Question

The rate constant for a reaction is 1.5 × 10^–7 sec^–1 at 50°C. What is the value of activation energy?

Options

• 2.2 × 10³ J mol^–1

• 2300 J mol^–1

• 2.2 × 10⁴ J mol^–1

• 220 J mol^–1

Answer 1

2.2 × 10⁴ J mol^–1 

Explanation:

`log K_2/K_1 = E_a/(2.303  R) [(T_2 - T_1)/(T_1 - T_2)]`

`log  ((4.5 xx 10^7))/((1.5 xx 10^7)) = E_a/(2.303  R) [(T_2 - T_1)/(T_1 - T_2)]`

or `log 3 = (E_a xx 50)/(2.303 xx 8.314 xx 19.147 xx 373 xx 232)`

E_a = 2011 J mol^–1 = 2.2 × 10⁴ J mol^–1