Question

Derive the rate equation for the rate constant of a first order reaction and show that the time required for the completion of half of the first order reaction is independent of initial concentration.

Derive an expression for half-life period in case of a first order reaction and show that it is independent of initial concentration.

Answer 1

Consider a first-order reaction

\[\ce{A -> Products}\]

The differential rate law is:

\[\ce{-\frac{d[A]}{dt} = k[A]}\]    ...(1)

where [A] is the concentration of A at time t and k is the rate constant.

\[\ce{\frac{d[A]}{[A]} = -k * dt}\]    ...(2)

Integrating Eq. (2) between limit [A] = [A]₀ at t = 0 and [A] = [A]_t at t = t

\[\int\limits_{[A]_0}^{[A]_t}\frac {d[A]}{[A]} = -k\int\limits_{0}^{t}dt\]

on integration

ln[A]_t ​− ln[A]₀ ​= −kt

\[\ce{ln \frac{[A]_t}{[A]_0} = -kt}\]    ...(3)

From Eq. (3),

\[\ce{k = \frac{1}{t} ln \frac{[A]_0}{[A]_t}}\]

= \[\ce{\frac{2.303}{t} log \frac{[A]_0}{[A]_t}}\]

= − kt    ...(3)

Eq. (4) is the integrated rate equation for a first-order reaction.

Half-Life (t_1/2​) of a First-Order Reaction

At t = t_1/2, \[\ce{[A]_t = \frac{[A]_0}{2}}\], Substituting into Eq. (3)

\[\ce{ln \frac{\frac{[A]_0}{2}}{[A]_0} = -kt_{1/2}}\]

\[\ce{ln \frac{1}{2} = -kt_{1/2}}\]

\[\ce{t_{1/2} = \frac{ln 2}{k}}\]

= \[\ce{\frac{0.693}{k}}\]     ...(5)

From Eq. (5), the half-life t_1/2 depends only on the rate constant k and is independent of the initial concentration [A]₀ for a first-order reaction.