Question

The activation energy of a certain reaction is 37.2 kJ/mol . At 20 ∘C, the rate constant is 0.0130 s−1. At what temperature would this reaction go twice as fast? Express your answer numerically in degrees Celsius

Answer 1

Answer: At 34°c

Step-by-step explanation:

Using The Arrhenius Equation:

k = Ae − Ea/RT

k represents rate constant

A represents frequency factor and is constant

R represents gas constant which is = 8.31J/K/mol

Ea represents the activation energy

T represents the absolute temperature.

By taking the natural log of both sides,

ln k = ln A- Ea/RT

Reactions at temperatures T1 and T2 can be written as;

ln k1= ln A− Ea/RT1

ln k2= ln A− Ea/RT2

Therefore,

ln(k1/k2) = −Ea/RT1 + Ea/RT2

Since k2=2k1 this becomes:

ln(1/2) = Ea/R*[1/T2 − 1/T1]

Theefore,

-0.693 = 37.2 x 10^3/8.31 * [ 1/T2 - 1/293]

1/T2 - 1/293 = -1.55 x 10^-4

1/T2 = -1.55 x 10^-4 + 34.13x 10^-4

1/T2 = 32.58 x 10^-4

Therefore T2 = 307K

T2 = 307 - 273 = 34 °c