Rate Law and Rate Constants

What is the Rate Law?

The rate law (also known as the rate equation) for a chemical reaction is an expression that provides a relationship between the rate of the reaction and the concentrations of the reactants participating in it.


For a reaction given by:

aA + bB → cC + dD

Where a, b, c, and d are the stoichiometric coefficients of the reactants or products, the rate equation for the reaction is given by:

Rate ∝ [A]x[B]y ⇒ Rate = k[A]x[B]y


  • [A] & [B] denote the concentrations of the reactants A and B.
  • x & y denote the partial reaction orders for reactants A & B (which may or may not be equal to their stoichiometric coefficients a & b).
  • The proportionality constant ‘k’ is the rate constant of the reaction.

It is important to note that the expression of the rate law for a specific reaction can only be determined experimentally. The rate law expression cannot be obtained from the balanced chemical equation (since the partial orders of the reactants are not necessarily equal to the stoichiometric coefficients).

Reaction Orders

The sum of the partial orders of the reactants in the rate law expression gives the overall order of the reaction.

If Rate = k[A]x[B]y ; overall order of the reaction (n) = x+y

The order of a reaction provides insight into the change in the rate of the reaction that can be expected by increasing the concentration of the reactants. For example:

  • If the reaction is a zero-order reaction, doubling the reactant concentration will have no effect on the reaction rate.
  • If the reaction is of the first order, doubling the reactant concentration will double the reaction rate.
  • In second-order reactions, doubling the concentration of the reactants will quadruple the overall reaction rate.
  • For third-order reactions, the overall rate increases by eight times when the reactant concentration is doubled.

Rate Constants

Rearranging the rate equation, the value of the rate constant ‘k’ is given by:

k = Rate/[A]x[B]y

Therefore, the units of k (assuming that concentration is represented in mol.L-1 or M and time is represented in seconds) can be calculated via the following equation.
k = (M.s-1)*(M-n) = M(1-n).s-1

The units of the rate constants for zero, first, second, and nth-order reactions are tabulated below.

Reaction Order Units of Rate Constant
0 M.s-1 (or) mol.L-1.s-1
1 s-1
2 M-1.s-1 (or) L.mol-1.s-1
n M1-n.s-1 (or) L(-1+n).mol(1-n).s-1

Differential Rate Equations

Differential rate laws are used to express the rate of a reaction in terms of the changes in reactant concentrations (d[R]) over a small interval of time (dt). Therefore, the differential form of the rate expression provided in the previous subsection is given by:

-d[R]/dt = k[A]x[B]y

Differential rate equations can be used to calculate the instantaneous rate of a reaction, which is the reaction rate under a very small-time interval. It can be noted that the ordinary rate law is a differential rate equation since it offers insight into the instantaneous rate of the reaction.

Integrated Rate Equations

Integrated rate equations express the concentration of the reactants in a chemical reaction as a function of time. Therefore, such rate equations can be employed to check how long it would take for a given percentage of the reactants to be consumed in a chemical reaction. It is important to note that reactions of different orders have different integrated rate equations.

Integrated Rate Equation for Zero-Order Reactions

The integrated rate equation for a zero-order reaction is given by:

kt = [R0] – [R] (or) k = ([R0] – [R])/t


  • [R0] is the initial concentration of the reactant (when t = 0)
  • [R] is the concentration of the reactant at time ‘t’
  • k is the rate constant

Integrated Rate Equation for First-Order Reactions

The integrated rate law for first-order reactions is:

kt = 2.303log([R0]/[R]) (or) k = (2.303/t)log([R0]/[R])

Integrated Rate Equation for Second-Order Reactions

For second-order reactions, the integrated rate equation is:

kt = (1/[R]) – (1/[R0])

Solved Examples on the Rate Law

Example 1

For the reaction given by 2NO + O2 → 2NO2, The rate equation is:

Rate = k[NO]2[O2]

Find the overall order of the reaction and the units of the rate constant.

The overall order of the reaction = sum of exponents of reactants in the rate equation = 2+1 = 3

The reaction is a third-order reaction. Units of rate constant for ‘nth’ order reaction = M(1-n).s-1

Therefore, units of rate constant for the third-order reaction = M(1-3).s-1 = M-2.s-1 = L2.mol-2.s-1

Example 2

For the first-order reaction given by 2N2O5 → 4NO2 + O2 the initial concentration of N2O5 was 0.1M (at a constant temperature of 300K). After 10 minutes, the concentration of N2O5 was found to be 0.01M. Find the rate constant of this reaction (at 300K).

From the integral rate equation of first-order reactions:

k = (2.303/t)log([R0]/[R])

Given, t = 10 mins = 600 s

Initial concentration, [R0] = 0.1M

Final concentration, [R] = 0.01M

Therefore, rate constant, k  = (2.303/600s)log(0.1M/0.01M) = 0.0038 s-1

The rate constant of this equation is 0.0038 s-1

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