The rate constant k for a reaction at 400 K is 2×10⁻³ s⁻¹. If the activation energy is 80 kJ/mol, what is the rate constant at 410 K? (R = 8.314 J/mol·K)
Correct: A
Use the Arrhenius equation: ln(k2/k1) = -Ea/R (1/T2 - 1/T1). Substitute values for T1=400K, T2=410K, Ea=80000 J/mol, k1=2×10⁻³. Solve for k2. The calculation yields ln(k2/2×10⁻³) ≈ 0.366, leading to k2 ≈ 3.15×10⁻³ s⁻¹.