The activation energy of a certain reaction is 46.2 kJ/mol . At 20 C, the rate c
ID: 523423 • Letter: T
Question
The activation energy of a certain reaction is 46.2 kJ/mol . At 20 C, the rate constant is 0.0130 s1. At what temperature would this reaction go twice as fast?
The activation energy of a certain reaction is 46.2 kJ/mol.At 20 oC, the rate constant is 0.0130 s 1 At what temperature would this reaction go twice as fast? Express your answer numerically in degrees Celsius T2 OC Submit Hints My Answers Give Up Review Part Incorrect, Try Again; 5 attempts remaining Part B Given that the initial rate constant is 0.0130 s 1 at an initial temperature of 20 °C, what would the rate constant be at a temperature of 100 C? Express your answer numerically in inverse seconds.Explanation / Answer
A)
T1 = 20.0 oC =(20.0+273)K = 293.0 K
K1 = 0.013 s-1
K2 = 2*K1 = 0.026 s-1
Ea = 46.2 KJ/mol = 46200.0 J/mol
use:
ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)
ln(0.026/0.013) = (46200.0/8.314)*(1/293.0 - 1/T2)
0.69 = 5556.89*(1/293.0 - 1/T2)
T2 = 304
= (304-273) oC
= 31 oC
Answer: 31 oC
B)
T1 = 20.0 oC =(20.0+273)K = 293.0 K
T2 = 100.0 oC =(100.0+273)K = 373.0 K
K1 = 0.013 s-1
Ea = 46.2 KJ/mol = 46200.0 J/mol
use:
ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)
ln(K2/0.013) = (46200.0/8.314)*(1/293.0 - 1/373.0)
ln(K2/0.013) = 5557*(7.32*10^-4)
K2 = 0.760 s-1
Answer: 0.760 s-1
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