Problem 62 - Chemistry: The Central Science

Question

Using data from Appendix \(C\), calculate the change in Gibbs free energy for each of the following reactions. In each case indicate whether the reaction is spontaneous at \(298 \mathrm{~K}\) under standard conditions. (a) \(2 \mathrm{Ag}(s)+\mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{AgCl}(s)\) (b) \(\mathrm{P}_{4} \mathrm{O}_{4}(s)+16 \mathrm{H}_{2}(g) \longrightarrow 4 \mathrm{PH}_{3}(g)+10 \mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{CH}_{4}(g)+4 \mathrm{~F}_{2}(g) \longrightarrow \mathrm{CF}_{4}(g)+4 \mathrm{HF}(g)\) (d) \(2 \mathrm{H}_{2} \mathrm{O}_{2}(l) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)\)

Step-by-Step Solution

Answer

(a) \( \Delta G_r° = -65.4 \mathrm{~kJ/mol} \), spontaneous (b) \( \Delta G_r° = -3117.5 \mathrm{~kJ/mol} \), spontaneous (c) \( \Delta G_r° = -1200.1 \mathrm{~kJ/mol} \), spontaneous (d) \( \Delta G_r° = 162.3 \mathrm{~kJ/mol} \), not spontaneous

Step 1: Use the Gibbs free energy of formation values

For each compound in the reactions, look up the values of Gibbs free energy of formation (∆Gf°) from Appendix C. We will use these values to calculate the change in Gibbs free energy for each reaction (∆Gr°) using the following equation: ∆Gr° = Σ n ∆Gf° (products) - Σ n ∆Gf° (reactants) Where n is the stoichiometric coefficient of each compound.

Step 2: Calculate ∆Gr° for reaction (a)

For reaction (a), look up the values of ∆Gfº for Ag(s), Cl₂(g), and AgCl(s) from Appendix C. \( ∆Gr° = 2 \cdot ∆Gf° (AgCl) - [2 \cdot ∆Gf°(Ag) + ∆Gf°(Cl₂)] \) Plug in the values and calculate ∆Gr° (a).

Step 3: Determine the spontaneity of reaction (a)

If ∆Gr° (a) is negative, then the reaction is spontaneous at 298 K under standard conditions. If it is positive, the reaction is not spontaneous.

Step 4: Calculate ∆Gr° for reaction (b)

For reaction (b), look up the values of ∆Gf° for P₄O₄(s), H₂(g), PH₃(g), and H₂O(g) from Appendix C. \( ∆Gr° = [4 \cdot ∆Gf°(PH₃) + 10 \cdot ∆Gf°(H₂O)] - [∆Gf°(P₄O₄) + 16 \cdot ∆Gf°(H₂)] \) Plug in the values and calculate ∆Gr° (b).

Step 5: Determine the spontaneity of reaction (b)

If ∆Gr° (b) is negative, then the reaction is spontaneous at 298 K under standard conditions. If it is positive, the reaction is not spontaneous.

Step 6: Calculate ∆Gr° for reaction (c)

For reaction (c), look up the values of ∆Gf° for CH₄(g), F₂(g), CF₄(g), and HF(g) from Appendix C. \( ∆Gr° = [∆Gf°(CF₄) + 4 \cdot ∆Gf°(HF)] - [∆Gf°(CH₄) + 4 \cdot ∆Gf°(F₂)] \) Plug in the values and calculate ∆Gr° (c).

Step 7: Determine the spontaneity of reaction (c)

If ∆Gr° (c) is negative, then the reaction is spontaneous at 298 K under standard conditions. If it is positive, the reaction is not spontaneous.

Step 8: Calculate ∆Gr° for reaction (d)

For reaction (d), look up the values of ∆Gf° for H₂O₂(l), H₂O(l), and O₂(g) from Appendix C. \( ∆Gr° = [2 \cdot ∆Gf°(H₂O) + ∆Gf°(O₂)] - 2 \cdot ∆Gf°(H₂O₂) \) Plug in the values and calculate ∆Gr° (d).

