Lesson 3: Spontaneity and Gibbs Free Energy

Part b: Gibbs Free Energy and Equilibrium

Part a: Predicting Spontaneity with Gibbs Free Energy
Part b: Gibbs Free Energy and Equilibrium

 

The Big Idea

The balance between reactants and products at equilibrium can be understood through Gibbs free energy. The sign and magnitude of ΔG° reveal whether a reaction favors products or reactants and how it relates to the equilibrium constant K.

 

Thermodynamic Model vs. Equilibrium Model of Spontaneity

We learned in Lesson 3a that the Gibbs free energy change (∆G) for a reaction provides a single indicator as to whether a reaction will  or will not occur. Spontaneous reactions have a Gibbs free energy change that is less than 0. The magnitude or size of this quantity provides a measure of the extent to which the reaction will be driven in the forward direction. A positive value of ∆G for a reaction indicates that it is not spontaneous. In such a case, the reverse reaction would be spontaneous. And finally, the ∆G value is 0 for reaction systems that have reached equilibrium.
 
The discussion in Lesson 3a referred to Gibbs free energy as reaction potential. A ∆G value has a sign (+ or -) and a magnitude (like 0.1 or 10 or 100). The sign of ∆G helps one to predict which direction a reaction system would proceed. And the magnitude helps us to predict the extent to which it proceeds. Previously in this Chemistry Tutorial, the ability to predict which direction a reaction proceeds was based on our Equilibrium model of reactions. A comparison of the reaction quotient (Q) to the equilibrium constant (K) allows us to predict if the reaction would proceed in the forward or the reverse direction. In this lesson, we will draw the connection between our Thermodynamic model of spontaneous reactions and our Equilibrium model.
 
 
 

Standard vs. Non-Standard ∆G

Gibbs free energy change values are most commonly available for standard conditions. Standard conditions include a temperature of 25.0°C, solution concentrations of 1.0 M, and gas pressures of 1 atm. Alterations in any of these parameters - temperature, concentration, and pressure - will alter the Gibbs free energy change value for the reaction. Using standard conditions allows scientists to have a reference point for discussing reactions and their reaction potential.
 
The symbol for standard Gibbs free energy change is ∆G°. With the superscripted ° symbol, this is sometimes pronounced as delta-G-zero or delta-G-naught.  While ∆G° represents the Gibbs free energy change for a specific set of conditions - standard conditions - the Gibbs free energy change can be determined for any set of non-standard conditions. This is important because most often the conditions aren’t standard. The symbol for Gibbs free energy change under non-standard conditions is ∆G, lacking the superscripted ° symbol.
 
The equation that relates the standard and the non-standard Gibbs free energy change values is

∆G = ∆G°  +  R•T•lnQ

where R = 8.314 J/mol/K, T = Kelvin temperature, and Q is the reaction quotient. (In the equation, ln represents a mathematical function known as the natural logarithm. You likely have a button on your scientific or graphing calculator for determining the natural logarithm of the numerical value for Q.) The reaction quotient includes concentrations and pressure values of aqueous-state solutions and gas-phase reactants and products. Its numerical value is determined using an expression of the same format as the equilibrium constant expression. This means product concentrations and pressures are in the numerator with reactant values in the denominator. Each concentration or pressure is raised to a power equal to the coefficient in the balanced chemical equation. Solid and liquid-state reactants and products are not included in the expression.

(If necessary, review Equilibrium Constant Expressions.)
 

  

 

Generic Expressions for Reaction Quotient

Consider a generic reaction of the form

a A(g)  +  b B(g)  →  c C(g)  +  d D(g) 

with all gas-phase reactants and products and with coefficients a, b, c, and d. The reaction quotient is determined by substituting partial pressure values into the following equation:

 
Consider the generic reaction of the form

a A(aq)  +  b B(aq)  →  c C(aq)  +  d D(aq) 

with all aqueous-state reactants and products and coefficients a, b, c, and d. The reaction quotient is determined by substituting concentration values into the following equation:

 
Finally, consider a generic reaction with a mix of states for reactants and products like

a A(g)  +  b B(aq)  →  c C(aq)  +  d D(s) 

The reaction quotient uses concentrations for aqueous-state substances, pressures for gas state substances, and omits solids and liquids from the expression. The reaction quotient expression is shown below:

 
 
 

Gibbs Free Energy and the Equilibrium Constant

For a system at equilibrium, our Thermodynamic model states that ∆G = 0. Our Equilibrium model states that the reaction quotient (Q) equals the equilibrium constant (K). We can apply these two ideas to the equation ∆G = ∆G°  +  R•T•lnQ and arrive at

At Equilibrium:      0 = ∆G°  +  R•T•lnK.

