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# 22.5: Gibbs Free Energy

Difficulty Level: At Grade Created by: CK-12

## Lesson Objectives

The student will:

• define Gibbs free energy.
• calculate Gibbs free energy given the enthalpy and entropy.
• use Gibbs free energy to predict spontaneity.

## Vocabulary

• Gibbs free energy

## Introduction

In the latter part of the 1800s, a Yale physics professor named J. Willard Gibbs published a paper that related the enthalpy of a system to its entropy. It is this relationship, according to Gibbs, that would allow chemists to determine the spontaneity of the system at a specific temperature. The relationship is known as free energy. In this final lesson of the chapter, we will explore Gibbs free energy.

## Gibbs Free Energy

Previously we have said that a system tended to be spontaneous if the enthalpy decreased or the entropy increased. However, there are systems that are spontaneous that do not follow that pattern. For instance, ice melts at room temperature. In this case, both the entropy and enthalpy increases. How do we know which will dominate a change? Gibbs free energy is defined as the maximum energy available to do useful work and can be determined by the combined effect of the change in the enthalpy of the system and change in the entropy of the reaction measured at a specific temperature.

## Gibbs Free Energy Equation

The definition of free energy is shown below.

\begin{align*}G = H - TS\end{align*}

where \begin{align*}H\end{align*} is the heat content (enthalpy) of a substance, \begin{align*}S\end{align*} is its entropy, and \begin{align*}T\end{align*} is the Kelvin temperature. However, since the only way we can know these values is to determine the change that takes place, the equation below is more useful, assuming that the temperature does not change.

\begin{align*}\triangle G = \triangle H - T\triangle S\end{align*}

This equation tells us that the change in free energy for an equation is equal to the change in enthalpy minus the change in entropy times the Kelvin temperature. The free energy available is the energy from the change in enthalpy of the bonds less the amount \begin{align*}T \triangle S\end{align*}. Or to look at it another way, the higher the temperature, the more the disorder, and the less available the energy becomes.

For Gibbs free energy, a spontaneous change is one where \begin{align*}\triangle G\end{align*} is negative. If a change takes place at low temperature and involves little change in entropy, \begin{align*}T\triangle S\end{align*} will be negligible and \begin{align*}\triangle G\end{align*} will be spontaneous for an exothermic change \begin{align*}(-\triangle H)\end{align*}. Combustion is a good example and is shown below.

\begin{align*}2\ \mathrm{C}_4\mathrm{H}_{10(g)} + 13 \ \mathrm{O}_{2(g)} \rightarrow 8 \ \mathrm{CO}_{2(g)} + 10 \ \mathrm{H}_2\mathrm{O}_{(g)} + \text{energy}\end{align*}

For this reaction, \begin{align*}\triangle H = -5315 \ \mathrm{kJ}\end{align*} while \begin{align*}\triangle S = 312 \ \mathrm{J/K}\end{align*}, so \begin{align*}\triangle G\end{align*} will definitely be negative. All combustion reactions are spontaneous at room temperature.

For highly endothermic changes (high positive \begin{align*}\triangle H\end{align*}), \begin{align*}\triangle G\end{align*} can only be negative (that is spontaneous) if \begin{align*}T\triangle S\end{align*} is large. This means that either the temperature is high or there is a large increase in entropy. An example of this is when solid carbon reacts with water:

\begin{align*}\mathrm{C}_{(s)} + \mathrm{H}_2\mathrm{O}_{(l)} + \text{energy} \rightarrow \mathrm{H}_{2(g)} + \mathrm{CO}_{(g)}\end{align*}

\begin{align*}\triangle S\end{align*} is positive because there is greater order in the solid carbon, which is being converted to the disorder of a gas. The temperature must be high for this reaction to occur (\begin{align*}1170 \ \mathrm{K}\end{align*} or higher). In fact, the reaction will stop or even reverse if the temperature decreases. Both conditions are met for \begin{align*}\triangle G\end{align*} to be negative for this endothermic reaction due to the increase in entropy and the high temperature.

