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.

Respuesta :

First you need to find the order of reaction.
Let the reaction follow a simple nth order rate law:
rate = kβˆ™[A]ⁿ

Half-life t₁₂ initial concentration [A]β‚€ and rate constant k for such a reaction are related as:
t₁₂ = (2ⁿ⁻¹ - 1) / ( (n - 1)βˆ™kβˆ™[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)βˆ™k)

K cancels out when you take the ratio of the two given half-lifes:
tβ‚β‚‚β‚β‚‚β‚Ž / tβ‚β‚‚β‚β‚β‚Ž = (K/[A]β‚€β‚β‚‚β‚ŽβΏβ»ΒΉ) / (K/[A]β‚€β‚β‚β‚ŽβΏβ»ΒΉ) = ([A]β‚€β‚β‚β‚Ž/[A]β‚€β‚β‚‚β‚Ž)ⁿ⁻¹
to find the exponent (n-1) take logarithm
ln(tβ‚β‚‚β‚β‚‚β‚Ž/tβ‚β‚‚β‚β‚β‚Ž) = ln(([A]β‚€β‚β‚β‚Ž/[A]β‚€β‚β‚‚β‚Ž)ⁿ⁻¹) = (n - 1)βˆ™ln([A]β‚€β‚β‚β‚Ž/[A]β‚€β‚β‚‚β‚Ž)
=>
n - 1 = ln(tβ‚β‚‚β‚β‚‚β‚Ž/tβ‚β‚‚β‚β‚β‚Ž) / ln([A]β‚€β‚β‚β‚Ž/[A]β‚€β‚β‚‚β‚Ž)
= ln(229s / 151s) / ln(0.297M / 0.196M )
= 1.00198...
β‰ˆ 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)βˆ™kβˆ™[A]₀²⁻¹ ) = 1/(kβˆ™[A]β‚€)
Hence
k = 1/(tβ‚β‚‚βˆ™[A]β‚€)
= 1/(151s βˆ™ 0.297M)
= 2.23Γ—10⁻² M⁻¹s⁻¹