devarshi-dt-logo

Question:

Rate constant k of a reaction varies with temperature according to the equation logk = constant − Ea/(2.303R × 1/T) where Ea is the energy of activation for the reaction. When a graph is plotted for logk vs 1/T a straight line with a slope −670k is obtained. The activation energy for this reaction will be: (R = 8.314 JK⁻¹mol⁻¹)

150.00 kJmol⁻¹

142.34 kJmol⁻¹

122.65 kJmol⁻¹

127.71 kJmol⁻¹

Solution:

Slope of the line = −Ea/(2.303R) = −670K
Ea = 2.303 × 8.314 (JK⁻¹mol⁻¹) × 670K = 127711.4 Jmol⁻¹ = 127.71 kJ mol⁻¹