Question

If the rate of a reaction increases by a factor of 2.8 when the temperature is...

If the rate of a reaction increases by a factor of 2.8 when the temperature is raised from 20 ∘C to 30 ∘C, what is the value of the activation energy in kJ/mol?

By what factor does the rate of this reaction increase when the temperature is raised from 130 ∘C to 140 ∘C?

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Homework Answers

Answer #1

1)

Given:

T1 = 20 oC

=(20+273)K

= 293 K

T2 = 30 oC

=(30+273)K

= 303 K

K2/K1 = 2.8/1

use:

ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)

ln(2.8/1) = ( Ea/8.314)*(1/293 - 1/303)

1.0296 = (Ea/8.314)*(1.126*10^-4)

Ea = 75997 J/mol

Ea = 76.0 KJ/mol

Answer: 76.0 KJ/mol

2)

Given:

T1 = 130 oC

=(130+273)K

= 403 K

T2 = 140 oC

=(140+273)K

= 413 K

Ea = 75.997 KJ/mol

= 75997 J/mol

use:

ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)

ln(K2/K1) = (75997.0/8.314)*(1/403 - 1/413.0)

ln(K2/K1) = 9141*(6.008*10^-5)

K2/K1 = 1.732

Answer: 1.732

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