1. false ΔG
lower, disorder
gas
a. negative
b. positive
c. negative
d. positive
e. negative
f. positive

2. never spontaneous
a. S > 0 because 1 mol gas produces 3 mol gas
c. ΔS > 0 because liquid turns to gas
d. ΔS > 0 because 1 mol liquid produces 1 mol liquid + ½mol gas
8) ΔS°rxn = [2 (81.2)] – [4(23.77 ) + 3( )] =
J/K
9) Plug the enthalpy and entropy values into ΔG° = ΔH° −TΔS° and you’ll find that ΔG° < 0 only when the temperature is above approximately 841 K ( 568 °C ); anything lower than this temperature makes ΔG°> 0 (non-spontaneous)

3. 10) a) ΔG°= – [(298 )( )]=
kJ/mol
b) No, it is not spontaneous at 25 °C
(ΔG°is a large positive value)
Set ΔG = 0:
0 = ΔH –TΔS ⇒ 0 = – T
T = K ⇐ This is T at equilibrium spontaneous at T > K
11) positive, positive
Spontaneous ABOVE the melting point of zero degrees Celsius.

4. 2C(s) + 2H2O(g) → CH4(g) + CO2(g)
C should be a reactant - first equation is good!
CO2 should be a product - second equation is good!
CH4 should be a product - third equation needs to be ‘flipped’
C(s)  + H2O(g)  →  CO(g) + H2(g)             ΔH = kJ
CO(g) + H2O(g)  → CO2(s)  +  H2(g)       ΔH = kJ 2
3H2(s)  + CO(g)  →  CH4(g) + H2O(g)        ΔH = kJ
Notice: when I ‘flipped’ the 3rd equation, I flipped the sign of ΔH
Apply ‘multipliers’ to arrive at the desired final equation

5. 2C(s) + 2H2O(g) → CH4(g) + CO2(g)
2C(s)  + 2H2O(g)  →  2CO(g) + 2H2(g)             ΔH = kJ
CO(g) + H2O(g)  → CO2(s)  +  H2(g)       ΔH = kJ
3H2(s)  + CO(g)  →  CH4(g) + H2O(g)        ΔH = kJ
2C(s)  + 2H2O(g) →  CH4(g)  +  CO2(g)
ΔH = kJ
intermediates → H2, CO
produced, then consumed

6. FeO(s) + CO(g) → Fe(s) + CO2(g)
Fe should be a product - second equation is good!
FeO(s)  + CO(g) →  Fe(s)  +  CO2(g)
FeO should be a reactant - third equation needs to be ‘flipped’
Why didn’t I deal with the first equation????
Everything in that equation was in other equations
Now I set the first equation up so everything in it will cancel
FLIP IT
FLIP IT REAL GOOD!!
2Fe3O4(s)  + CO2(g)  →  3Fe2O3(g) + CO(g)     ΔH = kJ
Fe2O3(s) + 3CO(g)  → 2Fe(s)  +  3CO2(g)       ΔH = kJ ⅙
3FeO(s)  + CO2(g)  →  Fe3O4(s) + CO(g)  ΔH = kJ ⅓
Apply ‘multipliers’ to arrive at the desired final equation

7. FeO(s) + CO(g) → Fe(s) + CO2(g)
⅓Fe3O4(s)  + ⅙CO2(g)  →  ½Fe2O3(g) + ⅙CO(g)      ΔH = kJ
½Fe2O3(s) + 1½CO(g)  → Fe(s)  +  1½CO2(g)       ΔH = kJ
FeO(s)  + ⅓CO2(g)  →  ⅓Fe3O4(s) + ⅓CO(g)  ΔH = kJ
FeO(s)  + CO(g) →  Fe(s)  +  CO2(g)
ΔH = - 11 kJ
intermediates → Fe2O3
produced, then consumed
catalysts → Fe3O4
consumed, then produced