Question

After 47.0 min, 20.0% of a compound has decomposed. What is the half-life of this reaction assuming first-order kinetics?

Answer #1

First calculate rate constant for the reaction following first-order kinetics by using the formula given below:

**K = 2.303/t log [A] _{o}/[A]_{o} -
x**

Here [A]_{o} = initial concentration of the reactant and
x = amount of reactant reacted in the time t

K = 2.303/47 log(100/80) = 2.303/47 log(1.25) = 0.0047
min^{-1}

Now calculate half life. (half life is time in which concentration of reactants reduced to half of initial concentration)

t_{1/2} = 2.303/K log[A]_{o}/([A]_{o}/2)
= 2.303/K log2 = 0.693 / K

**t _{1/2} = 0.693 / K = 0.693 / 0.0047 = 147 .4
min**

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