Ella_M
I am looking at the Ch 18 notes, and it says that when W>0 there is work done on the system. Does this mean that when W<0 there is work done BY the system? Does my question make sense?

Never mind! Just found my answer in the following notes.
Thanks!

Ella_M.....Glad you figured it out! The sign conventions are confusing but can be understood as follows. Energy (either in the form of heat or work) going into the the system (as in "endo" thermic) is assigned as positive. Conversely, energy is assigned as negative when it comes out of the system (as in "exo" thermic). Of course, the actual quantity (amount) of energy is always positive). The signs merely indicate the direction of the energy flow.

Hi Dr. Bob, I have a couple of questions regarding the exercises.

For question 37, I understand that if both delta S system and surrounding are positive, the reaction is spontaneous, and if both are negative, then the rxn is nonspontaneous. However, what if there is a system where Ssys is negative and Ssurr is positive (or vice versa), how would we determine spontaneity. Is it based on Suniv? And would the system's spontaneity be different based on temperature similar to question 41?

For question 41, since spontaneity is based on Suniv, how does the fact that delta H is positive pr negative factor in (beside it being used to calculate Suniv)? Like we've always learned that negative H is spontaneous, but with 41d, delta H is - but +Suniv so it's nonspontaneous.

For question 107, do you have any pointer to how we should approach it? I have listed all of the given (K1, T1, K2, T2) but I'm not sure how I could find delta S and H from them.

Thank you!

Mariana.....At first, one of the most confusing aspects of thermodynamics is the 2nd Law which relates spontaneity of a process in the system, e.g., a chemical reaction, to the overall entropy change of the universe. This is discussed in Section 18.5, example 18.3 (Tro, p 854), and the preceding paragraphs are especially useful, along with Eq 18.2: dS(surr) = -dH(sys)/T. (I'm using "d" for delta.) This shows that, if the reaction is exothermic, i.e., dH(sys) is negative, then the entropy of the surroundings increases. For the gas phase reactions in 14.37, you can estimate dS(sys) by looking at the change in moles of gas. More moles of gaseous products relates to an increase in entropy, i.e, dS(sys) is positive. See part c of Example 18.2 and then continue into Section 18.6 to see hows this all leads to the Gibbs free energy relationship, Eq 18.6: dG = dH - TdS.

Mariana.....The concept is related to #18.107 is "temperature dependence of the equilibrium constant" as per Tro, pp 875-876, and Eqs 18.18 and 18.19. Given K1 and K2 at temps T1 and T2, use Eq 18.19 to find dH(rxn). Then use this value in Eq 18.18, with one K and the corresponding T, to calculate dS(rxn).