1. THERMODYNAMICS – ENTROPY AND FREE ENERGY
Thermodynamics studies the energy of a system, how much work a system could produce, and how to predict the spontaneous changes a system will undergo
INTERNAL ENERGY (E) – The total energy of a system
Energy is conserved during all processes
3A-1 (of 22)

2. HERMANN VON HELMHOLTZ
Presented a mathematical argument for the Law of Conservation of Energy
FIRST LAW OF THERMODYNAMICS – The change in energy of a system is equal to heat that enters the system plus the work done on the system
ΔE = q + w
STATE FUNCTION – A property of a system that depends only on the current state of the system, not on the way in which the system acquired that state
Kinetics
Internal Energy is a state function, so ΔE is independent of the pathway of the change
Thermodynamics
3A-2 (of 22)

3. ENTHALPY (H) – The internal energy of a system plus the pressure-volume product of a system
H = E + pV
ΔH = ΔE + Δ(pV)
ΔH = ΔE + (Δp)V + p(ΔV)
At constant pressure, Δp = 0:
ΔH = ΔE + p(ΔV)
From the First Law:
ΔE = q + w
At constant pressure, w = -p(ΔV):
ΔE = q + -p(ΔV)
ΔE + p(ΔV) = q
At constant pressure the enthalpy change of a system is equal to the heat that enters the system
ΔH = qp
3A-3 (of 22)

4. 4NH3 (g) + 7O2 (g) → 4NO2 (g) + 6H2O (g)
Calculate the standard enthalpy change for
4NH3 (g) + 7O2 (g) → 4NO2 (g) + 6H2O (g)
Handout 5 of the class website (data at 298 K)
Thermodynamic data is tabulated under specified conditions called the STANDARD STATE, identified with a “º”, such as ΔHº
STANDARD STATE – Gaseous matter is at 1 atm, dissolved matter is 1 M, and elemental matter is in its natural state at room temperature and pressure
3A-4 (of 22)

5. 4NH3 (g) + 7O2 (g) → 4NO2 (g) + 6H2O (g)
Calculate the standard enthalpy change for
4NH3 (g) + 7O2 (g) → 4NO2 (g) + 6H2O (g)
Handout 5 of the class website (data at 298 K)
ENTHALPY OF FORMATION (ΔHf) – The enthalpy change for a formation reaction
FORMATION REACTION – A reaction that forms one mole of a substance from the elements composing it
3A-5 (of 22)

6. 4NH3 (g) + 7O2 (g) → 4NO2 (g) + 6H2O (g)
Calculate the standard enthalpy change for
4NH3 (g) + 7O2 (g) → 4NO2 (g) + 6H2O (g)
½N2 (g) + 3/2H2 (g) → NH3 (g)
½N2 (g) + O2 (g) → NO2 (g)
O2 (g) → O2 (g)
H2 (g) + ½O2 (g) → H2O (g)
3A-6 (of 22)

7. 4NH3 (g) + 7O2 (g) → 4NO2 (g) + 6H2O (g)
Calculate the standard enthalpy change for
4NH3 (g) + 7O2 (g) → 4NO2 (g) + 6H2O (g)
ΔHºrxn =
Σ(ΔHºf products)
- Σ(ΔHºf reactants)
4 mol NO2 (34 kJ/mol NO2)
+ 6 mol H2O (-242 kJ/mol H2O)
- 4 mol NH3 (-46 kJ/mol NH3)
- 7 mol O2 (0 kJ/mol O2)
= kJ
3A-7 (of 22)

8. 4NH3 (g) + 7O2 (g) → 4NO2 (g) + 6H2O (g)
Calculate the standard enthalpy change for
4NH3 (g) + 7O2 (g) → 4NO2 (g) + 6H2O (g)
EXTENSIVE PROPERTY – One that depends on the amount of chemical change
-1132 kJ
____________
4 mol NH3
= kJ
____________
mol NH3
3A-8 (of 22)

