Ch 18 – Thermodynamics and Equilibrium
Thermodynamics is the study of the relationship between heat and
other forms of energy that are involved in a chemical reaction.
Internal Energy (U)
 Internal energy is the sum of the kinetic and potential energies of the individual molecules.
 Both internal energy (U) and enthalpy (H) are state functions.
That is, they depend only on the present state, not on their history.
 A state function is completely determined by other state functions.
For example, U is determined by T and P.
1st Law of Thermodynamics
 U can be used to describe thermodynamic changes to a system due to a reaction.
U = Uf – Ui = q + w
where q is heat and w is work
 q is positive when heat is added to system, negative when heat is released
 w is positive when work is done to system, negative when work is done by the system
 For constant force, work is force times distance.
w = F(x)
 In a gravitational field, F = mg where m is mass and g is the gravitational constant, 9.8 m/s2.
 Work done by gravity is found by w = F(h) = mg(h), where (h) is change in height.
 For an expanding gas that is doing work by making a gas piston move upwards: w = −F(x)
 Volume of a cylinder is area times height.
V=A×x
V = A(x)
∆V
w = −F (
F
) = − (A)(V)
A
 Pressure is force over area: P = F/A. This gives us w = − P(∆V) for an expanding gas piston.
 Suppose a 1.0 kg weight pushes a gas piston downwards by 1.5 m, and 2.0 J are released as
heat due to friction during the process. Find the q, w, and U for the gas piston.
w = mg(h) = (1.0 kg)(9.8 m/s2)(1.5 m) = + 14.7 J (E is absorbed)
q = − 2.0 J (E is released)
U = q + w = (− 2.0) + 14.7 = + 12.7 J
Enthalpy
 H = U + PV
and
H = H f – Hi = U + (PV)
 If pressure is constant, then P = 0. This gives us H =U + P(V).
 U = q − P(V) can be substituted into the equation.
The result is:
H = q − P(V) + P(V) = qP
So, enthalpy is the heat of reaction at constant P.
 See Table 6.2, pg 246, and Appx C (pg A-8) for enthalpies of formation of substances.
 Standard enthalpy of reaction (Ho) is at 1 atm and 298 K.
It can be found by the difference between enthalpies of formation for products and reactants.
 Ho = ∑ 𝐧∆𝐇𝐟𝐨 (𝐩𝐫𝐨𝐝) − ∑ 𝐦∆𝐇𝐟𝐨 (𝐫𝐞𝐚𝐜𝐭) where n and m are stoichiometric coefficients.
Enthalpy for creating Urea from Ammonia and Carbon Dioxide
 2NH3(g) + CO2(g) ⇌ H2O(L) + (NH2)2CO(aq)
 Ho = [Hof (aq urea) + Hof(H2O(L))] − [2×Hof(NH3(g)) + Hof(CO2(g))]
 Ho = [(− 319.2) + (− 285.8)] – [2×(− 45.9) + (− 393.5)] = − 119.7 kJ per mol of urea
Entropy
 Entropy (S) is a measure of the randomness or disorder in a system.
 It measures how dispersed the energy of a system is among the possible energy states.
That is, it describes how randomly the energy is distributed.
 Measured in units of energy divided by temperature (J/K).
2nd Law of Thermodynamics
 The total combined entropy of a system and its surroundings
always increases for a spontaneous process. (Entropy Created) > 0
 A spontaneous process is one which randomizes (releases) energy and creates disorder.
 Heat flow involves entropy because it is a dispersal of energy.
It randomizes the E. Entropy associated with heat flow is q/T.
 For a spontaneous process: S = (entropy created) + q/T
and
S > q/T
 For a process that occurs only near eqm, such as if Q stays near K:
(entropy created) ~ 0
and
S = q/T
S = q/T, or S = H/T, for a phase change (melting, freezing, etc.)
 For a spontaneous process:
(entropy created) > 0 and
S > H/T
 H − TS < 0 for a spontaneous process
 H − TS > 0 is nonspontaneous, and H − TS = 0 is at equilibrium
3rd Law of Thermodynamics –
 A perfectly crystalline solid at absolute zero (0 K) has S = 0. That is, it has no disorder.
 Std Entropy (So) for an atom, molecule, or ion is measured at 1 atm and 298 K.
