Download Ch 20 Thermodynamics

Chem 2
Ch 20: Thermodynamics:
First law of thermodynamics: Law of conservation of energy: Energy can be neither
created nor destroyed.
∆E= q + w (q=heat, w=work, E=internal energy)
E univ= E sys + E surr
Heat gained by system is lost by surroundings and vice-versa.
Total energy of Universe is constant ∆E sys + ∆E surr=0=∆E univ
Spontaneous: It occurs without outside intervention.
Enthalpy=H=E+PV,E=internal energy, P=pressure, V=volume.
Entropy:
 S, A measure of molecular randomness or disorder.
 Thermodynamic function that describes number of arrangements that are
available to a system existing in a given state.
 Probability of occurrence of a particular arrangement(state) depends on the
number of ways(microstates) in which it can be arranged.
 S=k ln W, k= Boltzmann constant=R/Av number=1.38 x 10-23J/K.
 Un its of S are J/K
 More the microstates available, higher is the entropy. Ex: solid →liquid→gas
 Entropy is a state function depends only on present state and not its path.
 ∆ Ssys= Sfinal – Sinitial , ∆ Ssys > 0, as entropy increases,
 ∆ Ssys < 0, as entropy decreases.
 If there are n number of molecules, relative probability of finding all molecules
in one place is: (1/2)n
 ∆ Ssys= 5.76 J/K for 1 mol
 ∆ Ssys=qrev/T , T=temp, q=heat absorbed, rev=reversible process.
Entropy and Second Law of Thermodynamics:
Second Law of Thermo: In any spontaneous process there is always an increase in the
entropy of the universe. OR The entropy of the universe is increasing.
∆Suniv= ∆Ssys + ∆Ssurr
If ∆Suniv= +ve, process is spontaneous in direction written.
∆Suniv= -ve, process is spontaneous in the opposite direction.
∆Suniv= 0, process has no tendency to occur, system is at equilibrium.
Third Law of Thermodynamics: the entropy of a perfect crystal at 0 K is zero.
Standard molar Entropy: S0 has units J/mol.K
Standard states: 1 atm for gases, 1M for solutions, pure substances-solid, liquids.
1) S0 increases with temperature( as a substance is heated)
2) S0 increases as a substance changes from solid →liquid→gas, ∆ S0 vap>> ∆ S0 fus
3) Entropy of dissolved solid/liquid > entropy of pure solute, nature of solute and solvent
affect overall entropy change.
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4) Entropy of dissolved gas< Entropy of gas.
5) Entropy increases down a group.
6) Entropy increases with complexity.
Answer the following:
1. Will solid or gaseous CO2 have higher positional entropy?
2. Will N2 gas at 1 atm or 1.0 x 10-2 atm have higher positional entropy?
3. When solid sugar is dissolved in water to form a solution, what will be the sign of
entropy change?
4. When iodine vapors are condensed to form crystals, what will be the sign of
entropy change?
Standard entropy of the reaction: ∆S 0 rxn
 If number of moles increases, ∆S 0 rxn is +ve.
 If number of moles decreases, ∆S 0 rxn is -ve.
P.1. Write the reaction for combustion of 1 mol of propane and calculate the ∆S 0 rxn
Use S0 values given in the appendix B in your textbook.
Entropy changes in the surrounding:
 Exothermic change: Heat is lost by the system and gained by the surroundings.
q sys< 0 , q surr >0, ∆ Ssurr > 0
 Endothermic Change: Heat gained by the system is lost by the surroundings.
q sys> 0 , q surr <0, ∆ Ssurr < 0
 The change is entropy of the surroundings is α opposite charge in the heat of a
system 1/α temperature at which heat is transferred.
 ∆Ssurr = -q sys/T
 ∆Ssurr = -∆H sys/T
 Sign of ∆Ssurr depends on the direction of heat flow. When an exothermic
process occurs ∆Ssurr is + ve. Opposite is true for endothermic process. Nature
tends to seek the lowest possible energy.
 Magnitude of ∆Ssurr depends on the temperature. Transfer of heat at low
temperature produces a greater percent change in the randomness of the
surroundings.
P.2. Write an equation for the formation of ammonia.
∆S 0 sys=-197J/K. Calculate ∆S univ and state if the reaction occurs spontaneously at this
temperature.
P.3. Sb2 S3(s) + 3Fe(s)______ 2Sb (s) + 3FeS (s) ∆H=-125kJ
Sb4 O6(s) + 6C(s)______ 4Sb (s) + 6 CO (s) ∆H= 778 kJ
Calculate ∆Ssurr for both of these reactions at 25 C and 1 atm.
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The Entropy change and Equilibrium state:
At equilibrium ∆S univ =0, ∆S sys = -∆S surr

Neither forward, nor reverse reactions are spontaneous at equilibrium.