9. Why does this work?
NH4NO3 → N2O H2O
ΔHº = ?
N H2 + ³∕2O2
→ NH4NO3
ΔHºf NH4NO3 = kJ/mol
N2 + ½O2
→ N2O
ΔHºf N2O = 82 kJ/mol
H2 + ½O2
→ H2O
ΔHºf H2O = kJ/mol
NH4NO3
→ N H2 + ³∕2O2
ΔHº = kJ
N2 + ½O2
→ N2O
ΔHº = 82 kJ
2H2 + O2
→ 2H2O
ΔHº = kJ
NH4NO3 → N2O H2O
ΔHº = kJ
3A-9 (of 22)

10. Calculate the standard enthalpy change for
N H → N2H4
From Chem 1A Handout 8:
N ≡ N :
H – H →
H H
H – N – N – H
. .
+ 1 mol (941 kJ/mol)
= 941 kJ
= 864 kJ
= -160 kJ
= kJ
+ 2 mol (432 kJ/mol)
- 1 mol (160 kJ/mol)
- 4 mol (391 kJ/mol)
81 kJ
3A-10 (of 22)

11. RUDOLF CLAUSIUS
Proposed that there is always some internal energy that cannot be converted into work
He called this ENTROPY
( ΔT )
3A-11 (of 22)

12. A large amount of the internal energy of the blocks can be used for work
 a low amount cannot be used for work
 low entropy
A small amount of internal energy of the blocks can be used for work
 a high amount cannot be used for work
 high entropy
What is the spontaneous process?
3A-12 (of 22)

13. SECOND LAW OF THERMODYNAMICS – In any spontaneous process there is always an increase in total entropy
3A-13 (of 22)

14. LUDWIG BOLTZMANN
Proposed a statistical meaning for entropy
ENTROPY (S) – A measure of the number of arrangements available to a system in a given state
Boltzmann calculated the entropy of a system by
S = k log W
S = entropy
k = Boltzmann Constant (R/NA, or 1.38 x J/moleculeK)
W = number of ways particles can be arranged in a given state while keeping the total energy constant
3A-14 (of 22)

15. Boltzmann calculated the entropy of a system by
S = k log W
S = entropy
k = Boltzmann Constant (R/NA, or 1.38 x J/moleculeK)
W = number of ways particles can be arranged in a given state while keeping the total energy constant
3A-15 (of 22)

16. Calculate the entropies of the following samples of CO2 (g) and CO2 (s)
W = 20 x 19 = 380
W = 2 x 1 = 2
(1.38 x J/K) log 380
(1.38 x J/K) log 2
= 3.56 x J/K
= 4.15 x J/K
For a particular type of matter, the gaseous state has a higher entropy than the solid state
3A-16 (of 22)

17. FACTORS IN PREDICTING S FOR CHEMICAL SUBSTANCES
1) State of Matter
Sgas > Sliquid > Ssolid
More possible arrangements, higher entropy
2) Mass
Slarge mass > Slow mass
Ei = h2i2 i = 1,2,… 
______
8mX2   
A larger mass means closer spacing of energy levels, so more possible arrangements
3A-17 (of 22)

18. FACTORS IN PREDICTING S FOR CHEMICAL SUBSTANCES
3) molecular complexity
Sless elements < Smore elements
Molecules with different elements allows for more possible arrangements
Fa Fb Ha Hb
Fa – Fb and Ha – Hb
Ha – Fa and Hb – Fb or
Ha – Fb and Hb – Fa
Explain each difference in absolute entropy:
Cl2 (g) > Br2 (l)
Sgas is higher than Sliquid
Cl2 (g) > F2 (g)
S is higher for larger masses
C3H8 (g) > C2H6 (g)
S is higher for larger masses
CO (g) > N2 (g)
S is higher for more internal complexity
3A-18 (of 22)

