University of Washington Department of Chemistry Chemistry 452 Summer Quarter 2014

Lecture 8 7/11/14 ERD: 5.4-5.7 Devoe: 4.5

A. Entropy for Irreversible Processes. • In past lectures we showed how to calculate the enthalpy change for irreversible processes. Basically we composed a sequence of reversible steps which begin and end in the same states as the irreversible change. Because DH is a state function, the enthalpy change for the real, irreversible change is the same as the enthalpy change calculate for the sequence of reversible steps. • The strategy is the same for entropy calculations. To calculate the entropy for an δ q irreversible change we can still use dS = rev but we have to compose a T reversible path that starts and ends in the same states as the irreversible process. • It is very important that the properties of reversible and irreversible changes not be confused. A famous example is the entropy change for a reversible / irreversible adiabatic expansion. o By definition, for a reversible adiabatic expansion, qrev=0. Because δ q dS = rev we can assume ∆S=0 for such a process. T o For an irreversible adiabatic expansion we have qirrev=0. .But this does not mean the entropy change is zero. This can be seen by considering that

∆Uq=+rev w rev = q irrev + w irrev (8.1) which follows from the fact that U is a state function so ∆U is invariant with respect to whether a reversible or irreversible path is followed. o Now we have shown that the work performed by a gas which is undergoing a reversible is greater than the work performed if the same expansion is done reversibly:

−wwrev>− irrev (8.2) o .From equations 8.1 and 8.2 we conclude:

qqrev> irrev (8.3)

• So for a irreversible adiabatic expansion even though qirrev=0, the entropy change ∆S>0

B. Entropy Problems for Irreversible Processes

Example 1: Entropy Calculation for an Irreversible Vaporization

Suppose 18g of water at 298K are evaporated by falling on a iron crucible that is maintained at 423K. Calculat the entropy change for the water, the crucible and the l universe. For liquid water cJmoleKP =⋅75.4 / and for water vapor v cJmoleKP =⋅36.0 / . Also, assume ∆Hvap = 40. 68kJ / mole.

Solution: To calculate ∆S we break the transition from liquid water at T=298K to water vapor at 423K (i.e. in thermal contact with the crucible) into three steps

• Step 1: Heat liquid water from 298K to 373K

⎛⎞T2 ⎛⎞373 ∆=Sncwater P ln⎜⎟ =()( 1 moleJmoleK 75.4 / ⋅ ) ln⎜⎟ = 16.8 JK / ⎝⎠T1 ⎝⎠298

• Step 2: Vaporize the water at 373K

∆Hvap 40,/ 680Jmole ∆Swater ==b1moleg =1091./JK T 373K

• Step 3: Heat water vapor from373K to 423K

⎛⎞423 ∆=∆=SncTmoleJmoleKwater P ()(1 36.0 / ⋅ ) ln⎜⎟ = 4.40 JK / ⎝⎠373 • Total entropy change for the water:

∆Swater =+168./J K 1091 ./J K + 4 . 40J /K = 130 ./ 3J K

• To calculate the entropy change for the surroundings i.e. the crucible, we assume any heat that went into the water came out of the crucible. The heat transfeered from the crucible to the water was done so at constant pressure. The heat leaving the crucible is:

lv qHP=∆ crucible =−() ncTnHncT P ∆ step13 + ∆ vap + P ∆ step

=−(()(1mole 75.4 J / mole ⋅ K )( 373 K − 298 K )()( + 1 mole 40,680 J / mole ) +⋅b1molegb 36 J/ mole Kgb 423 K− 373 Kgh =-48,135J • To calculate the entropy chane of the crucible you now assume the hea calculated above is transferred at a constant temperature of 423K… ∆H −48, 135J ∆S ==crucible =−1138./JK crucible T 423K • The entropy change of the universe is:

∆∆∆Suniverse=+S waterS crucible = 1305./J K − 1138 ./J K = 16 . 7K /K

Example 2. Entropy of Mixing

Assume two ideal gases, initially in separate containers, are mixed at constant temperature. Assume their initial pressures are each 1 atm (i.e. when separate) and the final total pressure PT=PA+PB =1 atm. Calculate the entropy of mixing.

• For the gas A…

⎛⎞VP21AA ⎛⎞ ⎛⎞ P 2 A⎛⎞χ A ()1atm ∆=SnRnRnRnRAAln⎜⎟ = A ln ⎜⎟ =− A ln ⎜⎟ =− A ln ⎜⎟ ⎝⎠VP12AA ⎝⎠ ⎝⎠ P 1 A⎝⎠1 atm

n A where χ A = and where PP2 AATA==χχ(1 atm) nnAB+ • Similarly for gas B…

FV I F P I F P I F χ B 1atm I ∆SnR= ln2 B = nR ln1B =−nR ln2 B =−nR ln bg BBG J A G J B G J B G J H V1B K H P2B K H P1B K H 1atm K

nB where χ B = nnAB+ • Add the two results and divide by nA+nB to get the entropy of mixing per mole…

∆∆S AB+ S n A nB ∆S = =− R lnbgχχA − R ln bgB nnAB+ nnAB+ nnAB+

=−χχχχAABBRRlnb g − lnb g

The formula can be generalized to include any number of components. That is, N mixing of N gases together has an entropy change of ∆SR=− ∑ χχiilnbg. i=1 • Calculate the entropy change when 50 g each of oxygen O2, nitrogen N2, and hydrogen H2 are mixed at 1 atm pressure and 273K. Solution: • First calculate the moles of each gas mixed 50g molesO ==156. moles 2 32gmole/ 50g moles N ==179. moles 2 28gmole/ 50g moles H ==25moles 2 2gmole/ • Then calculate the mole fractions total moles== nT 156.. + 179 + 25= 28 . 35 moles 156. moles χ =≈006. O2 28. 35moles 179. moles χ =≈006. N2 28. 35moles 25moles χ =≈088. H2 28. 35moles • Calculate the final expression:

∆SR=−χχχχln + ln + χχ ln ejOO22di N 2 di N 2 HH 2 di 2 =−bgbgbgbg8. 31JmolesK / ⋅c 0 . 06 ln. 0 06 + 0 . 06 ln. 0 06 + 088 . ln. 088 h =−bgbg8./ 31JmolesK ⋅ − 0 . 34 − 011 . = 376 ./ JmolesK ⋅