 See Table 18.1 (pg 743) and Appx C (pg A-8).
 There is no “” or “f” in So because it is not measured the same way as Hof.
 It is actual total entropy, measured by integrating
from 0 Kelvin (where S = 0) up to the final T.
 Hof is based completely on chemical reactions at 298 K.
 Unlike Hof, values of So are nonzero (and positive) for elements.
 For a chemical reaction: So = ∑ 𝐧𝐒 𝐨 (𝐩𝐫𝐨𝐝) − ∑ 𝐦𝐒 𝐨 (𝐫𝐞𝐚𝐜𝐭)
where n and m are stoichiometric coefficients.



So (Entropy of Reaction)
 Entropy generally increases (So > 0) if any of the following occur:
o A molecule is broken into two or more smaller molecules.
(decomposition or elimination)
o There is an increase in moles of gas.
o The phase changes from solid to liquid or gas, or from liquid to gas.
(phase becomes less condensed)
 If the reverse of these conditions occur, then So < 0.
Ex 18.2 Is So positive or negative?
a. One mol glucose (s) ferments to two mol ethanol (liq) and two mol CO2 (g).
This a decomposition reaction and the moles of gas increase.
So > 0
b. 2NH3(g) + CO2(g) ⇌ H2O(L) + (NH2)2CO(aq)
The total number of molecules and the number of gas moles both decrease. So < 0
c. H2O(g) + CO (g) ⇌ CO2(g) + H2(g)
This cannot be determined easily because the total moles and the gas moles do not change.
The absolute value is a relatively small number, though (< 50 J/K).
Ex 18.3 Calculating So for a reaction (Use Table 18.1)
 2NH3(g) + CO2(g) ⇌ (NH3)2CO(aq) + H2O(L)
 So = [So(aq urea) + So(H2O(L))] − [2×So(NH3(g)) + So(CO2(g))]
1 kJ
 So = [(174) + (70)] − [2(193) + (214)] = (− 356 J/K)(1000 J) = − 0.356 kJ/K
Gibb’s Free Energy (G), named after J. Willard Gibbs, is found by G = H − TS
 For a chemical reaction, the change in free energy is Go = Ho − TSo.
 Go is also the maximum energy available to do useful work: Go = wmax.
 The degree sign symbol (o) is for standard state, which means a reaction involving
only liquids and/or solids has total P = 1 atm; each gas has partial P = 1 atm; and
each solute has concentration M = 1 mol/L. Also, standard state is at T = 298 K.
 If a reaction is spontaneous, then G < 0 (negative),
because H − TS < 0 if the process is spontaneous.
 Gof is the free energy of formation from elements in their reference forms at standard state.
 Note: Gof ≠ Hof − TSo does not work for individual substances because So is not Sof.
 For a reaction Go = ∑ 𝐧∆𝐆𝐟𝐨 (𝐩𝐫𝐨𝐝) − ∑ 𝐦∆𝐆𝐟𝐨 (𝐫𝐞𝐚𝐜𝐭) where n and m are stoich coeffs.
 Go can be found from either the summation equation or the enthalpy/entropy equation.
 See Table 18.2 (pg 747) and Appx C (pg A-8) for Gof values.
 Be sure S is in kJ/K when using G = H − TS equations where G and H are already in kJ.
Ex 18.4 Calculating Go from Ho and So for a reaction
 Use Ho and So values from Tables 6.2 and 18.1.
 N2(g) + 3H2(g) ⇌ 2NH3(g)
 Ho = [2(− 45.9)] – [0 + 3(0)] = − 91.8 kJ
1 kJ
 So = [2(192.7)] – [(191.6) + 3(130.6)] = (− 198.0 J/K)(1000 J) = − 0.198 kJ/K
 Go = Ho − TSo = (− 91.8 kJ) – (298 K)(− 0.198 kJ/K) = − 32.8 kJ
Ex 18.4 Calculating Go for a reaction by using Gof for each substance
 Use Go values from Table 18.2.
 C2H5OH(L) + 3O2(g) ⇌ 2CO2(g) + 3H2O(g)
 Go = [2(− 394.4) + 3(− 228.6)] – [(− 174.9) + (0.0)] = − 1299.7 kJ
Criterion for Spontaneity (This is similar to interpreting Kc)
 If Go is a negative number with a large magnitude, the reaction is spontaneous as written,
and the reactants transform almost completely to form products.