19. THIRD LAW OF THERMODYNAMICS – The entropy of a perfect crystal at 0 K is 0
(1.38 x J/K) log 1
(1.38 x J/K) log 16
= 0 J/K
= 1.66 x J/K
Because of the 3rd Law, all substances > 0 K have entropy values > 0
3A-19 (of 22)

20. CALCULATING ΔS FOR CHEMICAL CHANGES
Third Law Entropies of 1 mole of many substances in their standard states at 298 K (298Sº or just Sº) are found on Handout 5
Be (s) + O2 (g) + H2 (g) → Be(OH)2 (s)
3A-20 (of 22)

21. CALCULATING ΔS FOR CHEMICAL CHANGES
Third Law Entropies of 1 mole of many substances in their standard states at 298 K (298Sº or just Sº) are found on Handout 5
Sº values are not ΔSºf values,
Sº values are the absolute entropy (calculated from the Boltzmann equation) of 1 mole of each substance
All Sº values at temperatures > 0 K are positive
For a chemical reaction, a 298ΔSº can be calculated from Handout 5
298ΔSºreaction = Σ298Sºproducts - Σ298Sºreactants
Be (s) + O2 (g) + H2 (g) → Be(OH)2 (s)
NO!!!
3A-21 (of 22)

22. 2NiS (s) + 3O2 (g) → 2SO2 (g) + 2NiO (s)
Calculate 298ΔSº, the standard entropy change at 298 K, for
2NiS (s) + 3O2 (g) → 2SO2 (g) + 2NiO (s)
2 mol SO2 (248 J/molK SO2)
+ 2 mol NiO (38 J/molK NiO)
- 2 mol NiS (53 J/molK NiS)
- 3 mol O2 (205 J/molK O2)
= J/K
ΔS predicts whether a system is becoming more ordered or disordered
This says the system is becoming more ordered
3A-22 (of 22)

23. PREDICTING ΔS FOR CHEMICAL CHANGES
ΔS is + for reactions that have
1) more gaseous products than reactants
2) solid reactants dissolving
CaCO3 (s) → CaO (s) + CO2 (g)
Ag+ (aq) + Cl- (aq) → AgCl (s)
0 gas molecules → 1 gas molecule
dissolved ions precipitate
ΔS = +
ΔS = –
2SO2 (g) + O2 (g) → 2SO3 (g)
H+ (aq) + HCO3- (aq) → H2O (l) + CO2 (g)
3 gas molecules → 2 gas molecules
0 gas molecules → 1 gas molecule
ΔS = –
ΔS = +
3A-22 (of 22)

25. SPONTANEITY
SPONTANEOUS PROCESS – One that proceeds without outside intervention
SECOND LAW OF THERMODYNAMICS – In any spontaneous process there is always an increase in total entropy
For a spontaneous process:
ΔSuniv > 0
ΔSsys + ΔSsurr = ΔSuniv
HO 5
-ΔHsys
_________ T
3B-1 (of 19)

26. JOSIAH WILLARD GIBBS
Proposed a way to determine the maximum amount of internal energy that could be converted into work
GIBBS FREE ENERGY (G) – The maximum amount of internal energy of a system that can be converted into work
3B-2 (of 19)

27. G = H - TS
ΔGsys = ΔHsys - TΔSsys
-ΔGsys = -ΔHsys + ΔSsys
________ ________
T T
ΔSsurr
-ΔGsys = ΔSsurr + ΔSsys
________ T
ΔSuniv
3B-3 (of 19)

28. -ΔGsys = ΔSuniv
________ T
ΔSuniv > 0 for a spontaneous process 
ΔGsys < 0 for a spontaneous process
The change in Gibbs Free Energy (ΔG) indicates if a process is spontaneous or nonspontaneous
3B-4 (of 19)

