MOaD 12

Question 1

Entropy change formula from one temperture to another?

Answer 1

S2 = S1 + INT FUS(CP/T) + H/T?(FUS) + INT VAP,FUS(CP/T) + H/T(VAP) INT(VAP)CP/T

Question 2

Statistical entropy

Answer 2

KbinW

Question 3

Third law

Answer 3

The entropy (of all perfect crystals) at 0 K is zero”, or S0 = 0

At 0 K all quenchable energy has been quenched. In the case of a perfect crystal, all the atoms are in a regular uniform array, and the absence of disorder and thermal chaos suggests that such materials must have zero entropy as well.

Question 4

For example, if you cool CO very slowly, its entropy at 0 K is ~0, but if you cool it too rapidly its entropy is ~5 J K-1 mol-1, not zero. Why is this?

Question 5

Cooling slowly:

Answer 5

as the CO molecules solidify they have enough energy to equilibrate and settle in the lowest-energy configuration. This forms a perfect crystalline solid, with S0 = 0.

Answer 6

Cooling quickly: the CO molecules lose their thermal energy rapidly before they can rotate into the lowest energy configuration. So they freeze in a random arrangement. As there are 2 orientations and N molecules, W = 2N and S0 = NkBln2. For a mole, we have S0 = Rln2 ~ 5.8 J K-1 mol-1.

Question 7

S

Answer 7

, W = 2N and S0 = NkBln2.
for a mole:
For a mole, we have S0 = Rln2 ~ 5.8 J K-1 mol-1.

Question 8

entropy change

Answer 8

products - reactants

Question 9

S2

Answer 9

= rS 1 + intcp/t

Question 10

irreversibility

Answer 10

The surroundings don’t care what happens, they either supply the maximum amount of heat, qrev, (if the process takes place infinitely slowly) or less, q, if they can get away with it.

Question 11

S uni

Answer 11

ΔSuni = (qrev / T )sys + ( -q / T )surr

Question 12

if the q return from surrounding is less than the reaction then

Answer 12

q < qrev, ΔSuni will be positive, and this is the condition for a spontaneous (irreversible) change

Question 13

which from standard tables gives ΔrHo = -286 kJ mol-1 and ΔrSo = -163 J mol-1 K-1 at T = 298 K. So, do these values this make sense?

Answer 13

Experimentally we know the system gives out a lot of heat. And ΔHsys is negative (exothermic).

But the reaction is spontaneous, so we expect an increase in the overall entropy of the universe. The surroundings absorb the heat produced, therefore ΔSsurr = -ΔrH / T = 959 J mol-1 K-1. So although the entropy of our system (the reaction) actually decreased a modest amount, it is offset because the entropy of the surroundings increased hugely. Therefore ΔSuni = -163 + 959 = +796 J mol-1 K-1, which is positive. This is the condition for a spontaneous process!