(Part A) The activation energy of a certain reaction is 30.8 k J / m o l . At 22
ID: 819910 • Letter: #
Question
(Part A)The activation energy of a certain reaction is 30.8kJ/mol .
At 22 ?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 22 ?C ,
what would the rate constant be at a temperature of 180 ?C for the same reaction described in Part A?
(Part A)
The activation energy of a certain reaction is 30.8kJ/mol .
At 22 ?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 22 ?C ,
what would the rate constant be at a temperature of 180 ?C for the same reaction described in Part A?
Explanation / Answer
A)
Now using k = A e^(-Ea/RT)
taking log(base e);
-4.509 = ln A - 12.557
ln A = 8.048
A = 3130.49 s-1
thus for double the rate;
0.022 = ln (3130.49) - Ea/RT
0.022 = 8.048 - 30800/8.314xT
T = 461.57 K = 188.4 degree C
B) k1/k2 = e^ (-30800/8.314[1/295 - 1/453])
k1/k2 = 0.0125 s-1
k2 = 0.88 s-1
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