Question

Use the thermodynamicquantities given in Appendix C to calculate \(\Delta G^{\circ}, \Delta H^{\circ}\), and \(\Delta S^{\circ}\) for the reaction corresponding to roasting of \(\mathrm{PbO}\) in a CO atmosphere (Equation 23.5). Approximate the thermodynamic quantities of \(\mathrm{Pb}(I)\) using the thermodynamic properties of \(\mathrm{Pb}(s)\). Is this reaction spontaneous at \(25^{\circ} \mathrm{C}\) under standard conditions? Is it exothermic or endothermic?

Short answer

The calculated values for the reaction are \(\Delta G^{\circ} = -69,953.45 \text{J/mol}\), \(\Delta H^{\circ} = -65.7 \text{kJ/mol}\), and \(\Delta S^{\circ} = 14.3 \text{J/mol·K}\). Since \(\Delta G^{\circ} < 0\), the reaction is spontaneous under standard conditions at \(25^{\circ}\mathrm{C}\). Moreover, the negative value of \(\Delta H^{\circ}\) indicates that the reaction is exothermic.

Step-by-step solution

Calculate \(\Delta H^{\circ}\)

To calculate the change in enthalpy (\(\Delta H^{\circ}\)), use the following formula: \[\Delta H^{\circ} = \sum H^{\circ}_{products} - \sum H^{\circ}_{reactants}\] Using Appendix C, find the enthalpy of formation for the reactants and products. For PbO(s): -217.3 kJ/mol For CO(g): -110.5 kJ/mol For Pb(l): (Use the values for Pb(s)) 0 kJ/mol For CO2(g): -393.5 kJ/mol Now, substitute these values into the equation: \[\Delta H^{\circ} = (0 + (-393.5)) - ((-217.3) + (-110.5))\]

Compute \(\Delta H^{\circ}\)

Solve for \(\Delta H^{\circ}\): \[\Delta H^{\circ} = -393.5 - (-217.3) - (-110.5)\] \[\Delta H^{\circ} = -393.5 + 217.3 + 110.5\] \[\Delta H^{\circ} = -65.7 \text{kJ/mol}\] The negative value for \(\Delta H^{\circ}\) indicates that the reaction is exothermic.

Calculate \(\Delta S^{\circ}\)

To calculate the change in entropy (\(\Delta S^{\circ}\)), use the following formula: \[\Delta S^{\circ} = \sum S^{\circ}_{products} - \sum S^{\circ}_{reactants}\] Using Appendix C, find the entropy for the reactants and products. For PbO(s): 66.5 J/mol·K For CO(g): 197.7 J/mol·K For Pb(l): (Use the values for Pb(s)) 64.8 J/mol·K For CO2(g): 213.7 J/mol·K Now, substitute these values into the equation: \[\Delta S^{\circ} = (64.8 + 213.7) - (66.5 + 197.7)\]

Compute \(\Delta S^{\circ}\)

Solve for \(\Delta S^{\circ}\): \[\Delta S^{\circ} = 278.5 - 264.2 = 14.3 \text{J/mol·K}\]

Calculate \(\Delta G^{\circ}\)

To calculate the change in Gibbs free energy (\(\Delta G^{\circ}\)), use the following formula: \[\Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ}\] where \(T\) is the temperature in Kelvin. In this case, the temperature is 25°C, which is 298.15 K. Substitute the values for \(\Delta H^{\circ}\), \(\Delta S^{\circ}\), and \(T\) into the equation: \[\Delta G^{\circ} = -65.7 \times 10^3 \text{J/mol} - (298.15 \text{K} \times 14.3 \text{J/mol·K})\]

Compute \(\Delta G^{\circ}\)

Solve for \(\Delta G^{\circ}\): \[\Delta G^{\circ} = -65.7 \times 10^3 - (4253.45) = -69,953.45 \text{J/mol}\] Since \(\Delta G^{\circ} < 0\), the reaction is spontaneous under standard conditions at \(25^{\circ}\mathrm{C}\). The reaction is exothermic as the value of \(\Delta H^{\circ}\) is negative.

