Step 1: Calculate the ratio k₂/k₁ from first-order kinetics

For a first order reaction: t = (2.303/k) log([A]₀/[A])

At 298K, 10% completion (90% remains): t₁ = (2.303/k₁) log(100/90) = (2.303/k₁) log(10/9)

At 308K, 25% completion (75% remains): t₂ = (2.303/k₂) log(100/75) = (2.303/k₂) log(4/3)

Since t₁ = t₂: (2.303/k₁) log(10/9) = (2.303/k₂) log(4/3)

k₂/k₁ = log(4/3)/log(10/9)

log(4/3) = 0.125 and log(10/9) = 0.0458

k₂/k₁ = 0.125/0.0458 = 2.73

Step 2: Calculate Eₐ using Arrhenius equation

Using: log(k₂/k₁) = (Eₐ/2.303R)[1/T₁ - 1/T₂]

log(2.73) = 0.436

1/298 - 1/308 = (308-298)/($298 \times 308$) = 10/91784 = $1.089 \times 10$⁻⁴ K⁻¹

2.303 × R = $2.303 \times 8.314$ = 19.15 J mol⁻¹ K⁻¹

0.436 = (Eₐ/19.15) × $1.089 \times 10$⁻⁴

Eₐ = ($0.436 \times 19.15$)/($1.089 \times 10$⁻⁴)

Eₐ = 76700 J mol⁻¹ = 76.7 kJ mol⁻¹

Step 3: Calculate k at 318K

Using Arrhenius equation: k = A·e^(-Eₐ/RT)

At T = 318K:

Eₐ/RT = 76700/($8.314 \times 318$) = 76700/2643.85 = 29.02

k = $4 \times 10$¹⁰ × e^(-29.02)

k = $4 \times 10$¹⁰ × $2.58 \times 10$⁻¹³

k = $1.03 \times 10$⁻² s⁻¹

Alternatively using logarithmic form: log k = log A - Eₐ/(2.303RT) log k = 10.60 - 76700/($2.303 \times 8.314$ × 318) log k = 10.60 - 12.59 = -1.99 k = 10⁻¹·⁹⁹ = $1.03 \times 10$⁻² s⁻¹