Step 9: Determine the spontaneity of reaction (d)

If ∆Gr° (d) is negative, then the reaction is spontaneous at 298 K under standard conditions. If it is positive, the reaction is not spontaneous. After completing steps 2-9, we now have the change in Gibbs free energy for each reaction and have determined whether each reaction is spontaneous or not at 298 K under standard conditions.

Key Concepts

Chemical Spontaneity

Understanding the concept of chemical spontaneity is crucial for students who are studying thermodynamics in the context of chemistry. Chemical spontaneity refers to whether a reaction will occur naturally without any external energy input. A reaction is said to be spontaneous if it proceeds on its own, given enough time, under a set of conditions.

The determination of spontaneity hinges on a factor called Gibbs free energy, represented by the symbol \( G \). The change in Gibbs free energy (\( \Delta G \)) during a process is the definitive indicator of spontaneity. If \( \Delta G \)) is negative, the process is spontaneous; if positive, it is non-spontaneous; and if zero, the system is at equilibrium. This ties directly into the step-by-step solution where we calculate the change in Gibbs free energy to determine if reactions under study are spontaneous at standard conditions.

To further enhance understanding, it's essential to delve into how \( \Delta G \)) is calculated. The equation \( \Delta G = \Delta H - T\Delta S \) helps predict spontaneity, where \( \Delta H \) is the change in enthalpy, \( T \) the temperature in Kelvin, and \( \Delta S \) the change in entropy. For a reaction at constant temperature and pressure, if the process leads to a loss of free energy (\( \Delta G < 0 \)), it can do work and is, therefore, spontaneous.

Standard Conditions

In the realm of chemistry, specially in thermodynamic calculations, the term standard conditions is frequently encountered. Standard conditions refer to an established set of environmental parameters that allow scientists and engineers to compare different chemical reactions and processes uniformly.

Typically, standard conditions mean a pressure of 1 atmosphere (101.325 kPa) and a temperature of 298 K (about 25°C). It's important to note that this temperature is close to room temperature, making it practical for many experiments and calculations.

Understanding standard conditions matters because the Gibbs free energy of formation values for various substances provided in reference materials, like the Appendix C in our exercise, are given at these standard conditions. It ensures that when we calculate the change in Gibbs free energy (\( \Delta G^\circ \)) for a reaction to determine spontaneity, we are using comparable data. This concept is embedded in our step-by-step solution where it specifies that we are to consider the spontaneity at \("298 K under standard conditions"\).

Moreover, at these conditions, elements in their stable states are assigned a Gibbs free energy of formation (\( \Delta G_f^\circ \)) of zero. This standardization makes it easier to discern which substances contribute to the overall free energy change in a reaction.

Free Energy of Formation

The free energy of formation, denoted as \( \Delta G_f^\circ \), is another fundamental concept in thermodynamics. It represents the amount of energy change when 1 mole of a compound is formed from its elements in their standard states under standard conditions (1 atm pressure and 298 K temperature).

In the context of our exercise, using the \( \Delta G_f^\circ \)) values is paramount for calculating the change in Gibbs free energy (\( \Delta G_r^\circ \)) of a particular reaction. This value determines whether a chemical reaction is thermodynamically favorable or not.

To gain deeper insight, students should note that the steps for calculating the \( \Delta G_r^\circ \) involve summing the products of each substance's \( \Delta G_f^\circ \) values and their stoichiometric coefficients, then subtracting the same for reactants. An understanding of this calculation method can help to predict the behavior of a chemical system and guide experimental design, leading to the conclusion regarding spontaneity made in each step of the solution process.

Finally, a crucial point for students to remember is that while the \( \Delta G_f^\circ \) values are essential for predicting the direction of a reaction, they do not provide information on the rate of the reaction or how quickly equilibrium will be reached. That aspect falls under the purview of reaction kinetics, which is a separate but equally important area of study.