Untitled - Kelly Walsh High School

Kinetics 195
Quick Tip
Careful!
Remember that the k in the denominator, k 2, goes with the first temperature, T 2.
The two rate constant values, k 1 and k 2, are the values determined at two different
temperatures, T 1 and T 2. The temperatures must be in Kelvin. The units
on the rate constants will cancel, leaving a unitless ratio. R is 8.314 J/molK. The
activation energy will have units of joules/mol.
The variation in the rate constant at two different temperatures for the decomposition
of HI(g) to H 2(g) I 2(g) is given below. Calculate the activation energy.
T (K) k (L/mol s)
555 3.52 10 7
781 3.95 10 2
We will use the equation
ln k 1
k 2
T 1 781 K k 1 3.95 10 2 L/mols
T 2 555 K k 2 3.52 10 7 L/mols
R 8.314 J/molK E a ?
We could rearrange this equation and then enter the values or we can enter the
values and then rearrange the equation. Using the second method gives:
ln 3.95 102 L/mol s
3.52 10 7 L/mol s
Ea 1
¢
R T2 1
≤
T1 E a
8.314 J/mol K ¢
1
555 K
1
781 K ≤
Calculations containing logarithms and/or differences in reciprocals are very
sensitive to rounding. Be careful! not round intermediate values.