Question

Answer the following in brief.

Derive the integrated rate law for the first-order reaction.

Starting with the differential rate law equation, derive the integrated rate equation for a first order reaction.

Derive the integrated rate equation for a first order reaction.

Answer 1

Consider the first-order reaction,

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

The differential rate law is given by

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

Where [A] is the concentration of reactant at time t. Rearranging Eq. (1)

\[\ce{\frac{d[A]}{[A]} = -kdt}\]    ...(2)

Let [A]₀ be the initial concentration of the reactant A at time t = 0.

Suppose [A]_t is the concentration of A at time = t

The equation (2) is integrated between limits [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,

\[\ce{ln [A]{^{[A]_t}_{[A]_0}} = -k t^t_0}\]

Substitution of limits gives

ln[A]_t − ln[A]₀ = −kt

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

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

Converting ln to log₁₀, we write

\[\ce{k = \frac{2.303}{t} log_10 \frac{[A]_0}{[A]_t}}\]    ....(4)

Eq. (4) gives the integrated rate law for the first-order reactions.