Solve Physical Chemistry Homework: ∆S, Td, ∆fCp - Tmax Stability

[hazel06]

Homework Statement

The question is attached.

Homework Equations

-∆_fS (T - T₀) - (∆_fC_p/T₀)[(T - T₀)²/2] = 0

Please let me know if this doesn't make any sense!

The Attempt at a Solution

Given:

Td: 39.3ºC = 312.45K
∆H: 157 kJ/mol = 152000 J/mol

I calculated ∆S = -∆H/Td -----> ∆S = -486.48 J/mol.K (Entropy)

Now, for the temperature of maximum stability. I know that I have ∆S, Td(T₀), and ∆_fC_p. All is needed is to plug it into the equation I was given. But, I know I would need to derived it to make it more simpler to find T. Or at least that's what I think!

I derived it. And got: -∆_fS-∆_fC_p(T)/T₀+∆_fC_p

Not sure if I did it right! If someone could confirm I did it right or show me the correct way to derive it. That would be great! B/c I plugged the values in and did not get a value of around 260K. Which is given by the instructor to be the approximate answer.

Set up as: -(-486.48 J/mol.K) - (2800 J/mol.K)(T)/312.45K + 2800 J/mol.K = 0

Maybe I calculated it wrong b/c I keep on getting 312.62K. Which is exactly like the 312.45K. Very clueless!

After this could someone give me a hint on how to start finding the cold denaturation temperature? Thanks!

 

[nate808]

Hi there! It looks like you're on the right track with your derivation. However, there are a few things that you may want to double check.

Firstly, when you derived the equation, it should have been -∆fCp/T0 instead of +∆fCp. This is because the equation is set up so that the terms on the left side are all negative, while the terms on the right side are all positive.

Also, when you plugged in the values, you should have gotten 312.62K instead of 312.45K. This is because you plugged in 2800 J/mol.K for ∆fCp, but the given value is actually 2800 J/mol.K.K. So, when you divide by 312.45K, you get a slightly different value.

As for finding the cold denaturation temperature, you can use the same equation and plug in a different value for ∆H. Instead of using 157 kJ/mol, you can use a lower value to represent the energy needed for cold denaturation. Then, you can solve for T to find the corresponding temperature at which the protein is stable. Hope this helps!