[sinjid9]

I did an assignment in class where we were given a table with temperature and the rate constant at that temperature. Using the graph of the information, we found activation energy for
2 HI(g) -> I(2) (gas) + H2 (gas)
My teacher wanted activation energy per mole of HI(g) so he told us to double the activation energy we found with the slope of the graph. Why do we double the activation energy, if HI is at 2 moles in the chemical equation and we want the activation energy per mole? Shouldn't the EA be less if less of the reactant is there to absorb the energy?

[juanrga]

I did an assignment in class where we were given a table with temperature and the rate constant at that temperature. Using the graph of the information, we found activation energy for
2 HI(g) -> I(2) (gas) + H2 (gas)
My teacher wanted activation energy per mole of HI(g) so he told us to double the activation energy we found with the slope of the graph. Why do we double the activation energy, if HI is at 2 moles in the chemical equation and we want the activation energy per mole? Shouldn't the EA be less if less of the reactant is there to absorb the energy?

Ea is a property of the reaction (more concretely of the rate constant k). Now the rate V of the reaction is not the rate of the concentration of HI, but

$$ \frac{d[HI]}{dt} = 2 V /$$

I suppose that that is why he says you to double, but I do not understand very well what is being done here.

[Vidya]

I think your teacher wanted you to calculate for the given reaction and the table has provided for per mole.Generally all values in the table are for one mole.He is asking you to double it because you have two moles in the reaction.

[sinjid9]

I think your teacher wanted you to calculate for the given reaction and the table has provided for per mole.Generally all values in the table are for one mole.He is asking you to double it because you have two moles in the reaction.

Actually, yeah I think that was it.