Driving Forces and Gibbs Free Energy
Some reactions are spontaneous because they give off energy in the form of heat (H < 0). Others are spontaneous because they lead to an increase in the disorder of the system (S > 0). Calculations of H and S can be used to probe the driving force behind a particular reaction.
Practice Problem 5: Calculate H and S for the following reaction and decide in which direction each of these factors will drive the reaction. N_{2}(g) + 3 H_{2}(g) 2 NH_{3}(g) |
What happens when one of the potential driving forces behind a chemical reaction is favorable and the other is not? We can answer this question by defining a new quantity known as the Gibbs free energy (G) of the system, which reflects the balance between these forces.
The Gibbs free energy of a system at any moment in time is defined as the enthalpy of the system minus the product of the temperature times the entropy of the system.
G = H - TS
The Gibbs free energy of the system is a state function because it is defined in terms of thermodynamic properties that are state functions. The change in the Gibbs free energy of the system that occurs during a reaction is therefore equal to the change in the enthalpy of the system minus the change in the product of the temperature times the entropy of the system.
G = H - (TS)
If the reaction is run at constant temperature, this equation can be written as follows.
G = H - TS
The change in the free energy of a system that occurs during a reaction can be measured under any set of conditions. If the data are collected under standard-state conditions, the result is the standard-state free energy of reaction (G^{o}).
G^{o} = H^{o} - TS^{o}
The beauty of the equation defining the free energy of a system is its ability to determine the relative importance of the enthalpy and entropy terms as driving forces behind a particular reaction. The change in the free energy of the system that occurs during a reaction measures the balance between the two driving forces that determine whether a reaction is spontaneous. As we have seen, the enthalpy and entropy terms have different sign conventions.
Favorable | Unfavorable | |
H^{o} < 0 | H^{o} > 0 | |
S^{o} > 0 | S^{o} < 0 |
The entropy term is therefore subtracted from the enthalpy term when calculating G^{o} for a reaction.
Because of the way the free energy of the system is defined, G^{o} is negative for any reaction for which H^{o} is negative and S^{o} is positive. G^{o} is therefore negative for any reaction that is favored by both the enthalpy and entropy terms. We can therefore conclude that any reaction for which G^{o} is negative should be favorable, or spontaneous.
Favorable, or spontaneous reactions: | G^{o} < 0 |
Conversely, G^{o} is positive for any reaction for which H^{o} is positive and S^{o} is negative. Any reaction for which G^{o} is positive is therefore unfavorable.
Unfavorable, or non-spontaneous reactions: | G^{o} > 0 |
Reactions are classified as either exothermic (H < 0) or endothermic (H > 0) on the basis of whether they give off or absorb heat. Reactions can also be classified as exergonic (G < 0) or endergonic (G > 0) on the basis of whether the free energy of the system decreases or increases during the reaction.
When a reaction is favored by both enthalpy (H^{o} < 0) and entropy (S^{o} > 0), there is no need to calculate the value of G^{o} to decide whether the reaction should proceed. The same can be said for reactions favored by neither enthalpy (H^{o} > 0) nor entropy (S^{o} < 0). Free energy calculations become important for reactions favored by only one of these factors.
Practice Problem 6: Calculate H and S for the following reaction: NH_{4}NO_{3}(s) + H_{2}O(l) NH_{4}^{+} (aq) + NO_{3}^{-} (aq) Use the results of this calculation to determine the value of G^{o }for this reaction at 25^{o }C, and explain why NH_{4}NO_{3} spontaneously dissolves is water at room temperature. |
The Effect of Temperature on the Free Energy of a Reaction
The balance between the contributions from the enthalpy and entropy terms to the free energy of a reaction depends on the temperature at which the reaction is run.
Practice Problem 7: Use the values of H and S calculated in Practice Problem 5 to predict whether the following reaction is spontaneous at 25C: N_{2}(g) + 3 H_{2}(g) 2 NH_{3}(g) |
The equation used to define free energy suggests that the entropy term will become more important as the temperature increases.
G^{o} = H^{o} - TS^{o}
Since the entropy term is unfavorable, the reaction should become less favorable as the temperature increases.
Practice Problem 8: Predict whether the following reaction is still spontaneous at 500C: N_{2}(g) + 3 H_{2}(g) 2 NH_{3}(g) Assume that the values of H^{o} and S used in Practice Problem 7 are still valid at this temperature. |
Standard-State Free Energies of Reaction
G^{o} for a reaction can be calculated from tabulated standard-state free energy data. Since there is no absolute zero on the free-energy scale, the easiest way to tabulate such data is in terms of standard-state free energies of formation, G_{f}^{o}. As might be expected, the standard-state free energy of formation of a substance is the difference between the free energy of the substance and the free energies of its elements in their thermodynamically most stable states at 1 atm, all measurements being made under standard-state conditions.
