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Question
How is Half-life related to initial concentration for a second order reaction?
Numerical
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Solution
For a second-order reaction, the half-life is inversely proportional to the initial concentration of the reactant.
For a second-order reaction
Rate = k[A]2
The integrated rate law is
`1/[A] = 1/[A]_0 + kt`
At half-life t = t1/2, [A] = `[A]_0/2`
Substitute into the integrated rate law:
`1/([A]_0//2) = 1/[A]_0 + kt_(1//2)`
⇒ `2/[A]_0 = 1/[A]_0 + kt_(1//2)`
⇒ `kt_(1//2) = 1/[A]_0`
⇒ `t_(1//2) = 1/(k[A]_0)`
For a second order reaction `t_(1//2) prop 1/[A]_0`
This means:
As initial concentration increases, half-life decreases.
As initial concentration decreases, half-life increases.
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