29. ΔG = ΔH - TΔS
Spontaneous processes are favored by
a decrease in enthalpy (processes that are exothermic – ΔH negative)
an increase in entropy (processes that increase disorder – ΔS positive)
3B-5 (of 19)

30. CONTRIBUTIONS OF ΔH AND ΔS TO SPONTANEITY
ΔG = ΔH - TΔS
= (-) (+)
= (+) (+)
= (+) (-)
= (-) (-)
= (+)
= (?)
= (?)
= (-) ΔH - +
exothermic
endothermic ΔS + -
more disorder
more order ΔG - + ?
spontaneous
nonspontaneous
spontaneous at high T
spontaneous at low T
3B-6 (of 19)

31. Calculate the ΔGº at 298 K for the reaction
NH4Cl (s) ⇆ NH3 (g) + HCl (g)
298ΔGº
1 mol NH3 (-17 kJ/mol NH3)
+ 1 mol HCl (-95 kJ/mol HCl)
- 1 mol NH4Cl (-203 kJ/mol NH4Cl)
= 91 kJ
When starting with solid NH4Cl, 1 atm NH3, and 1 atm HCl, the reaction is nonspontaneous
 reverse reaction is spontaneous
3B-7 (of 19)

32. Calculate the ΔGº at 298 K for the reaction
NH4Cl (s) ⇆ NH3 (g) + HCl (g)
298 ΔHº
1 mol NH3 (-46 kJ/mol NH3)
+ 1 mol HCl (-92 kJ/mol HCl)
- 1 mol NH4Cl (-314 kJ/mol NH4Cl)
= kJ
298 ΔSº
1 mol NH3 (193 J/molK NH3)
+ 1 mol HCl (187 J/molK HCl)
- 1 mol NH4Cl (96 J/molK NH4Cl)
= J/K
298 ΔGº = 298 ΔHº - T 298 ΔSº
= 176 kJ - (298 K)(0.284 kJ/K)
= 91 kJ
3B-8 (of 19)

33. Calculate the ΔGº at 298 K for the reaction
NH4Cl (s) ⇆ NH3 (g) + HCl (g)
298 ΔHº
= kJ
Works against spontaneity
, and predominates
298 ΔSº
= J/K
Works for spontaneity
298 ΔGº = 298 ΔHº - T 298 ΔSº
= 176 kJ - (298 K)(0.284 kJ/K)
= 91 kJ
3B-9 (of 19)

34. Calculate the ΔHº, ΔGº, and ΔSº at 298 K for the reaction
2C2H6 (g) + O2 (g) ⇆ 2C2H5OH (l)
298 ΔHº
2 mol C2H5OH (-278 kJ/mol C2H5OH)
- 2 mol C2H6 (-84.7 kJ/mol C2H6)
- 1 mol O2 (0 kJ/mol O2)
= kJ
298 ΔGº
2 mol C2H5OH (-175 kJ/mol C2H5OH)
- 2 mol C2H6 (-32.9 kJ/mol C2H6)
- 1 mol O2 (0 kJ/mol O2)
= kJ
3B-10 (of 19)

35. Calculate the ΔHº, ΔGº, and ΔSº at 298 K for the reaction
2C2H6 (g) + O2 (g) ⇆ 2C2H5OH (l)
298 ΔSº
2 mol C2H5OH (161 J/molK C2H5OH)
- 2 mol C2H6 (229.5 J/molK C2H6)
- 1 mol O2 (0 J/molK O2)
= kJ
ΔGº = ΔHº - TΔSº
ΔGº - ΔHº = ΔSº
____________ -T
= kJ – ( kJ)
____________________________
-298 K
= kJ/K
3B-11 (of 19)

36. Calculate the ΔHº, ΔGº, and ΔSº at 298 K for the reaction
2C2H6 (g) + O2 (g) ⇆ 2C2H5OH (l)
ΔH and ΔS are temperature independent, so Handout 5 can be used to calculate a ΔHº and ΔSº at any temperature
However, ΔG depends on temperature, so the ΔGºf’s on Handout 5 can ONLY be used to calculate ΔGº at 298 K
 To calculate ΔGº at any temperature other than 298 K,
use ΔHº - TΔSº
3B-12 (of 19)