Free Energy: G= H-TS, H= enthalpy, T=temp in K, S= entropy.
 Also called as Gibbs Free Energy.
 Free energy change (∆G) is a measure of the spontaneity of a process and of the
useful energy available from it.
 ∆ G= ∆H-T∆S(constant temp)
 ∆Suniv = -∆G/T
 Process is spontaneous in the direction in which the free energy decreases.
(constant T and P)
 ∆S univ =0, at equilibrium
 ∆S univ > 0, for a spontaneous process.
 ∆S univ < 0, for a nonspontaneous process.
 ∆G=0, at equilibrium
 ∆G > 0, for a spontaneous process.
 ∆G< 0, for a nonspontaneous process.
Standard free energy of formation: ∆Gf0 is defined as a change in free energy that
accompanies the formation of 1 mole of that substance from its constituent elements with
all reactants and products in their standard states.
 Standard free energy of formation of an element in its standard state is zero.
 An equation coefficient x ∆Gf0 by the number.
 Reversing a reaction changes ∆Gf0
P.4. For the equation 2CH3OH (g) + 3 O2(g) _____ 2 CO 2(g) + 4 H2O (g)
Calculate ∆G0
-163
0
-394
-229
∆Gf0 kJ/mol
P.5. For the reaction
2 SO2 (g) + O2(g) ______ 2SO3(g) carried out at 25 C and
1 atm. Calculate ∆H0,∆S0,∆G0 -297
0
-396
∆Hf0 kJ/mol
248
205
257
S0 J/K.mol
P.6. Use ∆Gf0 values to calculate free energy change at 25 C for the following:
2NO (g) + O2(g) → 2NO2 (g)
P.7. For the equation 2CH3OH (g) + 3 O2(g) _____ 2 CO 2(g) + 4 H2O (g)
Calculate ∆G0
-163
0
-394
-229
∆Gf0 kJ/mol
P.8. Is the reaction C2H4 (g) + H2O (l) _______ C2H5OH (l) spontaneous under standard
conditions?
∆G and work a system can do:
 ∆G < 0, for a spontaneous process., ∆G=-wmax, maximum useful work obtainable
from the system.
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



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∆G > 0, for a spontaneous process., minimum work that must be done to the
system to make the process take place.
In any real process work is performed irreversibly and some of the free energy is
lost to the surroundings as heat.
You can never obtain maximum possible work.
You need more than minimum energy to start the process for a nonspontaneous
type.
At equilibrium a reaction can no longer do any work.
Effect of Temperature on Reaction spontaneity:
1. Reaction is spontaneous at all temperatures: ∆H < 0, ∆S >0, -T ∆S= -ve, ∆G =-ve
2. Reaction is nonspontaneous at all temperatures: ∆H > 0, ∆S <0, -T ∆S= +ve,∆G =+ve
3. Reaction is spontaneous at higher temperatures: ∆H > 0, ∆S >0, -T ∆S favors
spontaneity at high temp , ∆G =-ve and reaction is spontaneous.
4. Reaction is spontaneous at lower temperatures: ∆H < 0, ∆S <0, ∆H favors spontaneity
at low temp , Reaction occurs if -T ∆S< ∆H
P.9. An important reaction in the production of sulfuric acid is the oxidation of SO2(g) to
SO3(g):
2SO2(g) + O2(g)
2SO3(g)
At 298K, DG0 = -141.6kJ; DH0 = -198.4kJ; and DS0 = -187.9J/K
(a) Use the data to decide if this reaction is spontaneous at 250C, and predict how DG0
will change with increasing T.
(b) Assuming DH0 and DS0 are constant with increasing T, is the reaction spontaneous
at 900.0C?
Free Energy and Equilibrium:
G = G + RT ln(Q)
Q = reaction quotient from the law of mass action.
Equilibrium point occurs at the lowest value of free energy available to the reaction
system.
G = RT ln(K)
P.10. 4Fe(s) + 3 O2 (g) ↔ 2Fe2O3(s)
0
0
-826
Hf
27
205
90
S0
Calculate the equilibrium constant at 25 C.
P.11. The oxidation of SO2, which we considered in Sample Problem 20.6
2SO2(g) + O2(g) →
2SO3(g)
is too slow at 298K to be useful in the manufacture of sulfuric acid. To overcome this
low rate, the process is conducted at an elevated temperature.
(a) Calculate K at 298K and at 973K. (∆G0298 = -141.6kJ/mol of reaction as written
using DH0 and DS0 values at 973K. ∆G0973 = -12.12kJ/mol of reaction as written.)
(b) In experiments to determine the effect of temperature on reaction spontaneity, two
sealed containers are filled with 0.500atm of SO2, 0.0100atm of O2, and 0.100atm of
SO3 and kept at 250C and at 700.0C. In which direction, if any, will the reaction proceed
to reach equilibrium at each temperature?
(c) Calculate ∆G for the system in part (b) at each temperature.
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