Question

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.

Answer 1

The temperature is 39.1 ⁰C

Step-by-step explanation:

Applying Arrhenius equation which relates rate constants to the temperature

`ln((K_2)/(K_1)) = (E_a)/(R)[(1)/(T_1)-(1)/(T_2)]`

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

`ln((K_2)/(K_1)) = (E_a)/(R)[(1)/(T_1)-(1)/(T_2)]\\\\ln((0.724)/(0.531)) = (66,000)/(8.314)[(1)/(308.4)-(1)/(T_2)]\\\\0.31 = 7938.417[(1)/(308.4)-(1)/(T_2)]\\\\3.905 X 10^(-5) = (1)/(308.4)-(1)/(T_2)\\\\(1)/(T_2) = (1)/(308.4)- 3.905 X 10^(-5)\\\\(1)/(T_2) = 0.003243 - 0.00003905\\\\(1)/(T_2) = 0.003204\\\\{T_2} = 312.1 K`

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