Problem 58 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

Calculate the free energy change for the reaction: \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{HCl}(\mathrm{g})\) by using the following data at the reaction temperature of \(27^{\circ} \mathrm{C}\). Bond enthalpies \((\mathrm{kJ} / \mathrm{mol})\) : \(\mathrm{H}-\mathrm{H}=435 ; \mathrm{Cl}-\mathrm{Cl}=240 ; \mathrm{H}-\mathrm{Cl}=430\) Entropies (J/K-mol): \(\mathrm{H}_{2}=130 ; \mathrm{Cl}_{2}=222 ; \mathrm{HCl}=186\) (a) \(-185 \mathrm{~kJ}\) (b) \(-20 \mathrm{~kJ}\) (c) \(-179 \mathrm{~kJ}\) (d) \(-191 \mathrm{~kJ}\)

Step-by-Step Solution

Answer

The correct answer is \(\Delta G^\circ = -185 \,\text{kJ}\), which corresponds to option (a).

Step 1: Calculating the Enthalpy Change \(\Delta H^\circ\) of the Reaction

To calculate the enthalpy change, use the bond enthalpies given for the bonds broken and formed during the reaction. Bonds broken include one H-H bond and one Cl-Cl bond, while the bonds formed include two H-Cl bonds. The enthalpy change is the energy needed to break bonds minus the energy released when bonds are formed. \(\Delta H^\circ = (\text{Bonds broken}) - (\text{Bonds formed})\). The energy for bonds broken is \(\text{H-H} + \text{Cl-Cl} = 435 \,\text{kJ/mol} + 240 \,\text{kJ/mol}\). The energy for bonds formed is \(2 \times \text{H-Cl} = 2 \times 430 \,\text{kJ/mol}\). Hence, \(\Delta H^\circ = (435 + 240) - (2 \times 430)\).

Step 2: Calculating the Entropy Change \(\Delta S^\circ\) of the Reaction

The entropy change for the reaction is calculated by subtracting the sum of the entropies of the reactants from the sum of the entropies of the products. \(\Delta S^\circ = (\text{Products entropy}) - (\text{Reactants entropy})\). The entropy for reactants is \(\text{H}_2 + \text{Cl}_2 = 130 \,\text{J/K}\cdot\text{mol} + 222 \,\text{J/K}\cdot\text{mol}\). The entropy for products is \(2 \times \text{HCl} = 2 \times 186 \,\text{J/K}\cdot\text{mol}\). Hence, \(\Delta S^\circ = (2 \times 186) - (130 + 222)\).

Step 3: Convert Entropy Change to kJ/K-mol

Entropy is given in \(\text{J/K}\cdot\text{mol}\), but enthalpy is in \(\text{kJ/mol}\), so convert the entropy from J to kJ by dividing by 1000. \(\Delta S^\circ_{(\text{kJ/K}\cdot\text{mol})} = \Delta S^\circ_{(\text{J/K}\cdot\text{mol})} / 1000\).

Step 4: Calculating Gibbs Free Energy Change \(\Delta G^\circ\) Using Standard Free Energy Change Equation

Now use the formula for Gibbs free energy change: \(\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ\), where \(T\) is the temperature in Kelvin. First convert Celsius to Kelvin by adding 273.15. Then plug the values into the formula to find \(\Delta G^\circ\).

Step 5: Selecting the Correct Answer

After calculating the value of \(\Delta G^\circ\), select the answer choice that closest matches your calculated value from the provided options (a) through (d).

Key Concepts

Enthalpy Change

Enthalpy change, designated as \( \Delta H^\circ \), is a fundamental concept in chemistry that represents the heat content change during a chemical reaction. It is a state function, meaning its value is determined only by the current state of the system and not by how that state was reached. The enthalpy change for a reaction can be either positive (endothermic reaction, absorbing heat) or negative (exothermic reaction, releasing heat).

To calculate the \( \Delta H^\circ \) for a reaction, such as the formation of hydrogen chloride (HCl) from hydrogen (\( \mathrm{H}_2 \)) and chlorine (\( \mathrm{Cl}_2 \) gases), we consider the bond enthalpies of the reactants and products. Bond enthalpy refers to the energy required to break a particular type of bond between atoms in the gaseous state. When bonds are broken, energy is absorbed; when bonds are formed, energy is released. The overall \( \Delta H^\circ \) is the sum of energy needed to break bonds minus the energy released from forming bonds.

Bond Enthalpy

Bond enthalpy, or bond dissociation energy, is the energy needed to break one mole of a specific type of chemical bond in a gaseous substance. It is an indication of the bond strength; higher bond enthalpies mean stronger bonds. Each bond type, such as \( \mathrm{H}-\mathrm{H} \) or \( \mathrm{H}-\mathrm{Cl} \), has a characteristic bond enthalpy value associated with it.

To solve physical chemistry problems involving changes in enthalpy, it is essential to understand how to apply bond enthalpies. For example, when calculating the \( \Delta H^\circ \) in the reaction forming HCl, you must subtract the total bond enthalpies of bonds formed from the total bond enthalpies of bonds broken. If the resulting \( \Delta H^\circ \) is negative, the reaction releases energy, indicating it's exothermic.

Entropy Change

Entropy change, denoted as \( \Delta S^\circ \), measures the disorder or randomness of a system. Entropy is an extensive property that increases with the number of ways particles can be arranged while still maintaining the same energy state. In chemical thermodynamics, the entropy change of a reaction can predict the spontaneity of the process along with the Gibbs free energy change.

Calculating \( \Delta S^\circ \) involves the entropy of products minus the entropy of reactants. For instance, in our reaction forming HCl, we determine \( \Delta S^\circ \) by using the provided entropy values for \( \mathrm{H}_2 \), \( \mathrm{Cl}_2 \) and \( \mathrm{HCl} \). A positive \( \Delta S^\circ \) suggests increased disorder after a reaction, while a negative value suggests increased order.

Chemical Thermodynamics

Chemical thermodynamics encompasses the principles that govern the energy changes during chemical reactions and changes of state. The study integrates concepts of enthalpy, entropy, and temperature to predict reaction spontaneity, equilibrium, and system's heat exchanges.

Gibbs free energy change, \( \Delta G^\circ \), is a thermodynamic function derived from enthalpy and entropy changes. It reflects the maximum amount of work a thermodynamic system can perform at constant temperature and pressure. A negative value for \( \Delta G^\circ \) indicates a spontaneous reaction under specified conditions, whereas a positive value suggests non-spontaneity. The equation \( \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \)—where \( T \) is the temperature in Kelvin—couples all these variables to assess the feasibility of chemical processes.

Physical Chemistry Problems

Physical chemistry problems often require the integration of concepts like bond enthalpy, entropy, and Gibbs free energy to solve complex chemical reactions. A well-rounded understanding of these components allows students to approach these problems methodically and confidently.

When tackling problems such as calculating Gibbs free energy change, it is crucial to follow a step-by-step approach. Begin by computing bond enthalpies to find \( \Delta H^\circ \) and use the given entropies to calculate \( \Delta S^\circ \). Convert units where necessary to maintain consistency, for example, converting entropy units from J/K-mol to kJ/K-mol. Finally, use the equation \( \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \) to find the Gibbs free energy change. By mastering these steps, students can successfully solve even the most daunting physical chemistry problems.