Question

A certain reaction proceeds three times faster at 585K than it did at 535K. Calculate the activation energy, in KJ/mol, for this reaction.

a) 60 KJ/mol
b) 70 KJ/mol
c) 80 KJ/mol
d) 90 KJ/mol

Answer 1

Using the Arrhenius equation, we determine that the reaction proceeds three times faster at 585 K than at 535 K. The calculated activation energy for this reaction is approximately 70 kJ/mol(option b).

Step-by-step explanation:

To determine the activation energy for a reaction based on the given information, we can use the Arrhenius equation, which relates the rate of a chemical reaction to the temperature and the activation energy.

The Arrhenius equation in its logarithmic form is:

ln(k2/k1) = (Ea/R) * ((1/T1) - (1/T2))

where:
k1 = rate constant at temperature T1
k2 = rate constant at temperature T2
Ea = activation energy
R = universal gas constant (8.314 J/mol·K)
T1 = initial temperature in Kelvin
T2 = final temperature in Kelvin

Given that the reaction is three times faster at 585 K (k2) than at 535 K (k1), we can express this rate increase as k2 = 3k1. We can then plug the values into the Arrhenius equation:

ln(3) = (Ea/8.314) * ((1/535) - (1/585))

Solving for Ea, we find that the activation energy is approximately 70 kJ/mol, which corresponds to option (b).