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Derive an expression for the relation between half-life and rate constant for first-order reaction.

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Derive an expression for the relation between half-life and rate constant for first-order reaction.

With the help of integrated rate law, derive an expression for the half-life of a first order reaction.

Derivation
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Solution

The integrated rate law for the first-order reaction is

k = `2.303/t * log_10  [A]_0/[A]_t`

Where [A]0 is the initial concentration of reactant at t = 0. It falls to [A]t at time t after the start of the reaction.

The time required for [A]0 to become `[A]_0/2` is denoted as t1/2 or [A]t = `[A]_0/2` at t = t1/2

Putting this condition in the integrated rate law we write

k = `2.303/t_(1//2)  log_10  [A]_t/([A]_0/2)`

= `2.303/t_(1//2) log_10 2`

Substituting value of log102,

k = `2.303/t_(1//2) xx 0.3010`

∴ k = `0.693/t_(1//2)`

∴ `t_(1//2) = 0.693/k`

The half-life of a first-order reaction is independent of the initial reactant concentration.

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Chapter 6: Chemical Kinetics - Long answer questions

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SCERT Maharashtra Chemistry [English] Standard 12 Maharashtra State Board
Chapter 6 Chemical Kinetics
Long answer questions | Q 2 (i)
Nootan Chemistry [English] Class 12 ISC
Chapter 3 Chemical Kinetics
LONG ANSWER TYPE QUESTIONS | Q 5. ii. | Page 265

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