Detailed Study Notes on Gibbs Energy Change and Equilibrium Concepts

13-5 Gibbs Energy Change in a System of Variable Composition

• Standard Gibbs Energy Change $(ΔG^0)$: Defined as the difference in free energy between the products and reactants under standard conditions.

• Gibbs Energy Change $(ΔG)$: This is used in the context of not only standard conditions but also varying conditions in a chemical reaction.

• Equilibrium: The free energy at equilibrium is always lower than the free energy of pure reactants and products. The reaction proceeds spontaneously toward the minimum in free energy, which corresponds to equilibrium.

Free Energy Changes During a Reaction

• ### Example 1: Product Favoring Reaction

  • Relationship: $ΔG = ΔG^0 + RT ext{ln}Q$

• Interpretation: When $ΔG < 0$, the reaction is spontaneous in the forward direction, favoring products.

• ### Example 2: Reactant Favoring Reaction

  • Despite standard conditions the sign of $ΔG$ tells the direction of spontaneity.

Exterior Relation of Gibbs Free Energy to Equilibrium Constant

• At equilibrium, $ΔG = 0$ and reaction quotient $Q = K_{eq}$ (equilibrium constant).

• $ΔG^0 = –RT ext{ln}K_{eq}$

• Gibbs Energy Expression at non-equilibrium:

  • For a general reaction $aA + bB ⇆ cC + dD$:

    • $K_{eq} = rac{[C]^c[D]^d}{[A]^a[B]^b}$

    • With heterogeneous solutions: e.g., $2 Al(s) + 6 H^+(aq) ⇆ 2 Al^{3+}(aq) + 3 H_2(g)$

  • Note: Pure solids and liquids are excluded (they have a concentration of 1).

Interpreting Values of $ΔG^0$ and K

• The relationship can also be written as:
  $Δ_rG^0 = –RT ext{ln}K$

• In terms of reaction quotient:**
  $ΔG = ΔG^0 + RT ext{ln}Q$
  If the reaction is at equilibrium, then:
  $Δ_rG^0 = –RT ext{ln}K$

• Direction Prediction: When not at equilibrium, use the expression:
  $ΔG = RT ext{ln}(Q/K)$ to predict spontaneity.

Example Calculations of $K_{eq}$

• ### Example EOC #41

  • For the synthesis of gaseous methanol: $CO(g) + 2H<em>2(g) ⇆ CH</em>3OH(g)$

    • At 483 K, concentrations are as follows:

    • $[CO(g)] = 0.0911 M$

    • $[H_2(g)] = 0.0822 M$

    • $[CH_3OH(g)] = 0.00892 M$

    • Calculation of equilibrium constant $KP$ and $Δr G^0$ follows.

Example EOC #49

• Given reaction: $CH<em>3CO</em>2H(aq) + H<em>2O(l) ⇆ CH</em>3CO<em>2^{-}(aq) + H</em>3O^{+}(aq)$

  • Thermodynamic value: $Δ_rG^0 = 27.07 kJ mol^{-1}$ at $298 K$.

  • Analyze reaction's spontaneity at concentrations:

    • $[CH<em>3CO</em>2H] = 0.10 M, [CH<em>3CO</em>2^{-}] = 1.0 imes 10^{-3} M, [H_3O^{+}] = 1.0 imes 10^{-3} M$.

Cellular Energetics

• ATP Hydrolysis:

  • The hydrolysis of ATP $
    ightarrow$ ADP + $PO_4^{3-}$ with $ΔG^0 = -30.5 kJ/mol$.

  • Investigation into moles of ATP produced via glucose oxidation ($ΔG_{f}^{ heta}$ for glucose $= -917 kJ/mol$):

    • Energy released $= -2879 kJ$ produces approximately:

    • $ATP: rac{-2879 kJ/mol imes 1 mol ATP}{-30.5 kJ} = 94.4 mol ATP/mol glucose$

  • Efficiency of Cellular Machinery:

    • Yield = approximately 32%.

Example EOC #57

• At $298 K$, given:

  • $Δ_fG^{ heta}[CO(g)] = -137.2 kJ mol^{-1}$ and $K = 6.5 imes10^{11}$

  • Determine $ΔfG^{ heta}[COCl2(g)]$ accordingly; compare values from Appendix D.

Conservation Of Mass

• Lavoisier's studies on mercury(II) oxide, reaction: $HgO(s) ⇆ Hg(l) + rac{1}{2} O_2(g)$

  • At $25°C$, values include:

    • $ΔrH^{ heta} = +90.83 kJ mol^{-1}, ΔrG^{ heta}= +58.54 kJ mol^{-1}$.

  • A and B: Determine equilibrium pressure contributions and laboratory conditions for significant oxygen production rates.

Upcoming Topics

• 16-4: The Magnitude of an Equilibrium Constant.

• 16-5: Predicting the Direction of Chemical Change.

• 16-6: Le Châtelier’s Principle.

• 16-x: Relationships of $ΔG^{ heta}$ and $K$, with respect to temperature.

Relationships Regarding $ΔG^0$ and Temperature

a. $ΔG^0 = ΔH^0 - TΔS^0$
b. $ΔG^0 = – RT ext{ln}(K<em>{eq})$ c. $ln K</em>{eq} = − rac{ΔH^0}{RT} + rac{ΔS^0}{R}$

Van’t Hoff Equation

• Y-value Intercept: $ΔS^0/R$; Slope: $-ΔH^0/R$.

• Equation often used for shifts in reaction equilibrium with temperature changes:
  $ln K<em>2 - ln K</em>1 = - rac{ΔH^{ heta}}{R} \bigg( rac{ rac{1}{T<em>2} - rac{1}{T</em>1}}{1}\bigg)$

13-7 Coupled Reactions

• Driving a non-spontaneous reaction through coupling with a spontaneous one (e.g., smelting copper ore).

• Example:

  • $C(s) + rac{1}{2}O<em>2(g) + Cu</em>2O(s) → 2Cu(s) + CO(g)$, with relevant energies ::

    • Non-spontaneous: $ΔG^{ heta}_{673 K} = +125 kJ mol^{-1}$.

    • Spontaneous: $ΔG^{ heta}_{673 K} = -175 kJ mol^{-1}$ at standard pressure.

    • Coupled energy yields: total $ΔG^{ heta}_{673 K} = -50 kJ mol^{-1}$.

16-4: The Magnitude of the Equilibrium Constant

• Reaction determined by $K= [products]/[reactants]$. Indicates reaction favoring forward or reverse emissions.

  • Magnitude of K

  • Defaults: reaction favoring completion with $K ext{[large]}
    eq 10$.

  • Comparison basis:

    • Small: $ ext{K}< 10^{-3}$, Intermediate: $10^{-3} < K < 10^{3}$, Large: $K > 10^{3}$.