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\( \Delta G = \Delta H - T\Delta S \)
The Gibbs free energy equation determines the spontaneity of a chemical process at constant temperature and pressure. The change in free energy (ΔG) combines enthalpy change (ΔH) and entropy change (ΔS) at temperature T. When ΔG < 0, the process is spontaneous; when ΔG > 0, it is non-spontaneous; when ΔG = 0, the system is at equilibrium.
Key relationships:
Use this equation when predicting reaction spontaneity, calculating equilibrium constants (ΔG° = -RTlnK), or determining the temperature at which a reaction becomes spontaneous.
Gibbs free energy (ΔG) is a thermodynamic potential that measures the maximum reversible work that can be performed by a system at constant temperature and pressure. It combines enthalpy (ΔH) and entropy (ΔS) to predict the spontaneity of chemical processes. The change in Gibbs free energy determines whether a reaction will proceed spontaneously under given conditions. When ΔG is negative, the reaction is thermodynamically favorable; when positive, it is unfavorable; and when zero, the system is at equilibrium.
The fundamental equation for Gibbs free energy:
Where:
Key applications of Gibbs free energy include:
Measure of energy available to do work at constant T and P.
\( \Delta G = \Delta H - T\Delta S \)
Where ΔG = free energy change.
ΔG determines whether a reaction will occur without external input.
For a reaction with ΔH = -45 kJ/mol and ΔS = -120 J/K·mol at 298 K, what is the value of ΔG and is the reaction spontaneous?
Given: ΔH = -45 kJ/mol, ΔS = -120 J/K·mol, T = 298 K
First, convert ΔS to kJ: ΔS = -120 J/K·mol ÷ 1000 = -0.120 kJ/K·mol
Apply the Gibbs equation: ΔG = ΔH - TΔS
ΔG = -45 - (298)(-0.120) = -45 - (-35.76) = -45 + 35.76 = -9.24 kJ/mol
Since ΔG < 0, the reaction is spontaneous.
The answer is C) ΔG = -9.2 kJ/mol, Spontaneous.
This problem demonstrates the importance of unit consistency in thermodynamic calculations. We must ensure that ΔH and TΔS have the same units before subtraction. The negative ΔH (exothermic) tends to favor spontaneity, but the negative ΔS (decrease in disorder) opposes it. At 298 K, the enthalpy term dominates, making the reaction spontaneous. However, at higher temperatures, the TΔS term could become dominant and make the reaction non-spontaneous.
Gibbs Free Energy (ΔG): Energy available to do work at constant T and P
Spontaneous Process: Occurs without external energy input
Unit Conversion: 1 kJ = 1000 J
• ΔG < 0: Spontaneous
• ΔG > 0: Non-spontaneous
• ΔG = 0: Equilibrium
• Always check units before calculations
• Remember: ΔG = ΔH - TΔS
• Temperature must be in Kelvin
• Forgetting to convert units for ΔS
• Using Celsius instead of Kelvin
• Sign errors in the subtraction
A reaction has ΔH = +30 kJ/mol and ΔS = +100 J/K·mol. Calculate the temperature at which this reaction becomes spontaneous.
Step 1: Identify when the reaction becomes spontaneous
For spontaneity: ΔG < 0
At the boundary between spontaneous and non-spontaneous: ΔG = 0
Step 2: Set ΔG = 0 in the Gibbs equation
0 = ΔH - TΔS
Step 3: Solve for T
T = ΔH / ΔS
Step 4: Convert ΔS to kJ units
ΔS = +100 J/K·mol ÷ 1000 = +0.100 kJ/K·mol
Step 5: Calculate T
T = (+30 kJ/mol) / (+0.100 kJ/K·mol) = 300 K
Step 6: Interpret the result
The reaction is spontaneous when T > 300 K
At temperatures above 300 K, the TΔS term dominates over ΔH.
