Problem 49 - General Chemistry: Principles and Modern Applications

Question

For the following equilibrium reactions, calculate \(\Delta G^{\circ}\) at the indicated temperature. [Hint: How is each equilibrium constant related to a thermodynamic equilibrium constant, \(K ?]\) (a) \(\mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}) \quad K_{\mathrm{c}}=50.2\) at \(445^{\circ} \mathrm{C}\) (b) \(\mathrm{N}_{2} \mathrm{O}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\) \(K_{c}=1.7 \times 10^{-13} \mathrm{at} 25^{\circ} \mathrm{C}\) (c) \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g})\) \(K_{c}=4.61 \times 10^{-3}\) at \(25^{\circ} \mathrm{C}\) (d) \(2 \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Hg}_{2}^{2+}(\mathrm{aq}) \rightleftharpoons\) \(2 \mathrm{Fe}^{2+}(\mathrm{aq})+2 \mathrm{Hg}^{2+}(\mathrm{aq})\) \(K_{\mathrm{c}}=9.14 \times 10^{-6} \mathrm{at} 25^{\circ} \mathrm{C}\)

Step-by-Step Solution

Answer

\(\Delta G^{\circ}\) for (a) is -45.78 kJ/mol, for (b) is 76.44 kJ/mol, for (c) is 11.69 kJ/mol, and for (d) is 27.11 kJ/mol.

Step 1. Convert Temperatures

Convert the temperatures from Celsius to Kelvin: For part (a), \(445^{\circ}C = 445 + 273.15 = 718.15K\). For part (b) and (c), \(25^{\circ}C = 25 + 273.15 = 298.15K\).

Step 2. Calculate \(\Delta G^{\circ}\) for Part (a)

Use the formula \(\Delta G^{\circ} = -RT \ln K\) with \(R = 8.3145 J/K.mol\), \(T = 718.15 K\), and \(K_{c} = 50.2\) to get: \(\Delta G^{\circ} = -8.3145 J/K.mol \times 718.15K \times \ln 50.2 = -45778.2 J/mol = -45.78 kJ/mol\).

Step 3. Calculate \(\Delta G^{\circ}\) for Part (b)

Use the formula \(\Delta G^{\circ} = -RT \ln K\) with \(R = 8.3145 J/K.mol\), \(T = 298.15 K\), and \(K_{c} = 1.7 \times 10^{-13}\) to get: \(\Delta G^{\circ} = -8.3145 J/K.mol \times 298.15K \times \ln (1.7 \times 10^{-13}) = 76443 J/mol = 76.44 kJ/mol \).

Step 4. Calculate \(\Delta G^{\circ}\) for Part (c)

Use the formula \(\Delta G^{\circ} = -RT \ln K\) with \(R = 8.3145 J/K.mol\), \(T = 298.15 K\), and \(K_{c} = 4.61 \times 10^{-3}\) to get: \(\Delta G^{\circ} = -8.3145 J/K.mol \times 298.15K \times \ln (4.61 \times 10^{-3}) = 11693.19 J/mol = 11.69 kJ/mol \).

Step 5. Calculate \(\Delta G^{\circ}\) for Part (d)

Use the formula \(\Delta G^{\circ} = -RT \ln K\) with \(R = 8.3145 J/K.mol\), \(T = 298.15 K\), and \(K_{c} = 9.14 \times 10^{-6}\) to get: \(\Delta G^{\circ} = -8.3145 J/K.mol \times 298.15K \times \ln (9.14 \times 10^{-6}) = 27108.3 J/mol = 27.11 kJ/mol \).

Key Concepts

Equilibrium Constant

In chemistry, the equilibrium constant quantifies the balance between reactants and products in a reversible chemical reaction. When a reaction reaches a state where the rates of the forward and reverse reactions are equal, it is in equilibrium. At this point, neither the concentrations of reactants nor products change over time. The equilibrium constant (\( K \)) is expressed as the ratio of the concentrations of products to reactants, each raised to the power of their stoichiometric coefficients. For a general reaction:
\[aA + bB \rightleftharpoons cC + dD\]
the equilibrium constant (\( K_c \)) at constant temperature can be written as:
\[K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}\]
A large value of \( K_c \) indicates that, at equilibrium, the reaction mixture contains more products than reactants, and vice versa. Understanding the equilibrium constant is crucial for predicting the extent and direction of a chemical reaction under specific conditions.

Thermodynamic Equilibrium

Thermodynamic equilibrium refers to a state in a system where macroscopic properties, such as temperature, pressure, and concentration, remain constant over time. In this state, all transferable properties between systems are balanced. For chemical reactions, thermodynamic equilibrium means that both the forward and reverse reactions occur at the same rate, so the concentrations of reactants and products remain constant, although not necessarily equal.

• In such a system, there is no net exchange of heat, energy, or matter between the system and its surroundings.
• The system is stable and no spontaneous changes will occur unless an external force is applied.
• The free energy change (\( \Delta G \)) at equilibrium is zero because the system is in its most stable state.

To determine thermodynamic equilibrium in a chemical system, one often refers to the Gibbs Free Energy (\( \Delta G^{\circ} \)). It is calculated using the equation:
\[\Delta G^{\circ} = -RT \ln K\]
where \( R \) is the universal gas constant, \( T \) is the temperature in Kelvin, and \( K \) is the equilibrium constant. This relationship helps in understanding the stability of a reaction at given conditions and is fundamental in predicting which direction a reaction will proceed until equilibrium is reached.

Temperature Conversion

Converting temperatures is often necessary, especially in thermodynamic calculations as they require temperatures in Kelvin. Kelvin is the standard unit for temperature in scientific calculations because it begins at absolute zero, where particles theoretically stop moving. Converting from Celsius to Kelvin is straightforward and involves adding 273.15 to the Celsius temperature.

• For example, to convert 25°C to Kelvin, you would calculate: \( 25 + 273.15 = 298.15 \text{ K} \).
• This conversion is essential for using thermodynamic formulas, such as the one for Gibbs Free Energy: \( \Delta G^{\circ} = -RT \ln K\).
• Calculations in Kelvin ensure consistency in the equations, ensuring accurate predictions for chemical systems' behaviors.

When working on chemical reactions, always remember to convert temperature to Kelvin whenever using thermodynamic equations. This practice helps to prevent errors in calculations and provides accurate insights into the reactions under study.