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What is the half-life of a compound if 42% of a given sample of the compound decomposes in 60 minutes

Sagot :

The half-life of a compound is 76 years for 42% of a given sample of the compound decomposes in 60 minutes.

What is the expression for rate law for first order kinetics?

The rate law for first order kinetics is

k = [tex]\frac{2.303}{t}log\frac{a}{a-x}[/tex]

Where

k - rate constant

t - time passed by the sample

a - initial amount of the reactant

a-x - amount left after the decay process

What is expression for half-life of the compound?

The expression for the half-life of the compound is

[tex]t_{\frac{1}{2} }=\frac{0.693}{k}[/tex]

Where k is the rate constant

Calculation for the given compound:

It is given that

t = 60 minutes

a = 100 g

a-x = 100 g - 42 g = 58 g

(Since it is given that 42% of the given compound decomposes)

Then,

k = [tex]\frac{2.303}{60}log\frac{100}{58}[/tex]

  = 9.08 × [tex]10^{-3}[/tex] [tex]years^{-1}[/tex]

Then, the half-life of the compound is

[tex]t_{\frac{1}{2} }=\frac{0.693}{k}[/tex]

    = [tex]\frac{0.693}{9.08*10^{-3} }[/tex]

    = 76.3 years ≅ 76 years

Therefore, the half-life of the given compound is 76 years.

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