the rate constant doubles when the temperature is increased from 45.0 c to 73.0 c. what is the activation energy for this reaction? (r=8.314 j/kmol)?

The activation energy for this reaction is 22.6 kJ/ mol

Explanation:

Step 1: Data given

Rate constant doubles when Temperature goes from 45.0 °C to 73.0 °C

R = 8.314 J/K*mol

Step 2: Calculate the activation energy

Log (k2/k1) = Ea / 2.303R *((1/T1) - (1/T2))

⇒ with k1 = initial rate constant

⇒ with k2 = rate constant after doubled = 2k1

⇒ T1 = initial temperature = 45.0 °C = 318 Kelvin

⇒ T2 = Final temperature = 73.0 °C = 346 Kelvin

log (2) = Ea / (2.303*8.314) *((1/318) - (1/346))

log(2) = Ea / (2.303*8.314) * 0.00025448

Ea = 22649 J/mol = 22.6 kJ/mol

The activation energy for this reaction is 22.6 kJ/ mol

1 a

2 d

3 a

4 c

explanation: