April 18, 2024, 12:03:00 AM
Forum Rules: Read This Before Posting

### Topic: Help with pChem - Calculating K at different Temperatures  (Read 2464 times)

0 Members and 1 Guest are viewing this topic.

#### WalrusMaximus

• Very New Member
• Posts: 1
• Mole Snacks: +0/-0
##### Help with pChem - Calculating K at different Temperatures
« on: December 10, 2014, 11:20:46 PM »
I'm having trouble determining when to use: lnK = -(delta)G/RT vs dK/d(1/T) = (delta)H/R
My professor says I can use the first equation at any T but on homework the equation only works at 298K. He then said something like "the first equation tells you K at any T, the second tells you how K relates to T" what's the difference? Sounds like the same thing.
I appreciate any help.
If a specific example would help, I'm referring to this. http://puu.sh/dpPaj/5e1f6701f4.png

#### mjc123

• Chemist
• Sr. Member
• Posts: 2053
• Mole Snacks: +296/-12
##### Re: Help with pChem - Calculating K at different Temperatures
« Reply #1 on: December 12, 2014, 11:19:43 AM »
The first equation applies at any T, but you might only know ΔG° at 298K or some other specific temperature. If you know ΔH° (and assume it is temperature-independent), you can use the second equation to calculate K at any other temperature within reason.
Your second equation is not quite correct; it should be d(lnK)/d(1/T) = ΔH°/R

#### Plontaj

• Regular Member
• Posts: 29
• Mole Snacks: +4/-1
##### Re: Help with pChem - Calculating K at different Temperatures
« Reply #2 on: January 11, 2015, 05:47:50 AM »
Nope.
Quote
d(lnK)/d(1/T) = ΔH°/R

The correct form is: d(lnK)/d(1/T) = -ΔH°/R

Van't Hoff isobar: d(lnK)/d(1/T)= -ΔH°/R at constant pressure is a derivative of the temperature of equation: lnK = -ΔG°/(RT) = -ΔH°/(RT) + ΔS°/R . To calculate lnK from van't Hoff isobar you need integrate this equation and need to know the integral constant = ΔS°/R.

Derivation:
ΔG° = ΔH° - TΔS°

lnK=-ΔH°/(RT) + ΔS°/R

d(lnK)/d(T) = ΔH°/(RT^2)

d(lnK)/d(1/T) = -ΔH°/(R)