# Answer the following in brief. Derive the integrated rate law for first order reaction. - Chemistry

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

#### Solution

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.

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#### APPEARS IN

Balbharati Chemistry 12th Standard HSC Maharashtra State Board
Chapter 6 Chemical Kinetics
Exercise | Q 3.1 | Page 137