Problem 84 - Chemistry: The Central Science in SI Units, Global Edition

Question

The standard enthalpies of formation of \(\mathrm{ClO}\) and \(\mathrm{ClO}_{2}\) are 101 and \(102 \mathrm{~kJ} / \mathrm{mol}\), respectively. Using these data and the thermodynamic data in Appendix C, calculate the overall enthalpy change for each step in the following catalytic cycle: $$ \begin{array}{l} \mathrm{ClO}(g)+\mathrm{O}_{3}(g) \longrightarrow \mathrm{ClO}_{2}(g)+\mathrm{O}_{2}(g) \\ \mathrm{ClO}_{2}(g)+\mathrm{O}(g) \longrightarrow \mathrm{ClO}(g)+\mathrm{O}_{2}(g) \end{array} $$ What is the enthalpy change for the overall reaction that results from these two steps?

Step-by-Step Solution

Answer

The enthalpy changes for the two steps in the catalytic cycle are -141.7 kJ/mol and -250.2 kJ/mol. The overall enthalpy change for the reaction resulting from these two steps is -391.9 kJ/mol.

Step 1: Calculate the enthalpy change for the first step of the catalytic cycle

Using the standard enthalpies of formation and thermodynamic data, we can find the enthalpy change for the reaction: ClO(g) + O3(g) -> ClO2(g) + O2(g) Using the given information, the standard enthalpies of formation for the involved species are: ΔHf[ClO] = 101 kJ/mol ΔHf[ClO2] = 102 kJ/mol ΔHf[O3] = 142.7 kJ/mol (from Appendix C) ΔHf[O2] = 0 kJ/mol (element in its standard state) Now, we will use the formula for calculating the enthalpy change of the reaction: ΔHreaction = Σ(ΔHf[products]) - Σ(ΔHf[reactants]) For the first step: ΔH1 = (ΔHf[ClO2] + ΔHf[O2]) - (ΔHf[ClO] + ΔHf[O3]) ΔH1 = (102 + 0) - (101 + 142.7) = -141.7 kJ/mol

Step 2: Calculate the enthalpy change for the second step of the catalytic cycle

For the second step reaction: ClO2(g) + O(g) -> ClO(g) + O2(g) In this case, the standard enthalpies of formation for the involved species are: ΔHf[O] = 249.2 kJ/mol (from Appendix C) Again, using the formula for calculating the enthalpy change of the reaction: ΔHreaction = Σ(ΔHf[products]) - Σ(ΔHf[reactants]) For the second step: ΔH2 = (ΔHf[ClO] + ΔHf[O2]) - (ΔHf[ClO2] + ΔHf[O]) ΔH2 = (101 + 0) - (102 + 249.2) = -250.2 kJ/mol

Step 3: Calculate the overall enthalpy change for the catalytic cycle

Finally, we can find the overall enthalpy change for the catalytic cycle by adding the enthalpy changes of the two steps: ΔHoverall = ΔH1 + ΔH2 ΔHoverall = (-141.7 kJ/mol) + (-250.2 kJ/mol) = -391.9 kJ/mol The enthalpy change for the overall reaction that results from these two steps is -391.9 kJ/mol.

Key Concepts

Enthalpies of Formation

Understanding enthalpies of formation is key when dealing with chemical reactions and enthalpy changes. The enthalpy of formation, denoted as \(\Delta H_f\), refers to the energy change when one mole of a compound is formed from its elements in their standard states. Standard states typically mean at 1 pressure and 25°C, which helps ensure consistency in thermodynamic calculations.

In the exercise, the enthalpies of formation for \(\mathrm{ClO}\) and \(\mathrm{ClO}_2\) are given as 101 \(\mathrm{kJ/mol}\) and 102 \(\mathrm{kJ/mol}\) respectively. Such values provide insight into the stability of these molecules, as more energy (enthalpy) required to form them generally signals greater intrinsic stability.

• \(\mathrm{ClO}\): 101 \(\mathrm{kJ/mol}\)
• \(\mathrm{ClO}_2\): 102 \(\mathrm{kJ/mol}\)
• \(\mathrm{O}_3\): 142.7 \(\mathrm{kJ/mol}\)
• \(\mathrm{O}_2\): 0 \(\mathrm{kJ/mol}\), as it is in its elemental form

By using these standard enthalpies, we can determine the enthalpy change for reactions, such as each step of the catalytic cycle discussed in the problem.

Catalytic Cycle

A catalytic cycle refers to a sequence of chemical reactions where a catalyst facilitates the conversion of reactants to products, and is regenerated throughout the cycle. In this exercise, chlorine monoxide (ClO) serves as the catalyst.

The catalytic cycle involves two steps:

• \(\mathrm{ClO}(g) + \mathrm{O}_3(g) \rightarrow \mathrm{ClO}_2(g) + \mathrm{O}_2(g)\)
• \(\mathrm{ClO}_2(g) + \mathrm{O}(g) \rightarrow \mathrm{ClO}(g) + \mathrm{O}_2(g)\)

The overall result of the catalytic cycle is the transformation of ozone \((\mathrm{O}_3)\) and atomic oxygen \((\mathrm{O})\) into two molecules of molecular oxygen \((\mathrm{O}_2)\). It’s essential to note that the catalyst \(\mathrm{ClO}\) is not consumed but rather regenerated, allowing it to continue catalyzing more reactions. This is a common feature of catalytic processes: efficiency is enhanced as catalysts are not depleted in the reactions they facilitate.

Thermodynamic Data

Thermodynamic data provides essential physical information for understanding chemical processes, allowing us to predict how reactions proceed and calculate energy changes. For a reaction in a catalytic cycle, such data enables the calculation of individual and overall enthalpy changes, vital for determining the reaction's feasibility.

To compute the enthalpy change for each step in the given catalytic cycle, we make use of the standard enthalpies of formation for all reactants and products. The formula: \[\Delta H_{reaction} = \Sigma(\Delta H_f[\text{products}]) - \Sigma(\Delta H_f[\text{reactants}])\]permits these calculations by summing the enthalpies of products minus those for reactants.

In the given exercise:

• The first step results in an enthalpy change \(\Delta H_1 = -141.7 \mathrm{kJ/mol}\).
• The second step yields \(\Delta H_2 = -250.2 \mathrm{kJ/mol}\).

Adding these provides the overall enthalpy change for the catalytic cycle, which is \(-391.9 \mathrm{kJ/mol}\). By understanding these calculations and principles, students can robustly tackle complex chemical reactions.