Chapter 9: Entropy, Free Energy, and Equilibrium (Part 1) Study Notes

CHAPTER 9: Entropy, Free Energy, and Equilibrium (Part 1)

Spontaneous vs. Nonspontaneous Processes

  • Definition of Spontaneous Process: A process that occurs without needing to be driven by an external force.
      - Examples of spontaneous processes:
        - Ice melting at room temperature (RT)
        - A ball rolling downhill
        - Iron rusting at RT
        - Combustion of propane to form CO2 and H2O.
  • Definition of Nonspontaneous Process: A process that cannot occur under the given conditions without external intervention.
      - Example: Water freezing at RT, a ball rolling uphill, the conversion of rust back to iron at RT, the formation of propane (C3H8).
  • Conclusion: Nonspontaneous processes can occur if coupled with spontaneous processes.
  • Catalysts:
      - Catalysts speed up the rate of a reaction but do not affect the spontaneity of the reaction.

Factors Determining Spontaneity

  • The common assumption is that exothermic reactions (those that release heat) are spontaneous and endothermic reactions (those that absorb heat) are not. However:
      - Examples of Endothermic Spontaneous Processes:
        - Ice melting above 0°C
        - Water evaporating
        - NaCl dissolving in water.
  • Conclusion: Enthalpy alone cannot determine spontaneity.

Entropy (S)

  • Definition: Entropy is a measure of the dispersion of energy within a system.
      - It measures how spread out energy is and is also often described in terms of disorder within a system.
  • Spontaneous Energy Dispersion: Energy tends to disperse from localized areas to being more spread out unless hindered.
      - As ice melts, the ordered arrangement (solid state) of water molecules becomes less organized (liquid state), leading to an increase in entropy (ΔS > 0).
Specific Examples of Entropy Change:
  • Melting of Ice:
      - Transition from solid (highly ordered) to liquid (less ordered) results in an increase in entropy.
      - ΔS > 0.
  • Evaporation of Water:
      - Water molecules transition from a somewhat disordered liquid state to a highly disordered vapor state, increasing entropy.
      - ΔS > 0.
  • Dissolution of NaCl in Water:
      - NaCl transitions from a highly ordered solid state to a more disordered solution in water.
      - ΔS > 0.
Entropy Changes During Phase Changes:
  • Transition from Solid to Liquid to Gas: In these transitions, entropy generally increases.
      - Example: I_2(g)
    ightarrow I_2(s) results in a decrease in entropy (ΔS < 0).   - Example: 2N_2O(g) ightarrow 2N_2(g) + O_2(g) results in an increase in entropy (ΔS > 0).

Laws of Thermodynamics

  • First Law of Thermodynamics: Energy cannot be created or destroyed; it can only be converted from one form to another.
  • Second Law of Thermodynamics: For a process to be spontaneous, the total change in entropy of the universe (ΔS_univ) must be positive:
      - extΔS<em>extuniv=extΔS</em>extsys+extΔSextsurrext{ΔS}<em>{ ext{univ}} = ext{ΔS}</em>{ ext{sys}} + ext{ΔS}_{ ext{surr}}
  • Third Law of Thermodynamics: The entropy of a perfect crystalline substance is zero at absolute zero (0 K).

Gibb’s Free Energy (G)

  • Definition: Gibb’s Free Energy is the energy available for doing useful work in a system.
      - Introduced by J. Willard Gibbs (1839-1903).
  • Equation for Gibb's Free Energy:
      - G=H−TSG = H - TS
      - Criteria for spontaneity: extΔG0=extΔH0−TextΔS0ext{ΔG}^0 = ext{ΔH}^0 - T ext{ΔS}^0
      - Where:
        - GG = Free Energy
        - HH = Enthalpy
        - TT = Temperature (in Kelvin)
        - SS = Entropy.
Calculating ΔS for a Reaction
  • Formula:
      - extΔS<em>extsystem=S</em>extfinal−Sextinitialext{ΔS}<em>{ ext{system}} = S</em>{ ext{final}} - S_{ ext{initial}}
      - extΔSextrxn0=Snextproducts0−Snextreactants0ext{ΔS}^0_{ ext{rxn}} = S^0_{n_{ ext{products}}} - S^0_{n_{ ext{reactants}}}
  • Example: For the reaction ext{CaCO}3(s) ightarrow ext{CaO}(s) + ext{CO}_2(g)   - Given values:     - S0(extCaCO3)=92.9extJ/KmolS^0( ext{CaCO}_3) = 92.9 ext{ J/K mol}     - S0(extCaO)=39.8extJ/KmolS^0( ext{CaO}) = 39.8 ext{ J/K mol}     - S0(extCO2)=213.6extJ/KmolS^0( ext{CO}_2) = 213.6 ext{ J/K mol}   - Calculation:     - extΔS0</em>extrxn=(39.8+213.6)−92.9=160.5extJ/Kext{ΔS}^0</em>{ ext{rxn}} = (39.8 + 213.6) - 92.9 = 160.5 ext{ J/K}

