The activation energy of a certain reaction is 46 6 kJ/mol At 21 °C , the rate c
ID: 551884 • Letter: T
Question
The activation energy of a certain reaction is 46 6 kJ/mol At 21 °C , the rate constant is 0.0190s.At what temperature in degrees Celsius would this reaction go twice as fast? Express your answer with the appropriate units Hints 2 TValue Units Submit My Answers Give Up Part B Given that the initial rate constant is 0.0190s at an initial temperature of 21 °C , what would the rate constant be at a temperature of 100. C for the same reaction described in Part A? Express your answer with the appropriate units Hints IA 2 kValue Units Submit My Answers Give UpExplanation / Answer
A)
we have:
T1 = 21 oC
=(21+273)K
= 294 K
K2/K1 = 2
Ea = 46.6 KJ/mol
= 46600 J/mol
we have below equation to be used:
ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)
ln(2) = (46600.0/8.314)*(1/294.0 - 1/T2)
0.6931 = 5605.0036*(1/294.0 - 1/T2)
T2 = 305 K
= (305-273) oC
= 32 oC
Answer: 32 oC
B)we have:
T1 = 21 oC
=(21+273)K
= 294 K
T2 = 100 oC
=(100+273)K
= 373 K
K1 = 1.9*10^-2 s-1
Ea = 46.6 KJ/mol
= 46600 J/mol
we have below equation to be used:
ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)
ln(K2/1.9*10^-2) = (46600.0/8.314)*(1/294.0 - 1/373.0)
ln(K2/1.9*10^-2) = 5605*(7.204*10^-4)
K2 = 1.08 s-1
Answer: 1.08 s-1
Feel free to comment below if you have any doubts or if this answer do not work
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