This problem illustrates how temperature affects the spontaneity of reactions. Here we have an endothermic reaction (ΔH > 0) with an increase in entropy (ΔS > 0). At low temperatures, the enthalpy term dominates, making ΔG positive. As temperature increases, the TΔS term grows larger until it overcomes the positive ΔH, making ΔG negative. This is typical of reactions that become spontaneous at high temperatures, such as many decomposition reactions.
Endothermic Reaction: Absorbs heat (ΔH > 0)
Entropy Increase: Greater disorder (ΔS > 0)
Temperature Dependence: Effect of T on spontaneity
• Set ΔG = 0 to find transition temperature
• T = ΔH/ΔS for phase transition
• High T favors entropy-driven processes
• Remember to set ΔG = 0 for transition points
• Check which term dominates at different temperatures
• Units must be consistent in division
• Forgetting to set ΔG = 0 for transition temperature
• Incorrectly solving the algebraic equation
• Forgetting to convert units properly
The hydrolysis of ATP to ADP has ΔG° = -30.5 kJ/mol under standard conditions. If the cellular concentrations are [ATP] = 5 mM, [ADP] = 0.5 mM, and [Pi] = 10 mM, calculate the actual ΔG for this reaction at 37°C. The reaction is: ATP + H₂O → ADP + Pi
Step 1: Write the reaction quotient expression
Q = [ADP][Pi] / [ATP]
Step 2: Convert concentrations to M
[ATP] = 5 mM = 0.005 M, [ADP] = 0.5 mM = 0.0005 M, [Pi] = 10 mM = 0.010 M
Step 3: Calculate Q
Q = (0.0005)(0.010) / (0.005) = 0.000005 / 0.005 = 0.001
Step 4: Convert temperature to Kelvin
T = 37°C + 273 = 310 K
Step 5: Use the non-standard Gibbs equation
ΔG = ΔG° + RT ln(Q)
Step 6: Substitute values (R = 8.314 J/K·mol)
ΔG = -30500 J/mol + (8.314 J/K·mol)(310 K) ln(0.001)
ΔG = -30500 + (2577.34) ln(0.001)
ΔG = -30500 + (2577.34)(-6.908)
ΔG = -30500 - 17800 = -48300 J/mol = -48.3 kJ/mol
This problem demonstrates the biological significance of Gibbs free energy. The standard free energy (ΔG°) is measured under standard conditions (1 M for all species), but cellular conditions are very different. The actual free energy (ΔG) depends on the real concentrations in the cell. Here, the high ATP concentration relative to ADP makes the reaction even more favorable than under standard conditions. This is crucial for understanding how biological systems maintain energy flow.
Standard Free Energy (ΔG°): Free energy change under standard conditions
Actual Free Energy (ΔG): Free energy change under actual conditions
Reaction Quotient (Q): Ratio of product to reactant activities
• ΔG = ΔG° + RT ln(Q)
• Standard conditions: 1 M, 1 bar, 298 K
• R = 8.314 J/K·mol
• Convert all concentrations to M before calculations
• Temperature must be in Kelvin
• Pay attention to signs in logarithmic calculations
• Forgetting to convert units for concentrations
• Using Celsius instead of Kelvin
• Sign errors in logarithmic calculations
For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔG° = -33.0 kJ/mol at 298 K. Calculate the equilibrium constant (K) for this reaction. How does this relate to the spontaneity of the reaction?
Step 1: Use the relationship between ΔG° and K
ΔG° = -RT ln(K)
Step 2: Rearrange to solve for K
ln(K) = -ΔG° / RT
K = e^(-ΔG° / RT)
Step 3: Convert ΔG° to J/mol
ΔG° = -33.0 kJ/mol = -33000 J/mol
Step 4: Substitute values (R = 8.314 J/K·mol, T = 298 K)
K = e^(-(-33000) / (8.314 × 298))
K = e^(33000 / 2477.572)
K = e^(13.32)
K ≈ 6.1 × 10⁵
Step 5: Relate to spontaneity
Since ΔG° < 0, the reaction is spontaneous under standard conditions, which corresponds to K > 1. The large value of K indicates that products are heavily favored at equilibrium.
