# Chapter16 Thermodynamics12 by J1XyExu

VIEWS: 15 PAGES: 20

• pg 1
```									      Chapter 16
Thermodynamics
Entropy, Free Energy, and
Equilibrium
Enthalpy, Entropy and
Spontaneous Processes
 Although an exothermic process (-∆H) favors the spontaneity
of a chemical reaction, it doesn’t guarantee that it happens.
 The melting of ice is an endothermic process yet it occurs
spontaneously at certain temperatures.
 The other thermodynamic quantity that affects the spontaneity
of a process is the change in entropy (∆S). A positive entropy
favors a spontaneous process.
 Entropy is a measure of the randomness of a system. The more
random the system, the higher the entropy.
 Ice melts spontaneously because of the transition from a highly
ordered state of a crystal to a less ordered state of a liquid.
Spontaneous Processes
 A spontaneous process once started proceeds without
any external influence.
 An example would be a match once lit continues to
burn even out any outside influence.
 A low activation energy favors a spontaneous process.
Activation
Reactant            Energy

Energy
Released                Product
 The reverse of a spontaneous reaction is always non-
spontaneous.
 Some spontaneous reactions occur extremely slowly.
Problems
16.1 Which of the following processes are spontaneous? Which are
non-spontaneous?
(a) Diffusion of perfume molecules from one side of the
room to the other.
(b) Heat flow from a cold object to a hot object.
(c) Decomposition of rust (Fe2O3.H2O) to iron metal,
oxygen, and water.
(d) Decomposition of solid CaCO3 to solid CaO and
gaseous CO2 at 25oC and 1 atm pressure (Kp = 1.4 x 10-23)
16.2 Predict the sign of ∆S in the system for each of the following
processes:
(a) H2O(g)  H2O(l)
(b) I2 (g)  2I(g)
(c) CaCO3(s)  CaO(s) + CO2(g)
(d) Ag+(aq) + Br-(aq)  AgBr(s)
Entropy and Probability
 The tendency of a process to increase in entropy is due to the
fact that a more disordered state is more probable and
consequently more achievable.
 Austrian physicist Ludwig Boltzmann quantified the relationship
between probability and entropy with the equation:
S = k ln W
where S is the entropy of the system, W is the number of possible
ways a system can be arranged, and k is the Boltzmann constant
equal to the gas constant divided by Avogadro’s number (R/NA).
 For gases expanding to a larger volume at constant temperature
and given the initial and final volume ∆S is equal to:
Vfinal
∆S = nR ln ----------
Vinitial
Problem
16.4 Which state has the higher entropy? Explain
in terms of probability.
(a) A perfectly ordered crystal of solid nitrous
oxide or a disordered crystal in which the
molecules are oriented randomly.
(b) Quartz glass or quartz crystal.
(c) 1 mol of N2 gas at STP or 1 mol of N2 gas at
273 K in a volume of 11.2 L.
(d) 1 mol of N2 gas at STP or 1 mol of N2 gas at
273 K and 0.25 atm.
Entropy and Temperature
 The higher the temperature the greater the kinetic energy
of particles, the higher the degree of randomness, the
higher the entropy.
 The 3rd law of thermodynamics states that the entropy of
a perfectly ordered crystalline solid at absolute 0 (0oK) is
zero.
 As temperature of a solid rises, entropy steadily increases
as the vibrational energy can be distributed in many more
ways.
 At the melting point, entropy jumps vertically since the
randomness in liquids are much higher than solids.
 The same changes occur going from liquid to gaseous
state.
Standard Molar Entropies and Standard
Entropies of Reaction
 Standard Molar Entropy is defined as:
The entropy of 1 mole of pure substance at a pressure
of 1 atm and at a specified temperature usually 25oC.
 The standard molar entropy is used as basis for the
calculation of the standard entropy of reactions.
Problem
16.5 Calculate the standard entropy at 25oC for the
decomposition of calcium carbonate:
CaCO3(s)  CaO(s) + CO2(g)
Entropy and the 2nd Law of Thermodynamics
 The 2nd law of thermodynamics states that in any spontaneous
process the total entropy of the system and its surrounding always
increases.
 Though the 1st law of thermodynamics tells us which direction of
a chemical reaction is favored energy wise, it doesn’t tell us
whether the reaction is spontaneous. The 2nd law does. It states
that the total entropy change is equal to the change in entropy of
the system and the change of entropy of the surroundings
∆Stotal = ∆Ssystem + ∆Ssurroundings
 For a spontaneous process, ∆Stotal is +, for a non-spontaneous it’s
-, and for a process in equilibrium it’s equal to 0.
 The change in entropy of the surroundings is proportional to      -
∆H but is inversely proportional to the absolute temperature.
∆H
∆Ssurroundings = - -----
T
Problem
16.6 By determining the sign of ∆Stotal, show
whether the decomposition of CaCO3 is
spontaneous under the standard conditions at
25oC.
CaCO3(s)  CaO(s) + CO2(g)
Gibb’s Free Energy
 To calculate for the thermodynamic properties of a system, the
∆Ssurroundings have to be determined which can be complicated.
 A way around this is by using the Gibb’s free energy (G): G = H
- TS. Since H and S are state functions, G IS also a state function.
