15.63 How to know what A is

[PranithaPrasad]

Can someone explain how you would know what A is in these types of problems? In the solutions manual for this problem at the end of the set up for the problem, they put in -0.59 and I'm assuming that this is from ln A from the equation we're supposed to use, but I'm not sure how we would go about soling for A

[Austin Ho 1E]

Not sure what you mean? You don't need A at all in this problem. Given that ln(k₂/k₁)=E_a/R * (1/T₁ - 1/T₂), you just need to plug in values. Additionally, A is just the pre-exponential constant in the Arrhenius equation, so you could find it.

[Lindsay H 2B]

I don't think A has anything to do with the 0.59 in the solutions manual because A is eliminated when subtracting ln(k2)-ln(k1). However, I also don't understand why the 0.59 is there, could someone explain this part?

[Ashley Davis 1I]

Not sure what you mean? You don't need A at all in this problem. Given that ln(k₂/k₁)=E_a/R * (1/T₁ - 1/T₂), you just need to plug in values. Additionally, A is just the pre-exponential constant in the Arrhenius equation, so you could find it.

What do you mean we could find it? How? How would we find A?

[Ashley Davis 1I]

Update: I saw there's a chart with values for A depending on the reaction. For this reaction (CO2 + OH- → HCO3-), A would be 1.5 x 10^10 L/mol*s. But even then, ln(1.5 x 10^10) does not equal -0.59. So why is -0.59 included in the solutions manual?

[Jana Sun 1I]

Oh wow I didn't even notice the -0.59 in the solutions manual until now. I didn't use it at all in my calculations and I still go the same answer (probably because our answer ended up being so large that the -0.59 wouldn't have made a difference). But now I'm really curious about the -0.59. Does anyone know where this comes from?

[Joshua Hughes 1L]

I believe it meant to say that it is equal to .59 and not a minus .59. when you calculate it out Ln(k'/k) is .5937

[Charles Ang 1E]

if you think about it, If A=1, then Ln(A) should be 0. I'm not sure if this is correct, but its the way I think of it.