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AP Chemistry Flashcards: Free Energy Of Dissolution

Study Free Energy Of Dissolution in AP Chemistry with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

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What this deck covers

This deck focuses on Free Energy Of Dissolution, giving you a quick way to review the definitions, rules, and examples that matter most for AP Chemistry.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

AP Chemistry Flashcards: Free Energy Of Dissolution

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QUESTION

Which quantity becomes zero at the dissolution equilibrium point?

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ANSWER

△G\triangle G△G. Free energy equals zero when forward and reverse rates are equal.

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Flashcard 1: Which quantity becomes zero at the dissolution equilibrium point?

Answer: △G\triangle G△G. Free energy equals zero when forward and reverse rates are equal.

Flashcard 2: Identify the factor that primarily affects △S\triangle S△S in dissolution.

Answer: Molecular disorder. Dissolution typically increases randomness as molecules spread out.

Flashcard 3: Determine the spontaneity: △H=5 kJ/mol\triangle H = 5 \text{ kJ/mol}△H=5 kJ/mol, T=298 KT = 298 \text{ K}T=298 K, △S=30 J/mol\cdotpK\triangle S = 30 \text{ J/mol·K}△S=30 J/mol\cdotpK.

Answer: Spontaneous. △G=5000−298(30)=5000−8940=−3940\triangle G = 5000 - 298(30) = 5000 - 8940 = -3940△G=5000−298(30)=5000−8940=−3940 J/mol, so spontaneous.

Flashcard 4: What does a positive △S\triangle S△S indicate about entropy in dissolution?

Answer: Entropy increases. Positive △S\triangle S△S means the system becomes more disordered.

Flashcard 5: Determine the spontaneity: △H=5 kJ/mol\triangle H = 5 \text{ kJ/mol}△H=5 kJ/mol, T=298 KT = 298 \text{ K}T=298 K, △S=30 J/mol\cdotpK\triangle S = 30 \text{ J/mol·K}△S=30 J/mol\cdotpK.

Answer: Spontaneous. △G=5000−298(30)=5000−8940=−3940\triangle G = 5000 - 298(30) = 5000 - 8940 = -3940△G=5000−298(30)=5000−8940=−3940 J/mol, so spontaneous.

Flashcard 6: State the effect on △G\triangle G△G if both △H\triangle H△H and △S\triangle S△S are positive.

Answer: Depends on TTT. Sign depends on whether T△ST\triangle ST△S is greater or less than △H\triangle H△H.

Flashcard 7: Calculate △H\triangle H△H given △G=−3 kJ/mol\triangle G = -3 \text{ kJ/mol}△G=−3 kJ/mol, T=310 KT = 310 \text{ K}T=310 K, △S=80 J/mol\cdotpK\triangle S = 80 \text{ J/mol·K}△S=80 J/mol\cdotpK.

Answer: △H=21.8 kJ/mol\triangle H = 21.8 \text{ kJ/mol}△H=21.8 kJ/mol. △H=△G+T△S=−3000+310(80)=21800\triangle H = \triangle G + T\triangle S = -3000 + 310(80) = 21800△H=△G+T△S=−3000+310(80)=21800 J/mol.

Flashcard 8: Identify the type of dissolution with △S>0\triangle S > 0△S>0 and △H<0\triangle H < 0△H<0.

Answer: Spontaneous dissolution. Both favorable enthalpy and entropy make △G\triangle G△G negative at all temperatures.

Flashcard 9: State the condition for a dissolution to be spontaneous at all temperatures.

Answer: △H<0\triangle H < 0△H<0, △S>0\triangle S > 0△S>0. Exothermic enthalpy and positive entropy ensure spontaneity.

Flashcard 10: What is indicated by a positive △G\triangle G△G in dissolution?

Answer: Non-spontaneous process. Positive free energy means the process is thermodynamically unfavorable.

Flashcard 11: What determines the spontaneity of a dissolution process?

Answer: Sign of △G\triangle G△G. Negative △G\triangle G△G means spontaneous, positive means non-spontaneous.

Flashcard 12: How does the dissolution of a gas in a liquid typically affect ΔS\Delta SΔS?

Answer: ΔS\Delta SΔS decreases. Gas molecules become more ordered when dissolved in liquid.

Flashcard 13: What is the relationship between △H\triangle H△H and △S\triangle S△S in spontaneous dissolution?

Answer: △H<T△S\triangle H < T\triangle S△H<T△S. For spontaneity, the entropy term must overcome unfavorable enthalpy.

Flashcard 14: Calculate △G\triangle G△G given △H=10 kJ/mol\triangle H = 10 \text{ kJ/mol}△H=10 kJ/mol, T=298 KT = 298 \text{ K}T=298 K, △S=50 J/mol\cdotpK\triangle S = 50 \text{ J/mol·K}△S=50 J/mol\cdotpK.

Answer: △G=−4.9 kJ/mol\triangle G = -4.9 \text{ kJ/mol}△G=−4.9 kJ/mol. △G=10000−298(50)=10000−14900=−4900\triangle G = 10000 - 298(50) = 10000 - 14900 = -4900△G=10000−298(50)=10000−14900=−4900 J/mol.

