A first order reaction has a rate constant of 0.531 sec-1 at 35.4 oC. If the activation energy is 66 kJ, calculate the temperature in oC at which the rate constant is 0.724 sec-1. Use 1 decimal place.

The temperature is 39.1 ⁰C

Explanation:

Applying  Arrhenius equation which relates rate constants to the temperature

where;

Ea is the activation energy = 66 kJ

R is gas constant = 8.314 J/k.mol

T₁ is the initial gas temperature = 35.4°C = 308.4K

T₂ is the final gas temperature  = ?

K₁ is rate constant at T₁ = 0.531

K₂ is rate constant at T₂ = 0.724

T₂(°C) = 312.1 -273 = 39.1 ⁰C

Therefore, the temperature at which the rate constant is 0.724 s⁻¹ is 39.1°C

thermal energy is the answer