If a plot of ln(k) versus 1/T yields a line with slope -Ea/R and the slope is -6.0 x 10^3 K, what is Ea approximately?

• A. About 49.9 kJ/mol
• B. About 0.0499 kJ/mol
• C. About 499 kJ/mol
• D. About 6.0 kJ/mol

Answer: (A)

The relationship ln k versus 1/T is linear because of the Arrhenius equation, ln k = ln A − Ea/(RT). The slope of this line is −Ea/R, so Ea equals the negative of the slope times R. Here the slope is −6.0 × 10^3 K, and using R = 8.314 J/mol·K gives Ea = −(−6.0 × 10^3) × 8.314 J/mol ≈ 4.9884 × 10^4 J/mol ≈ 49.9 kJ/mol. So the activation energy is about 50 kJ/mol.