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