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Lecture March 13

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Chapter 14 - Gibbs Free Energy

Remember: Chapter 14 HW and Quiz dueTuesday, March 26th

All regrade requests must besubmitted by Friday, March 15th!

Chapter 16 Sec. 16.9 and 16.10

Summary of Entropy

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! Suniverse = ! Sreaction - ! H/T

! Ssys + ! Ssurr

Eq 14.8

Rearranging:-T ! Suniv = ! H - T ! Sreact

! G OR ! G = ! H - T ! S ! G < 0 reaction spontaneous

! G > 0 reaction not spontaneous! G = 0 reaction is in equilibrium Note: signs

inverted from ! S!!

Calculating ! Go - Part II

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Use the standard values of formation! Go to calculate ! Go for a reaction

CaCO 3(s) " CaO(s) + CO 2(g)

Analogous to!

Ho

offormation

! Go = #! Go(products) - #! Go(reactants)! Go = [(-603.3) + (-394.4)] - (-1128.8)

= +131.1 kJ

Positive so reaction is NOT spontaneous at 25 oC.

T = 298.15 KPressure =1 atm

Temperature Dependence of ! G

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CaCO 3(s) " CaO(s) + CO 2(g)

! Go = ! Ho - T ! So

! Go > 0 Reaction NOT spontaneous (at 25 oC)If you assume that ! S and ! H at di $ erent temperatures

can be estimated by their values at 25 oC, then cancalculate T when reaction becomes spontaneous "

Set ! G =0 = ! Ho - T ! So

T = ! H/ ! S = (178.5)/(.1588) =1124 Kso T >1124 K reaction is spontaneous

Same Tas

before

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Analyzing a System inTerms of Sign of ! H and ! S

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!

exothermic ! spontaneous at

LOW T! + Always

spontaneous

+ endo ! Never spontaneous

+ +spontaneous at

HIGH T

! H ! S ! G = ! H -T ! S

Qualitative Example

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Given the reaction can determine the sign

of ! S

Given that the reaction is exothermic

Consider the reaction2NO(g) + 2H 2(g) " N2(g) + 2H 2O(g)

4 moles of gas to 3 moles of gas -

Reaction will be spontaneous ( ! G

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AgCl at Non-StandardConditions

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What if [Ag + ] = 1.0 x 10 -6 and [Cl - ] = 2.2 x 10 -6

Ag+ (aq) + Cl - (aq) AgCl(s)

Q = 1/(1.0x10 -6 )(2.2x10 -6 ) = 4.5 x 10 11

Note: reverse of K sp

! G = -53.0 + (.008315)(373)(26.84)= +30.2 kJ

Using T= 100. oC lnQ = 26.84

NOT Spontaneous - no ppt

Quiz of the Day

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Use the accompanying thermodynamic data for theformation of CCl 4(g) via the hypothetical reaction of

C(s) + 2Cl 2 (g) CCl 4(g)

!

Ho

(kJ/mol) So

(J/mol-K)CCl4 (g) -102.9 309.7C(s) 5.7Cl2(g) 223.0

If you are at standardpressures of 1 atm, whatis ! Go for this reaction at

50.0 oC?Ans: ! Go = ! Ho - T ! So ! Ho = -102.9 kJ

! So = 309.7 - {5.7+ 2 (223.0)} =-142.0 J/K

! G = -102.9 - (323.15)(-.1420)= -102.0 + 45.9 = -57.0 kJ