By rearranging the equation, we can derive an equation that relates the standard Gibbs free energy change (∆G°) for a reaction to the equilibrium constant.

∆G°  =  - R•T•lnK

 

With a few more steps of algebra (divide each side by -R•T; for both sides, use the inverse of the natural logarithm function), we can derive an equation for calculating an equilibrium constant for a reaction from the standard Gibbs free energy change.

K = e^-∆G°/RT

(In math, e is the natural exponential function. There’s a button on your calculator for that also.) In Examples 1 and 2, we will see how to use the equation to calculate the equilibrium constant (K) for a reaction from a given value of the standard Gibbs free energy change.
 

 
 

Example 1

For the decomposition reaction:

N₂O₄(g)   ⇌   2 NO₂(g)

The standard ∆G° value is +4.73 kJ/mol. Determine the equilibrium constant for the reaction at 25.0°C.
 
 

Solution:

The known information for this problem includes:

The unknown is K. The equation relating the unknown to the known values is

 

K = e^-∆G°/RT

The exponent on the right side of the equation is -∆G°/RT. The value of this exponent is

-∆G°/RT = -1*(+4730 J/mol) / (8.314 J/mol/K * 298.15 K) = -1.908166 ...

This exponent can be used to determine the value of K:

K = e^{-1.908166 ...}

K = 0.148
(rounded from 0.1483521 ...)

 

 
 
 

Example 1 Commentary

There are two important quantities in Example 1 that correlate to each other. One is an equilibrium quantity - K. The other is a thermodynamic quantity - ∆G°. Three claims can be made regarding these two numbers:

1. The value of K is less than 1.0. The Equilibrium model views this an indicator that there is a reversible system with an equilibrium position that lies on the reactant side. At equilibrium, there is mostly reactants and very little products in the system.
2. The ∆G° value is positive. The Thermodynamic model views this as an indicator that the reaction is not spontaneous in the forward direction at standard conditions.  On the other hand, the reverse reaction is spontaneous at standard conditions.
3. The ∆G° value is positive. The Thermodynamic model views this as an indicator of a very low reaction potential. It is not expected that reactants would react away and turn into products.

 
The Equilibrium model and the Thermodynamic model are making the same claims in a different language. Both models are claiming that this system has little ability to produce NO₂ from the given reactant.
 
 
 

Example 2

For the synthesis reaction:
   
The standard ∆G° value is -33.3 kJ/mol. Determine the equilibrium constant for the reaction at 25.0°C.
 
 

Solution:

The known information for this problem includes:
 
 
The unknown is K. The equation relating the unknown to the known values is
   
The exponent on the right side of the equation is -∆G°/RT. The value of this exponent is
   
This exponent can be used to determine the value of K:
   
 
 

Example 2 Commentary

The Equilibrium model and the Thermodynamic model lead to three correlated claims.

1. The value of K is much greater than 1.0. The Equilibrium model views this an indicator that there is a reversible system with an equilibrium position that lies on the product side at standard conditions. At equilibrium, there is mostly products and very little reactants in the system.
2. The ∆G° value is negative. The Thermodynamic model views this as an indicator that the reaction is spontaneous in the forward direction at standard conditions. 
3. The ∆G° value is negative. The Thermodynamic model views this as an indicator of a high reaction potential. It is expected that reactants would react away and turn into products (at standard conditions). We would think of this as a reaction that goes to completion.

 
Once again, the Equilibrium model and the Thermodynamic model are making identical claims. Both models are claiming that NH₃ will be produced by this reaction system.
   
 

Pressure and Concentration Effects

As is evident in the equations above, the value of Q is dependent upon the concentration of solute in solutions and the partial pressure of any gas-phase reactants and products. As the concentrations and partial pressures of reactants and products change, the Q changes, and so does the ∆G value for the reaction. Using the equation ∆G = ∆G°  +  R•T•lnQ, the effects of pressure and concentration can be quantified. This is shown in Examples 3, 4, and 5.
 