Key concepts

Enthalpy

Enthalpy, symbolized as \( H \), is a concept in thermodynamics representing the total heat content of a system. It is an essential quantity because it helps us determine the energy changes occurring during a chemical reaction. Imagine it as the account balance of heat which tells us if we are gaining or losing energy during the reaction process. The change in enthalpy \( \Delta H \) is calculated using the formula:

• \( \Delta H^{\circ} = \sum H^{\circ}_{\text{products}} - \sum H^{\circ}_{\text{reactants}} \)

This equation indicates that we subtract the total enthalpy of the reactants from the total enthalpy of the products. If \( \Delta H \) is negative, the reaction releases heat, qualifying it as exothermic. On the other hand, a positive \( \Delta H \) means the reaction absorbs heat, making it endothermic. In our specific example, the negative \( \Delta H^{\circ} = -65.7 \text{kJ/mol} \) shows that energy is released, confirming an exothermic reaction.

Entropy

Entropy, symbolized as \( S \), is a measure of randomness or disorder in a system. It's an abstract concept related to how spread out or scrambled the energy of a system is. Higher entropy means greater disorder, while lower entropy indicates more order. The change in entropy \( \Delta S \) of a reaction is calculated using:

• \( \Delta S^{\circ} = \sum S^{\circ}_{\text{products}} - \sum S^{\circ}_{\text{reactants}} \)

This formula tells us that by subtracting the total entropy of reactants from that of the products, we can find out how the disorder changes throughout the reaction. A positive \( \Delta S \) suggests that the product molecules are more disordered compared to the reactants. In our example, the positive \( \Delta S^{\circ} = 14.3 \text{J/mol·K} \) indicates that randomness has indeed increased.

Gibbs Free Energy

Gibbs Free Energy, represented by \( G \), merges enthalpy and entropy into a single value that reveals the spontaneity of a process. It's the magical thermodynamic quantity that helps us determine whether a reaction can occur without external intervention. The equation for calculating the change in Gibbs Free Energy \( \Delta G \) is:

• \( \Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} \)

Where \( T \) is the temperature in Kelvin. If \( \Delta G \) is negative, the reaction is spontaneous, meaning it can proceed on its own. In our case, with \( \Delta G^{\circ} = -69,953.45 \text{J/mol} \), the reaction is spontaneous under standard conditions at \( 25^{\circ}\mathrm{C} \). This counts as more thermodynamic insight confirming that our system energetically prefers the reaction to move forward.

Spontaneity

Spontaneity in a chemical reaction refers to the ability of the process to proceed without any help from external forces or energy. Thermodynamics gives us \( \Delta G \) as the metric for determining spontaneity. A reaction having a negative \( \Delta G \) signifies spontaneity, whereas a positive \( \Delta G \) suggests the need for additional energy to proceed. Spontaneity doesn’t imply speed; a spontaneous reaction can either be fast or slow. In our example, since the \( \Delta G^{\circ} \) is approximately \(-70 \text{kJ/mol} \), we know the reaction will proceed by itself at room temperature. Embracing the concept of spontaneity helps us in predicting and understanding the nature of chemical reactions in practical environments.

Exothermic reaction

An exothermic reaction is one that releases energy, usually in the form of heat. This type of reaction causes the surrounding temperature to increase as the system sheds some of its energy outward. Exothermic processes are often recognized by the negative sign of \( \Delta H \). Such reactions are common in everyday life, ranging from the simple combustion of fuels to complex biochemical reactions in our bodies.In our particular example, the reaction between \( \mathrm{PbO} \) and \( \text{CO} \) produces \( \text{Pb} \) and \( \text{CO}_2 \) and results in \( \Delta H^{\circ} = -65.7 \text{kJ/mol} \), clearly marking the reaction as exothermic. This means the chemical process releases energy into the environment, thus aligning with the universal principle of systems moving towards lower energy and greater stability. Understanding this concept allows us to harness such reactions effectively in industrial and scientific applications.