Answered

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Molecular iodine,
I2(g)
dissociates into iodine atoms at 625 K with a first-order rate constant of
0.271s-1
What is the half-life for this reaction?

Sagot :

jufsanabriasa

Answer:

Half-life for this reaction is 2.56s

Explanation:

The general expression in a reaction that follows first-order is:

Ln[A] = -kt + ln[A]₀

Where [A] is concentration of reactant after time t,

k is rate constant = 0.271s⁻¹

[A]₀ is initial concentration of reactant.

Half-life is defined as the time required to decrease the initial concentration of the reactant (I2 in this case) halved.

If [A]₀ = 1

[A] = 1/2

Solving the equation:

Ln[1/2] = -0.271s⁻¹*t + ln[1]

Ln[1/2] = -0.271s⁻¹*t + 0

Ln[1/2] = -0.271s⁻¹t

Ln 2 = 0.271s⁻¹

2.56s = t

Half-life for this reaction is 2.56s

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