 
 

Example 3

For the decomposition reaction:
   
The standard ∆G° value is +4.73 kJ/mol. Determine the reaction quotient and the ∆G value at 25.0°C and partial pressure of 1.00 atm and 0.0001 atm, respectively for N₂O₄ and NO₂.
 

Solution:

The given information is:
 
The goal is to determine the ∆G value. The reaction quotient must first be determined. We know that Q = P_NO2² / P_N2O4. By substitution, we can determine the value of Q.
   
Now the equation ∆G = ∆G°  +  R•T•lnQ can be used to determine the value of ∆G.
   
The reaction is spontaneous for these conditions.
 
 
 

Example 4

For the same decomposition reaction:
   
The standard ∆G° value is +4.73 kJ/mol. Determine the reaction quotient and the ∆G value at 25.0°C, after a bit of reacting of the N₂O₄ to form NO₂ - partial pressure of 0.950 atm and 0.1001 atm, respectively for N₂O₄ and NO₂.
 
 

Solution:

The given information is:
 
The goal is to determine the ∆G value. The reaction quotient must first be determined. We know that Q = P_NO2² / P_N2O4. By substitution, we can determine the value of Q.
   
 
Now the equation ∆G = ∆G°  +  R•T•lnQ can be used to determine the value of ∆G.
   
The reaction is also spontaneous for these conditions.
 
 
 

Example 5

Once more for the same decomposition reaction:
   
The standard ∆G° value is +4.73 kJ/mol. Determine the reaction quotient and the ∆G value at 25.0°C, after a bit more reacting of the N₂O₄ to form NO₂ - partial pressure of 0.800 atm and 0.4001 atm, respectively for N₂O₄ and NO₂.
 
 

Solution:

The given information is:
 
The goal is to determine the ∆G value. The reaction quotient must first be determined. We know that Q = P_NO2² / P_N2O4. By substitution, we can determine the value of Q.
   
Now the equation ∆G = ∆G°  +  R•T•lnQ can be used to determine the value of ∆G.
   
The reaction is NOT spontaneous for these conditions.
 
 
 
 

The Gibbs Free Energy Change Changes as a Reaction Proceeds

As shown in the above calculations, the value of the Gibbs free energy change for a reaction depends on the amounts of the reactants and products that are present. The standard value, ∆G°, never changes since it is based on a single set of fixed conditions. But the non-standard value does change since it is based upon amounts and the amounts of reactants and products change during the course of a reaction. In effect, the ∆G value provides an at-this-moment indicator of the reaction potential.
 
The ∆G is the difference in free energy between reactants and products. For spontaneous reactions, this ∆G is less than 0. That is, the free energy of products is less than the free energy of reactants. But as the amounts of reactants and products change, the difference in Gibbs free energy changes. The bar charts below demonstrate this change in ∆G for a reaction that proceeds in the forward direction. Once the difference in Gibbs free energy is 0 (∆G = 0), the reaction is at an equilibrium.
   
The value of ∆G varies from the standard value (∆G°) by an amount equal to R•T•lnQ. The R is a constant value. The reaction is assumed to occur at constant T. But the Q changes since the concentrations and pressures are changing as the reaction proceeds in the forward direction.
 
 
 

The Behavior of a Reactant Loaded N₂O₄-NO₂ System

Let’s consider the N₂O₄-NO₂ reaction system used in some of our earlier example problems.
     
The standard ∆G° value for this reaction is +4.73 kJ/mol. In Example 1, we determined the K value for this system to be 0.148. Let’s suppose we load the system up with all N₂O₄ and use our equation ∆G = ∆G°  +  R•T•lnQ to determine ∆G values for the various points along the path towards equilibrium. From calculation to calculation, we will change the N₂O₄ pressure by -0.020 atm. Because of the stoichiometry of the equation, the partial pressure of NO₂ will change by twice this amount (by +0.040 atm). We will use a spreadsheet to assist with the calculations. The results are shown in the table below. Note that we have inserted a line (in red) to represent the equilibrium condition in which Q = K and ∆G = 0.
   
This data illustrate a variety of concepts associated with the Thermodynamic and Equilibrium models.

