Integrated And Differential Rate Equations

As we know that the Differential rate equation is directly related to the rate laws, while the Integrated rate equations are obtained by integrating the differential rate equations. The integrated rate law equation expresses the concentration of the reactant as a function of time.

The order and rate constant of the reaction can be found from the integrated rate law equation. The Integrated rate law equation is different for different orders of reactions.

Using experimental data, one can find out the value of rate constant "k" and predict the order of reaction. 

Now, we calculate the integrated rate equations for zero, first and second order reactions one by one in detail;

Integrated and Differential Rate Equations

(1) Integrated And Differential Rate Equations For Zero Order Reaction

A zero order reaction is defined as a reaction in which the order of reaction is equal to one.

OR 

"A zero order reaction takes place at a constant/fixed rate, independent of the reactant's concentration/initial concentration". 

Zero Order reactions are rarely observed in nature. Consider a general equation of zero order reaction,

                 A  ➝  B

Rate of reaction is given as,

           ➩ rA = -d[A]/dt = d[B]/dt = k[A]⁰

                   ➩  -d[A]/dt = k 

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

Equation(1) is known as "Differential Rate Equation For Zero Order Reaction". 

Now, we integrate the above equation to obtain the integrated rate equation for zero order reaction;

                    ➩ -d[A] = k dt 

         Integrate both sides, we get;

                    ➩ ∫-d[A] = k∫dt

Boundary conditions are;

                    At t =0 then [A] = [A]

                    At, t = t₁ then [A] = [A]₁

After applying these limits and evaluate the above integration then we have,

                ➩ -( [A]₁– [A]) = k (t₁–0)

                ➩ [A]₁ –[A]₀ = - k.t₁

                ➩ { [A]₁= - k.t₁+ [A]₀}

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

                                  OR

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

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

Here, equation (2) is known as "Integrated Rate Equation For Zero Order Reaction". We can see that the above equation looks like a straight line equation.

On comparing the above equation with a straight line equation,

              ➩ Y = m.x + c

  Then, 

        ➛ Slope, m = –k

        ➛ Intercept, c = [A]₀

(2) Integrated And Differential Rate Equations For First Order Reaction

A reaction whose reaction's order is equal to one is called first order reaction. 

OR 

"A first order reaction is defined as a reaction in which the rate of reaction depends on the first power of the reactant's concentration".

(the concentration of only one reactant molecule, whose exponent is equal to one, affects the rate of reaction)

For Example 

Natural and artificial radioactive decay of unstable nuclei, thermal decomposition of nitrogen pentoxide and hydrolysis of hydrogen peroxide etc. are examples of first order reaction. 

Now let's look at the Integrated rate equations and differential rate equations for the first order reaction. 

Consider a general first order reaction,

                           A ➝ B

Rate of reaction is given as,

                   ➩ rA = -d[A]/dt = k[A]¹

                   ➩ { -d[A]/dt = k[A]¹ } 

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

Here, equation (1) is called "Differential Rate Equation For First Order Reaction". To obtain an integrated rate equation, we integrate the above equation(1) across the boundary conditions.

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

                     ➩ (1/[A]) d[A] = -k dt

                      Integrating both side,

                     ➩ ∫(1/[A]) d[A] = ∫-k dt

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

Boundary conditions are

                  At t=0 then [A] = [A]₀

                  At t=t₁ then [A] = [A]₁

           And,

                ∵∫(1/[A]) d[A] = ln([A])

Evaluate the above integration over the boundary conditions and we get,

        ➩ ln([A]₁) - ln([A]₀) = - k(t₁-0)

        ➩ { ln([A]₁) = -kt₁ + ln([A]₀) } 

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

                              OR

        ➩ { k = (1/t) ln([A]₀/[A]₁) } [ say, t₁=t ]

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

                               OR

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

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

The above equation(3) is known as "Integrated Rate Equation For First Order Reaction". If we observe the above equation (equation-3), we find that this equation looks like a straight line equation.

On comparing the above equation(3) with a straight line equation,

                    ➩ Y = m.x + c

             Then, 

                 ➛ Slope, m = –k

                 ➛ Intercept, c = ln([A]₀)

Equation(3) can also be written in one more way,

              ➩ { [A]₁ = [A]₀ exp(-kt) }

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

We can rearrange the above equation(4) in a new form;

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

              ➩ { log₁₀([A]₀/[A]₁) = kt/2.303}

                                        --------------------(7) 

(3) Integrated And Differential Rate Equations For Second Order Reaction

A reaction whose reaction's order is equal to two, is called second order reaction 

OR

"A second order reaction is defined as a reaction in which the rate of reaction depends on the second power of the reactant's concentration".(The concentration of only one reactant molecule, whose exponent is equal to two or two different individual reactant molecules whose exponent is equal to one, affects the rate of reaction)

For Example 

Thermal decomposition of Nitrous oxide (N₂O), decomposition of NO₂, decomposition of hydrogen iodide, hydrolysis of esters by alkali (saponification reaction) etc. are examples of second order reaction. 

