Question

Using data from Appendix \(C\), calculate \(\Delta G^{\circ}\) for the following reactions. Indicate whether each reaction is spontaneous under standard conditions. (a) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\) (b) \(\mathrm{NO}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g) \longrightarrow 3 \mathrm{NO}(g)\) (c) \(6 \mathrm{Cl}_{2}(g)+2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s) \longrightarrow 4 \mathrm{FeCl}_{3}(s)+3 \mathrm{O}_{2}(g)\) (d) \(\mathrm{SO}_{2}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{S}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)\)

Short answer

(a) \(\Delta G^{\circ} = -140.52 \, \mathrm{kJ/mol}\), the reaction is spontaneous. (b) \(\Delta G^{\circ} = 106.10 \, \mathrm{kJ/mol}\), the reaction is not spontaneous. (c) \(\Delta G^{\circ} = -117.04 \, \mathrm{kJ/mol}\), the reaction is spontaneous. (d) \(\Delta G^{\circ} = -174.33 \, \mathrm{kJ/mol}\), the reaction is spontaneous.

Step-by-step solution

(a) Given reaction, Gibbs free energy values, and Formula

The reaction is: \[ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g) \] The standard Gibbs free energy values (\(\Delta G^{\circ}_{f}\)) are: \( \Delta G^{\circ}_{f}(\mathrm{SO}_{2}) = -300.13 \, \mathrm{kJ/mol} \) \( \Delta G^{\circ}_{f}(\mathrm{O}_{2}) = 0 \, \mathrm{kJ/mol} \) \( \Delta G^{\circ}_{f}(\mathrm{SO}_{3}) = -370.39 \, \mathrm{kJ/mol} \) We will apply the formula for \(\Delta G^{\circ}\).

(a) Calculation of \(\Delta G^{\circ}\) for the reaction

Using the formula, we have: \[ \Delta G^{\circ} = [2(-370.39) - (2(-300.13) + 0)] \, \mathrm{kJ/mol} \] Simplifying the expression: \[ \Delta G^{\circ} = [-740.78 + 600.26] \, \mathrm{kJ/mol} = -140.52 \, \mathrm{kJ/mol} \] Since \(\Delta G^{\circ}<0\), the reaction is spontaneous under standard conditions.

(b) Given reaction, Gibbs free energy values, and Formula

The reaction is: \[ \mathrm{NO}_{2}(g)+\mathrm{N}_{2}\mathrm{O}(g) \longrightarrow 3\mathrm{NO}(g) \] The standard Gibbs free energy values (\(\Delta G^{\circ}_{f}\)) are: \( \Delta G^{\circ}_{f}(\mathrm{NO}_{2}) = 51.30 \, \mathrm{kJ/mol} \) \( \Delta G^{\circ}_{f}(\mathrm{N}_{2}\mathrm{O}) = 104.20 \, \mathrm{kJ/mol} \) \( \Delta G^{\circ}_{f}(\mathrm{NO}) = 87.20 \, \mathrm{kJ/mol} \) We will apply the formula for \(\Delta G^{\circ}\).

(b) Calculation of \(\Delta G^{\circ}\) for the reaction

Using the formula, we have: \[ \Delta G^{\circ} = [3(87.20) - (51.30 + 104.20)] \, \mathrm{kJ/mol} \] Simplifying the expression: \[ \Delta G^{\circ} = [261.60 - 155.50] \, \mathrm{kJ/mol} = 106.10 \, \mathrm{kJ/mol} \] Since \(\Delta G^{\circ}>0\), the reaction is not spontaneous under standard conditions.

(c) Given reaction, Gibbs free energy values, and Formula

The reaction is: \[ 6 \mathrm{Cl}_{2}(g)+2 \mathrm{Fe}_{2}\mathrm{O}_{3}(s) \longrightarrow 4\mathrm{FeCl}_{3}(s)+3\mathrm{O}_{2}(g) \] The standard Gibbs free energy values (\(\Delta G^{\circ}_{f}\)) are: \( \Delta G^{\circ}_{f}(\mathrm{Cl}_{2}) = 0 \, \mathrm{kJ/mol} \) \( \Delta G^{\circ}_{f}(\mathrm{Fe}_{2}\mathrm{O}_{3}) = -740.40 \, \mathrm{kJ/mol} \) \( \Delta G^{\circ}_{f}(\mathrm{FeCl}_{3}) = -399.46 \, \mathrm{kJ/mol} \) \( \Delta G^{\circ}_{f}(\mathrm{O}_{2}) = 0 \, \mathrm{kJ/mol} \) We will apply the formula for \(\Delta G^{\circ}\).

(c) Calculation of \(\Delta G^{\circ}\) for the reaction

Using the formula, we have: \[ \Delta G^{\circ} = [4(-399.46) + 3(0) - (6(0) + 2(-740.40))] \, \mathrm{kJ/mol} \] Simplifying the expression: \[ \Delta G^{\circ} = [-1597.84 + 1480.80] \, \mathrm{kJ/mol} = -117.04 \, \mathrm{kJ/mol} \] Since \(\Delta G^{\circ}<0\), the reaction is spontaneous under standard conditions.

