[chelovek]

I'm a bit confused here. The problem I'm working on gives me the standard entropy of solid lead at 298K as 64.8jk/mol. I'm given an equation for the heat capacity Cp of solid lead (based on T), and an equation for the heat capacity of liquid lead (also based on T). I'm also given the melting point of lead (600k) and the heat of fusion.

The question is "What is the standard entropy of liquid lead at 770K"

What exactly does "standard entropy" mean? Is it the integral from absolute 0 to T, of Cp/T dt? I tried to verify that by calculating the Cp of solid lead at 298, divided by T, integrated from 0 to 298, but this didn't give me 64.8. In fact, I can't integrate at 0 since it's undefined...so, how exactly is standard entropy calculated? OR, am I confusing "standard entropy" with "entropy at standard state"?

[Tflarr]

I'm not sure if this is exactly what your looking for but it might help:

http://www.chemicool.com/definition/standard_entropy.html

Hope it helps

[Hunt]

I think you need the entropy of the liquid at P = 1 bar ( standard state ) and T = 770 K. So take the change in entropy for a process of a solid lead melting to liquid.

 S(770) = S(298) + \int_{298}^{T_f} [ C_{p}(solid) \frac {dT}{T}] + \frac { \Delta H(fusion) }{T_f} + \int_{T_f}^{770} C_{p}(liquid) \frac {dT}{T} 

T_f = freezing / melting pt
S(298) = 64.8 KJ

ofcourse here i assume T_f < 770 K