Question

Answer the following in brief.

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

Answer 1

Consider the first-order reaction,

A → product

The differential rate law is given by

rate = -`"d[A]"/"dt" = "k[A]"` ...(1)

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

`"d[A]"/"[A]" = -"k"` dt   .....(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,

In`["A"]_(["A"]_0)^(["A"]_"t") = -"k"  "t"_0^"t"`

Substitution of limits gives

`"In"["A"]_"t" - "In"["A"]_0 = -"kt"`

or `"In"["A"]_"t"/["A"]_0 = -"kt"`  ...(3)

or k = `1/"t" "In"["A"]_0/["A"]_"t"`

Converting ln to log₁₀, we write

k = `2.303/"t" "log"_10["A"]_0/["A"]_"t"`   ...(4)

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