## Final Van't Hoff Eqn

$\ln K = -\frac{\Delta H^{\circ}}{RT} + \frac{\Delta S^{\circ}}{R}$

jillian1k
Posts: 54
Joined: Sat Jul 22, 2017 3:00 am

### Final Van't Hoff Eqn

I was going over the notes from lecture on friday and came across the van't hoff eqn derivation. I'm confused about whether we're supposed to use this eqn -ln(K2/K1) = -(deltaH*/RT2) + (deltaH*/R) or this one -ln (K2/K1) = -deltaH*/R(1/T2-1/T1). If it's the latter, then my question would be why can we eliminate ∆S° from the eqn, but not ∆H°? I realize there would be no way to use the eqn without one of these variables, but I'm confused why we can get rid of ∆S° and not ∆H° from the final eqn even though they are both constant. Sorry if this is obvious and I just missed a step in lecture!

Joyce Lee 1C
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Joined: Fri Sep 29, 2017 7:03 am
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### Re: Final Van't Hoff Eqn

at T1:
ln(K1) = (-ΔH°/RT1) + (ΔS°/R)
at T2:
ln(K2) = (-ΔH°/RT2) + (ΔS°/R)
if you assume that ΔS° is constant and subtract ln(K1) from ln(K2), the (ΔS°/R) cancels out
ln(K2) - ln(K1) = (-ΔH°/RT2) + (ΔS°/R) - [(-ΔH°/RT1) + (ΔS°/R)] = (-ΔH°/RT2) + (ΔS°/R) + (ΔH°/RT1) - (ΔS°/R) =
(-ΔH°/RT2) + (ΔH°/RT1)
you're left with:
ln(K2/K1) = (-ΔH°/RT2) + (ΔH°/RT1) = (-ΔH°/R)[(1/T2) - (1/T1)]

Tiffany Dao 1A
Posts: 32
Joined: Fri Sep 29, 2017 7:05 am

### Re: Final Van't Hoff Eqn

In lecture, I believe the reasoning was because the standard entropy is assumed to be constant and the same for both temperatures, so they would cancel out when you derive the equation. Also, since the R is a constant. If you look at the explanation above you'll see this. The enthalpies, however, do not cancel because the denominator is different due to the different temperatures. Therefore they can't be removed, but you are right that they are both assumed to be constant.