Example:

In the production of ammonia at \begin{align*}25^\circ\text{C}\end{align*}, the entropy was found to be \begin{align*}-198.0 \ \mathrm{J/K} \cdot \mathrm{mol}\end{align*}. Calculate the Gibbs free energy for the production of ammonia.

\begin{align*}\mathrm{N}_{2(g)} + 3 \ \mathrm{H}_{2(g)} \rightarrow 2 \ \mathrm{NH}_{3(g)} \ \ \ \ \ \triangle H = -93.0 \ \text{kJ/mol}\end{align*}

Solution:

\begin{align*}\triangle H = -93.0\ \text{kJ/mol}\end{align*}
\begin{align*}T = 25^\circ\mathrm{C} + 273.15 = 298.15 \ \text{K}\end{align*}
\begin{align*}\triangle S = -198.0 \ \text{J/K} \cdot \text{mol} = -0.198 \ \text{kJ/K} \cdot \text{mol} \end{align*}
\begin{align*}\triangle G = \triangle H - T\triangle S\end{align*}
\begin{align*}\triangle G = (-93.0\ \text{kJ/mol}) - (298.15\ \text{K})(-0.198 \ \text{kJ/K} \cdot \text{mol}) \end{align*}
\begin{align*}\triangle G = (-93.0\ \text{kJ/mol}) - (-59.0\ \text{kJ/mol}) = -34.0\ \text{kJ/mol}\end{align*}

This video shows an example of how to plug values into Gibbs Free Energy equation (7f): http://www.youtube.com/watch?v=ECjH1ErqzRU (3:14).

## The Sign of ΔG and Spontaneity

The sign of \begin{align*}\triangle G\end{align*} indicates spontaneity. If the sign of \begin{align*}\triangle G\end{align*} is positive, the reaction is non-spontaneous; if \begin{align*}\triangle G\end{align*} is negative, the reaction is spontaneous. If \begin{align*}\triangle G\end{align*} is zero, the reaction is at equilibrium. In the production of ammonia, \begin{align*}\triangle G\end{align*} was found to be \begin{align*}-34.0 \ \mathrm{kJ/mol}\end{align*} at \begin{align*}25^\circ\mathrm{C}\end{align*} or \begin{align*}298.15 \ \mathrm{K}\end{align*}. Therefore, at this temperature the production of ammonia is a spontaneous process. What would happen if the temperature were increased?

Example:

Calculate the Gibbs free energy for the production of ammonia at \begin{align*}200^\circ\mathrm{C}\end{align*}, the entropy was found to be \begin{align*}-198.0 \ \mathrm{J/K} \cdot \mathrm{mol}\end{align*}.

\begin{align*}\mathrm{N}_{2(g)} + 3 \ \mathrm{H}_{2(g)} \rightarrow 2 \ \mathrm{NH}_{3(g)} \ \ \ \ \ \triangle H = -93.0 \ \text{kJ/mol}\end{align*}

Solution:

\begin{align*}\triangle H = -93.0 \ \text{kJ/mol}\end{align*}
\begin{align*}T = 200^\circ\mathrm{C} + 273.15 = 473.15 \ \text{K}\end{align*}
\begin{align*}\triangle S = -198.0 \ \text{J/K} \cdot \text{mol} = -0.198 \ \text{kJ/K} \cdot \text{mol} \end{align*}
\begin{align*}\triangle G = \triangle H - T\triangle S\end{align*}
\begin{align*}\triangle G = (-93.0\ \text{kJ/mol}) - (473.15\ \text{K}) (-0.198\ \text{kJ/K} \cdot \ \text{mol}) \end{align*}
\begin{align*}\triangle G = (-93.0\ \text{kJ/mol}) - (-93.7\ \text{kJ/mol}) = 0.7 \ \text{kJ/mol})\end{align*}

At \begin{align*} 200^\circ\mathrm{C}\end{align*}, \begin{align*}\triangle G\end{align*} is positive, so the reaction is now non-spontaneous. Gibbs free energy is temperature dependent. At high temperatures, some systems that are spontaneous will become non-spontaneous (as is the case with \begin{align*}\mathrm{NH}_{3(g)}\end{align*} formation), and some systems that are non-spontaneous will become spontaneous. Table below summarizes the conditions that relate \begin{align*}\triangle H, \ \triangle S, \ \triangle G\end{align*}, and temperature to spontaneity.

Summary for Gibbs Free Energy
\begin{align*} \triangle H\end{align*} \begin{align*}\triangle S\end{align*} \begin{align*}\triangle G\end{align*} Spontaneity
Positive Positive Positive at low temperatures Negative at high temperatures Spontaneous at high temperatures
Negative Positive Negative Spontaneous at all temperatures
Negative Negative Negative at low temperatures Positive at high temperatures Spontaneous at low temperatures
Positive Negative Positive Non-spontaneous at all temps

Example:

Calculate \begin{align*}\triangle G\end{align*} for the reaction \begin{align*}\mathrm{Cu}_{(s)} + \mathrm{H}_2\mathrm{O}_{(g)} \rightarrow \mathrm{CuO}_{(s)} + \mathrm{H}_{2(g)}\end{align*} when \begin{align*}\triangle H = 84.5 \ \mathrm{kJ/mol}\end{align*}, \begin{align*}\triangle S = -48.7 \ \mathrm{J/K} \cdot \mathrm{mol}\end{align*}, and \begin{align*}T = 150^\circ\mathrm{C}\end{align*}.

