Question

A reaction has a half-life of 1 min. The time required for 99.9% completion of the reaction is ______ min. (Round off to the nearest integer).

[Use In 2 = 0.69, In 10 = 2.3]

Answer 1

A reaction has a half-life of 1 min. The time required for 99.9% completion of the reaction is 10 min.

Explanation:

If a reaction is 99.9% complete, it means that only 0.1% of the reactant remains.

If we assume the initial concentration of the reactant A0 is 100 units, then at 99.9% completion, the remaining concentration A is

A = A₀ − X = 100 − 99.9 = 0.1

The half-life T_1/2 of a first-order reaction is related to the rate constant k by the formula:

\[\ce{T_{1/2} = \frac{0.693}{k}}\]

Given T_1/2 = 1 min, we can rearrange this to find k:

\[\ce{k = \frac{0.693}{T_{1/2}} = \frac{0.693}{1} = 0.693 min^{-1}}\]

The first-order kinetics equation is given by

\[\ce{k = \frac{1}{T} ln (\frac{A_0}{A})}\]

Here, A₀ = 100 and A = 0.1. Substituting these values into the equation

\[\ce{0.693 = \frac{1}{T} ln (\frac{100}{0.1})}\]

Calculate \[\ce{\frac{100}{0.1} = 1000}\], so we need to find ln(1000)

ln(1000) = ln(10³) = 3 ln(10)

Using ln(10) ≈ 2.3:

ln(1000) ≈ 3 × 2.3 = 6.9

Now substituting ln(1000) back into the equation

\[\ce{0.693 = \frac{6.9}{T}}\]

Rearranging to solve for T

\[\ce{T = \frac{6.9}{0.693}}\] ≈ 9.95 min

Rounding off 9.95 to the nearest integer gives us 10 min.