Question

Rate constant k of a reaction varies with temperature according to the equation

log k = constant $–\frac{E_{a}}{2.303RT}$. When a graph is plotted for log k versus $\frac{1}{T}$ a straight line with a slope – 5632 is obtained. The energy of activation for this reaction is –

• A. 127.67 kJ mol^–1
• B. 107.84 kJ mol^–1
• C. 86 kJ mol^–1
• D. 246.8 kJ mol^–1

Answer 1

Answer: (B)

107.84 kJ mol^–1

Answer 2

Slope = $–\frac{E_{a}}{2.303R}$ = – 5632

Ž E_a = 5632 × 2.303 × 8.314 = 107.8 kJ/mol