Chemistry 102: Kinetics: frequency factor and temperature

Question

The compound 1,1-difluoroethane decomposes at elevated temperatures to give fluoroethylene and hydrogen fluoride according to the equation:

CH3​CHF2​(g)CH2​CHF(g)+HF(g)

Part 1). At 460K, the rate constant for this reaction is k=5.8 x 10^-6 s^-1 and the activation energy is E_a = 265 kJ/mol. What is the frequency factor for this reaction?

Part 2). At what temperature would the reaction proceed with a rate four times faster than 460k?

Answer 1

k = Ae^-E/RT A is the "frequency factor"

A = ke^E/RT

A = 5.8 x 10^-6 x e^{265000 J/mol/8.314 J/mole-K/460K} = 7.181 x 10²⁴

We can solve for t for a k = 4 x 5.8 x 10^-6: since k is proportional to the rate.

-E/RT = ln(k/A)

RT/E = -1/ln(k/A)

T = -E/(Rln(k/A)) plugging in A from part (1)

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