Order of a reaction is an important ·parameter for every chemical reaction. It is always determined experimentally and cannot be written from the balanced chemical equation. It refers to the number of reacting particles whose concentration terms determine the reaction rate.

It may be defined as, “the sum of powers or exponents to which the concentration terms are raised in the rate law expression.”

For example, for a hypothetical reaction,

aA + bB à products

if the rate law expression for this reactions is,

Rate = k [A]^{m} [B]^{n}

The order of the above reaction is equal to (m + n). These powers or exponents, i.e., m and n have no relation to the stoichiometric coefficients a and b of the balanced chemical equation. Order of the reaction with respect to A is m and that with respect to B is n. If the sum of the power is equal to one, the reaction is called first order reaction. If the sum of the powers is two or three, the reaction is second order or third order reaction respectively. The order of a reaction can also be zero or fractional.

For illustration, a few examples are given below:

(i) Nitrous oxide (N_{2}O) decomposes as:

2N_{2}O à 2N_{2}(g) + O_{2}(g)

The rate law expression for this reaction is

Rate = k [N_{2}O];Order = 1

Therefore, it is a .first order reaction.

(ii) For the reaction H_{2}(g) + I_{2}(g) à 2HI(g); the rate law is

Rate = k[H2][I2]; Order = 1 + 1 = 2

So, it is a second order reaction.

(iii) The reaction between nitric oxide and oxygen is an example of third order reaction.

2NO(g) + O_{2} (g) à 2NO_{2}(g)

The rate law is

Rate = k[NO]2[O2] ; Order = 2 + 1 =3

(iv) The reaction between hydrogen and bromine to form hydrogen bromide has fractional order

H_{2}(g) + Br_{2}(g) à 2HBr(g);

The rate law is

Rate = k [H2] [Br2] ½ , order = 1 + 1/2 = 3/2

(v) Decomposition of ammonia over platinum or gold is a zero order reaction

Rate = k [NH3]0 ; order = zero

In the above decomposition reaction, metal acts a catalyst.

**UNITS OF RATE CONSTANT FOR REACTIONS OF DIFFERENT ORDERS
**

The rate constant k has different units for reactions having different orders.

(i) For zero order reactions,

Rate = dx / dt = k [A]0

mol dm^{ -3} / sec = k . 1 or k = mold m ^{-3} s^{-1}

Thus, the units of rate constant for zero order reaction are: mol dm-3 s^{-1}

iii) For first order reactions, Rate = dx / dt = k [A]^{2}

mol dm ^{-3} / sec = k ( mol dm ^{-3}) ^{2}

k = 1 / mol dm ^{-3} sec = dm^{3} mol ^{-1} sec^{ -1}

The units of the rate constant for second order reaction are: dm3 moi-1 sec-1

(iv) For third order reactions,

Rate = dx / dt = k [A]^{3}

mol dm / sec ^{-3} = k (mol dm ^{-3})^{3}

k = 1 / sec mol^{2} dm ^{-6} = dm^{6} mol ^{-2} sec ^{-1}

The units of rate constant for third order reaction are: dm6 mol-2 sec-1

**INTEGRATED RATE EQUATIONS**

The integrated rate equations are different for the reactions of different reaction orders. Let us illustrate this method by applying it to zero order and first order reactions.

** **

**1. ZERO ORDER REACTION**

A reaction is zero order if its rate is independent of the concentration of reactants or a zero order reaction means that the rate of reaction is proportional to zero power of the concentration of reactants. For a zero order reaction of the form R à P

The rate of reaction , rate = – d [R] / dt = k [R]_{0}

As any quantity raised to power zero is unity

Rate = – d [R] / dt = k x 1

d[R] = – kdt

Integrating both sides, we get

[R] = – kt + Iwhere, I is the constant of integration

At t = 0, [R] = [R]_{0}, where [R]_{0} is initial concentration of reactant.

Substituting this value in equation (20.4), we get

[R] = – kt + [R]_{0 }or kt = [R]

_{0}– [R]

K = 1 / t [ [R]_{0} – [R]]

**SIGNIFICANCE OF INTEGRATED RATE EQUATION**

(i) All zero order reactions obey the equation.

(ii) The value of k can be evaluated if [R]_{0} and [R] at time t are known.

