Lesson 16.2 Free Energy
Suggested Reading
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Zumdahl Chapter 16 Sections 16.4, 16.6 - 16.9
Essential Question
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What is the relationship between spontaneity and free energy?
Learning Objectives:.
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Relate free energy to spontaneity.
Calculate the standard free energy change for a chemical reaction.
Show the relationship between free energy and the equilibrium
constant.
Predict the direction of a reaction using the concept of free energy.
Describe the free energy change with respect to equilibrium for both
spontaneous and non-spontaneous reaction.
Introduction
In the previous lesson you learned that the quantity ∆H - T∆S can serve as
a criterion for the spontaneity of a reaction at constant temperature and
pressure. If the value of this quantity is negative the reaction is
spontaneous, if positive the reaction is non-spontaneous, and if equal the
reaction is at equilibrium. It is convenient to define a new thermodynamic
quantity in terms of H and S that will be directly useful as a criterion of
spontaneity. For this purpose, American physicist J. Willard Gibbs (1839 1903) introduced the concept of free energy, G, which is a thermodynamic
quantity defined by the equation G = H - TS, or ∆G = ∆H - T∆S This
quantity gives a direct criterion for spontaneity. As a reaction proceeds at a
given T and P, reactants form products and the enthalpy and entropy
changes. These changes result in a change in free energy. The following
rules are useful in predicting the spontaneity of a reaction.
1.
2.
3.
When ∆G is a large negative number (more negative than -10 kJ), the
reaction is spontaneous as written, and reactants transform almost
entirely to products at equilibrium (K > 1).
When ∆G is a large positive number (larger than about 10 kJ), the
reaction is non-spontaneous as written, and reactants do not give
significant amounts of products at equilibrium (K < 1).
When ∆G has a small negative or positive value (less than about 10
kJ), the reaction gives an equilibrium mixture with significant amounts of
both reactant and products.
Standard Free Energy Change
Recall that for the purposes of tabulating thermodynamic data, certain
standard states are choses, which are indicated by a superscript degree sign
on the symbol of the quantity. The standard states are as follows:
Pure liquids and solids: 1 atm pressure, 298 K
Gases: 1 atm partial pressure, 298 K
Solutions: 1 M concentration, 298 K
The standard free energy change, ∆G∘, is the free energy change that occurs
when reactants in their standard states are converted to products in their
standard states.
∆G∘ = ∆H∘ - T∆S∘
Thus if we know the standard enthalpy and entropy changes for a reaction we can calculate
the standard free energy change.
Example: Calculating ∆G∘ = ∆H∘ - T∆S∘
What is the standard free energy change for the Haber process at 25 C.
N2(g) + 3H2(g) → 2NH3(g)
Solution:
Calculate ∆H∘ and ∆S∘, then substitute these values into ∆G∘ = ∆H∘ - T∆S∘.
You may find it helpful to use a table to organize your data.
Reaction
N2(g)
3H2(g)
2NH3
Enthalpy (kJ)
0
3x0=0
2 x (-45.9) = -91.8
Entropy (J/K)
191.5
3 x 130.6 = 391.8
2 x 193 = 386
Now calculate each quantity.
∆H∘ = -91.8 - (0) = -91.8 kJ
∆S∘ = 386 - (191.5 + 391.8) = -197. J/K
Substitute into equation for free energy, using units of kJ/K for S.
∆G∘ = -91.8 kJ - (298 K)(-.197 kJ/K) = -33.1 kJ
Thus, this reaction is deemed spontaneous under these conditions.
Standard Free Energies of Formation
The standard free energy of formation, ∆G∘f, of a substance is defined
similarly to standard enthalpy of formation. That is, ∆G∘f is the free energy
change that occurs when 1 mole of substance is formed from its elements
in their stablest states (reference forms) at 1 atm and a specified
temperature (usually 298 K).
For example, the standard free energy of formation of NH3(g) is the free
energy change for the reaction 1/2N2(g) + 3/2H2(g) → NH3(g). The
reactants, each at 1 atm, are converted to 1 mol of product at 1 atm. In the
previous example, we found that the ∆G∘ for two moles of ammonia is -33.1
kJ, so the ∆G∘f for 1 mole should equal -33.1kJ / 2 = -16.6 kJ.
As in the case of standard enthalpies of formation, the standard free
energies of formation of the elements in their stablest states are assigned a
value of 0. Also, as with standard enthalpies, ∆G∘ can be found
from ∆G∘f using the equation below. Standard energies of formation can be
found alongside values for standard enthalpies of formation and standard
entropies in tables of thermodynamic data.
Example: Calculating ∆G∘ from Standard Free Energies of Formation
Calculate ∆G∘ for the combustion of 1 mol of ethanol at 25 deg C.
C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g)
Solution:
Obtain the values for ∆G∘f from a table of thermodynamic data, and then
calculate ∆G∘ from ∆G∘f using the equation above. Always double check to
make sure your equations are balanced!
C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g)
∆G∘f kJ: -174.8
0
2 x -394.4
3 x -228.6
Substitute and solve.
∆G∘ = [2(-394.4) + 3( -228.6) - (-174.8)]kJ = -1299.8 kJ
Since ∆G∘ is a large negative number we can predict that this reaction is
spontaneous.
Free Energy Change During a Reaction
For any spontaneous reaction, the total free energy of the substances in
the reaction mixture decreases (exergonic reaction) as the reaction
proceeds. For a non-spontaneous reaction, free energy increases during a
reaction (endergonic reaction).
This is analogous to what occurs during exothermic and endothermic
reactions.
Free Energy and Equilibrium Constants
One of the most important results of chemical thermodynamics is an
equation relating the standard free energy change for a reaction to the
equilibrium constant.
If we know the equilibrium constant, Keq, for a chemical change (or if we
can determine the equilibrium constant), we can calculate the standard
state free energy change, Go, for the reaction using the equation:
In this equation
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R = 8.314 J mol-1 K-1 or 0.008314 kJ mol-1 K-1.
T is the temperature on the Kelvin scale.
Keq is the equilibrium constant at the temperature T.
This equation also take the form
Go = -2.303 RT log K
If you want the free energy change for a reaction involving reactants and
products in nonstandard states ( G), you can obtain it from the standard
free energy change using this equation:
G=
Go + RT ln Q
where Q is the reaction quotient.
Example: Calculating K from the Standard Free Energy Change
Find the value of the equilibrium constant K at 298 K for the reaction
The standard free energy change equals -13.6 kJ.
Using Go = -2.303 RT log K, we can rearrange for K as follows
You can understand and use Go as a criterion for spontaneity in terms of
its relationship with K. When the equilibrium constant is greater than one,
the equilibrium mixture is mostly products, Go is negative, so the reaction
is spontaneous. When the equilibrium constant is less than one, the
equilibrium mixture is mostly reactants, Go is positive, so the reaction is
non-spontaneous. When K is near 0, Go is near 0 and there are
appreciable amounts of both reactants and products.
For a given chemical reaction, the minimum free energy of the system is
found at the equilibrium position as shown in the diagram to the right.
Change in Free Energy with
Temperature
If you must calculate the change in temperature at a temperature other than
298 K, you can use a simple method that gives approximate results. Precise
calculation are complicated, so this is the preferred method under these
circumstances. In this method you assume ∆H∘ and ∆S∘ are approximately
constant with temperature and use
∆G∘T = ∆H∘ - T∆S∘ (approximation)
The following table summarizes the effect of temperature on spontaneity.
Homework: Book questions pg. 783 questions 45, 47,
51, 61, 63