Question

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

Answer 1

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 = $\frac{2.303}{t}log\frac{a}{a-x}$

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

$t_{\frac{1}{2} }=\frac{0.693}{k}$

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 = $\frac{2.303}{60}log\frac{100}{58}$

  = 9.08 × $10^{-3}$ $years^{-1}$

Then, the half-life of the compound is

$t_{\frac{1}{2} }=\frac{0.693}{k}$

    = $\frac{0.693}{9.08*10^{-3} }$

    = 76.3 years ≅ 76 years

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

Learn more about rate law for first order kinetics here:

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