On plotting a graph of [R] vs. t we get a straight line as shown in the figure below. The slope of the line is given as slope = – k ‘

The intercept on the concentration axis = [R]_{0}

Rate = k [NH3]0 ; order = zero

In the above decomposition reaction, metal acts a catalyst.

**UNITS OF RATE CONSTANT FOR REACTIONS OF DIFFERENT ORDERS
**

The rate constant k has different units for reactions having different orders.

(i) For zero order reactions,

Rate = dx / dt = k [A]0

mol dm^{ -3} / sec = k . 1 or k = mold m ^{-3} s^{-1}

Thus, the units of rate constant for zero order reaction are: mol dm-3 s^{-1}

iii) For first order reactions, Rate = dx / dt = k [A]^{2}

mol dm ^{-3} / sec = k ( mol dm ^{-3}) ^{2}

k = 1 / mol dm ^{-3} sec = dm^{3} mol ^{-1} sec^{ -1}

The units of the rate constant for second order reaction are: dm3 moi-1 sec-1

(iv) For third order reactions,

Rate = dx / dt = k [A]^{3}

mol dm / sec ^{-3} = k (mol dm ^{-3})^{3}

k = 1 / sec mol^{2} dm ^{-6} = dm^{6} mol ^{-2} sec ^{-1}

The units of rate constant for third order reaction are: dm6 mol-2 sec-1

**INTEGRATED RATE EQUATIONS**

The integrated rate equations are different for the reactions of different reaction orders. Let us illustrate this method by applying it to zero order and first order reactions.

**1. ZERO ORDER REACTION**

A reaction is zero order if its rate is independent of the concentration of reactants or a zero order reaction means that the rate of reaction is proportional to zero power of the concentration of reactants. For a zero order reaction of the form R à P

The rate of reaction , rate = – d [R] / dt = k [R]_{0}

As any quantity raised to power zero is unity

Rate = – d [R] / dt = k x 1

d[R] = – kdt

Integrating both sides, we get

[R] = – kt + I

where, I is the constant of integration

At t = 0, [R] = [R]_{0}, where [R]_{0} is initial concentration of reactant.

Substituting this value in equation (20.4), we get

[R] = – kt + [R]_{0 }or kt = [R]

_{0}– [R]

K = 1 / t [ [R]_{0} – [R]]

**SIGNIFICANCE OF INTEGRATED RATE EQUATION**

(i) All zero order reactions obey the equation.

(ii) The value of k can be evaluated if [R]_{0} and [R] at time t are known.

On plotting a graph of [R] vs. t we get a straight line as shown in the figure below. The slope of the line is given as

slope = – k ‘

The intercept on the concentration axis = [R]

**2. FIRST ORDER REACTION**

Let us assume a simple hypothetical first order reaction represented as,

R P

If the initial concentration of R is [R]_{0}, k is the rate constant and [R] is cone. at time t then the differential form of this first order reaction will be

-d [R] /dt = k [R]

Rearranging the equation, we get

-d [R] / [R] = Kdt

On integrating equation (20.6),

[ Integral of d [R] / [R] is In [R]

To get the value of constant k let us put the value of t= 0 in Eqn. (20.7), then

-In [R]0 = k x 0 + constant

Constant = – In [R]_{0}

Substituting this value in Eqn. (20.7), we get

– In [R] = k t- In [R]_{0}

or k t =In [R]_{0} – In [R]

** **

** SIGNICANCE OF INTEGRATED EQUATION **

The integrated rate equation is also used to evaluate the value of k by graphical method. Equation (20.9) can be written as

Kt / 2.303 = log [R]_{0} / [R] or kt / 2.303 = log [R]_{0} –log [R]

or log [R] = -K / 2.303 T + log [R]_{0}

If we plot log [R]_{0} / [R] vs time or log [R]o vs time the following graphs would be obtained.

In both the plots i.e., Fig. 20.19 (a) and 20.19 (b) we get straight lines whose slope would be given as

Slope = – k / 2.303 ( fig. 20.19(a))

Slope = k / 2.303 ( fig.20.19 9b))

From the value of slope, the value of k can be calculated. The intercept made on the Y -axis would be log [R]0.

All natural and artificial radioactive decay of nuclei takes place by first order reaction and follow first order kinetics