37. Calculate ΔHº, ΔSº, and ΔGº at 308 K for
2SO2 (g) + O2 (g) → 2SO3 (g)
298ΔHº = 308ΔHº
(constant at all temperatures)
2 mol SO3 (-396 kJ/mol SO3)
- 2 mol SO2 (-297 kJ/mol SO2)
- 1 mol O2 (0 kJ/mol O2)
= kJ
∴ reaction is exothermic
3B-13 (of 19)

38. Calculate ΔHº, ΔSº, and ΔGº at 308 K for
2SO2 (g) + O2 (g) → 2SO3 (g)
298ΔSº = 308ΔSº
(constant at all temperatures)
2 mol SO3 (257 J/molK SO3)
- 2 mol SO2 (248 J/molK SO2)
- 1 mol O2 (205 J/molK O2)
= J/K
∴ reaction becomes more ordered
3B-14 (of 19)

39. Calculate ΔHº, ΔSº, and ΔGº at 308 K for
2SO2 (g) + O2 (g) → 2SO3 (g)
298ΔGº ≠ 308ΔGº
(not constant at all temperatures)
308 ΔGº = ΔHº - TΔSº
= kJ - (308 K)( kJ/K)
= kJ
∴ when all reactants and products are present and start in their standard states, the forward reaction is spontaneous at 308 K
3B-15 (of 19)

40. For the reaction: H2O (s) ⇆ H2O (l)
Find ΔGº at 20.0ºC
Calculated from Handout 5:
298 ΔHºfus = 6,038 J
298 ΔSºfus = J/K
293.2ΔGº = ΔHº - T ΔSº
= 6,038 J - (293.2 K)(22.1 J/K)
= J
reaction is spontaneous
 forward reaction is spontaneous
H2O (s) melts at 20.0ºC
3B-16 (of 19)

41. For the reaction: H2O (s) ⇆ H2O (l)
Find ΔGº at -20.0ºC
Calculated from Handout 5:
298 ΔHºfus = 6,038 J
298 ΔSºfus = J/K
253.2ΔGº = ΔHº - T ΔSº
= 6,038 J - (253.2 K)(22.1 J/K)
= 442 J
reaction is nonspontaneous
 reverse reaction is spontaneous
H2O (l) freezes at -20.0ºC
Notice ΔGº is different at the two different temperatures: -442 J and +442 J
3B-17 (of 19)

42. For the reaction: H2O (s) ⇆ H2O (l)
Find ΔGº at 0.0ºC
Calculated from Handout 5:
298 ΔHºfus = 6,038 J
298 ΔSºfus = J/K
273.2ΔGº = ΔHº - T ΔSº
= 6,038 J - (273.2 K)(22.1 J/K)
= 0 J
The temperature of a phase change (like melting) is when the two phases are in equilibrium
For any process at equilibrium, ΔG = 0
(if it is a standard state reaction like this, then ΔGº = 0)
3B-18 (of 19)

43. For the reaction: H2O (l) ⇆ H2O (g)
Find the normal boiling point of water
A phase change temperature is an equilibrium between 2 phases, so ΔG = 0
“Normal” means vapor pressure is 1 atm when boiling ∴ standard state, so ΔGº = 0
Calculated from Handout 5:
298 ΔHºvap = 44,400 J
298 ΔSºvap = 119 J/K
TΔGº = ΔHº - TΔSº
0 = ΔHº - TeqΔSº
-ΔHº = - TeqΔSº
ΔHº = Teq
_____
ΔSº
= 44,400 J
___________
119 J/K
= K
= ºC
3B-19 (of 19)

