Question

For a reaction under study, a graph of ln k versus 1/T (K−1) was prepared. The line of best fit to the data was determined to be: y = -8.09 × 103 x + 31.5 . What is the activation energy, E a, for this reaction in kJ/mol?

Answer 1

To do this you need the general equation for this graph that is the following:

K = Ae^-Ea/RT

If you graph ln k vs 1/T so, adjusting the expression at this:

lnK = lnA - Ea/RT

the equation you have can be interpreted like this:

y = lnK

lnA = 31.5

-Ea/R = -8.09x10³

using R as 8.3144 J/mol K Ea can be solved:

Ea = 8.09x10³ R

Ea = 8.09x10³ * 8.3144

Ea = 67263.496 J/mol and in kJ/mol it would be:

Ea = 67263.496 / 1000

Ea = 67.26 kJ/mol

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