Define Gibbs free energy .

Gibbs free energy ( G ) is the energy that accounts for both the entropy change and the enthalpy change of a reaction, used to determine reaction feasibility.

State the Gibbs equation and give the units of each term.

Δ G θ = Δ H reaction θ - T Δ S system θ Δ G θ : kJ mol -1 Δ H reaction θ : kJ mol -1 T : K Δ S system θ : J K -1 mol -1 (must be converted to kJ K -1 mol -1 )

True or False? When using the Gibbs equation, Δ S system θ must be converted from J K -1 mol -1 to kJ K -1 mol -1 by dividing by 1000.

True. The units of Δ H θ and Δ G θ are kJ mol -1 , so Δ S system θ must be divided by 1000 to ensure consistent units in the calculation.

Calculate Δ G θ for the reaction: 2NaHCO 3 (s) → Na 2 CO 3 (s) + H 2 O (l) + CO 2 (g) at 298 K. Δ H θ = +135 kJ mol -1 Δ S θ = +344 J K -1 mol -1

Δ S θ = +344 ÷ 1000 = +0.344 kJ K -1 mol -1 Δ G θ = +135 - (298 × 0.344) Δ G θ = +32.49 kJ mol -1

The Gibbs equation can be rearranged to find temperature: T = (Δ H reaction θ - .......... ) ÷ ..........

The Gibbs equation can be rearranged to find temperature: T = (Δ H reaction θ - Δ G θ ) ÷ Δ S system θ

True or False? The Gibbs free energy change accounts for both enthalpy and entropy changes, making it more reliable than entropy alone for determining feasibility.

True. Using entropy change alone is inaccurate because it ignores the enthalpy contribution. Δ G θ combines both to give a complete picture of feasibility.

Calculate Δ G θ for: CH 3 OH (l) + HBr (g) → CH 3 Br (g) + H 2 O (l) at 298 K. Δ H r θ = -47 kJ mol -1 Δ S system θ = -23.0 J K -1 mol -1

Δ S system θ = -23.0 ÷ 1000 = -0.023 kJ K -1 mol -1 Δ G θ = -47 - (298 × -0.023) Δ G θ = -40.1 kJ mol -1

In the Gibbs equation Δ G θ = Δ H reaction θ - T Δ S system θ , the symbol T represents .......... measured in .......... .

In the Gibbs equation Δ G θ = Δ H reaction θ - T Δ S system θ , the symbol T represents temperature measured in kelvin (K) .

Why is it insufficient to use only entropy change to predict whether a reaction is feasible?

Feasibility depends on both entropy and enthalpy changes. A reaction with a negative entropy change could still be feasible if it is sufficiently exothermic. The Gibbs equation combines both terms to give an accurate prediction.

State the condition for a reaction to be feasible in terms of Δ G θ .

A reaction is feasible when Δ G θ is negative (less than zero). When Δ G θ is positive, the reaction is not feasible.

True or False? An exothermic reaction with a positive entropy change is feasible at all temperatures.

True. When Δ H θ is negative and Δ S θ is positive, both terms in Δ G θ = Δ H θ - T Δ S θ are negative at any temperature, so Δ G θ is always negative.

Define feasible reaction .

A feasible reaction is one in which Δ G θ is negative, indicating the reaction is thermodynamically likely to occur spontaneously.

For an endothermic reaction with a positive Δ S θ , at what temperatures is the reaction feasible?

At high temperatures . The - T Δ S θ term becomes sufficiently negative to overcome the positive Δ H θ , making Δ G θ negative and the reaction feasible.

True or False? An endothermic reaction with a negative entropy change is never feasible.

True. When Δ H θ is positive and Δ S θ is negative, both terms in Δ G θ are positive at any temperature. Δ G θ is always positive, so the reaction is never feasible.

For an exothermic reaction with a negative Δ S θ , the reaction is feasible at .......... temperatures because the - T Δ S θ term does not .......... the negative Δ H θ .

For an exothermic reaction with a negative Δ S θ , the reaction is feasible at low temperatures because the - T Δ S θ term does not overcome the negative Δ H θ .

Calculate Δ G θ for: 2Ca (s) + O 2 (g) → 2CaO (s) at 298 K. Δ H θ = -635.5 kJ mol -1 Δ S system θ = -207.0 J K -1 mol -1

Δ S θ = -207.0 ÷ 1000 = -0.207 kJ K -1 mol -1 Δ G θ = -635.5 - (298 × -0.207) Δ G θ = -573.8 kJ mol -1 (feasible)

When Δ H θ is positive and Δ S θ is positive, Δ G θ is negative only at .......... temperatures where T Δ S θ .......... Δ H θ .

When Δ H θ is positive and Δ S θ is positive, Δ G θ is negative only at high temperatures where T Δ S θ exceeds Δ H θ .

Explain why an exothermic reaction with a negative entropy change becomes less feasible at high temperatures.

At high temperatures , the - T Δ S θ term becomes large and positive (since Δ S θ is negative). This positive term can overcome the negative Δ H θ , making Δ G θ positive and the reaction not feasible .

A LevelChemistryCambridge (CIE)Flashcards23. Chemical EnergeticsGibbs Free Energy Change, ΔG

Gibbs Free Energy Change, ΔG (Cambridge (CIE) A Level Chemistry): Flashcards

Exam code: 9701

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Define Gibbs free energy.

