22.5: Gibbs Free Energy
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.



G=H−TS


where



△G=△H−T△S


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
For Gibbs free energy, a spontaneous change is one where



2 C4H10(g)+13 O2(g)→8 CO2(g)+10 H2O(g)+energy


For this reaction,
For highly endothermic changes (high positive



C(s)+H2O(l)+energy→H2(g)+CO(g)


Example:
In the production of ammonia at



N2(g)+3 H2(g)→2 NH3(g) △H=−93.0 kJ/mol


Solution:



△H=−93.0 kJ/mol





T=25∘C+273.15=298.15 K





△S=−198.0 J/K⋅mol=−0.198 kJ/K⋅mol





△G=△H−T△S





△G=(−93.0 kJ/mol)−(298.15 K)(−0.198 kJ/K⋅mol)





△G=(−93.0 kJ/mol)−(−59.0 kJ/mol)=−34.0 kJ/mol


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
Example:
Calculate the Gibbs free energy for the production of ammonia at



N2(g)+3 H2(g)→2 NH3(g) △H=−93.0 kJ/mol


Solution:



△H=−93.0 kJ/mol





T=200∘C+273.15=473.15 K





△S=−198.0 J/K⋅mol=−0.198 kJ/K⋅mol





△G=△H−T△S





△G=(−93.0 kJ/mol)−(473.15 K)(−0.198 kJ/K⋅ mol)





△G=(−93.0 kJ/mol)−(−93.7 kJ/mol)=0.7 kJ/mol)


At
\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  Nonspontaneous 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 nonspontaneous 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 nonspontaneous. 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 nonspontaneous.
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 nonspontaneity. If the sign of \begin{align*}\triangle G\end{align*} is positive, the reaction is nonspontaneous; 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
 Define Gibbs free energy.
 Summarize the conditions of spontaneity according to Gibbs free energy equation.
 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?
 The reaction will always be spontaneous.
 The reaction will always be nonspontaneous.
 The reaction will be spontaneous only at high temperatures.
 The reaction will be spontaneous only at high temperatures.
 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?
 The reaction will always be spontaneous.
 The reaction will always be nonspontaneous.
 The reaction will be spontaneous only at high temperatures.
 The reaction will be spontaneous only at low temperatures.
 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*}?
 \begin{align*}68.7 \ \mathrm{kJ/mol}\end{align*}
 \begin{align*}106 \ \mathrm{kJ/mol}\end{align*}
 \begin{align*}3.32 \times 10^3 \ \mathrm{kJ/mol}\end{align*}
 \begin{align*}4.08 \times 10^4 \ \mathrm{kJ/mol}\end{align*}
 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*}?
 \begin{align*}1.132 \times 10^5 \ \mathrm{kJ/mol}\end{align*}
 \begin{align*}5.590 \times 10^3 \ \mathrm{kJ/mol}\end{align*}
 \begin{align*}311.1 \ \mathrm{kJ/mol}\end{align*}
 \begin{align*}203.5 \ \mathrm{kJ/mol}\end{align*}
 Which of the following regarding reaction spontaneity is true?
 A reaction with a positive \begin{align*}\triangle S^o\end{align*} will always be spontaneous.
 A reaction with a negative \begin{align*}\triangle H^o\end{align*} will always be spontaneous.
 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.
 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.
 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.
 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:
 \begin{align*}\triangle H\end{align*}
 \begin{align*}\triangle S\end{align*}
 \begin{align*}\triangle G\end{align*} at \begin{align*}100^\circ\mathrm{C}\end{align*}
 Is the system spontaneous or nonspontaneous?
\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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