So the change in free energy is: ∆G = ∆H - T ∆S
If ∆G < 0, the reaction is spontaneous.
If ∆G > 0, the reaction is non-spontaneous.
If ∆G = 0, the reaction is in equilibrium.
∆H   ∆S    ∆G                  Results                        Examples
-    +      -   Spontaneous at all temperatures.        2NO2(g)  N2(g) + 2O2(g)
-    -    - or + Spontaneous at low temperatures,       N2(g) + 3H2(g)  2NH3(g)
Non-spontaneous at high temperatures
+    -      +   Non-spontaneous at all temperatures.        3O2(g)  2O3(g)
+    +    - or + Non-spontaneous at low temperatures,   2HgO(s)  2Hg(l) + O2(g)
Spontaneous at high temperatures
Problem
16.7 Consider the decomposition of N2O4:
N2O4(g)  2NO2(g) ∆Ho = 55.3kJ, ∆So = +175.7 J/K
(a) Is the reaction spontaneous under the standard
conditions at 25oC?
(b) Estimate the temperature at which the reaction
becomes spontaneous.
16.9 What are the signs (+, -, or 0) of ∆H, ∆S, ∆G for
the following spontaneous reaction of A atoms (red)
and B atoms (black)?
Standard Free-Energy Changes for Reactions
 Just like the other thermodynamic quantities, free
energy is temperature, state and concentration
dependent.
 The standard conditions are for:
1. Solids, liquids, and gases in pure form and at 1 atm.
2. Solutions at 1M concentrations
3. Temperature at 25oC.
 Just like enthalpy, the value of ∆G depends on the
stoichiometry of the reaction.
 In the calculation of the standard free energy change,
the standard enthalpy and standard entropy must be
used.
∆Go = ∆Ho - T ∆So
Problem
16.10 Consider the thermal decomposition of CaCO3:
CaCO3(s)  CaO(s) + CO2(g)
(a) Use the data in Appendix B to calculate the standard free-energy change
for this reaction at 25oC?
(b) Will a mixture of solid CaCO3, solid CaO, and gaseous CO2 at 1 atm
pressure react spontaneously at 25oC to produce more CaO and CO2?
(c) Assuming that ∆Ho and ∆So are independent of temperature, estimate the
temperature at which the reaction becomes spontaneous.

16.11 Consider the following endothermic reaction of AB2 molecules:

(a) What is the sign (+, -, or 0) of ∆So for the reaction?
(b) Is the reaction more likely to be spontaneous at high or low temperatures?
Standard Free Energies of Formation
 The standard free energies of formation is the change in free
energy when 1 mole of substance is formed from its elements in
their standard states.
 Calculations with ∆Gof is similar to ∆Hof . The most stable form
of the element has a ∆Gof = 0.
 Substances with a ∆Gof < 0 or (-) are stable and have little
tendency to decompose. Conversely substances with a ∆Gof > 0 or
(+) are unstable and may spontaneously decompose.
 The ∆Go of substances can be calculated from given ∆Gof using
the equation:
∆Go = ∆Gof products - ∆Gof reactants
 A high (+) ∆Go indicates that the substance can not be easily
synthesized at standard temperatures and pressures.
Problem
16.12 (a) Using values of ∆Gof in Appendix B,
calculate the standard free energy change for the
reaction of calcium carbide (CaC2) with water. Might
this reaction be used for the synthesis of acetylene
(HC≡CH, or C2H2)?
CaC2(s) + 2H2O(l)  C2H2(g) + Ca(OH)2(s)

(b) Is it possible to synthesize acetylene from solid
graphite and gaseous H2 at 25oC and 1 atm pressure?
Free-Energy Changes and Composition of
the Reaction Mixture
 The free energy at non-standard conditions depends on the
composition of the reaction mixture.
 The following equation is used for the calculation of ∆G at non-
standard pressures and concentrations:
∆G = ∆Go + RT ln Q
where ∆G is the free energy at non-standard conditions, ∆Go is
the standard free energy, R is the gas constant, T is the
temperature in oK, and Q is the concentration quotient QC or QP.
 For the reaction: A(g) + B(g) ↔ C(g) + D(g)
[C][D]             PC PD
QC = ---------     QP = ---------
[A][B]             PA PB
Problems
16.13 Calculate ∆G for the formation of ethylene (C2H4)
from carbon and hydrogen at 25oC when the partial
pressures are 100 atm H2 and 0.10 atm C2H4.

2C(s) + 2H2(g)  C2H4(g)     ∆Go = 68.1 kJ

Is the reaction spontaneous in the forward or reverse
direction?
Free-Energy and Chemical Equilibrium
 The value of the expression RT ln Q determines whether a
reaction is spontaneous or not.
 A large Q (much more products than reactants) means that RT
ln Q is a very large (+) value and no matter what the value of
∆Go, ∆G will be (+) so the forward reaction will be non-
spontaneous.
 At equilibrium Q = K and ∆G = 0 so the free energy chemical
equilibrium equation becomes:
0 = ∆Go + RT ln K or ∆Go = -RT ln K
where K can be Kc or Kp
∆Go         ln K        K
∆Go < 0     ln K > 0   K>1           Mainly products
∆Go > 0     ln K < 0   K<1           Mainly reactants
∆Go = 0     ln K = 0   K=1      Products and reactants have
comparable amounts
Problems
16.15 Given the data in Appendix B, calculate Kp at
25oC for the reaction:
CaCO3(s) ↔ CaO(s) + CO2(g)

16.16 Use data in Appendix B to calculate the vapor
pressure of water at 25oC.

16.17 At 25oC, Kw for the dissociation of water is
1.0 x 10-14. Calculate ∆Go for the reaction:
2H2O(l) ↔ H3O+(aq) + OH-(aq).

```
To top