Is the reaction spontaneous or non-spontaneous at \begin{align*}150^\circ\mathrm{C}\end{align*}.

Solution:

\begin{align*}\triangle H = 84.5 \ \text{kJ/mol}\end{align*}
\begin{align*}T = 150^\circ\mathrm{C} + 273.15 = 423.15 \ \text{K}\end{align*}
\begin{align*}\triangle S = -48.7 \ \text{kJ/K} \cdot \mathrm{mol} = -0.0487 \ \text{kJ/K} \cdot \mathrm{mol}\end{align*}
\begin{align*}\triangle G = \triangle H - T \triangle S\end{align*}
\begin{align*}\triangle G = (84.5 \ \text{kJ/mol}) - (423.15 \ \text{K})(-0.0487 \ \text{kJ/K} \cdot \mathrm{mol})\end{align*}
\begin{align*}\triangle G = (84.5 \ \text{kJ/mol}) - (-20.6 \ \text{kJ/mol}) = 105.1 \ \text{kJ/mol}\end{align*}

\begin{align*}\triangle G\end{align*} is positive so the reaction is non-spontaneous. From looking at Table below, \begin{align*}\triangle H\end{align*} is positive and \begin{align*}\triangle S\end{align*} is negative, so \begin{align*}\triangle G\end{align*} will always be positive and non-spontaneous.

A blackboard example of using Gibbs free energy equation to determine whether a reaction is spontaneous (7f) is available at http://www.youtube.com/watch?v=sG1ZAdYi13A (9:57).

## Lesson Summary

• Gibbs free energy equation: \begin{align*}\triangle G = \triangle H - T\triangle S\end{align*}.
• The sign of \begin{align*}\triangle G\end{align*} indicates spontaneity or non-spontaneity. If the sign of \begin{align*}\triangle G\end{align*} is positive, the reaction is non-spontaneous; if \begin{align*}\triangle G\end{align*} is negative, the reaction is spontaneous; if \begin{align*}\triangle G\end{align*} is zero, the reaction is at equilibrium.

## Further Reading / Supplemental Links

The learner.org website allows users to view streaming videos of the Annenberg series of chemistry videos. You are required to register before you can watch the videos but there is no charge. The website has one video that relates to this lesson called “The Driving Forces.”

