High School

A researcher raises the temperature from 46.5 to 64.4°C and finds that the rate of the reaction doubles. What is the activation energy (in kJ/mol) for this reaction?

Options:
A) 23.5 kJ/mol
B) 46.5 kJ/mol
C) 54.8 kJ/mol
D) 69.2 kJ/mol

Answer :

Final answer:

The activation energy for the reaction can be calculated using the Arrhenius equation, but the given information is not sufficient to perform the calculation and find the correct option.

Explanation:

This question is about determining the activation energy of a chemical reaction using the Arrhenius equation, which is: k = Ae^-Ea/RT. In this equation, k is the rate constant, A is the frequency factor, Ea is the activation energy, R is the ideal gas constant (8.314 J/mol/K), and T is the temperature in Kelvin. Given that the rate of the reaction doubles when the temperature raised from 46.5 to 64.4 °C, we can set up two equations using the Arrhenius equation (one for the initial rate and one for the final rate) and solve for Ea. However, without the exact values for k and A, we can not perform the calculation and select the correct option among the ones given (A) 23.5 kJ/mol, B) 46.5 kJ/mol, C) 54.8 kJ/mol, D) 69.2 kJ/mol).

Learn more about Activation Energy here:

https://brainly.com/question/28384644

#SPJ11