Question

A reactant decomposes with a half-life of 137 s when its initial concentration is 0.165 m. when the initial concentration is 0.722 m, this same reactant decomposes with the same half-life of 137 s.

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(229s / 151s) / ln(0.297M / 0.196M )
= 1.00198...
≈ 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/(151s ∙ 0.297M)
= 2.23×10⁻² M⁻¹s⁻¹