Question

A reactant decomposes with a half-life of 153 s when its initial concentration is 0.269 M. When the initial concentration is 0.568 M, this same reactant decomposes with the same half-life of 153 s. It is a first order reaction. I am unsure how to do this. SOS

Answer 1

First you need to find the order of reaction.
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(151s / 151s) / ln(0.269M / 0.568M )
= 0

=>
n = 0

With known order n we can compute k from given half-life and initial concentration.
For a zero order reaction half-life is given by:
t₁₂ = [A]₀ / 2k
Hence
k = [A]₀/2t
= 0.269/2*153
= 8.79×10⁻4 M⁻¹s⁻¹