Gibbs Free Energy Sample Problems

Gibbs Free Energy: Sample Problems and Detailed Explanations

Gibbs Free Energy (ΔG), a thermodynamic potential, measures the maximum reversible work that may be performed by a thermodynamic system at a constant temperature and pressure. Understanding Gibbs Free Energy is crucial in chemistry, particularly for predicting the spontaneity of chemical reactions and determining the equilibrium constant. This article will delve into several sample problems demonstrating various applications of Gibbs Free Energy calculations, providing detailed explanations and highlighting important concepts along the way. We'll cover calculations involving standard free energy changes, non-standard conditions, and the relationship between Gibbs Free Energy and the equilibrium constant.

Understanding the Fundamentals

Before tackling the problems, let's refresh our understanding of the key equation:

ΔG = ΔH - TΔS

Where:

• ΔG is the change in Gibbs Free Energy (in Joules or kJ)
• ΔH is the change in enthalpy (heat content) (in Joules or kJ)
• T is the temperature in Kelvin (K)
• ΔS is the change in entropy (disorder) (in Joules/Kelvin or kJ/Kelvin)

A negative ΔG indicates a spontaneous reaction (favors product formation), while a positive ΔG indicates a non-spontaneous reaction (favors reactants). A ΔG of zero signifies a system at equilibrium.

Sample Problem 1: Calculating ΔG° from ΔH° and ΔS°

Problem: The standard enthalpy change (ΔH°) for the reaction:

2H₂(g) + O₂(g) → 2H₂O(l)

is -571.6 kJ/mol. The standard entropy change (ΔS°) is -326.6 J/mol·K. Calculate the standard Gibbs Free Energy change (ΔG°) at 298 K. Is this reaction spontaneous under standard conditions?

Solution:

1. Convert units: Ensure all units are consistent. Convert ΔS° from J/mol·K to kJ/mol·K by dividing by 1000: ΔS° = -0.3266 kJ/mol·K

2. Apply the Gibbs Free Energy equation:

ΔG° = ΔH° - TΔS° ΔG° = -571.6 kJ/mol - (298 K)(-0.3266 kJ/mol·K) ΔG° = -571.6 kJ/mol + 97.3 kJ/mol ΔG° = -474.3 kJ/mol

3. Interpret the result: Since ΔG° is negative (-474.3 kJ/mol), the reaction is spontaneous under standard conditions (298 K and 1 atm pressure).

Sample Problem 2: Calculating the Equilibrium Constant (K) from ΔG°

Problem: The standard Gibbs Free Energy change (ΔG°) for the reaction:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

is -33.0 kJ/mol at 298 K. Calculate the equilibrium constant (K) for this reaction at 298 K.

Solution:

The relationship between ΔG° and K is given by:

ΔG° = -RTlnK

Where:

• R is the ideal gas constant (8.314 J/mol·K)
• T is the temperature in Kelvin (298 K)
• lnK is the natural logarithm of the equilibrium constant

1. Convert units: Convert ΔG° from kJ/mol to J/mol by multiplying by 1000: ΔG° = -33000 J/mol

2. Solve for lnK:

-33000 J/mol = -(8.314 J/mol·K)(298 K)lnK lnK = 13.3

3. Solve for K:

K = e¹³.³ ≈ 6.0 x 10⁵

Therefore, the equilibrium constant (K) for the reaction at 298 K is approximately 6.0 x 10⁵. This large value indicates that the formation of ammonia is strongly favored at equilibrium under standard conditions.

Sample Problem 3: Non-Standard Conditions – Using the Equation ΔG = ΔG° + RTlnQ

Problem: Consider the reaction from Problem 2: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) (ΔG° = -33.0 kJ/mol at 298 K). Calculate the Gibbs Free Energy change (ΔG) at 298 K when the partial pressures are: P(N₂) = 10 atm, P(H₂) = 5 atm, and P(NH₃) = 0.1 atm.