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Cards in this collection (18)

Define Gibbs free energy.
Gibbs free energy (G) is the energy that accounts for both the entropy change and the enthalpy change of a reaction, used to determine reaction feasibility.
State the Gibbs equation and give the units of each term.
ΔG^θ = ΔH_reaction^θ - TΔS_system^θ
ΔG^θ: kJ mol^-1
ΔH_reaction^θ: kJ mol^-1
T: K
ΔS_system^θ: J K^-1 mol^-1 (must be converted to kJ K^-1 mol^-1)
True or False?
When using the Gibbs equation, ΔS_system^θ must be converted from J K^-1 mol^-1 to kJ K^-1 mol^-1 by dividing by 1000.
True.
The units of ΔH^θ and ΔG^θ are kJ mol^-1, so ΔS_system^θ must be divided by 1000 to ensure consistent units in the calculation.
Calculate ΔG^θ for the reaction: 2NaHCO₃ (s) → Na₂CO₃ (s) + H₂O (l) + CO₂ (g) at 298 K.
ΔH^θ = +135 kJ mol^-1 ΔS^θ = +344 J K^-1 mol^-1
ΔS^θ = +344 ÷ 1000 = +0.344 kJ K^-1 mol^-1
ΔG^θ = +135 - (298 × 0.344)
ΔG^θ = +32.49 kJ mol^-1
The Gibbs equation can be rearranged to find temperature:
T = (ΔH_reaction^θ - ..........) ÷ ..........
The Gibbs equation can be rearranged to find temperature:
T = (ΔH_reaction^θ - ΔG^θ) ÷ ΔS_system^θ
True or False?
The Gibbs free energy change accounts for both enthalpy and entropy changes, making it more reliable than entropy alone for determining feasibility.
True.
Using entropy change alone is inaccurate because it ignores the enthalpy contribution. ΔG^θ combines both to give a complete picture of feasibility.
Calculate ΔG^θ for: CH₃OH (l) + HBr (g) → CH₃Br (g) + H₂O (l) at 298 K.
ΔH_r^θ = -47 kJ mol^-1 ΔS_system^θ = -23.0 J K^-1 mol^-1
ΔS_system^θ = -23.0 ÷ 1000 = -0.023 kJ K^-1 mol^-1
ΔG^θ = -47 - (298 × -0.023)
ΔG^θ = -40.1 kJ mol^-1
In the Gibbs equation ΔG^θ = ΔH_reaction^θ - TΔS_system^θ, the symbol T represents .......... measured in .......... .
In the Gibbs equation ΔG^θ = ΔH_reaction^θ - TΔS_system^θ, the symbol T represents temperature measured in kelvin (K).
Why is it insufficient to use only entropy change to predict whether a reaction is feasible?
Feasibility depends on both entropy and enthalpy changes. A reaction with a negative entropy change could still be feasible if it is sufficiently exothermic. The Gibbs equation combines both terms to give an accurate prediction.
State the condition for a reaction to be feasible in terms of ΔG^θ.
A reaction is feasible when ΔG^θ is negative (less than zero). When ΔG^θ is positive, the reaction is not feasible.
True or False?
An exothermic reaction with a positive entropy change is feasible at all temperatures.
True.
When ΔH^θ is negative and ΔS^θ is positive, both terms in ΔG^θ = ΔH^θ - TΔS^θ are negative at any temperature, so ΔG^θ is always negative.
Define feasible reaction.
A feasible reaction is one in which ΔG^θ is negative, indicating the reaction is thermodynamically likely to occur spontaneously.
For an endothermic reaction with a positive ΔS^θ, at what temperatures is the reaction feasible?
At high temperatures. The -TΔS^θ term becomes sufficiently negative to overcome the positive ΔH^θ, making ΔG^θ negative and the reaction feasible.
True or False?
An endothermic reaction with a negative entropy change is never feasible.
True.
When ΔH^θ is positive and ΔS^θ is negative, both terms in ΔG^θ are positive at any temperature. ΔG^θ is always positive, so the reaction is never feasible.
For an exothermic reaction with a negative ΔS^θ, the reaction is feasible at .......... temperatures because the -TΔS^θ term does not .......... the negative ΔH^θ.
For an exothermic reaction with a negative ΔS^θ, the reaction is feasible at low temperatures because the -TΔS^θ term does not overcome the negative ΔH^θ.
Calculate ΔG^θ for: 2Ca (s) + O₂ (g) → 2CaO (s) at 298 K.
ΔH^θ = -635.5 kJ mol^-1 ΔS_system^θ = -207.0 J K^-1 mol^-1
ΔS^θ = -207.0 ÷ 1000 = -0.207 kJ K^-1 mol^-1
ΔG^θ = -635.5 - (298 × -0.207)
ΔG^θ = -573.8 kJ mol^-1 (feasible)
When ΔH^θ is positive and ΔS^θ is positive, ΔG^θ is negative only at .......... temperatures where TΔS^θ .......... ΔH^θ.
When ΔH^θ is positive and ΔS^θ is positive, ΔG^θ is negative only at high temperatures where TΔS^θ exceeds ΔH^θ.
Explain why an exothermic reaction with a negative entropy change becomes less feasible at high temperatures.
At high temperatures, the -TΔS^θ term becomes large and positive (since ΔS^θ is negative). This positive term can overcome the negative ΔH^θ, making ΔG^θ positive and the reaction not feasible.

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