Half Life Of A Reaction: 1st, 2nd and 3rd Order Reaction

Usually we calculate the half-life of a reaction to find the reaction's order. Let us understand what is the half-life of the reaction. "The time taken for the concentration of a given reactant to reach half of its initial concentration (if A is a reactant then, [A] = 1/2 [A]o) is called as the half-life of a reaction". Half-life is denoted by 't½'.

Half-Life Of Reaction

Half-life is a period of reaction is defined as the time during which the concentration of a reactant is reduced to half of its initial concentration.

The half-life of a reaction 't½' occurs when (consider 'A' as a reactant molecule) ;

➩ [A] = 1/2 [A]₀  Or  [A]₀/2

Remember that the formula for the half-life of a reaction varies along with changing the order of the reaction. And half-life is usually measured in 'Seconds'. By determining the half-life as a function of initial concentration, one can find the reaction's order and specific reaction rate. 

Now Let's Calculate The Half-life For Different Order Of Reaction.

(1)Zero order reaction

The integrated rate equation for zero order reaction ( where the rate of reaction is independent of the reactant's concentration) is as follows;

          ➩ k = ([A]₀- [A]₁) / t   

                                -----------------(1)

If t = t½, Then [A]₁= 1/2 [A]₀

Thus, equation(1) becomes;

             ➩ k = ([A]₀ - 1/2 [A]₀) / t½

             ➩ k = (2[A]₀-[A]o) / 2t½

             ➩ k = [A]₀ / 2t½

             ➩ { t½ = [A]₀ / 2k }

                             ------------------(2)

Here, equation (2) denotes the half-life for zero order reaction. As shown in the above equation, the half-life of the zero-order reaction depends on the initial concentration of reactant molecule and changes with the change in the initial concentration of the reactant molecule.

☆ [A]₀⇈ as a result t½ ⇈ 

(2)First order reaction

The integrated rate equation for first order reaction ( where the rate of reaction depends on the first power of the reactant's concentration) is as follows;

          ➩ k = (1/t ) ln([A]₀/[A]₁) 

          ➩ k = (2.303/t ) log([A]₀/[A]₁)

                                         -----------------(3)

If t = t½, Then [A]₁= 1/2 [A]₀

Thus, equation(3) becomes;

      ➩ k = (2.303 / t½) log₁₀{([A]o/([A]o/2)}

      ➩ k = (2.303 / t½) log₁₀(2[A]o/([A]o)

      ➩ k = (2.303 / t½) log₁₀(2)

                             [ ∵ log₁₀(2) = 0.3010 ]  

      ➩ k = (2.303/t½ ) (0.3010)

      ➩ { k = 0.693 / t½ }

                   OR

      ➩ { t½ = 0.693 / k }

                          ----------------(4)

Here, equation (4) denotes the half-life for first order reaction. As shown in the above equation, the half-life of the first-order reaction does not depend on the initial concentration of reactant molecule. Thus, if we change the initial concentration of the reactant molecule, the half-life of the first-order reaction remains unaffected. It is a constant value.

(3)Second order reaction

The integrated rate equation for second order reaction ( where the rate of reaction depends on the second power of the reactant's concentration or the concentration of the two different reactants) is as follows;

          ➩ k = (1/[A]₁-1/[A]₀) / t

                               -------------------(5)

If t = t½, Then [A]₁= 1/2 [A]₀

Thus, equation(1) becomes;

             ➩ k = {1/([A]₀/2) - 1/[A]₀} / t½

             ➩ k = {2/[A]₀- 1/[A]₀}/ t½

             ➩ k = {1/[A]₀} / t½

             ➩ { k = 1 /[A]₀t½ }

                            OR

             ➩ { t½ = 1 /[A]₀k }

                              ------------------(6)

Here, equation (6) denotes the half-life for second order reaction. As shown in the above equation, the half-life of the second-order reaction is inversely proportional to the initial concentration of reactant molecule. Thus if we increase the initial concentration of the reactant molecule, the half-life of the second-order reaction decreases.

☆ [A]₀⇊ as a result t½ ⇊   

Numerical Problem

QUE- Show That The Time Required To Complete The Reaction Upto 99.9% Of First Order Reaction Is 10 Times The Half-life Of That Reaction?? 

Solution 

  To complete the reaction upto 99.9%, 

  That means, 

                  [A]₁ = [A]₀ - 0.999 [A]₀

  Integrated rate equation for first order, 

            ➩ k = (2.303/t ) log₁₀([A]₀/[A]₁)

       ➩ k = (2.303/t ) log₁₀([A]₀/[A]₀- 0.999[A]₀) 

            ➩ k = (2.303/t ) log₁₀[1/(1- 0.999)]

            ➩ k = (2.303/t ) log₁₀[1000]

            ➩ k = (2.303/t ) × 3

            ➩ k = 6.909 / t 

            ➩ t = 6.909 / k

                        --------------(1) 

And, The half life of a first order reaction, 

                 ➩ t½ = 0.693 / k 

                                 --------------(2)

Now dividing equation (1) by equation (2) ;

          ➩ t / t½ = (6.909 / k) / (0.693 / k) 

          ➩ t / t½ = 6.909/0.693 

          ➩ t / t½ = 10

          ➩ [ t = 10 (t½ ) ]

                       ➝ Hence proved

!!Hope you have found this article helpful!!

Let me know what you think about HALF LIFE OF REACTIONS!!

!!Feel free to comment if you have any queries!!