Problem 1: First-Order Half-Life from k

Given: k = $6.93 \times 10^{-3}$ $s^{-1}$

$t_{1/2} = \frac{0.693}{6.93 \times 10^{-3}} = \frac{0.693}{0.00693} = 100 \text{ s}$

Problem 2: Time for Specific % Completion

Same reaction (k = $6.93 \times 10^{-3}$ $s^{-1}$, t_{1}/_{2} = 100 s):

75% completion: 25% remains = (1/2)^{2} → 2 half-lives = 200 s

87.5% completion: 12.5% remains = (1/2)^{3} → 3 half-lives = 300 s

93.75% completion: 6.25% remains = (1/2)^{4} → 4 half-lives = 400 s

99.9% completion: 0.1% remains ≈ (1/2)^10 → 10 half-lives = 1000 s

Log formula verification (75%): $t = \frac{2.303}{6.93 \times 10^{-3}} \times \log\frac{1}{0.25} = 332.3 \times \log 4 = 332.3 \times 0.602 = 200 \text{ s} \checkmark$

Problem 3: Zero-Order Half-Life

k = 0.01 mol/(L·s), [A]_{0} = 0.5 M:

$t_{1/2} = \frac{0.5}{2 \times 0.01} = 25 \text{ s}$

Note: If [A]{0} is doubled to 1.0 M, t{1}/_{2} = 1.0/(0.02) = 50 s (doubles).

Problem 4: Finding k from Incomplete Reaction Data

For first-order: 60% complete in 40 min (40% = [A]/[A]_{0} remaining):

$k = \frac{2.303}{40} \log\frac{1}{0.40} = \frac{2.303}{40} \times \log 2.5 = \frac{2.303 \times 0.398}{40} = 0.02291 \text{ min}^{-1}$

$t_{1/2} = \frac{0.693}{0.02291} = 30.25 \text{ min}$

Problem 5: Radioactive Decay

t_{1}/_{2} = 5 years. Fraction remaining after 20 years:

n = 20/5 = 4 half-lives. Fraction = (1/2)^{4} = 1/16 = 6.25% remains.