± The Arrhenius Equation Part A The Arrhenius equation shows the relationship be
ID: 551347 • Letter: #
Question
± The Arrhenius Equation Part A The Arrhenius equation shows the relationship between the rate constant k and the temperature T in kelvins and is typically written as The activation energy of a certain reaction is 46.0 kJ/mol. At 24 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 where R is the gas constant (8.314 J/mol K), A is a constant called the frequency factor, and Es is the activation energy for the reaction. However, a more practical form of this equation is Th= 56.5 which is mathmatically equivalent to Submit My Answers Give Up ln (4-4 Incorrect; Try Again; 2 attempts remaining where ki and k2 are the rate constants for a single reaction at two different absolute temperatures (T1 and T2) Part B Given that the initial rate constant is 0.01 90s 1 at an initial temperature of 24C , what would the rate constant be at a temperature of 180 ° C for the same reaction described in Part A? Express your answer with the appropriate units. Hints Value UUnits Submit My Answers Give Up vide Feedba ContinueExplanation / Answer
A)
we have:
T1 = 24 oC
=(24+273)K
= 297 K
K2/K1 = 2
Ea = 46 KJ/mol
= 46000 J/mol
we have below equation to be used:
ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)
ln(2/1) = (46000.0/8.314)*(1/297.0 - 1/T2)
0.6931 = 5532.8362*(1/297.0 - 1/T2)
T2 = 308 K
= (308-273) oC
= 35 oC
Answer: 35 oC
B)
we have:
T1 = 24 oC
=(24+273)K
= 297 K
T2 = 180 oC
=(180+273)K
= 453 K
K1 = 1.90*10^-2 s-1
Ea = 46 KJ/mol
= 46000 J/mol
we have below equation to be used:
ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)
ln(K2/1.90*10^-2) = (46000.0/8.314)*(1/297.0 - 1/453.0)
ln(K2/1.90*10^-2) = 5533*(1.159*10^-3)
K2 = 11.6 s-1
Answer: 11.6 s-1
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