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Spontaneity, Entropy and Free Energy

     This part of thermodynamics predicts whether a process is spontaneous, and how much energy is available from the process.  A spontaneous process is one which occurs without outside intervention.  The speed at which it occurs is governed by kinetics.  Spontaneous processes don't necessarily occur quickly, they just occur on their own.
     There are several factors which influence spontaneity.  One is the energy involved, DH.  Although processes which are exothermic are often spontaneous, endothermic processes can also be spontaneous.  For example, ice melts at a temperature above 0oC, even though the process requires heat.  This example also illustrates that temperature is a factor.  Some processes are spontaneous at high temperatures, other at lower temperatures.  Lastly, entropy must be considered.  Entropy is a measure of randomness or disorder.  The second law of thermodynamics states that in any spontaneous process, the entropy of the universe increases.  The universe includes both the system's disorder and the disorder of the surroundings.  Thus, a spontaneous process can occur, in which the system becomes more ordered, only if the entropy increase in the surroundings is greater.  For example, when ice freezes below 0oC, the liquid water goes to a more ordered solid state.  That is, the system loses entropy.  However, as the ice freezes, heat is given off the surroundings.  This heat flow increases random motion of molecules in the surroundings and increases the entropy of the surroundings.  Since the process is spontaneous below 0oC, we know the increase in entropy of the surroundings must be greater than the decrease in entropy of the water molecules.  The symbol for entropy change is DS.
    The factors of entropy change, temperature and enthalpy change have been combined to provide a measure of spontaneity.  This thermodynamic function,  DG, the Gibbs free energy change, not only predicts if the process is spontaneous, but also tells us how much excess energy is available to do work.  If the process is not spontaneous, it tells us how much work is needed to make the process happen.
DG= DH - TDS

     If DG is negative, the process is spontaneous.  If DG is positive, the process is not spontaneous.
     Tables of standard enthalpies of formation (DHfo), free energies of formation (DGfo) and entropy values (So) are listed in the appendix of the text.  Note that entropy values are absolute, so you must calculate entropy changes using the following relationship:

DSo = (sum of Soproducts) – (sum of Soreactants)

  The following table summarizes the relationship between temperature, enthalpy change and entropy change and spontaneity.
 

enthalpy change 
entropy change
spontaneity
(–) exothermic
(+) increase
spontaneous at all T
(–) exothermic
(–) decrease
spontaneous at lower T
(+) endothermic
(+) increase
spontaneous at higher T
(+) endothermic
(–) decrease
never spontaneous
• Consider the reaction:  CaCO3(s)  -->CaO(s)  +  CO2(g)  at 25oC.  Calculate DGo using the tables in the appendix of your textbook.  Is the process spontaneous at this temperature?  Is it spontaneous at all temperatures?  If not, at what temperature does it become spontaneous?
 

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The Relationship Between DG and Keq

     When a chemical reaction occurs, the reaction proceeds until it reaches equilibrium.  At equilibrium, DG = 0.  The value of DG changes during the course of the reaction as the concentrations (or pressures) of products and reactants change.  If initial conditions are not standard, i.e., with all P=1 atm and all concentrations = 1.00M, or the temperature is not 298K, you can calculate DG using the following equation:

DG=DGo + RT lnQ       where R = 8.314 J/mol-K, Q = reaction quotient
Calculation of Equilibrium Constants
     At equilibrium, DG = 0, and Q = K, so the above equation becomes:
DG = 0 = DGo + RT lnK; or
DGo = –RT lnK

    Reactions with large negative values of DGo have very large vales of K, showing that the driving force for the reaction to proceed to the right is large.  Reactions with positive values of DGo have small values of K, indicating the reaction is favored to the left.

• Calculate DHo, DSo, DGo and K at 298K for:  H2(g) + Cl2(g) <--> 2 HCl(g)
 

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Copyright ©1998 Beverly J. Volicer and Steven F. Tello, UMass Lowell.  You may freely edit these pages  for use in a non-profit, educational setting.  Please include this copyright notice on all pages.