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ok so there are following three steps
A+F ---> AF Activation energy: +2 KJ/mol Heat of Rx= -.1 KJ/mol
AF+H -> AFM+ TM Activation E: +10KJ/mol Heat of Rx= +5.2 "
AFM+ F->AF+ TM Activation E: +15 " Heat of Rx= -6 "
I have to find which one is the rate-determining step : one thats slowest. This stuff is from the beginning of equlibrium. And for potential energy diagram which step should I use?

2006-11-23 18:04:35 · 3 answers · asked by passion flower 1 in Science & Mathematics Chemistry

3 answers

The second one:
AF + H --> AFM + TM

Activation E: +10KJ/mol is moderate
Heat of Rx= +5.2 KJ/mol is positive, making this an endothermic reaction (requires the input of heat or energy to accomplish).

See the site below:

2006-11-23 18:42:53 · answer #1 · answered by Richard 7 · 28 1

According to the Arrhenius equation the rate constant of a reaction and the activation energy are related as follows:

k= Ae^(-Ea/RT)

Assuming that A is the same for the reactions (otherwise you can't say anything), the higher the Ea (activation energy), the slower the reaction, regardless if it is exothermic or endothermic.
So the rate determining step is the last one.

For the potential energy diagram depends on what you want to show. You can use all reactions for a detailed diagram or you can plot each step separately or only the rate limiting step or the total reaction.

2006-11-24 06:24:36 · answer #2 · answered by bellerophon 6 · 0 0

None of them. The energies are not implicated in the reaction rates, which are more determined by temperature, concentration, and the possible presence of catalysts.

2006-11-24 02:22:28 · answer #3 · answered by Anonymous · 0 1

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