Calculating Gibb’s Free Energy (ΔG°)

  • Method 1:
      - Example Reaction: ext{CH}_4(g) + 2 ext{O}_2(g)
    ightarrow ext{CO}_2(g) + 2 ext{H}_2 ext{O}(g)
      - Values from reaction:
        - Enthalpies: extΔH0=(−74.9)+2(0)−(−393.5)−(2(−241.8))=−802.2extkJext{ΔH}^0 = (-74.9) + 2(0) - (-393.5) - (2(-241.8)) = -802.2 ext{ kJ}
        - Entropies: extΔS0=213.7+(2imes188.7)−(186.1+(2imes205.5))=−5.0extJ/Kext{ΔS}^0 = 213.7 + (2 imes 188.7) - (186.1 + (2 imes 205.5)) = -5.0 ext{ J/K}
        - Free Energy: extΔG0=extΔH0−TextΔS0=−802.2−(298)(−0.005)=−800.7extkJext{ΔG}^0 = ext{ΔH}^0 - T ext{ΔS}^0 = -802.2 - (298)(-0.005) = -800.7 ext{ kJ}

  • Method 2: Using standard state free energy values:
      - Reaction: ext{CH}_4(g) + 2 ext{O}_2(g)
    ightarrow ext{CO}_2(g) + 2 ext{H}_2 ext{O}(g)
      - Values:
        - extΔG0ext{ΔG}^0:
          - extCH4=−50.8extkJ/molext{CH}_4 = -50.8 ext{ kJ/mol}
          - extO2=0ext{O}_2 = 0
          - extCO2=−394.4extkJ/molext{CO}_2 = -394.4 ext{ kJ/mol}
          - extH2extO=−228.6extkJ/molext{H}_2 ext{O} = -228.6 ext{ kJ/mol}
      - Calculation:
        - extΔG0=[−394.4+2(−228.6)]−[−50.8+0]=−800.8extkJext{ΔG}^0 = [-394.4 + 2(-228.6)] - [-50.8 + 0] = -800.8 ext{ kJ}

Evaluation of ΔG°
  • If ext{ΔG}^0 < 0: Reaction is spontaneous in the forward direction.
  • If ext{ΔG}^0 > 0: Reaction is not spontaneous in the forward direction. Work/energy must be applied for the forward direction to occur; the reverse reaction is spontaneous.
  • If extΔG0=0ext{ΔG}^0 = 0: Reaction is at equilibrium.
Temperature Effects on ΔG°
  • The relationship between enthalpy, entropy, and free energy at various temperatures influences spontaneity:
      - If extΔH<0ext{ΔH} < 0 and extΔS>0ext{ΔS} > 0: Spontaneous at all temperatures.
      - If ext{ΔH} > 0 and extΔS<0ext{ΔS} < 0: Nonspontaneous at all temperatures.   - If extΔH<0ext{ΔH} < 0 and extΔS<0ext{ΔS} < 0: Spontaneous at low temperatures, nonspontaneous at high temperatures.   - If extΔH>0ext{ΔH} > 0 and ext{ΔS} > 0: Nonspontaneous at low temperatures, spontaneous at high temperatures.

Calculating Reaction Temperature for Decomposition

  • Example Evaluation: Calculating the temperature at which mercury(II) oxide decomposes to oxygen gas (at 1 atm):
        - Reaction: 2 ext{HgO} (s)
    ightarrow 2 ext{Hg} (g) + ext{O}_2 (g)
        - Enthalpy (ΔH°): -90.8 kJ (for HgO), 61.3 kJ for gaseous products; total = 0 kJ.
        - Entropy values for analysis: 70.3 J/K (HgO), 174.9 J/K (Hg), and 205.0 J/K (O2).
        - Summing gives ΔH° = 304.2 kJ, ΔS° = 414.2 J/K (or 0.4142 kJ/K).
        - For the reaction: extΔG0=ΔH°−TΔS°=0ext{ΔG}^0 = ΔH° - TΔS° = 0
        - Setting ΔG° = 0 to solve for T gives: T=ΔH°ΔS°=00.4142=734.4KT = \frac{ΔH°}{ΔS°} = \frac{0}{0.4142} = 734.4 K