 If Go is a large positive number, the reaction is not spontaneous as written,
and the reactants do not form significant amounts of products.
(The reaction may be spontaneous in reverse.)
 If Go has a small magnitude (pos. or neg.), the reaction gives an equilibrium mixture
containing significant amounts of both reactants and products.
Thermodynamic Eqm Constant (K)
 Includes gas concentrations expressed as the partial pressure (P, in atm),
and liquid solute concentrations expressed as molarity (M).
 K includes both gas pressures and solute molarities.
 Solids and pure liquids are still omitted.
 If a reaction includes only gases, then K = Kp.
 If a reaction includes only aqueous solutes, then K = Kc, Ka, Kb, Ksp, or Kf.
Ex 18.7 Expressions for K
[(NH2 )2 CO(aq) ]
a. 2NH3(g) + CO2(g) ⇌ H2O(L) + (NH2)2CO(aq)
K=
b. AgCl(s) ⇌ Ag+1(aq) + Cl-1(aq)
K = Ksp = [Ag+1][ Cl-1]
PCO2 P2NH3
Thermodynamic Reaction Quotient (Q)
 Q has the same form as K, but uses interim concentrations: [A]i.
 G at nonstandard conditions is found by
G = Go + (RT)ln[Q]
R = 8.31 J/(K-mol) is the universal gas constant. The result for G is in Joules.
Relating G to K
 What if Q = K?
Then, no changes in concentration are spontaneous, which means G = 0.
Implies 0 = Go + (RT)ln[K]
or
Go = − (RT)ln[K]
 Also:
K = exp[−
∆𝐆𝐨
𝐑𝐓
] = 𝐞[− ∆𝐆
𝐨 /𝐑𝐓]
𝐐
 G = − (RT)ln[K] + (RT)ln[Q] = (RT)ln[ ]
𝐊
Ex 18.8 Find K at 298K from Go
1000 J
 Go = (− 13.6 kJ) (
1 kJ
) = − 13,600 J for 2NH3(g) + CO2(g) ⇌ H2O(L) + (NH2)2CO(aq)
 K = exp(− Go/RT) = exp[
− (−13,600 J)
(8.31
J
)( 298 K)
K
] = exp[+ 5.49] = 242
Ex 18.9 Calculate Ksp from Go for AgCl(s) ⇌ Ag+1(aq) + Cl-1(aq)
1000 J
 Go = [(77.1) + (−131.3)] – [−109.8] = (+ 55.6 kJ) (
 K = exp(− Go/RT) = exp[
− (+ 55,600 J)
(8.31
J
)( 298 K)
K
1 kJ
] = exp[− 22.45] = 1.8 x 10−10
 This value matches the value in Table 17.2 of the text.



) = + 55,600 J

GTo (standard state concentrations and partial pressures, but at temperatures other than 298 K)
 GTo = Ho − TSo (Assumes H and S to be approximately constant as T changes.)
H S
− +
+ −
G
Always G < 0
Always G > 0
+
+
G < 0 if T >
−
−
G < 0 if T <
Spontaneity
Reaction is spontaneous at all T’s.
Reaction is not spontaneous at any T.
∆H
Reaction is spontaneous only at T’s above that value.
∆S
∆H
Reaction is spontaneous only at T’s below that value.
∆S
Ex 18.10 Find GTo and K at 1000 oC for CaCO3(s) ⇌ CaO(s) + CO2(g)
1 kJ
 Ho = + 178.3 kJ
So = (+ 159.0 J/K)(1000 J) = + 0.159 kJ/K
 Go1273 K = Ho − TSo = +178.3 kJ – (1273 K)(0.159 kJ/K) = − 24.1 kJ = − 24,100 J
 K1273 K = exp[
− ∆Go
1273K
RT
] = exp[
− (− 24,100 J)
(8.31
J
× 1273 K)
K
] = exp[+ 2.278] = 9.8
 Expression: K1273 K = PCO2 = 9.8 atm
 Also, since Ho > 0 and So > 0, Go is negative at high temperatures.
G < 0 at T >
∆H
∆H
∆S
∆S
=
(+178.3 kJ)
(0.159
kJ
)
K
= 1121 K
Reaction is spontaneous only at T > 1121 K