Free Energy and Equilibrium

Free Energy and Equilibrium

• This lecture connects thermodynamics and kinetics.

Standard vs. Non-Standard States

• Reactants and products are only in their standard states for an instant during a reaction.
• Standard states for reactions in solution are 1 mol/L concentrations.
• For gases, standard state is 1 atmosphere of pressure.
• $\Delta G$ in standard states predicts the direction of reaction at equilibrium.
• $\Delta G$ in non-standard states predicts the direction of reaction given specific reaction conditions.

Equation for Non-Standard Gibbs Free Energy

• $\Delta G = \Delta G^{\circ} + RT \log Q$
  • $\Delta G$: Gibbs free energy under non-standard conditions.
  • $\Delta G^{\circ}$: Gibbs free energy under standard conditions.
  • R: Ideal gas constant (8.314 J/mol*K).
  • T: Temperature in Kelvin.
  • Q: Reaction quotient (proportional to product concentrations over reactant concentrations, each raised to their stoichiometric powers).

Equilibrium Conditions

• At equilibrium, Q becomes K (equilibrium constant).
• At equilibrium, $\Delta G = 0$, meaning the change in entropy of the universe is zero.
• Therefore, $\Delta G^{\circ} = -RT \log K$
  • This equation links thermodynamics ($\Delta G$) and kinetics (K).

Relationship Between K and $\Delta G$

• K approaches 1: Significant amounts of both reactants and products are present at equilibrium.
• K = 1: $\Delta G = 0$. Products and reactants are equally favored at equilibrium.
• K > 1: $\Delta G$ is negative. Products are favored (forward reaction is spontaneous).
• K < 1: $\Delta G$ is positive. Reactants are favored (reverse reaction is spontaneous).

Calculating K from Calorimetry

• Sometimes it's experimentally impossible to make the kinetics measurements to get K, but easy to measure $\Delta G$ with thermodynamic experiments.
• Conversely, if kinetics experiments give K, $\Delta G$ can be back-calculated.

Interpreting $\Delta G$ and K

• K is proportional to product concentrations at equilibrium over reactant concentrations at equilibrium.
• K > 1 means the numerator (products) is much bigger than the denominator (reactants).
• K < 1 means the denominator (reactants) is bigger than the numerator (products), favoring the reverse reaction.

Free Energy and Reaction Direction

• A negative $\Delta G$ indicates a forward reaction.
• The value of Q can place the system on either side of $\Delta G^\circ$ on a free energy diagram, influencing the initial reaction direction until equilibrium is reached.

Example Problem 1: Calculating $\Delta G^\circ$ and Kp

• Reaction: $PCl<em>5(g) \rightleftharpoons PCl</em>3(g) + Cl_2(g)$
• Given: $\Delta G_f^\circ$ values for reactants and products at 25°C.
• Steps:
  • Calculate $\Delta G^\circ$ for the reaction using the formula:
    $\Delta G^\circ<em>{rxn} = \Sigma \Delta G</em>f^\circ (products) - \Sigma \Delta G_f^\circ (reactants)$
  • Use $\Delta G^\circ = -RT \log K$ to solve for Kp.
    • $log K = - \frac{\Delta G^\circ}{RT}$
    • $K = e^{-\frac{\Delta G^\circ}{RT}}$
• Solution:
  • $\Delta G^\circ = +39 \text{ kJ}$
  • $Kp = 1.5 \times 10^{-7}$
• Interpretation:
  • Reaction lies strongly in the reverse direction at equilibrium because $\Delta G^\circ$ is positive and Kp << 1.

Example Problem 2: Calculating $\Delta G$ under Non-Standard Conditions

• Using the same reaction as above, calculate $\Delta G$ given initial partial pressures.
• Use the equation: $\Delta G = \Delta G^\circ + RT \log Q$
• Calculate Q using initial partial pressures:
  • $Q = \frac{P<em>{PCl</em>3} \times P<em>{Cl</em>2}}{P<em>{PCl</em>5}}$
• Solution:
  • $Q = 37$
  • $\Delta G = +48 \text{ kJ}$
• Interpretation:
  • The reaction will proceed in reverse under these conditions because Q > K and $\Delta G$ is positive.

Example Problem 3: Ozone Depletion

• Reaction: $NO(g) + O<em>3(g) \rightarrow NO</em>2(g) + O_2(g)$
• Use data from appendix three to calculate $\Delta G^{\circ}$ and K at 298 K.
• $\Delta G^\circ<em>{rxn} = \Sigma \Delta G</em>f^\circ (products) - \Sigma \Delta G_f^\circ (reactants)$
• Solution:
  • $\Delta G^\circ = -198.3 \text{ kJ}$
  • $K = 5.76 \times 10^{34}$
• Interpretation:
  • The forward reaction is highly favored because $\Delta G^{\circ}$ is large and negative, and K is very large.

Example Problem 4: Autoionization of Water

• Reaction: $H_2O(l) \rightleftharpoons H^+(aq) + OH^-(aq)$
• K = $1 \times 10^{-14}$ at 25°C.
• Calculate $\Delta G^{\circ}$ using $\Delta G^\circ = -RT \log K$
• Solution:
  • $\Delta G^\circ = +79.9 \text{ kJ/mol}$
• Interpretation:
  • The autoionization of water is not favored forward because $\Delta G^{\circ}$ is large and positive and K is very small.
• At 40°C, $\Delta G^{\circ}$ = +81.1 kJ/mol, showing the reaction remains unfavorable forward.

Predicting Signs of Thermodynamic Quantities Without Math

• For the reaction: $H_2(g) \rightarrow 2H(g)$
• Predict the signs of $\Delta H$ and $\Delta S$ and spontaneity at high or low temperatures.
• $\Delta S$ is positive because one mole of gas goes to two moles of gas, increasing disorder.
• $\Delta H$ is endothermic (positive) because it takes energy to break stable H-H bonds.
  • High temperatures would favor spontaneity.

Rubber Band Entropy

• Relaxed rubber has higher entropy, stretched rubber has lower entropy because polymer chains align.
• Contraction is spontaneous.

Nursery Rhyme and Second Law of Thermodynamics

• Humpty Dumpty illustrates the second law. It is easy to break an egg (increase entropy) but impossible to perfectly restore it (decrease entropy).

Environmental Studies Excerpt

• Discusses the misconceptions the general public has about energy.
• The first law of thermodynamics, says that energy is conserved.
• The second law says that energy changes from a more useful to a less useful form every time it's converted with the first law.
• The second law is really why we have an energy crisis.