Now let's look at the Integrated rate equations and differential rate equations for the second order reaction. But in second order reaction, two different reactant molecules react with each other or single molecule react with itself and convert into products. So keep in mind that two different cases exist here.

Case -1

Consider the general equation of second-order reaction where only one reactant molecule affects the rate of reaction;

                   2A ➝ products

Rate of reaction is given as,

               ➩ rA = -d[A]/dt = k[A]²

               ➩ { -d[A]/dt = k[A]²} 

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

Here, equation (1) is called "Differential Rate Equation For Second Order Reaction". And to obtain an integrated rate equation, we integrate the above equation(1) across the boundary conditions.

               ➩ -d[A]/dt = k[A]²

               ➩ (1/[A]²)d[A] = -k dt

           Integrating both side,

               ➩ ∫(1/[A]²) d[A] = ∫-k dt

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

Boundary conditions are

            At t=0 then [A] = [A]₀

            At t=t₁ then [A] = [A]₁

And,

         ∵∫(1/[A]²) d[A] = - 1/[A]

Evaluate the above integration over the boundary conditions and we get,

           ➩ (1/[A]₁) - (1/[A]₀) = k(t₁-0)

           ➩ { (1/[A]₁) = kt₁ + (1/[A]₀)}

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

The above equation(3) is known as "Integrated Rate Equation For Second Order Reaction". If we observe the above equation, we find that this equation looks like a straight line equation.

Equation(3) can also be written in one more way (say t₁=t), 

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

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

Case -2

Consider the general equation of second-order reaction where two different reactant molecules affect the rate of reaction,

                  A + B ➝ products

Rate of reaction is given as;

      ➩ rA = -d[A]/dt = -d[B]/dt = k[A][B]

      ➩ { -d[A]/dt = -d[B]/dt = k[A][B]}

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

Here, equation (5) is called Differential rate equation for second order reaction. In this case, both the initial concentrations are not equal to each other (means [A]₀≠ [B]₀). 

Let 'x' be the concentration of each reactant species react at time 't' then the boundary conditions are-

       At t=0 then [A]=a=[A]₀ and [B]=b=[B]₀

        At t=t then [A]= a-x and [B]=b-x

Then the expression of the rate law becomes,

          ➩ -d[A]/dt = -d[B]/dt = - d(x)/dt 

          ➩ {-d(x)/dt = k(a-x)(b-x) }

After rearrangement,

            ➩ 1/(a-x)(b-x) dx = - kdt

        Integrating both sides we get,

            ➩∫1/(a-x)(b-x) dx = ∫- kdt

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

Integrating the above equation over the time limit '0 to t' and concentration limit '0 to x'. And by using method of partial fraction,

∵∫1/(a-x)(b-x) dx ＝ ∫1/(b-a) {1/(a-x)- 1/(b-x)} ＝  

1/ (b-a) {ln(a-x) – ln(b-x)}

Thus the above integration is easily solved after putting the integration limit and we get,

      ➩1/(b-a) { ln(a/a-x) - ln(b/b-x) } = kt

             Applying rules of logarithm,

                ∵ (b-x)=[B] and (a-x)=[A]

            as well as, a= [A]₀ and B= [B]₀

Thus,

      ➩1/([B]₀-[A]₀) ln([A]₀[B] /[B]₀[A]) = kt

      ➩{ln([A]₀[B] /[B]₀[A]) = k([B]₀- [A]₀)t}

                         OR

  ➩ { ln([A]/[B]) = k([B]₀- [A]₀)t + ln([B]₀/[A]₀)}

                                                 ------------------------(7)

The above equation(7) is known as "Integrated Rate Equation For Second Order Reaction". If we observe the above equation, we find that this equation looks like a straight line equation. Thus, here also a linear relationship exist between "ln([A]/[B])" and "Time, t". 

On comparing the above equation with a straight line equation,

                    ➩ Y = m.x + c

      Then, 

           ➛ Slope, m = k.([B]₀-[A]₀)

           ➛ Intercept, c = ln([B]₀/[A]₀)

Hope you have found this article helpful!!

Let me know what you think about these rate equations. Feel free to comment if you have any queries.!!