suppose V0, Vt and V are the volumes of standard alkali at the beginning, after a time t and after the completion of the reaction.

It will be noted that infinite reading (V) is generally taken after heating the reaction mixture for some minutes or after keeping the reaction mixture at the experimental temperature for a long time.

V = Volume of alkali equivalent to:

  • Acetic acid discharged after the completion of the reaction and
  • HCl present

Vt = Volume of alkali equivalent to:

  • Acetic acid generated at the time t and
  • HCl present

V0 = Volume of the alkali equal to only volume HCl.

As the concentration of HCl is constant all through the experiments,

[A]0 (that is, Initial concentration of ester)

α (V - V0) and [A]t  (that is, concentration of ester remaining unreached at t) α (V - V0)

We can compute pseudo-first order rate constant for the acid hydrolysis of ethyl acetate by employing the given modified form of:

K't/2.303 = log [A]o/[A]t = log (V - V0)/(V - V0)

Here k' is the pseudo first order rate constant.

B) Inversion of Sucrose:

The hydrolysis of sucrose to prepare glucose and fructose in the presence of mineral acid is identical to the acid hydrolysis of ester to the extent that the reaction kinetics is concerned.

C12H22O11 + H2O + H3O+ → C6H12O6 + C6H12O6 + H3O+

Sucrose                                  Glucose      Fructose

Sucrose turns the plane-polarized light to the right, (that is, it is dextro-rotatory). Glucose as well turns the plane-polarized light to the light, whereas fructose turns it to the left (that is, it is laevo-rotatory). On completion of the reaction, the reaction mixture is laevo-rotatory, as the angle of rotation is more for fructose than for glucose. To beginning with, the reaction mixture is dextrorotatory because of sucrose. Therefore, the completion of reaction (that is, infinite reading) is marked by the change in the sign of rotation. Because of this reason, the reaction is termed as inversion of sucrose.

If r0, rt and r are, the angles of rotation at the starting, after time t; and after completion of the reaction, then the pseudo first order rate constant (k') for the inversion of sucrose is represented by:

K' = 2.303/t log (ro - r)/(rt - r)

The equation above is a modified form of (= 2.303/t log [A]o/[A]t), here [A]o is proportional to (r0 - r) and [A]t and is proportional to (r0 - r).

Determining the Order of Reaction:

In order to represent or write the rate law, we should know the order of reaction with respect to each and every reactant. In this part, we shall illustrate a few methods for finding out the order of reaction.

1) Method of Initial Rates:

The instantaneous rate of reaction extrapolated to the instant if the reagents were just mixed is known as the initial rate of the reaction. Let us consider the reaction:

A + B → Products

Suppose that the rate of reaction be represented as:

v = k[A]m [B]n

Here the reaction is mth order in A and nth order in B. The rate constant for the reaction is k. We have to get the initial rates from at least two experiments in which the initial concentrations of A (a1 and a2) are different whereas the initial concentration of B (b1,) is constant

Rate in Experiment I, v1 = ka1m b1n

Rate in Experiment II, v2 = ka2m b1n

From the ratio (v1/v2), we can compute m, as a1 and a2 are well-known.

 (Rate in Experiment I/Rate in Experiment II) = v1/v2 = (ka1m b1n/ka2m b1n)

                                                                                    = (a1/a2) m

Taking logarithms we can represent,

log (v1/v2) = m log (a1 - a2)

Extrapolation is the procedure of extending a curve up to a desired x or y coordinate to acquire the corresponding y or x value.

Initial reaction rate could be graphically arrived at via plotting the concentration of a reactant against time. The tangent to the concentration curve is drawn at the beginning of the reaction and its slope is computed. The negative of the slope value is the initial rate.

Likewise, the rate is computed for one more experiment in which the initial concentration of A is a2 and the initial concentration of B is b2.

Therefore, rate in experiment III, v3 = k a2m b2n

(Rate in experiment II/rate in experiment III) = v2/v3 = (ka2m b1n/k a2m b2n)

                                                                                     = (b1/b2)n

Taking logarithms we can write,

log (v2/v3) = n log (b1 - b2)

As v2, v3, b1 and b2 are known, 'n' can be computed. The overall reaction order = m + n.

Fast Reactions:

Most of the reactions are of fast which ordinary experimental methods are inadequate to make measurement of reaction rates or the rate constants. These reactions are known as fast reactions and the half-life periods of fast reactions are less than 10-2s. Some of the special methods employed for measuring the constants of fast reactions are:

  • Flash photolysis
  • Flow methods
  • Relaxation methods

Flow Methods:

Two methods are available under flow method. In continuous flow method (figure below) the reacting solutions or gases are taken in separate containers (A and B) and are allowed to flow via the mixing chamber (C) to an observation tube (D). At different points all along the observation tube, the composition of the mixture is found out by some physical procedures.

