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

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(205.138)] = -548.1 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)

10) a) ΔG°= 491.18 – [(298 )(0.19767)]= + 432.27 kJ/mol b) No, it is not spontaneous at 25 °C (ΔG°is a large positive value) Set ΔG = 0: 0 = ΔH –TΔS ⇒ 0 = 491.18 – 0.19767T T = 2484.8 K ⇐ This is T at equilibrium . . . spontaneous at T > 2484.8 K 11) positive, positive Spontaneous ABOVE the melting point of zero degrees Celsius.

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 = +131.3 kJ CO(g) + H2O(g)  → CO2(s)  +  H2(g)       ΔH = - 41.2 kJ 2 3H2(s)  + CO(g)  →  CH4(g) + H2O(g)        ΔH = -206.1 kJ Notice: when I ‘flipped’ the 3rd equation, I flipped the sign of ΔH Apply ‘multipliers’ to arrive at the desired final equation

2C(s) + 2H2O(g) → CH4(g) + CO2(g) 2C(s)  + 2H2O(g)  →  2CO(g) + 2H2(g)             ΔH = +262.6 kJ CO(g) + H2O(g)  → CO2(s)  +  H2(g)       ΔH = - 41.2 kJ 3H2(s)  + CO(g)  →  CH4(g) + H2O(g)        ΔH = -206.1 kJ 2C(s)  + 2H2O(g) →  CH4(g)  +  CO2(g) ΔH = + 15.3 kJ intermediates → H2, CO produced, then consumed

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 = +47.0 kJ Fe2O3(s) + 3CO(g)  → 2Fe(s)  +  3CO2(g)       ΔH = - 25.0 kJ ⅙ ½ 3FeO(s)  + CO2(g)  →  Fe3O4(s) + CO(g)  ΔH = -19.0 kJ ⅓ Apply ‘multipliers’ to arrive at the desired final equation

FeO(s) + CO(g) → Fe(s) + CO2(g) ⅓Fe3O4(s)  + ⅙CO2(g)  →  ½Fe2O3(g) + ⅙CO(g)      ΔH = +7.83 kJ ½Fe2O3(s) + 1½CO(g)  → Fe(s)  +  1½CO2(g)       ΔH = - 12.5 kJ FeO(s)  + ⅓CO2(g)  →  ⅓Fe3O4(s) + ⅓CO(g)  ΔH = -6.33 kJ FeO(s)  + CO(g) →  Fe(s)  +  CO2(g) ΔH = - 11 kJ intermediates → Fe2O3 produced, then consumed catalysts → Fe3O4 consumed, then produced