Question

Derive an integrated rate law expression for first order reaction: A → B + C

Answer 1

Consider first order reaction, A → B + C

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_(["A"]_0)^(["A"]_"t") ("d"["A"])/(["A"]) = - "k" int_0^"t"`dt

On integration,

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

Substitution of limits gives

ln[A]_t − ln[A]₀ = –kt

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

or k = `1/"t"` ln `["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.