Relationship Between ΔG° and ΔG

  • Standard State Conditions: ΔG° refers to reactions under standard state conditions (1M, 1 atm, etc.).
  • Nonstandard State Conditions: ΔG applies to reactions under nonstandard conditions and is related to ΔG° by:
      - ΔG=ΔG°+RTextlnQΔG = ΔG° + RT ext{ln}Q
      - Where Q = reaction quotient.
Example Calculation of ΔG
  • Given the reaction: ext{H}2(g) + ext{I}_2(g) ightarrow 2 ext{HI}(g)   - ΔG° = 2.60 kJ at 25 °C.   - Given starting conditions:     - P</em>extH<em>2=3.5extatm,P</em>extI<em>2=1.5extatm,P</em>extHI=1.75extatmP</em>{ ext{H}<em>2} = 3.5 ext{ atm}, P</em>{ ext{I}<em>2} = 1.5 ext{ atm}, P</em>{ ext{HI}} = 1.75 ext{ atm}
      - Calculation of ΔG:
      - ΔG=2.60+(0.008314)(298)extln(0.5833)=1.26extkJΔG = 2.60 + (0.008314)(298) ext{ln}(0.5833) = 1.26 ext{ kJ}
      - Conclusion: Since ΔG is positive, the reaction is not spontaneous as written. The reverse reaction will therefore have ΔG negative.
Example Calculation of ΔG with Nonstandard Conditions
  • Reaction Under Review: 2 ext{NO}(g) + ext{O}_2(g)
    ightarrow 2 ext{NO}_2(g)
      - Given:
        - Values:
          - ΔG°: 87.6 kJ/mole (for NO2), 51.3 kJ (for NO)
      - Calculation:
      - Determine ΔG°:
        - ΔG°=(2imes51.3)−[(2imes87.6)+0]=102.6−175.2=−72.6extkJΔG° = (2 imes 51.3) - [(2 imes 87.6) + 0] = 102.6 - 175.2 = -72.6 ext{ kJ}
      - Final ΔG Calculation:
      - Using nonstandard condition Q with pressure values:
      - Q=(0.100)2(0.100)(0.100)=4000Q = \frac{(0.100)^2}{(0.100)(0.100)} = 4000
      - ΔG=ΔG°+RTextlnQ=−72.6+(0.008314)(298)extln(4000)=−52.1extkJΔG = ΔG° + RT ext{ln}Q = -72.6 + (0.008314)(298) ext{ln}(4000) = -52.1 ext{ kJ}

Relationship Between ΔG°, ΔG, and K

  • When a reaction is at equilibrium:
      - ΔG=0ΔG = 0 and Q=KQ = K
      - Therefore, 0=ΔG°+RTextlnK0 = ΔG° + RT ext{ln}K
      - Rearranging gives: ΔG°=−RTextlnKΔG° = -RT ext{ln}K
Calculation of Kp
  • Example Reaction: 2 ext{H}_2O(l)
    ightleftharpoons 2 ext{H}_2(g) + ext{O}_2(g) at 25°C:
      - Given:
        - ΔG°=−237.2extkJ/moleΔG° = -237.2 ext{ kJ/mole}
      - Calculation for ΔG°:
        - ΔG°=(2imes0)+0−2(−237.2)=474.4extkJΔG° = (2 imes 0) + 0 - 2(-237.2) = 474.4 ext{ kJ}
      - Rearranging gives: extKp=e−474.4RText{K}_p = e^{\frac{-474.4}{RT}} leading to extKpextapproximately7imes10−84ext{K}_p ext{ approximately } 7 imes 10^{-84}
Vapor Pressure Calculation of Iodine
  • Phase Transition: I_2(s)
    ightleftharpoons I_2(g)
      - Given Values:
        - ΔG°=19.3extkJ/moleΔG° = 19.3 ext{ kJ/mole}
      - Calculation of vapor pressure of iodine at 298K:
        - Q=Kp=PI2=extvaporpressureofI2Q = K_p = P_{I_2} = ext{vapor pressure of } I_2
        - From ΔG°:19.3=0−(0.008314)(298)extlnKΔG°: 19.3 = 0 - (0.008314)(298) ext{ln}K giving K=0.000414extatmK = 0.000414 ext{ atm}
Vapor Pressure at Different Temperature
  • At 400K:
      - Given:
        - ΔH°=62.25extkJ/moleΔH° = 62.25 ext{ kJ/mole}
        - ΔS°=260.57−116.7=143.87extJ/KΔS° = 260.57 - 116.7 = 143.87 ext{ J/K}
      - Calculation:
        - ΔG°=62.25−(400)(0.14387)=4.702extkJΔG° = 62.25 - (400)(0.14387) = 4.702 ext{ kJ}
      - Applying:
        - ΔG°=−RTextlnKΔG° = -RT ext{ln}K leads to further calculations.

Conclusion

  • Understanding the balance between entropy, free energy, temperature, and reaction spontaneity equips one with deep insights into thermodynamics and its applications to chemical reactions.