Kinetics, Equilibrium, and Thermodynamics Review

Tying Together Kinetics, Equilibrium & \Delta G

Temperature and Reaction Speed

• Reactions generally speed up with increasing temperature.
• Activation Energy (E_a): Minimum energy required for a reaction to occur.
• Low Temperature: Low kinetic energy (KE), small fraction of molecules possess sufficient energy to overcome the activation energy barrier.
• High Temperature: High KE, larger fraction of molecules have enough energy to surpass the activation energy barrier.
• Temperature is directly related to kinetic energy.
• E_a represents the energy barrier from reactants to products.

Arrhenius Equation

• The Arrhenius equation describes the relationship between temperature and the rate constant (k).
• k = A e^{-\frac{E_a}{RT}}
  • k: Rate constant
  • A: Frequency factor (collision frequency and other factors)
  • E_a: Activation energy
  • R: Ideal gas constant
  • T: Temperature (in Kelvin)
• Exponential Factor (e^{-\frac{E_a}{RT}}) represents the fraction of molecules with sufficient energy to surmount the activation barrier.
• Using two different temperatures to determine E_a:
  • \ln\left(\frac{k1}{k2}\right) = -\frac{Ea}{R} \left(\frac{1}{T1} - \frac{1}{T_2}\right)

Example Calculation of Activation Energy

• Reaction: NO2(g) + CO(g) \rightarrow NO(g) + CO2(g)
• Given:
  • At 701 K, k_1 = 2.57 M^{-1}s^{-1}
  • At 895 K, k_2 = 567 M^{-1}s^{-1}
• Calculation:
  • \ln\left(\frac{567}{2.57}\right) = \frac{-E_a}{8.314 J/(mol \cdot K)} \left(\frac{1}{895K} - \frac{1}{701K}\right)
  • \ln(220.623) = \frac{-E_a}{8.314} (0.0011173 - 0.0014265)
  • 5.39445 = \frac{-E_a}{8.314} (-0.0003092)
  • E_a = \frac{5.39445}{0.0003092/8.314} = 145110.1 J/mol
• Result:
  • E_a = 145.1 \times 10^3 J/mol \approx 145 kJ/mol

Reaction Coordinate Diagram

• Illustrates the potential energy (PE) changes during a reaction.
• Shows the energy profile as reactants transition to products.
• Transition State: The highest energy point on the diagram, representing the activated complex.
• The "ball over the hill" analogy: overcoming the activation energy barrier.

Reaction Coordinate Diagram and Reaction Mechanism

• Reaction Coordinate Diagrams (RCD) reflect the rate law.
• Each elementary reaction step is depicted with a distinct activation energy.
• Large E_a: Slower reaction step.
• Small E_a: Quicker reaction step.
• Intermediates exist between transition states.
• The rate-limiting step has the highest activation energy; the transition state for this step is the rate-limiting transition state.

Equilibria and Gibbs Free Energy

• Exergonic Reaction: \Delta G < 0, favors products (K > 1).
• Endergonic Reaction: \Delta G > 0, favors reactants (K < 1).
• Equilibrium Reaction: \Delta G = 0, K = 1.
• \Delta G = \Delta H - T\Delta S
• RCD Changes with Temperature: Only considers the temperature's influence on \Delta G, NOT the \Delta E of transition states.

Relationship Between K and \Delta G

• \Delta G = -RT \ln(K)
• At standard conditions, free energies of all reactants and products are equal when K=1.
• If K = 1, \Delta G = 0
• If K > 1, \Delta G < 0, products are favored.
• If K < 1, \Delta G > 0, reactants are favored.

Example Calculation of Equilibrium Constant

• Reaction: N2O4(g) \rightleftharpoons 2NO_2(g)
• Given:
  • \Delta Gf(NO2) = 51.3 kJ/mol
  • \Delta Gf(N2O_4) = 99.8 kJ/mol
• Calculation:
  • \Delta G{rxn}^{\circ} = \sum n \Delta Gf(products) - \sum n \Delta G_f(reactants)
  • \Delta G_{rxn}^{\circ} = [2(51.3) - 99.8] kJ/mol = 2.8 kJ/mol
  • \Delta G_{rxn}^{\circ} = -RT \ln K
  • \ln K = -\frac{\Delta G_{rxn}^{\circ}}{RT}
  • \ln K = -\frac{2.8 \times 10^3 J/mol}{8.314 J/(mol \cdot K) \cdot 298 K} = -1.13
  • K = e^{-1.13} = 0.3

The van't Hoff Equation

• Applies to reactions not at standard conditions.
• \Delta G^{\circ} = -RT \ln(K)
• \Delta G = \Delta H - T \Delta S
• Rearranging:
  • \ln(K) = -\frac{\Delta H}{RT} + \frac{\Delta S}{R}
• van't Hoff Equation:
  • \ln \left( \frac{K2}{K1} \right) = -\frac{\Delta H}{R} \left( \frac{1}{T2} - \frac{1}{T1} \right)
• Plotting \ln K vs. \frac{1}{T}: Slope is -\frac{\Delta H}{R}, Y-intercept is \frac{\Delta S}{R}.
• The sign of \Delta H dictates the slope's sign.
• See example 10.15.