Question

In: Chemistry

A particular reactant decomposes with a half-life of 119 s when its initial concentration is 0.324...

A particular reactant decomposes with a half-life of 119 s when its initial concentration is 0.324 M. The same reactant decomposes with a half-life of 233 s when its initial concentration is 0.165 M.

Determine the reaction order?

What is the value and unit of the rate constant for this reaction?

Solutions

Expert Solution

Let the reaction follow a simple nth order rate law:
rate = k∙[A]ⁿ

Half-life t₁₂ initial concentration [A]₀ and rate constant k for such a reaction are related as:
t₁₂ = (2ⁿ⁻¹ - 1) / ( (n - 1)∙k∙[A]₀ⁿ⁻¹ )
except the particular case of first order reactions, i.e. n=1, in which half-life does not depend on initial concentration:
t₁₂ = ln(2)/k

Apparently your reaction is not a first order reaction. When you combine the constant factors in the relation above to a constant K, you can see that half-life of a non-first order reaction is inversely proportional to initial concentration raised to the power (n-1):
t₁₂ = K/[A]₀ⁿ⁻¹
with K=(2ⁿ⁻¹ - 1)/((n - 1)∙k)

K cancels out when you take the ratio of the two given half-lifes:
t₁₂₍₂₎ / t₁₂₍₁₎ = (K/[A]₀₍₂₎ⁿ⁻¹) / (K/[A]₀₍₁₎ⁿ⁻¹) = ([A]₀₍₁₎/[A]₀₍₂₎)ⁿ⁻¹
to find the exponent (n-1) take logarithm
ln(t₁₂₍₂₎/t₁₂₍₁₎) = ln(([A]₀₍₁₎/[A]₀₍₂₎)ⁿ⁻¹) = (n - 1)∙ln([A]₀₍₁₎/[A]₀₍₂₎)
=>
n - 1 = ln(t₁₂₍₂₎/t₁₂₍₁₎) / ln([A]₀₍₁₎/[A]₀₍₂₎)
= ln(233s / 119s) / ln(0.324M / 0.165M )

= 0.995

≈ 1
=>
n = 2

With known order n we can compute k from given half-life and initial concentration.
For a second order reaction half-life is given by:
t₁₂ = (2²⁻¹ - 1) / ( (2 - 1)∙k∙[A]₀²⁻¹ ) = 1/(k∙[A]₀)
Hence
k = 1/(t₁₂∙[A]₀)
= 1/(119s ∙ 0.324M)
= 2.593×10⁻² M⁻¹s⁻¹

= 0.02593 M-1s-1


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