You must already know that calculating the rate of a reaction is extremely important to understand the reaction. But it also necessary to infer the rate law of a reaction to find out the order of the reaction. But what is the order of a reaction you ask? Let’s find out below and also learn about the different types of reactions.

### Suggested Videos

## Rate Law

We know that the rate law is the expression in which reaction rate is given in terms of 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.

Consider a general reaction, aA + bB → cC + dD where a, b, c, d are the stoichiometric coefficients of the reactants and the products. Therefore, the rate law for the above reaction is,

Rate \( \propto \) [A]* ^{x}* [B]

^{y}where *x *and *y *may or may not be equal to the stoichiometric coefficients of the reactants. Therefore, the rate of the reaction is equal to k [A]* ^{x}* [B]

*where k is the rate constant.*

^{y},∴ -d[R]/dt = k [A]* ^{x}* [B]

^{y }## Order of Reaction

The order of a reaction is the sum of the powers of the concentrations of the reactants in the rate law expression**.** In the above general reaction, *x* and *y* are the powers. The sum of them will give the order of the reaction. Order of a reaction can be 0, 1, 2, 3 and even a fraction. A zero-order reaction means that the rate of the reaction is *independent* of the concentration of reactants.

**Download Integrated Rate Equations Cheat Sheet by clicking on the button below**

**Browse more Topics under Chemical Kinetics**

- Collision Theory of Chemical Reactions
- Rate of a Chemical Reaction
- Pseudo First Order Reaction
- Factors Influencing Rate of a Reaction
- Temperature Dependence of the Rate of a Reaction

## Zero Order Reaction

So we already know that in a zero order reaction, the rate is independent of the concentration of the reactants. Thus, it means the sum of the powers of the concentrations is zero. It can only be zero when the all the powers are zero. Consider a reaction, R → P. Therefore, the rate law of this reaction is,

Rate \( \propto \) [R]^{0}

∴ Rate = -d[R]/dt = k[R]^{0 }= k × 1

∴ Rate = -d[R]/dt = k

∴ d[R] = -kdt

Integrating both sides, [R] = -kt + I……………….(I)

where I is the constant if integration. At t = 0, the concentration of the reactant R = [R]_{0}. Where [R]_{0} is the initial concentration of the reactant. Substituting this value in the equation I,

[R]_{0} = -k × 0 + I_{ }= I

Substituting this value of I in the equation (I), we get

[R] = -kt + [R]_{0 }………………(II)

Comparing equation II with the equation of a straight line y= mx +c, if we plot [R] against t, we get a straight line with slope = -k and intercept =** **[R]_{0}_{ }

(Source: askiitians.com)

Therefore, on simplifying equation II, we get

k ={[R]_{0} – [R]}/t……………..(III)

### Example of Zero Order Reaction

Zero-order reactions are very uncommon but they occur under certain condition. An example of a zero-order reaction is decomposition of ammonia, 2NH_{3 }→ N_{2} + 3H_{2}** **

Rate = k[NH_{3}]^{0} = k

## First Order Reaction

In this type of reaction, the sum of the powers of concentrations of reactants in rate law is equal to 1, that is the rate of the reaction is proportional to the first power of the concentration of the reactant. Consider the reaction R → P again. Therefore, the rate law for this reaction is,

Rate \( \propto \) [R]

We know that [R] = -kt + [R]_{0 }( from equation II). Taking log of both sides, we get

ln[R] = -kt + ln[R]_{0 }………………………….(IV)

∴ ln[R]/[R]_{0 }= -kt …………………………….(V)

∴ k = (1/t) ln [R]_{0 }/[R]** **………………………(VI)

Now, consider equation II again. At time t_{1} and time t_{2}, the equation II will be [R]_{1} = -kt_{1} + [R]_{0}** _{ }**and [R]

_{2}= -kt

_{2}+ [R]

_{0}

**respectively, where [R]**

_{ }_{1}and [R]

_{2}are concentrations of the reactants at time t

_{1 }and t

_{2}

**respectively. Subtracting second equation from first one, we get**

ln [R]_{1}– ln[R]_{2} = -kt_{1} – (- kt_{2} )

∴ ln[R]_{1} /[R]_{2 }= k (t_{2} – t_{1})

∴ k = [1/(t_{2} – t_{1})] ln[R]_{1} /[R]_{2}

Now, taking antilog of both sides of equation V, we get [R] = [R]_{0}e^{-kt}

Comparing this equation with equation of a straight line y = mx + c, if we plot ln [R] against t, we get a straight line with slope = -k and intercept = ln[R]_{0}

(Source: nonsibihighschool.org)

On removing natural logarithm from equation VI, the first-order reaction can also be written as,

k = 2.303/t log[R]_{0 }/[R]** **…………..(VII)

If we plot a graph of log[R]_{0 }/[R] against t, we get slope = k/2.303

### Example of First Order Reaction

An example of a first-order reaction is the hydrogenation of ethene.

C_{2}H_{4} + H_{2} → C_{2}H_{6}

Therefore the rate of reaction for the above is k [C_{2}H_{4}]. Hence, equations III and VII are the equations of rate constants of zero and first order reactions respectively. We can find rate constants, initial and final concentrations and the time taken for the reaction to occur using these reactions.

## A Solved Question for You

Q: The order of a reaction is:

a) can never be zero b) can never be a fraction

c) must be a whole number only d) can be a whole number or a fraction or zero

Solution: d) can be a whole number or a fraction or zero. It depends on the dependency of the rate of reaction on the reactants. If the rate is independent of the reactants, then the order of reaction is zero. Therefore, the rate law of a zero order reaction would be rate α [R]^{0 }where [R] is the concentration of the reactant.