Homework 16D: Thermodynamics and Electrochemistry Study Guide

Relationship Between Gibbs Free Energy and the Equilibrium Constant

• The relationship between the standard Gibbs Free Energy change ($\Delta G^\circ$) and the equilibrium constant ($K$) of a chemical reaction is defined by the following expression:   $\Delta G^\circ = -RT \ln(K)$

  • $\Delta G^\circ$ is the Gibbs Free Energy change under standard conditions.

  • $R$ is the ideal gas constant, typically valued at $8.314\,J\,mol^{-1}\,K^{-1}$.

  • $T$ is the absolute temperature in Kelvin ($K$).

  • $\ln(K)$ is the natural logarithm of the equilibrium constant.

• Application: Determining $\Delta G$ from an Equilibrium Constant (Problem 1)

  • Reaction: $CO_2 + H_2 \rightleftharpoons CO + H_2O$

  • Given Equilibrium Constant ($K$): $2$

  • Temperature ($T$): $1120^\circ C$.

  • Conversion to Kelvin: $T = 1120 + 273.15 = 1393.15\,K$.

  • Calculation: $\Delta G^\circ = -(8.314\,J\,mol^{-1}\,K^{-1}) \times (1393.15\,K) \times \ln(2)$.

  • This calculation yields the energy change in Joules, which can be converted to kilojoules ($kJ$) by dividing by $1000$.

• Application: Determining the Equilibrium Constant from $\Delta G$ (Problem 2)

  • Reaction: $2Cu_2O + 2NO \rightarrow 4CuO + N_2$

  • Given $\Delta G^\circ$: $-389\,kJ$ at $25^\circ C$.

  • Temperature ($T$): $25 + 273.15 = 298.15\,K$.

  • Preparation: Convert $\Delta G^\circ$ to Joules: $-389000\,J$.

  • Rearranging the formula for $K$:     $\ln(K) = \frac{-\Delta G^\circ}{RT}$     $K = e^{\frac{-\Delta G^\circ}{RT}}$

  • Substitution: $K = e^{\frac{-(-389000\,J)}{(8.314\,J\,mol^{-1}\,K^{-1} \times 298.15\,K)}}$.

Relationship Between Gibbs Free Energy and Electrochemistry

• The work performed by an electrochemical cell can be directly related to the change in Gibbs Free Energy. The formula connecting the two is:   $\Delta G^\circ = -nFE^\circ_{cell}$

  • $n$ represents the number of moles of electrons transferred in the balanced redox equation.

  • $F$ is Faraday's constant, which is approximately $96485\,C\,mol^{-1}$.

  • $E^\circ_{cell}$ is the standard cell potential (electromotive force) measured in Volts ($V$).

• Calculating Free Energy Change for a Voltaic Cell (Problem 3)

  • Cell Notation: $Zn | Zn^{2+} || I_2 | I^{-1}$.

  • Anode (Oxidation): $Zn \rightarrow Zn^{2+} + 2e^-$

  • Cathode (Reduction): $I_2 + 2e^- \rightarrow 2I^-$

  • Total Electrons Transferred ($n$): $2$.

  • Standard Reduction Potentials:

  • $E^\circ_{red}(Zn^{2+}/Zn) = -0.76\,V$

  • $E^\circ_{red}(I_2/I^-) = +0.54\,V$

  • Cell Potential Calculation: $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$

  • $E^\circ_{cell} = 0.54\,V - (-0.76\,V) = 1.30\,V$

  • Calculation of $\Delta G^\circ$:     $\Delta G^\circ = -(2) \times (96485\,C\,mol^{-1}) \times (1.30\,V)$

  • Result: The value will be in Joules (since $1\,J = 1\,C \times V$).

• Determining Unknown Reduction Potentials (Problem 4)

  • Reaction: $Zn + M^{2+} \rightarrow Zn^{2+} + M$

  • Given $\Delta G^\circ$: $-212\,kJ$ ($-212000\,J$).

  • Number of electrons ($n$): $2$ (based on the charge change of $Zn$ to $Zn^{2+}$ and $M^{2+}$ to $M$).

  • First, find total $E^\circ_{cell}$:     $E^\circ_{cell} = \frac{-\Delta G^∘}{nF} = \frac{-(-212000\,J)}{(2 \times 96485\,C\,mol^{-1})}$

  • Solve for the reduction potential of $M^{2+}$:     $E^\circ_{cell} = E^\circ_{red}(\text{cathode}) - E^\circ_{red}(\text{anode})$     $E^\circ_{cell} = E^\circ_{red}(M^{2+}/M) - E^\circ_{red}(Zn^{2+}/Zn)$     $E^\circ_{red}(M^{2+}/M) = E^\circ_{cell} + E^\circ_{red}(Zn^{2+}/Zn)$

  • Substituting the standard potential for Zinc ($-0.76\,V$) allows for the identification of the reduction potential for $M^{2+}$.

Thermodynamic Calculations of Gibbs Free Energy

• The thermodynamic definition of Gibbs Free Energy combines enthalpy ($H$), temperature ($T$), and entropy ($S$):   $\Delta G = \Delta H - T\Delta S$

• Spontaneity and Calculation (Problem 5)

  • Parameters:

  • $\Delta H_{system} = 145\,kJ$

  • $T = 293\,K$

  • $\Delta S_{system} = 192\,J/K$

  • Adjustment for Units: Convert $\Delta S$ to $kJ/K$: $0.192\,kJ/K$.

  • Substitution: $\Delta G = 145\,kJ - (293\,K \times 0.192\,kJ/K)$.

  • Spontaneity Criteria:

  • If \Delta G < 0, the process is spontaneous.

  • If \Delta G > 0, the process is nonspontaneous.

  • If $\Delta G = 0$, the system is at equilibrium.

• Calculating Threshold Temperature (Problem 6)

  • Parameters:

  • $\Delta G = -34.7\,kJ$

  • $\Delta H = -28.8\,kJ$

  • $\Delta S = 22.2\,J/K$ ($0.0222\,kJ/K$).

  • Rearranging for Temperature ($T$):     $\Delta G = \Delta H - T\Delta S$     $T\Delta S = \Delta H - \Delta G$     $T = \frac{\Delta H - \Delta G}{\Delta S}$

  • Substitution: $T = \frac{-28.8\,kJ - (-34.7\,kJ)}{0.0222\,kJ/K}$.

  • This allows for finding the temperature at which the specific free energy change of $-34.7\,kJ$ occurs.

Summary of Mathematical Constants and Conversions

• Temperature Conversion:

  • $\text{Temperature in Kelvin (K)} = \text{Temperature in Celsius (}^{\circ}\text{C)} + 273.15$

• Energy Constants:

  • Ideal Gas Constant ($R$): $8.314\,J\,mol^{-1}\,K^{-1}$

  • Faraday's Constant ($F$): $96485\,C\,mol^{-1}$

• Unit Prefixes:

  • $1\,kJ = 1000\,J$

  • $\Delta S$ values are frequently provided in $J/K$ and must be converted to $kJ/K$ if $\Delta H$ is in $kJ$ to ensure dimensional consistency in the equation $\Delta G = \Delta H - T\Delta S$.