This question connects thermodynamics with equilibrium chemistry. The relationship ΔG° = -RT ln(K) shows that standard free energy change determines the position of equilibrium. A negative ΔG° corresponds to K > 1 (products favored), while a positive ΔG° corresponds to K < 1 (reactants favored). This relationship is fundamental in understanding how thermodynamic properties influence chemical equilibria. The Haber process for ammonia synthesis has a large equilibrium constant at low temperatures, which explains why high ammonia yields are thermodynamically favorable.
Standard Free Energy (ΔG°): Free energy change under standard conditions
Equilibrium Constant (K): Ratio of concentrations at equilibrium
Relationship: ΔG° = -RT ln(K)
• ΔG° < 0 → K > 1 (products favored)
• ΔG° > 0 → K < 1 (reactants favored)
• ΔG° = 0 → K = 1 (equal concentrations)
• Remember to convert ΔG° to J/mol when using R = 8.314
• Use exponential function to solve for K
• Large negative ΔG° leads to large K
• Forgetting to convert ΔG° to J/mol
• Sign errors in the exponential calculation
• Forgetting the relationship between ΔG° and K
Which of the following statements about Gibbs free energy is FALSE?
Let's examine each option:
A) TRUE - A negative ΔG indicates thermodynamic favorability
B) FALSE - ΔG indicates thermodynamic favorability but not the rate of reaction. Kinetics determines how fast a reaction occurs, while thermodynamics determines whether it's favorable
C) TRUE - At equilibrium, ΔG = 0 because forward and reverse reactions occur at equal rates
D) TRUE - The TΔS term in ΔG = ΔH - TΔS explains why temperature affects spontaneity
The answer is B) ΔG can predict whether a reaction will occur quickly.
This question highlights one of the most important distinctions in chemistry: thermodynamics versus kinetics. Thermodynamics (ΔG) tells us whether a reaction is energetically favorable but says nothing about how fast it occurs. Kinetics deals with reaction rates. For example, diamond is thermodynamically unstable relative to graphite (ΔG < 0 for diamond → graphite), but the conversion is extremely slow due to high activation energy barriers. This distinction is crucial for understanding chemical processes.
Thermodynamics: Study of energy changes and equilibrium
Kinetics: Study of reaction rates and mechanisms
Activation Energy: Energy barrier for reaction to occur
• Thermodynamics ≠ Kinetics
• ΔG predicts spontaneity, not rate
• Both are needed for complete understanding
• Always distinguish between thermodynamic and kinetic factors
• Remember that stable doesn't mean fast
• Consider both energy and rate in real systems
• Confusing thermodynamic favorability with reaction rate
• Thinking stable means unreactive
• Forgetting the kinetics vs thermodynamics distinction
Q: Why is Gibbs free energy called "free" energy?
A: The term "free" in Gibbs free energy refers to the portion of the system's energy that is available to do useful work at constant temperature and pressure. Not all of the internal energy of a system can be converted to work—some energy is always lost to entropy. The Gibbs free energy represents the maximum amount of energy that can be extracted from a system to perform non-pressure-volume work. The "free" energy is the part that is "free" to be used for purposes other than maintaining the system's temperature and pressure.
Q: How does Gibbs free energy relate to chemical equilibrium and reaction rates?
A: Gibbs free energy determines the position of chemical equilibrium through the relationship ΔG° = -RT ln(K), where K is the equilibrium constant. However, it's important to distinguish between thermodynamics and kinetics. While ΔG predicts whether a reaction is favorable (thermodynamics), it doesn't determine how fast the reaction occurs (kinetics). A reaction with a large negative ΔG may still be very slow if it has a high activation energy barrier. At equilibrium, ΔG = 0, meaning the forward and reverse reactions occur at equal rates, but both may be extremely slow. This distinction between thermodynamic favorability and kinetic accessibility is fundamental in chemistry.