II. Calculation of activation energy (5points) Arrhenius equation can be represe
ID: 536089 • Letter: I
Question
II. Calculation of activation energy (5points)
Arrhenius equation can be represented as
lnk = -Ea/(RT) + lnA
where k is the rate constant, Ea the activation energy, T the absolute temperature, R the gas constant (8.314 J/molK), and A the collision frequency factor among molecules.
Plotting of the set of data between lnk and 1/T reveals a straight line with its slope being Ea/R. Therefore, the activation energy, Ea, can be obtained because Excel plotting will give you the value for the slope and the gas constant R is already known.
The following table is assumed to show the rate constants for the above reaction:
Temperature, X(°c) k, Y (s-1)
189.7 0.0000252
198.9 0.0000525
230.3 0.0006300
251.2 0.0031600
(II-1) Using these data, calculate the activation energy for the reaction. Show both the regenerated data table and the plot from which you obtain the activation energy.
(II-2) What is the value of the rate constant at 430.0 K?
(Hint: Use anyone of the rate constants and temperatures from the given table and the following equation):
ln(k1/k2) = Ea/R(1/T2 – 1/T1)
Explanation / Answer
the data on temperature need to be converted into K, by T (K)= t(deg.c)+273
from Arhenius equation, lnK= lnKo-Ea/RT, Ea= activation energy , K= rate constant , T= temperature in K, Ko= Frequency factor,
so a plot of lnK vs 1/T gives a straight line whose slope is Ea/R, where R= 8.314 J/mole.K. the plot along with the data points are shown below
from the plot, the equation of best fit is lnK= -19038*1/T+30.51
from the best fit, Ea/R= slope= 19038, Ea= 19038*8.314 joules/mole =158292 J/mole
at 430K
lnK = -19038/430+30.51 = -13.76
K= 1.052*10-6/sec