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Chapter 18: Problem 84

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?

Short Answer

Expert verified
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 by step solution

01

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
02

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
03

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.

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Most popular questions from this chapter

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