Flashcard 15: What is the sign of △G\triangle G△G when dissolution is at equilibrium?

Answer: △G=0\triangle G = 0△G=0. Zero free energy indicates the system has reached equilibrium.

Flashcard 16: Which factor affects the sign of △G\triangle G△G at constant pressure and temperature?

Answer: Enthalpy and entropy. Both enthalpy and entropy contributions determine free energy sign.

Flashcard 17: Calculate △G\triangle G△G with △H=20 kJ/mol\triangle H = 20 \text{ kJ/mol}△H=20 kJ/mol, T=350 KT = 350 \text{ K}T=350 K, △S=70 J/mol\cdotpK\triangle S = 70 \text{ J/mol·K}△S=70 J/mol\cdotpK.

Answer: △G=−4.5 kJ/mol\triangle G = -4.5 \text{ kJ/mol}△G=−4.5 kJ/mol. △G=20000−350(70)=20000−24500=−4500\triangle G = 20000 - 350(70) = 20000 - 24500 = -4500△G=20000−350(70)=20000−24500=−4500 J/mol.

Flashcard 18: What effect does increased pressure have on gas solubility in liquids?

Answer: Increases solubility. Henry's law: gas solubility is directly proportional to pressure.

Flashcard 19: What does a positive △S\triangle S△S indicate about entropy in dissolution?

Answer: Entropy increases. Positive △S\triangle S△S means the system becomes more disordered.

Flashcard 20: Determine the spontaneity: ΔH=5 kJ/mol\Delta H = 5 \text{ kJ/mol}ΔH=5 kJ/mol, T=298 KT = 298 \text{ K}T=298 K, ΔS=30 J/mol\cdotpK\Delta S = 30 \text{ J/mol·K}ΔS=30 J/mol\cdotpK.

Answer: Spontaneous. ΔG=5000−298(30)=5000−8940=−3940\Delta G = 5000 - 298(30) = 5000 - 8940 = -3940ΔG=5000−298(30)=5000−8940=−3940 J/mol, so spontaneous.

Flashcard 21: Which parameter is zero for a process at equilibrium?

Answer: △G\triangle G△G. At equilibrium, the free energy change equals zero.

Flashcard 22: Calculate the enthalpy change: ΔG=−5 kJ/mol\Delta G = -5 \text{ kJ/mol}ΔG=−5 kJ/mol, T=298 KT = 298 \text{ K}T=298 K, ΔS=20 J/mol\cdotpK\Delta S = 20 \text{ J/mol·K}ΔS=20 J/mol\cdotpK.

Answer: ΔH=1.96 kJ/mol\Delta H = 1.96 \text{ kJ/mol}ΔH=1.96 kJ/mol. ΔH=ΔG+TΔS=−5000+298(20)=1960\Delta H = \Delta G + T \Delta S = -5000 + 298(20) = 1960ΔH=ΔG+TΔS=−5000+298(20)=1960 J/mol.

Flashcard 23: Identify the factor that primarily affects △S\triangle S△S in dissolution.

Answer: Molecular disorder. Dissolution typically increases randomness as molecules spread out.

Flashcard 24: State the sign of △G\triangle G△G for a spontaneous dissolution.

Answer: △G<0\triangle G < 0△G<0. Negative free energy indicates a thermodynamically favorable process.

Flashcard 25: What is the formula for Gibbs free energy change?

Answer: △G=△H−T△S\triangle G = \triangle H - T\triangle S△G=△H−T△S. Fundamental thermodynamic equation relating free energy to enthalpy and entropy.

Flashcard 26: What is the sign of △G\triangle G△G when △H<0\triangle H < 0△H<0 and △S>0\triangle S > 0△S>0?

Answer: △G<0\triangle G < 0△G<0. Both terms favor spontaneity, making △G\triangle G△G always negative.

Flashcard 27: Calculate △G\triangle G△G when △H=15 kJ/mol\triangle H = 15 \text{ kJ/mol}△H=15 kJ/mol, T=300 KT = 300 \text{ K}T=300 K, △S=60 J/mol\cdotpK\triangle S = 60 \text{ J/mol·K}△S=60 J/mol\cdotpK.

Answer: △G=−3 kJ/mol\triangle G = -3 \text{ kJ/mol}△G=−3 kJ/mol. △G=15000−300(60)=15000−18000=−3000\triangle G = 15000 - 300(60) = 15000 - 18000 = -3000△G=15000−300(60)=15000−18000=−3000 J/mol.

Flashcard 28: What is the significance of △G=0\triangle G = 0△G=0 in dissolution?

Answer: Equilibrium state. Zero free energy indicates the system is at thermodynamic equilibrium.

Flashcard 29: What does a negative △S\triangle S△S for dissolution indicate?

Answer: Entropy decreases. Negative entropy change indicates decreased molecular disorder.

Flashcard 30: What is the unit of Gibbs free energy change?

Answer: Joules (J) or kilojoules (kJ). Energy units for thermodynamic quantities in the SI system.