Given:

Half-life \(t_{1/2} = 1.7\) hours, [A]₀ = 100%, [A]ₜ = 100 - 20 = 80%

To find:

Time for 20% of the reactant to react, \(t\)

Formulas:

i. \(t_{1/2} = \frac{0.693}{k}\)

ii. \(t = \frac{2.303}{k} \log_{10} \frac{[A]_0}{[A]_t}\)

Calculation:

\(t_{1/2} = \frac{0.693}{k}\)

\(k = \frac{0.693}{t_{1/2}} = \frac{0.693}{1.7 \, \text{h}} = 0.4076 \, \text{h}^{-1}\)

\(t = \frac{2.303}{k} \log_{10} \frac{100}{80}\)

\(t = \frac{2.303}{0.4076 \, \text{h}^{-1}} \log_{10} \frac{100}{80} = 0.5475 \, \text{h} \times \frac{60 \, \text{min}}{1 \, \text{h}} = 32.9 \, \text{min}\)

The time required for 20% of the reaction to react is 32.9 minutes.