## Review Questions

1. Define Gibbs free energy.
2. Summarize the conditions of spontaneity according to Gibbs free energy equation.
3. For the reaction \begin{align*}\mathrm{C}_2\mathrm{H}_5\mathrm{OH}_{(l)} \rightarrow \mathrm{C}_2\mathrm{H}_5\mathrm{OH}_{(g)}\end{align*}, \begin{align*}\triangle S = 122.0 \ \mathrm{J/K} \times \mathrm{mol}\end{align*} and \begin{align*}\triangle H = 42.59 \ \mathrm{kJ/mol}\end{align*} at \begin{align*}25^\circ\mathrm{C}\end{align*}. Which of the following statements is true?
1. The reaction will always be spontaneous.
2. The reaction will always be non-spontaneous.
3. The reaction will be spontaneous only at high temperatures.
4. The reaction will be spontaneous only at high temperatures.
4. For the reaction \begin{align*}\mathrm{C}_6\mathrm{H}_{6(l)} + 3 \ \mathrm{H}_{2(g)} \rightarrow \mathrm{C}_6\mathrm{H}_{12(l)}\end{align*}, \begin{align*}\triangle S = -101.6 \ \mathrm{J/K} \times \mathrm{mol}\end{align*} and \begin{align*}\triangle H = -205.4 \ \mathrm{kJ/mol}\end{align*} at \begin{align*}25^\circ\mathrm{C}\end{align*}. Which of the following statements is true?
1. The reaction will always be spontaneous.
2. The reaction will always be non-spontaneous.
3. The reaction will be spontaneous only at high temperatures.
4. The reaction will be spontaneous only at low temperatures.
5. For the reaction \begin{align*}\mathrm{COCl}_{2(g)} \rightarrow \mathrm{CO}_{(g)} + \mathrm{Cl}_{2(g)}\end{align*}, \begin{align*}\triangle H = 109.6 \ \mathrm{kJ/mol}\end{align*} and \begin{align*}\triangle S = 137.1 \ \mathrm{J/K} \times \mathrm{mol}\end{align*}. What is the value of \begin{align*}\triangle G\end{align*} at \begin{align*}25.0^\circ\mathrm{C}\end{align*}?
1. \begin{align*}68.7 \ \mathrm{kJ/mol}\end{align*}
2. \begin{align*}106 \ \mathrm{kJ/mol}\end{align*}
3. \begin{align*}-3.32 \times 10^3 \ \mathrm{kJ/mol}\end{align*}
4. \begin{align*}-4.08 \times 10^4 \ \mathrm{kJ/mol}\end{align*}
6. Hydrazine, \begin{align*}\mathrm{N}_2\mathrm{H}_{4(l)}\end{align*}, has an important use in the space industry as rocket fuel. The preparation of hydrazine is: \begin{align*}\mathrm{N}_2\mathrm{O}_{(g)} + 3 \ \mathrm{H}_{2(g)} \rightarrow \mathrm{N}_2\mathrm{H}_{4(l)} + \mathrm{H}_2\mathrm{O}_{(l)} \ (\triangle H = -317.0 \ \text{kJ})\end{align*}. If the value of \begin{align*}\triangle S\end{align*} is \begin{align*}-393.8 \ \mathrm{J/K} \cdot \mathrm{mol}\end{align*} at \begin{align*}15^\circ\mathrm{C}\end{align*}, what is the value of \begin{align*}\triangle G\end{align*}?
1. \begin{align*}1.132 \times 10^5 \ \mathrm{kJ/mol}\end{align*}
2. \begin{align*}5.590 \times 10^3 \ \mathrm{kJ/mol}\end{align*}
3. \begin{align*}-311.1 \ \mathrm{kJ/mol}\end{align*}
4. \begin{align*}-203.5 \ \mathrm{kJ/mol}\end{align*}
7. Which of the following regarding reaction spontaneity is true?
1. A reaction with a positive \begin{align*}\triangle S^o\end{align*} will always be spontaneous.
2. A reaction with a negative \begin{align*}\triangle H^o\end{align*} will always be spontaneous.
3. A reaction with a positive \begin{align*}\triangle S^o\end{align*} and a negative \begin{align*}\triangle H^o\end{align*} will always be spontaneous.
4. A reaction with a negative \begin{align*}\triangle S^o\end{align*} and a negative \begin{align*}\triangle H^o\end{align*} will always be spontaneous.
5. A reaction with a positive \begin{align*}\triangle S^o\end{align*} and a positive \begin{align*} \triangle H^o\end{align*} will always be spontaneous.
8. For the reaction \begin{align*}\mathrm{CuO}_{(s)} + \mathrm{H}_{2(g)} \rightarrow \mathrm{Cu}_{(s)} + \mathrm{H}_2\mathrm{O}_{(g)}\end{align*}, use the data provided in Table below to find the values of:
1. \begin{align*}\triangle H\end{align*}
2. \begin{align*}\triangle S\end{align*}
3. \begin{align*}\triangle G\end{align*} at \begin{align*}100^\circ\mathrm{C}\end{align*}
4. Is the system spontaneous or non-spontaneous?
\begin{align*}\triangle H\end{align*} \begin{align*}\triangle S\end{align*}
\begin{align*}\mathrm{CuO}_{(s)}\end{align*} \begin{align*}-155.2 \ \mathrm{kJ/mol}\end{align*} \begin{align*}43.5 \ \mathrm{J/K} \cdot \mathrm{mol}\end{align*}
\begin{align*}\mathrm{H}_{2(g)}\end{align*} \begin{align*}0 \ \mathrm{kJ/mol}\end{align*} \begin{align*}131.0 \ \mathrm{J/K} \cdot \mathrm{mol}\end{align*}
\begin{align*}\mathrm{Cu}_{(s)}\end{align*} \begin{align*}0 \ \mathrm{kJ/mol}\end{align*} \begin{align*}33.3 \ \mathrm{J/K} \cdot \mathrm{mol}\end{align*}
\begin{align*}\mathrm{H}_2\mathrm{O}_{(g)}\end{align*} \begin{align*}-241.8 \ \mathrm{kJ/mol}\end{align*} \begin{align*}188.7 \ \mathrm{J/K} \cdot \mathrm{mol}\end{align*}

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CK.SCI.ENG.SE.2.Chemistry.22.5