1. Initially, ∆G is negative. The reaction is spontaneous in the forward direction.
2. The value of Q is initially less than the value of K. The reaction proceeds in the forward direction to reach equilibrium.
3. As products are produced and reactants are reacted away, the sign on ∆G remains negative, but its magnitude is decreasing. It’s approaching 0, the value for the equilibrium condition.
4. As products are produced and reactants are reacted away, the value of Q increases. It is approaching the value of K.
5. Once equilibrium is reached (red line in the table), the ∆G is 0 and Q = K.
6. The spreadsheet continues to perform calculations, but the grey-shaded values in the table would not be observed. Once ∆G equals 0, the reaction no longer proceeds in the forward direction. The system has attained an equilibrium.

 
 
 

The Behavior of a Product Loaded N₂O₄-NO₂ System

Let’s consider the same N₂O₄- NO₂ reaction system.
  But this time, let’s load the system with products and have our system perform the same calculations. From calculation to calculation, we will change the NO₂ pressure by -0.080 atm. Because of the stoichiometry of the equation, the partial pressure of N₂O₄ will change by one-half this amount (by +0.040 atm). Our spreadsheet produces the values shown below. We have inserted a red line to represent the equilibrium position.
   
Let’s use our Thermodynamic and Equilibrium models to perform the same analysis of this data.

1. Initially, ∆G is positive. This indicates that the reaction is not spontaneous in the forward direction. But this also means that the reaction is spontaneous in the reverse direction.
2. The value of Q is initially greater than the value of K. The reaction proceeds in the reverse direction to reach equilibrium.
3. As N₂O₄ is produced and the NO₂ is reacted away, the sign on ∆G remains positive, but its magnitude is decreasing. It’s approaching 0, the value for the equilibrium condition.
4. As N₂O₄ is produced and the NO₂ is reacted away, the value of Q decreases. It is approaching the value of K.
5. Once equilibrium is reached (red line in the table), the ∆G is 0 and Q = K
6. The spreadsheet continues to perform calculations, but the grey-shaded values in the table would not be observed. Once ∆G equals 0, the reaction no longer proceeds in the forward direction. The system has attained an equilibrium.

 
 
 
 
 

Before You Leave - Practice and Reinforcement

Now that you've done the reading, take some time to strengthen your understanding and to put the ideas into practice. Here's some suggestions.
 

• The Check Your Understanding section below includes questions with answers and explanations. It provides a great chance to self-assess your understanding.
• Download our Study Card on Thermodynamic and Equilibrium Model. Save it to a safe location and use it as a review tool.

 
 
 

Check Your Understanding of Gibbs Free Energy and Equilibrium

Use the following questions to assess your understanding of the relationship between the thermodynamic and the equilibrium models of spontaneity. Tap the Check Answer buttons when ready.
 
1. Use the equation ∆G = ∆G°  +  R•T•lnQ to answer part a and part b.
a. When Q = 1, how does the value of ∆G compare to the value of ∆G°?
 

b. What is the value of Q for standard conditions (i.e., 25°C, partial pressures of 1.00 atm, and concentrations of 1.00 M)?
 
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2. Write the expression for reaction quotient for the following reaction:

AgCl(s)    ⇌    Ag⁺(aq)  +  Cl^-(aq)

 

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3. Write the expression for reaction quotient for the following reaction:

2 C₄H₁₀(l)  +  13 O₂(g)    ⇌    8 CO₂(g)  +  10 H₂O(g)

 

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4. For the decomposition of water

2 H₂O(l)    ⇌    2 H₂ (g)  +  O₂(g)

the value of ∆G° is +474 kJ/mol. Determine the equilibrium constant for the reaction at 25.0°C.
 

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5. For the reaction

2 SO₂(g)    +   O₂(g)    ⇌    2 SO₃(g)

the ∆G° value is -142 kJ/mol. Determine the equilibrium constant for the reaction at 25.0°C.
 

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6. Consider the reaction

2 CO(g)    +   O₂(g)    ⇌    2 CO₂(g)

 
for which the ∆G° value is -514 kJ/mol.
a. Determine the equilibrium constant for the reaction at 25.0°C.
 

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b. If the partial pressures of CO, O₂, and CO₂ are 0.752 atm, 0.829 atm, and 1.340 atm respectively, what is the value of the reaction quotient?
 
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c. Compare the reaction quotient to the equilibrium constant. In which direction does the reaction proceed to reach equilibrium?
 
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d. Calculate the ∆G value for these conditions at 25.0°C.
 
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