Data Set K: 0.054 T:10.0C K:0.152 T:20.0C K:0.398 T:30.0C K:0.980 T:40.0C Using
ID: 635946 • Letter: D
Question
Data Set
K: 0.054 T:10.0C
K:0.152 T:20.0C
K:0.398 T:30.0C
K:0.980 T:40.0C
Using A and Ea, answer the following questions: A=3.04x10^-13 Ea(kj/mol)= -0.765
1) Find the value of k at 50.0degrees celsius. Report your answer to 3 decimal places. T=323K
2) Find the temperature where k will be 100x greater than the k at 20degrees celsius (from your data set). report your answer to 1 decimal place in celsius.
3) A catalysts is added and the rate constant increases by 100x at 20degrees celsius. find the new activation energy. report your answer in kj/mol to one decimal place. use the k at 20degrees celsius from your data set.
Explanation / Answer
Governing equation:
ln k=ln A - Ea/(RT)
K:rate constant
A: Arrhenius constant
Ea: activation energy
R: gas constant (0.008314 KJ/mol.K)
T: temperature in Kelvin
Putting values in the formula
ln k= ln (3.04×10^(-13) ) -(-.765)/(.008314×323)
On solving
K=4.042×10^[-13]
Calculation of temperature using formula:
ln(k 2 /k 1)= -Ea/R ×{1/T 2 - 1/T 1)
Putting k 2 =100 k 1 and other values
ln(100)= .765/.008314 ×{1/T 2 -1/293)
So T 2 =18.7 kelvin
As k=100×0.152=15.2
Calculating Ea using first formula
ln15.2=ln(3.04× 10^[-13] )- Ea/(.008314×293)
On solving
Ea =-76.8 KJ/mol
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