Question

• Assignment Score: 990/1100 Resources Ex Give Up? Feedback Try Again Question 11 of 11 > Attempt 3 Calculate the activation energy, E, in kilojoules per mole for a reaction at 74.0 °C that has a rate constant of 0.237s and a frequency factor of 4.50 x 10's! E = 9.2343452 x10-9 J/mol No Pw

Answer 1

Step 1: Explanation:

Activation Energy can be calculated from Arrhenius equation

k=Ae^−Ea/RT

Where, k = Rate constant, A = Frequency factor , Ea = Activation energy, R = Gas constant

T = Absolute Temperature

Step 2: Extract the data from question

k = 0.237 s^-1

A = 4.50 × 10¹¹ s^-1

Ea = we need to calculate

R = 8.314 J / mol.K

T = 74 °C = (74+273.15)K = 347.15 K

Step 3: Calculation of activation Energy

k=Ae^−Ea/RT

By substituting the above value

0.237 s^-1 = 4.50 × 10¹¹ s^-1 × e^−Ea/RT

=> ( 0.237 s^-1 / 4.50 × 10¹¹ s^-1 ) = e^−Ea/RT

Taking ln both side of equation to remove the expoenential

=> ln (( 0.237 s^-1 / 4.50 × 10¹¹ s^-1 )) = -Ea / RT

=> -28.27220856 =   -Ea / ( 8.314 J / mol.K ) × 347.15 K

=> -28.27220856 =   -Ea / 2886.2051 J /mol

=> -28.27220856 × 2886.2051 J /mol =   -Ea

on cancelling negative sign both sides we got

Ea = 81599.39 J/mol

Hence activation Energy in kJ/mol

we know, 1 kJ = 1000 J

Ea =( 81599.39 J/mol ) × (1 kJ / 1000 J ) = 81.6 kJ /mol