If a temperature increase from 12.0 ∘c to 22.0 ∘c doubles the rate constant for a reaction, what is the value of the activation barrier for the reaction?

48,501 J/mol

Explanation:

1) Action barrier = activation energy = Ea

2) Data:

i) T₁ = 12°C = 12 + 273.15 K = 285.15K

ii) T₂ = 22°C = 22 + 273.15 K = 295.15 K

iii) rate constant = k: k₂ / k₁ = 2

iv) Ea = ?

3) Formula:

Arrhenius' law gives the relationship between the constant of reaction and the temperature:

4) Solution

By arranging the formula, you get:

㏑[k₂/k₁] =Ea/R [1/T₁ - 1/T₂]

Replace k₂ = 2k₁; T₁ = 285.15; and T₂ = 295.15

ln[2] = Ea/8.314 J/K mol × [1/285.15 - 1/295.15]K

Ea = ln [2] × 8.314 J/K mol / [1.18818×10⁻⁴K] = 48,501 J/mol

while cooking a piece of 1/2 pound beef steak, a cast iron pan (2.4kg in weight) was heated 260 celsius.