Question

Part A:

The activation energy of a certain reaction is 50.0 kJ/mol . At 25 ^∘C , the rate constant is 0.0110s⁻¹. At what temperature in degrees Celsius would this reaction go twice as fast?

Express your answer with the appropriate units.

Part B:

Given that the initial rate constant is 0.0110s⁻¹ at an initial temperature of 25  ^∘C , what would the rate constant be at a temperature of 140.  ^∘C for the same reaction described in Part A?

Express your answer with the appropriate units.

Answer 1

A)
Given:
T1 = 25 oC
=(25+273)K
= 298 K
K1 = 1.10*10^-2 s-1
K2 = 2*K1 = 2.20*10^-2 s-1
Ea = 50 KJ/mol
= 50000 J/mol

use:
ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)
ln(2.2*10^-2/1.1*10^-2) = (50000/8.314)*(1/298.0 - 1/T2)
0.6931 = 6013.9524*(1/298.0 - 1/T2)
T2 = 309 K
= (309-273) oC
= 36 oC
Answer: 36 oC

B)
Given:
T1 = 25 oC
=(25+273)K
= 298 K
T2 = 140 oC
=(140+273)K
= 413 K
K1 = 1.1*10^-2 s-1
Ea = 50 KJ/mol
= 50000 J/mol

use:
ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)
ln(K2/1.1*10^-2) = (50000.0/8.314)*(1/298.0 - 1/413.0)
ln(K2/1.1*10^-2) = 6014*(9.344*10^-4)
K2 = 3.03 s-1
Answer: 3.03 s-1