The rate constant can be computed by employing any one of the three rates named above.

v1 = 3.0 × 10-3 Ms-1 = k (0.10M) (0.10 M)2

k = (3.0 x 1-3)/(0.10)3 M-2 s-1 = 3.0 M-2 s-1

1261_Apparatus for Continuous Flow Method.jpg

Fig: Apparatus for Continuous flow method

Care should be taken in applying the procedure or techniques of initial rates. For complex reactions such as the formation of HBr, the product as well influences the rate. The procedure of initial rates is applicable to only simple reactions.

A) Clock Reactions:

In the simplicity of certain reactions, the time taken for the colour change of the reaction mixture can be employed for measuring the initial rate. Such self- indicating reactions are termed as clock reactions. For illustration, the kinetics of the given reaction can be monitored as a clock reaction. A reaction mixture is made by employing potassium perdisulphate and potassium iodide in a higher concentration and sodium thiosulphate in much lower concentration. A group of (starch is as well present in the reaction mixture. The reaction mixture colourless in the starting and it turns blue after some time.

2KI + K2S2O8 → 2K2SO4 + I2



The time, 'Δt', between the mixing of reactants and the appearance of blue colour noted. The blue colour forms because of the discharge of free iodine, after sodium thiosulphate (that is, present in less concentration) is consumed fully as per the reaction:

2 Na2S2O0 + I2 → 2NaI + Na2S4O6



The order of reaction with respect to KI (m) and the order of the reaction with respect to K2S2O8 (n) can be computed by employing the given formula:

log (1/Δt) = m log [KI] + n log [K2S2O8] + constant

A plot of log (1/Δt) against log [KI] is prepared by employing Δt values acquired by varying [KI] and keeping [K2S2O8] constant. The slope of the straight line provides m. Likewise n is acquired from the slope of the straight line got by plotting log (1/Δt) against [K2S2O8]. For the second plot, Δt is acquired by varying [K2S2O8] and keeping [KI] constant.

Experimentally it has been determined that m = 1 and n = 1

Therefore, -d[K2S2O8] = k [KI] [K2S2O8]

For clock reactions, the two more illustrations are represented below:

  • Acid catalyzed iodination of acetone
  • Saponification of ester (by employing phenolphthalein indicator).

B) Trial and Error Method:

We can find out the order of reaction:

a) By replacing the experimental data into equations {(K't/2.303) = log [A]0/[A]t}, {1/[A]t + 1/[A]o = kt and [A]t = [A]0 - kt

b) By graphical technique using plots like log [A] against t, 1/[A] against t and [A] against t. The order of the reaction is one, two or zero based on

  • Which of the equations gives rise to a constant value for k or
  • Which of the plots gives a straight lineΔ

C) Half-Life Method

The half-lives are found out by using various initial concentrations of the reactant. Whenever the half-life is independent of initial concentration, the reaction is of first order. If the half-life is inversely proportional to the power of initial concentration, then the reaction is of second order. If the half-life is directly proportional to the first power of initial concentration, the reaction is Zeroth order.

In common, half-life period (t) is proportional to [A]1-n where [A]0 is the initial concentration of the reactant and n is the order of the reaction.

If the half-life periods are t1 and t2 corresponding to the initial concentrations

[A]1 and [A]2 of a reactant, then

t2/t1 = ([A]2/[A]1)1-n = ([A]2/[A]1)n-1

n - 1 = log (t2/t1)/{log[A]1/[A]2}

n = 1 + log (t2/t1)/{log[A]1/[A]2}

D) Isolation Method:

In case of reaction containing more than one reactant, the rate law can be simplified whenever the concentrations of all reactants apart from one are taken in surplus. The reaction rate then based on the reactant present in lesser quantity. The order of the reaction is found out by one of the methods given above. This is equivalent to the order in the reactant present in lesser quantity. This method is repeated in turn with each of the reactants being in lesser amount whereas others are in surplus. The procedure is known as Vant Hoff's isolation method.

For illustration, consider the reaction,

A + B → Products

For which the rate law is,

Rate = k [A]m [B]n

In the first set of experiments B is in surplus as compared to A, in such a way that the rate depends only on A.

Therefore, the rate - k1[A]m

Here, k1 = k (Initial concentration of B)n

As B is in surplus as compared to A, the concentration of B approximately remains a constant during the experiment. From the rate, the measurements, m can be determined by utilizing graphical method.

In the second set of experiments, [A] is much big as compared to [B]. The rate measurements are made and employing the rate law described below, n is computed.

Rate = k2[B]n

Here, K2 = k (Initial concentration of A]m

The overall order of the reaction is m + n.

Relaxation Method:

For studying the reactions, which are complete in 10-5 s or even less, one of the problems encountered is that the time of mixing the reactants must be much smaller than the time taken for the completion of the reaction. To ignore this problem, relaxation techniques have been developed. In a relaxation method, we disturb a reaction at equilibrium by employing an external influence like unexpected variation of pressure or temperature (termed as pressure jump or temperature jump). The system is stated to be perturbed from its equilibrium position. 

The perturbed system then returns to the latest equilibrium position. Relaxation refers to the passage of a perturbed system to the new equilibrium. The concentration of the perturbed system is recorded at different time intervals, by employing appropriate physical methods. From such measurements, it is possible to compute the rate constants.

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