45. THERMODYNAMIC FUNCTIONS
(1) Enthalpy (H) – The total energy plus pV product of a system
Change in Enthalpy (ΔH) – Indicates if a process is exothermic or endothermic at constant pressure
(2) Entropy (S) – The energy of a system that cannot be converted to work, related to the number of arrangements available to a system in a given state
Change in Entropy (ΔSsys) – Indicates if a process is becoming more ordered or more disordered
Change in Entropy (ΔSuniv) – Indicates if a process is spontaneous or nonspontaneous
(3) Gibbs Free Energy (G) – The maximum energy of a system that can be converted into work
Change in Gibbs Free Energy (ΔG) – Indicates if a process is spontaneous or nonspontaneous
3C-1 (of 13)

46. PREDICTING ΔG FOR NON-STANDARD STATE CHEMICAL REACTIONS
ΔGº predicts spontaneity when reactants and products are all present in the container and in their standard state (1 M or 1 atm)
ΔG predicts spontaneity when reactants and products are not in their standard state
ΔG = ΔGº + RT ln Q
3C-2 (of 13)

47. The standard free energy change at 298 K is -514 kJ for the reaction
2CO (g) + O2 (g) → 2CO2 (g)
If a reaction vessel contains atm CO, atm O2, and 3.0 atm CO2, calculate the free energy change for the reaction at 298 K
ΔG = ΔGº + RT ln Q
298ΔG = 298ΔGº + RTln Q
Q = pCO22
__________
pCO2pO2
= (3.0)2
_________________
(0.010)2(0.020)
= x 106
298ΔG =
-514,000 J
+ (8.314 J/K)(298 K) ln 4.5 x 106
= ,000 J
3C-3 (of 13)

48. HIO2 (aq) + H2O (l) ⇆ H3O+ (aq) + IO2- (aq)
(a) Calculate the standard free energy change at 298 K for the reaction
HIO2 (aq) + H2O (l) ⇆ H3O+ (aq) + IO2- (aq)
298ΔGº = 1 mol H3O+ (-237 kJ/mol H3O+) + 1 mol IO2- (-23 kJ/mol IO2-)
- 1 mol HIO2 (-55 kJ/mol HIO2) mol H2O (-237 kJ/mol H2O)
= 32 kJ
This means that the reverse reaction is spontaneous in a solution that is 1 M HIO2, 1 M H3O+, and 1 M IO2-
3C-4 (of 13)

49. HIO2 (aq) + H2O (l) ⇆ H3O+ (aq) + IO2- (aq)
(b) If a solution is M HIO2, 1.00 x 10-7 M H3O+, and M IO2-, calculate the free energy change at 298 K for the reaction
HIO2 (aq) + H2O (l) ⇆ H3O+ (aq) + IO2- (aq)
ΔG = ΔGº + RT ln Q
298ΔG = 298ΔGº + RTln Q
Q = [H3O+][IO2-]
_______________
[HIO2]
= (1.00 x 10-7)(0.250)
_______________________
(0.500)
= x 10-8
298ΔG =
32,000 J
+ (8.314 J/K)(298 K) ln 5.00 x 10-8
= -10,000 J
This means that the forward reaction is spontaneous in a solution that is M HIO2, 1.00 x 10-7 M H3O+, and M IO2-
3C-5 (of 13)

50. THE MOST IMPORTANT REASON FOR KNOWING ΔGº
For a reaction at equilibrium
ΔG = 0
ΔG = ΔGº + RT ln Q
0 = ΔGº + RT ln Q
0 = ΔGº + RT ln Keq
-RT ln Keq = ΔGº
ΔGº = -RT ln Keq
Knowing ΔGº is valuable because it is related to a reaction’s equilibrium constant
3C-6 (of 13)

51. 4NO (g) ⇆ 2N2O (g) + O2 (g)
The 298ΔGº for the above reaction is kJ. Calculate Keq at 298 K.
ΔGº = -RT ln Keq
ΔGº = ln Keq
_____
-RT
e-ΔGº/RT = Keq
Keq = e-(-28,200 J)/(8.314 J/K)(298 K)
= x 104
3C-7 (of 13)

