Question

Derive integrated rate equation for the first order reaction.

Answer 1

Consider a general first order reaction

R → P

The differential rate equation for given reaction can be written as

Rate = \(- \frac{d[R]}{dt} = K[R]^1\)

Rearrange above equation.

\(\frac {d[R]}{[R]} = - K \times dt\)

Integrating on both sides of the given equation

\(\int \frac{d[R]}{[R] }= -k \int d t\)

\(n[R] = −Kt + I\)      .....(1)

Where I is Integration constant

At t = 0 the concentration of reactant [R] = [R]​₀ where [R]₀​ is initial concentration of reactant

Substituting in equation (1) we get

\(ln[R]_0 = (−K × 0) + I\)

\(ln[R]_0 = I \)     ......(2)

Substitute I value in equation (1)

\(ln[R] = −Kt + ln[R]_0\)​

\(Kt = ln[R]_0​ − lnR\)

\(Kt= ln\frac{[R]_0}{[R]}\)

\(K= \frac 1tln \frac{[R]_0}{[R]}\)

\(K = \frac{2.303}t \log \frac{[R]_0}{[R]}\)

Answer 2

The rate of reaction is directly proportional to the first power of the concentration of the reactant ‘R’. 

Consider a first order reaction

R → P

In [R] = -Kt +1 .....(1) 

Where I → Integration constant 

To find I, where = 0, [R] = [R]₀ 

In [R]₀ = -K × 0 + 1 

I = In [R]₀ 

Substituting in equ (1) we get 

In [R]₀ – In [R] = Kt