Interpreting Standard-State Free Energy of Reaction Data
We are now ready to ask the obvious question: What does the value of G^{o} tell us about the following reaction?
N_{2}(g) + 3 H_{2}(g) 2 NH_{3}(g) G^{o} = -32.96 kJ
By definition, the value of G^{o} for a reaction measures the difference between the free energies of the reactants and products when all components of the reaction are present at standard-state conditions.
G^{o} therefore describes this reaction only when all three components are present at 1 atm pressure.
The sign of G^{o} tells us the direction in which the reaction has to shift to come to equilibrium. The fact that G^{o} is negative for this reaction at 25^{o}C means that a system under standard-state conditions at this temperature would have to shift to the right, converting some of the reactants into products, before it can reach equilibrium. The magnitude of G^{o} for a reaction tells us how far the standard state is from equilibrium. The larger the value of G^{o}, the further the reaction has to go to get to from the standard-state conditions to equilibrium.
Assume, for example, that we start with the following reaction under standard-state conditions, as shown in the figure below.
N_{2}(g) + 3 H_{2}(g) 2 NH_{3}(g)
The value of G at that moment in time will be equal to the standard-state free energy for this reaction, G^{o}.
When Q_{p} = 1: G = G^{o}
As the reaction gradually shifts to the right, converting N_{2} and H_{2} into NH_{3}, the value of G for the reaction will decrease. If we could find some way to harness the tendency of this reaction to come to equilibrium, we could get the reaction to do work. The free energy of a reaction at any moment in time is therefore said to be a measure of the energy available to do work.
The Relationship Between Free Energy and Equilibrium Constants
When a reaction leaves the standard state because of a change in the ratio of the concentrations of the products to the reactants, we have to describe the system in terms of non-standard-state free energies of reaction. The difference between G^{o} and G for a reaction is important. There is only one value of G^{o} for a reaction at a given temperature, but there are an infinite number of possible values of G.
The figure below shows the relationship between G for the following reaction and the logarithm to the base e of the reaction quotient for the reaction between N_{2} and H_{2 }to form NH_{3}.
N_{2}(g) + 3 H_{2}(g) 2 NH_{3}(g)
Data on the left side of this figure correspond to relatively small values of Q_{p}. They therefore describe systems in which there is far more reactant than product. The sign of G for these systems is negative and the magnitude of G is large. The system is therefore relatively far from equilibrium and the reaction must shift to the right to reach equilibrium.
Data on the far right side of this figure describe systems in which there is more product than reactant. The sign of G is now positive and the magnitude of G is moderately large. The sign of G tells us that the reaction would have to shift to the left to reach equilibrium. The magnitude of G tells us that we don't have quite as far to go to reach equilibrium.
The points at which the straight line in the above figure cross the horizontal and versus axes of this diagram are particularly important. The straight line crosses the vertical axis when the reaction quotient for the system is equal to 1. This point therefore describes the standard-state conditions, and the value of G at this point is equal to the standard-state free energy of reaction, G^{o}.
When Q_{p} = 1: G = G^{o}
The point at which the straight line crosses the horizontal axis describes a system for which G is equal to zero. Because there is no driving force behind the reaction, the system must be at equilibrium.
When Q_{p} = K_{p}: G = 0
The relationship between the free energy of reaction at any moment in time (G) and the standard-state free energy of reaction (G^{o}) is described by the following equation.
G = G^{o} + RT ln Q
In this equation, R is the ideal gas constant in units of J/mol-K, T is the temperature in kelvin, ln represents a logarithm to the base e, and Q is the reaction quotient at that moment in time.
As we have seen, the driving force behind a chemical reaction is zero (G = 0) when the reaction is at equilibrium (Q = K).
0 = G^{o} + RT ln K
We can therefore solve this equation for the relationship between G^{o} and K.
G^{o} = - RT ln K
This equation allows us to calculate the equilibrium constant for any reaction from the standard-state free energy of reaction, or vice versa.
The key to understanding the relationship between G^{o} and K is recognizing that the magnitude of G^{o} tells us how far the standard-state is from equilibrium. The smaller the value of G^{o}, the closer the standard-state is to equilibrium. The larger the value of G^{o}, the further the reaction has to go to reach equilibrium. The relationship between G^{o} and the equilibrium constant for a chemical reaction is illustrated by the data in the table below.
