Answer to Question #179782 in General Chemistry for Tzuyu

Question

A first order reaction has a rate constant of 4.6x10^-2/sec and 8.10x10^-2/sec at 0°C and 20°C, respectively.

What is the value of the activation energy?

Answer 1

ln (k2/k1) = Ea/R (1/T1-1/T2)

R = 8.3145

For the reaction to go twice as fast, k2 = 2 k1, so (k2/k1) = 2. Then,

(4.6X10^-2 J/mol / 8.1 × 10^-2 J/molK) = Ea/8.3145 × (1/273-1/293)

0.5679 = Ea/8.3145 × (0.00366 - 0.00341)

0.5679× 8.3145 = 0.00025Ea

0.00025 Ea = 4.72180455

Ea = 18887.2182kJ