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Published byΠανκρατιος Ζερβός Modified over 5 years ago
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false ΔG lower, disorder gas a. negative b. positive c. negative d. positive e. negative f. positive
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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)
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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.
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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
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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
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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
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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
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