Enthalpy Entropy Gibb’s Free Energy Practice Problems, Exercises of Chemistry

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Enthalpy/Entropy/

Gibb’s Free Energy

Practice Problems

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2(g)

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• Calculate;
• ∆S
• ∆H f
• ∆H rxn
• ∆G at 20˚C
• Critical Temperature
• ∆T if this reaction were used to change the temperature of 500ml of water.

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• 1. Determine, just by looking at this equation whether ∆S is positive or negative. POSITIVE – although it is 9 molecules going to 8, there is a liquid going to gas and phase changes are more important usually.
• 2. Calculate ∆S using the table provided. Compound S˚ J /mol*K ∆ Hf kJ /mol HCN (^) (l) 94.1 151. O2 (g) 205.0 0 H 2 O (^) (g) 189.0 - 242. CO2 (g) 214 - 393. N2 (g) 192 0

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• 2. Calculate ∆S using the table provided. Compound S˚ J /mol*K ∆ Hf kJ /mol HCN (^) (l) 94.1 151. O2 (g) 205.0 0 H 2 O (^) (g) 189.0 - 242. CO2 (g) 214 - 393. N2 (g) 192 0

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• 2. Calculate ∆S using the table provided.
• +216. J / mol*K
• Is this result consistent with your expectations? Compound S˚ J /mol*K ∆ Hf kJ /mol HCN (^) (l) 94.1 151. O2 (g) 205.0 0 H 2 O (^) (g) 189.0 - 242. CO2 (g) 214 - 393. N2 (g) 192 0

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2(g)

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• 3. Calculate ∆H f using the table provided. Compound S˚ J /mol*K ∆ Hf kJ /mol HCN (^) (l) 94.1 151. O2 (g) 205.0 0 H 2 O (^) (g) 189.0 - 242. CO2 (g) 214 - 393. N2 (g) 192 0

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• 4. Calculate ∆H rxn using bond enthalpy using the table below Bond Bond Enthalpy kJ/mol H-C 413 C≡N 891 O=O 498 H-O 463 C=O 804 N≡N 945

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• 4. Calculate ∆H rxn using bond enthalpy using the table below
• 4(413) + 4(891) + 5(498) = +7,
• 4(463) + 8(804) + 2(945) = - 10,
• ∆H = - 2268 kJ/mol Bond Bond Enthalpy kJ/mol H-C 413 C≡N 891 O=O 498 H-O 463 C=O 804 N≡N 945

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• 5. Which value would be more accurate to use: ∆H f or ∆H rxn ? Explain why
• ∆H f = - 2662 kJ/mol
• ∆H rxn = - 2268 kJ/mol
• Heats of formation values are more specific to standard conditions and are more accurate than averaged bond enthalpy values.

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• 6. Under what conditions would this reaction be spontaneous, given the ∆S value from #2 and the ∆H f value from #3.
• ∆S = +216. J / mol*K
• ∆ H f = - 2662 kJ/mol Δ H → Δ S ↓ Exothermic (-) Endothermic (+) Increasing Entropy (+) Always Spontaneous at any temperature Only spontaneous above critical temperature Decreasing Entropy (-) Only spontaneous below critical temperature Never Spontaneous (but reverse is)

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• 8. What is the critical temperature in which this reaction become spontaneous or is no longer spontaneous?
• 9. Is this value consistent with your predictions from #7? Explain why.

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• 8. What is the critical temperature in which this reaction become spontaneous or is no longer spontaneous?
• 0 = - 2662 – (T*0.2166) = - 12,290 K
• 9. Is this value consistent with your predictions from #7? Explain why.
• Yes, because negative kelvin values are impossible. Thus, any and all temperature values would result in a negative ∆G. That’s why when ∆S is positive and ∆H is negative, the reaction is always spontaneous.