52. HF (aq) + H2O (l) ⇆ H3O+ (aq) + F- (aq)
The Keq (Ka) for the above reaction is 7.2 x 10-4 at 298 K. Calculate 298ΔGº.
ΔGº = -RT ln Keq
298ΔGº = -(8.314 J/K)(298 K) ln (7.2 x 10-4)
298ΔGº = 18,000 J
3C-8 (of 13)

53. RELATIONSHIP BETWEEN ΔGº AND Keq
The sign of ΔGº indicates the magnitude of the Keq
Keq
>1
<1 1
ΔGº = -RT ln (>1)
ΔGº = -RT ln (<1)
ΔGº = -RT ln (1)
ln (>1) = +
ln (<1) = –
ln (1) = 0
ΔGº – +
∴ if ΔGº is – it means Keq > 1
∴ if ΔGº is + it means Keq < 1
∴ if ΔGº is 0 it means Keq = 1
Why?
ΔGº means all chemicals are 1 atm or 1M
∴ Q = 1
If Q = 1 and Keq > 1
, then the forward reaction must be spontaneous, so ΔGº = –
3C-9 (of 13)

54. Find 298ΔGº and Keq at 298 K for the reaction
N2O4 (g) ⇆ 2NO2 (g)
298ΔGº = 2 mol NO2 (52 kJ/mol NO2) mol N2O4 (98 kJ/mol N2O4) = 6 kJ
e-ΔGº/RT = Keq
Keq = e-(6,000 J)/(8.314 J/K)(298 K) =
3C-10 (of 13)

55. HNO2 (aq) + H2O (l) ⇆ H3O+ (aq) + NO2- (aq)
Find ΔGº and Ka at 298 K for the reaction
HNO2 (aq) + H2O (l) ⇆ H3O+ (aq) + NO2- (aq)
298ΔGº =
1 mol H3O+ (-237 kJ/mol H3O+)
+ 1 mol NO2- (-35 kJ/mol NO2-)
- 1 mol HNO2 (-54 kJ/mol HNO2)
- 1 mol H2O (-237 kJ/mol H2O)
= 19 kJ
e-298ΔGº/RT = Keq
= e-(19,000 J)/(8.314 J/K)(298 K)
= x 10-4
3C-11 (of 13)

56. HNO2 (aq) + H2O (l) ⇆ H3O+ (aq) + NO2- (aq)
Find ΔGº and Ka at 398 K for the reaction
HNO2 (aq) + H2O (l) ⇆ H3O+ (aq) + NO2- (aq)
298ΔHº =
1 mol H3O+ (-286 kJ/mol H3O+)
+ 1 mol NO2- (-174 kJ/mol NO2-)
- 1 mol HNO2 (-207 kJ/mol HNO2)
- 1 mol H2O (-286 kJ/mol H2O)
= 33 kJ
298ΔSº =
1 mol H3O+ (70 J/molK H3O+)
+ 1 mol NO2- (156 J/molK NO2-)
- 1 mol HNO2 (146 J/molK HNO2)
- 1 mol H2O (70 J/molK H2O)
= 10. J/K
398ΔGº = 33 kJ - (398 K)(0.010 kJ/K)
= 29 kJ
3C-12 (of 13)

57. HNO2 (aq) + H2O (l) ⇆ H3O+ (aq) + NO2- (aq)
Find ΔGº and Ka at 398 K for the reaction
HNO2 (aq) + H2O (l) ⇆ H3O+ (aq) + NO2- (aq)
Keq = e-ΔGº/RT
= e-398ΔGº/R(398 K)
= e-(29,000 J)/(8.314 J/K)(398 K)
= 1.6 x 10-4
3C-13 (of 13)