Solution:

We'll use the equation:

ΔG = ΔG° + RTlnQ

Where Q is the reaction quotient, calculated using the partial pressures:

Q = (P(NH₃))²/ (P(N₂) * (P(H₂))³) = (0.1)² / (10 * 5³) = 8 x 10⁻⁶

1. Substitute values:

ΔG = -33000 J/mol + (8.314 J/mol·K)(298 K)ln(8 x 10⁻⁶) ΔG = -33000 J/mol + (2477.572 J/mol)(-11.34) ΔG = -33000 J/mol - 28062 J/mol ΔG = -61062 J/mol = -61.1 kJ/mol

Therefore, under these non-standard conditions, the reaction is still spontaneous (ΔG is negative), but the free energy change is less negative than under standard conditions, indicating a smaller driving force for product formation.

Sample Problem 4: Determining Spontaneity at Different Temperatures

Problem: A certain reaction has ΔH° = +25 kJ/mol and ΔS° = +100 J/mol·K. Determine whether the reaction is spontaneous at 25°C (298 K) and at 100°C (373 K).

Solution:

We need to calculate ΔG at both temperatures using the equation ΔG = ΔH° - TΔS°:

At 298 K:

ΔG = 25000 J/mol - (298 K)(0.100 kJ/mol·K) = 25000 J/mol - 29800 J/mol = -4800 J/mol

The reaction is spontaneous at 298 K (ΔG is negative).

At 373 K:

ΔG = 25000 J/mol - (373 K)(0.100 kJ/mol·K) = 25000 J/mol - 37300 J/mol = -12300 J/mol

The reaction is spontaneous at 373 K (ΔG is negative). In this case, the positive ΔH° is outweighed by the positive TΔS° term at both temperatures. This suggests that the increase in entropy (disorder) is the primary driving force for the spontaneity of this reaction.

Sample Problem 5: Calculating ΔG° from Equilibrium Constant Data

Problem: The equilibrium constant (K) for the reaction:

A(g) + B(g) ⇌ C(g)

is 10 at 298 K. Calculate the standard Gibbs Free Energy change (ΔG°) for this reaction.

Solution:

We can use the equation: ΔG° = -RTlnK

1. Substitute values:

ΔG° = -(8.314 J/mol·K)(298 K)ln(10) ΔG° ≈ -5705 J/mol ≈ -5.7 kJ/mol

The standard Gibbs Free Energy change is approximately -5.7 kJ/mol, indicating a spontaneous reaction under standard conditions.

Frequently Asked Questions (FAQ)

Q1: What is the difference between ΔG and ΔG°?

A1: ΔG represents the Gibbs Free Energy change under any conditions (non-standard conditions), while ΔG° specifically refers to the Gibbs Free Energy change under standard conditions (1 atm pressure, 1 M concentration for solutions, 298 K).

Q2: Can a reaction with a positive ΔH and a positive ΔS still be spontaneous?

A2: Yes, if the TΔS term is larger than the ΔH term, resulting in a negative ΔG. This is more likely at higher temperatures.

Q3: What are the limitations of using Gibbs Free Energy to predict spontaneity?

A3: Gibbs Free Energy predicts thermodynamic spontaneity, but it doesn't provide information about the rate of a reaction. A spontaneous reaction might be incredibly slow in practice. It also assumes equilibrium conditions are achieved.

Q4: How does Gibbs Free Energy relate to cell potentials in electrochemistry?

A4: The change in Gibbs Free Energy is directly related to the cell potential (E) through the equation: ΔG = -nFE, where n is the number of moles of electrons transferred and F is Faraday's constant.

Conclusion

Gibbs Free Energy is a powerful tool for predicting the spontaneity and equilibrium position of chemical reactions. By understanding the fundamental equation and its applications under both standard and non-standard conditions, we can analyze and interpret the thermodynamic feasibility of various chemical processes. The sample problems presented here illustrate a range of calculations, emphasizing the importance of unit consistency and the interpretation of results in the context of thermodynamic principles. Mastering these concepts forms a crucial foundation for advanced studies in chemistry and related fields. Remember to always consider the context of the problem, paying close attention to the given values and the necessary conversions to ensure accurate results.