(d) Given reaction, Gibbs free energy values, and Formula

The reaction is: \[ \mathrm{SO}_{2}(g)+2\mathrm{H}_{2}(g) \longrightarrow \mathrm{S}(s)+2\mathrm{H}_{2}\mathrm{O}(g) \] The standard Gibbs free energy values (\(\Delta G^{\circ}_{f}\)) are: \( \Delta G^{\circ}_{f}(\mathrm{SO}_{2}) = -300.13 \, \mathrm{kJ/mol} \) \( \Delta G^{\circ}_{f}(\mathrm{H}_{2}) = 0 \, \mathrm{kJ/mol} \) \( \Delta G^{\circ}_{f}(\mathrm{S}) = 0 \, \mathrm{kJ/mol} \) \( \Delta G^{\circ}_{f}(\mathrm{H}_{2}\mathrm{O}) = -237.23 \, \mathrm{kJ/mol} \) We will apply the formula for \(\Delta G^{\circ}\).

(d) Calculation of \(\Delta G^{\circ}\) for the reaction

Using the formula, we have: \[ \Delta G^{\circ} = [0 + 2(-237.23) - (-300.13 + 2(0))] \, \mathrm{kJ/mol} \] Simplifying the expression: \[ \Delta G^{\circ} = [-474.46 + 300.13] \, \mathrm{kJ/mol} = -174.33 \, \mathrm{kJ/mol} \] Since \(\Delta G^{\circ}<0\), the reaction is spontaneous under standard conditions.

Key concepts

Understanding Thermodynamics in Gibbs Free Energy Calculations

Thermodynamics, a branch of physics, investigates the relationships between heat, work, temperature, and energy. In the context of chemical reactions, thermodynamics allows us to predict whether a process will occur spontaneously through the concept of Gibbs free energy (\( \triangle G \)). Gibbs free energy is a thermodynamic potential that combines the concepts of enthalpy (\( H \)), entropy (\( S \)), and temperature (\( T \)) in the equation \( \triangle G = \triangle H - T\triangle S \). A negative value of \( \triangle G \) indicates a spontaneous process under constant pressure and temperature, which means the reaction can occur without any external energy input. Conversely, a positive \( \triangle G \) value suggests that the reaction is non-spontaneous under the same conditions and requires additional energy to proceed. Zero \( \triangle G \) indicates a system at equilibrium, meaning the forward and reverse reactions occur at the same rate. Understanding Gibbs free energy is paramount for students of chemistry, as it provides critical insights into the feasibility of chemical reactions and the energy changes involved.

To make thermodynamics more accessible for learners, providing examples with real-life applications helps in understanding abstract concepts. For example, the energy released during the combustion of fuel in an engine or the energy required for the synthesis of biochemical compounds can be analyzed through Gibbs free energy calculations. To enhance clarity, educators can break down each term in the Gibbs free energy equation and explore how changes in temperature or entropy affect the spontaneity of reactions.

Spontaneous Reactions and Gibbs Free Energy

Spontaneous reactions are processes that proceed on their own without the need for continuous external energy. It's a common misconception that 'spontaneous' implies that a reaction occurs quickly. In reality, spontaneity refers to the direction in which a process naturally proceeds, and it can be either rapid or extremely slow. For example, the rusting of iron is a spontaneous process, yet it occurs over a lengthy period.

The sign of the Gibbs free energy change (\( \triangle G \)) in a reaction determines spontaneity. If \( \triangle G < 0 \), the process is spontaneous; if \( \triangle G > 0 \), it is non-spontaneous and requires external energy; and if \( \triangle G = 0 \), the system is at equilibrium. To deepen students' understanding, educators might use analogies, such as comparing spontaneous reactions to a ball rolling downhill without any push needed. In contrast, non-spontaneous reactions are like pushing a ball uphill, requiring effort. Furthermore, explaining that the rate of a reaction (how fast it occurs) is not the same as spontaneity can help students differentiate between kinetics and thermodynamics—two key aspects of chemical reactions.

Standard Conditions in Gibbs Free Energy Calculations

Standard conditions, usually denoted by the degree sign (\( ^{\text{o}} \) ), refer to a specific set of conditions under which the measurements of thermodynamic properties are reported and compared. For Gibbs free energy changes, standard conditions mean a pressure of 1 bar (or approximately 1 atmosphere) for gases, a concentration of 1 mol/L for solutions, and pure substances in their most stable forms at 1 bar and 25°C (298.15K). These conditions are not to be confused with standard temperature and pressure (STP), which is defined as 0°C and 1 atmosphere.

The concept of standard conditions is critical because it provides a reference point that enables scientists and students to compare the thermodynamic properties of reactions consistently. When students calculate Gibbs free energy changes under standard conditions, as in the exercise above, they are able to assess the spontaneity of reactions without additional variables that could complicate their understanding. By emphasizing the importance of these conditions, we can help students realize that the \( \triangle G \) values they calculate are not universal for all situations but pertain to those specific conditions. Therefore, it's crucial to always check the conditions under which a reaction occurs to determine the correct spontaneity and feasibility of the process.