Part A The activation energy of a certain reaction is 38.2 kJ/mol . At 26 C , th
ID: 877573 • Letter: P
Question
Part A
The activation energy of a certain reaction is 38.2 kJ/mol . At 26 C , the rate constant is 0.0120s1. At what temperature in degrees Celsius would this reaction go twice as fast?
Express your answer with the appropriate units.
k
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Part B
Given that the initial rate constant is 0.0120s1 at an initial temperature of 26 C , what would the rate constant be at a temperature of 200 C for the same reaction described in Part A?
Express your answer with the appropriate units.
s
T2 =k
Explanation / Answer
From equation
ln(K2/K1)= (Ea/R)*(1/T1-1/T2)
Ea= activation energy = 38.2 Kj/mol= 38.2X1000 J/Mol= 38200j/mol
it is given that the reactionat T2 is twice fast at T1 (26 Deg.C)
T2 has to be calculate given T1 =26+273= 299 K
ln2= (382200/8.314)*(1/299-1/T2)
1/299- 1/T2= ln2/4594.66 =0.000151
1/T2= 1/299-0.000151
T2=313K=40deg.c
given that K2/K1=2
K2= 2* K1=2*0.0120 =0.0240s-1
b) At temperature 200 deg.c, the rate constant can be dtermined from
ln(K2/K1)= (Ea/R)* (1/T1- 1/T2)
Where T1= 26+ 273= 299 K
T2= 200 +273 =473K
Ea= 38.2Kj/Mol= 38200 J/Mol
R= 8.314 J/Mol
ln(k2/K1)= (38200/8.314) *(1/299-1/473) =5.65
K2/ K1 = e5.65= 285.11
K2= 285*0.0120=3.42s-1