Values of G^{o} and K for Common Reactions at 25^{o}C
Reaction | G^{o }(kJ) | K | |||||
2 SO_{3}(g) | 2 SO_{2}(g) + O_{2}(g) | 141.7 | 1.4 x 10^{-25} | ||||
H_{2}O(l) | H^{+}(aq) + OH^{-}(aq) | 79.9 | 1.0 x 10^{-14} | ||||
AgCl(s) + H_{2}O | Ag^{+}(aq) + Cl^{-}(aq) | 55.6 | 1.8 x 10^{-10} | ||||
HOAc(aq) + H_{2}O | H^{+}(aq) + OAc^{-}(aq) | 27.1 | 1.8 x 10^{-5} | ||||
N_{2}(g) + 3 H_{2}(g) | 2 NH_{3}(g) | -32.9 | 5.8 x 10^{5} | ||||
HCl(aq) + H_{2}O | H^{+}(aq) + Cl^{-}(aq) | -34.2 | 1 x 10^{6} | ||||
Cu^{2+}(aq) + 4 NH_{3}(aq) | Cu(NH_{3})_{4}^{2+}(aq) | -76.0 | 2.1 x 10^{13} | ||||
Zn(s) + Cu^{2+}(aq) | Zn^{2+}(aq) + Cu(s) | -211.8 | 1.4 x 10^{37} |
Practice Problem 9: Use the value of G^{o} obtained in Practice Problem 7 to calculate the equilibrium constant for the following reaction at 25C: N_{2}(g) + 3 H_{2}(g) 2 NH_{3}(g) |
The equilibrium constant for a reaction can be expressed in two ways: K_{c} and K_{p}. We can write equilibrium constant expressions in terms of the partial pressures of the reactants and products, or in terms of their concentrations in units of moles per liter.
For gas-phase reactions the equilibrium constant obtained from G^{o} is based on the partial pressures of the gases (K_{p}). For reactions in solution, the equilibrium constant that comes from the calculation is based on concentrations (K_{c}).
Practice Problem 10: Use the following standard-state free energy of formation data to calculate the acid-dissociation equilibrium constant (K_{a}) at for formic acid: Compound G_{f}^{o}(kJ/mol) HCO_{2}(aq) -372.3 H^{+}(aq) 0.00 HCO_{2}^{-}(aq) -351.0 |
The Temperature Dependence of Equilibrium Constants
Equilibrium constants are not strictly constant because they change with temperature. We are now ready to understand why.
The standard-state free energy of reaction is a measure of how far the standard-state is from equilibrium.
G^{o} = - RT ln K
But the magnitude of G^{o} depends on the temperature of the reaction.
G^{o} = H^{o} - TS^{o}
As a result, the equilibrium constant must depend on the temperature of the reaction.
A good example of this phenomenon is the reaction in which NO_{2} dimerizes to form N_{2}O_{4}.
2 NO_{2}(g) N_{2}O_{4}(g)
This reaction is favored by enthalpy because it forms a new bond, which makes the system more stable. The reaction is not favored by entropy because it leads to a decrease in the disorder of the system.
NO_{2} is a brown gas and N_{2}O_{4} is colorless. We can therefore monitor the extent to which NO_{2} dimerizes to form N_{2}O_{4} by examining the intensity of the brown color in a sealed tube of this gas. What should happen to the equilibrium between NO_{2} and N_{2}O_{4} as the temperature is lowered?
For the sake of argument, let's assume that there is no significant change in either H^{o} or S^{o} as the system is cooled. The contribution to the free energy of the reaction from the enthalpy term is therefore constant, but the contribution from the entropy term becomes smaller as the temperature is lowered.
G^{o} = H^{o} - TS^{o}
As the tube is cooled, and the entropy term becomes less important, the net effect is a shift in the equilibrium toward the right. The figure below shows what happens to the intensity of the brown color when a sealed tube containing NO_{2} gas is immersed in liquid nitrogen. There is a drastic decrease in the amount of NO_{2} in the tube as it is cooled to -196^{o}C.
Practice Problem 11: Use values of H^{o} and S^{o}for the following reaction at 25C to estimate the equilibrium constant for this reaction at the temperature of boiling water (100C), ice(0C), a dry ice-acetone bath (-78C), and liquid nitrogen (-196C): 2 NO_{2}(g) N_{2}O_{4}(g) |
The Relationship Between Free Energy and Cell Potentials
The value of G for a reaction at any moment in time tells us two things. The sign of G tells us in what direction the reaction has to shift to reach equilibrium. The magnitude of G tells us how far the reaction is from equilibrium at that moment.
The potential of an electrochemical cell is a measure of how far an oxidation-reduction reaction is from equilibrium. The Nernst equation describes the relationship between the cell potential at any moment in time and the standard-state cell potential.
Let's rearrange this equation as follows.
nFE = nFE^{o} - RT ln Q
We can now compare it with the equation used to describe the relationship between the free energy of reaction at any moment in time and the standard-state free energy of reaction.
G = G^{o} + RT ln Q
These equations are similar because the Nernst equation is a special case of the more general free energy relationship. We can convert one of these equations to the other by taking advantage of the following relationships between the free energy of a reaction and the cell potential of the reaction when it is run as an electrochemical cell.
Practice Problem 12: Use the relationship between G^{o} and E^{o} for an electrochemical reaction to derive the relationship between the standard-state cell potential and the equilibrium constant for the reaction. |