Question

The reaction rate doubles on raising the reaction temperature from 298K to 308K. What is the activation energy of this reaction? Assume Ea is temperature independent in this range.

a) 20 kJ/mol
b) 40 kJ/mol
c) 60 kJ/mol
d) 80 kJ/mol

Answer 1

The activation energy of this reaction is approximately 57.3 kJ/mol.

Step-by-step explanation:

According to the question, the reaction rate doubles when the temperature is increased from 298K to 308K. If the activation energy (Ea) is temperature independent, we can use the Arrhenius equation to find the Ea value. The Arrhenius equation is given by:

k = A imes e^(-Ea/RT)

where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin.

Since the reaction rate doubles, we can set up the equation:

k_2 = 2 imes k_1

Substituting the values into the Arrhenius equation:

A imes e^(-Ea_2/RT) = 2 imes A imes e^(-Ea_1/RT)

Dividing both sides by A and taking the natural logarithm:

-Ea_2/RT = ln(2) -Ea_1/RT

Simplifying:

Ea_2 - Ea_1 = R imes T imes ln(2)

Since Ea is temperature independent, Ea_1 = Ea_2, so:

Ea = R imes T imes ln(2)

Substituting the given values: Ea = 8.314 J/mol K imes 308 K imes ln(2)

Calculating the value: Ea ≈ 57.3 kJ/mol

Therefore, the activation energy of this reaction is approximately 57.3 kJ/mol, which is closest to option (c) 60 kJ/mol.