Question

Calculate \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\) for each reaction: (a) \(2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NOCl}(g) ; K=1.58 \times 10^{7}\) (b) \(\mathrm{Cu}_{2} \mathrm{~S}(s)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{Cu}(s)+\mathrm{SO}_{2}(g) ; K=3.25 \times 10^{37}\)

Short answer

Reaction (a) \( ΔG^{\text{°}} ≈ -41139 \text{ J/mol} \) Reaction (b) \( ΔG^{\text{°}} ≈ -213601 \text{ J/mol} \)

Step-by-step solution

- Understand the relationship

To calculate the standard Gibbs free energy change (\text {Δ}G^{\text{°}}), use the equation: \( ΔG^{\text{°}} = -RT \times \text{ln} K \) where R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin, and K is the equilibrium constant.

Step 2A - Plug in values for reaction (a)

For the reaction 2 NO(g) + Cl_2(g) \rightleftharpoons 2 NOCl(g): \( K = 1.58 \times 10^{7} \) Plug in the values: \( \text{Δ}G^{\text{°}} = - (8.314 \text{ J/mol·K}) \times (298 \text{ K}) \times \text{ln} (1.58 \times 10^{7}) \)

Step 3A - Calculate for reaction (a)

First, calculate the natural logarithm: \( \text{ln}(1.58 \times 10^{7}) ≈ 16.58 \) Next, calculate ΔG^{\text{°}}: \( ΔG^{\text{°}} = - (8.314) \times (298) \times (16.58) ≈ -41139 \text{ J/mol} \)

Step 2B - Plug in values for reaction (b)

For the reaction \mathrm{Cu}_2 \text{S} + O_2 \rightleftharpoons 2 Cu + SO_2: \(K = 3.25 \times 10^{37} \) Plug in the values: \( ΔG^{\text{°}} = - (8.314 \text{ J/mol·K}) \times (298 \text{ K}) \times \text{ln} (3.25 \times 10^{37}) \)

Step 3B - Calculate for reaction (b)

First, calculate the natural logarithm: \( \text{ln}(3.25 \times 10^{37}) ≈ 86.18 \) Next, calculate ΔG^{\text{°}}: \( ΔG^{\text{°}} = - (8.314) \times (298) \times (86.18) ≈ - 213601 \text{ J/mol} \)

Key concepts

Gibbs free energy

Gibbs free energy, often represented as \( \text{Δ}G \), is a thermodynamic quantity that combines enthalpy and entropy to predict the feasibility of a reaction. In simple terms, it helps us determine whether a reaction will proceed spontaneously. A negative \( \text{Δ}G \) means the reaction is spontaneous, while a positive \( \text{Δ}G \) indicates non-spontaneity. The standard Gibbs free energy change, represented as \( \text{Δ}G^{\text{°}} \), is calculated under standard conditions, such as 1 atm pressure and 298 K temperature. The equation used is \[ \text{Δ}G^{\text{°}} = -RT \times \text{ln} K \], where:

• R is the gas constant, equal to 8.314 J/mol·K.
• T is the temperature in Kelvin.
• K is the equilibrium constant for the reaction.

This equation bridges thermodynamics and reaction equilibria by relating the free energy change to the equilibrium constant.

Equilibrium constant

The equilibrium constant (K) is a dimensionless number that provides information about the proportion of reactants and products at equilibrium. For a given reaction, it is derived from the concentrations of the products divided by the concentrations of the reactants, each raised to the power of their coefficients in the balanced equation. A high K value (e.g., much greater than 1) indicates that the reaction favors product formation. In contrast, a low K value (e.g., much less than 1) suggests the reaction favors the reactants. Importantly, K is temperature-dependent — it can change when the temperature changes, which makes it essential to specify the temperature (usually 298 K) when providing an equilibrium constant for calculations like Gibbs free energy.

Thermodynamics

Thermodynamics is the study of energy transformation and exchanges within physical systems. It comprises several laws that govern these processes. For Gibbs free energy calculations, the first and second laws of thermodynamics are particularly relevant.

• First Law: Energy cannot be created or destroyed, only transformed. This law is essential for understanding the conservation of energy in reactions.
• Second Law: The entropy of an isolated system always increases over time. Entropy, a measure of disorder, plays a key role in determining the spontaneity of reactions. When calculating Gibbs free energy, we consider both enthalpy (heat content) and entropy to determine the feasibility of a reaction. Together, these principles underscore why reactions occur and how much energy is involved.

Natural logarithm

The natural logarithm, denoted as ln, is a mathematical function that is the inverse of the exponential function (\(e\)). Unlike regular logarithms that use base 10, the natural logarithm uses base \(e\) (approximately 2.718). In Gibbs free energy calculations, we use the natural logarithm to relate the equilibrium constant (K) to the free energy change: \[ \text{Δ}G^{\text{°}} = -RT \times \text{ln} K \]. Here:

• ln(K) translates the equilibrium constant into a value that can be linearly related to the free energy change.
• This relationship simplifies the mathematical manipulation of exponential growth common in reaction kinetics and equilibria.

Understanding ln(K) in this context helps us appreciate why large changes in equilibrium constants cause significant changes in Gibbs free energy.

Standard conditions

Standard conditions refer to a set of predefined conditions typically used in thermodynamic calculations to allow consistent comparisons and interpretations. These conditions usually include:

• Temperature: 298 K (25°C).
• Pressure: 1 atm for gases.
• Concentration: 1 M for solutions.

Using standard conditions allows us to derive standard values like \( \text{Δ}G^{\text{°}} \), which are essential for comparing different reactions under the same reference state. This standardization is crucial because real-world reactions often occur under varying conditions, making it challenging to compare results directly. With standard conditions, we have a common benchmark for predicting reaction behavior and calculating associated thermodynamic quantities like Gibbs free energy.