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------------------------------------------------ Consider the following reaction:

\[ \text{H}_2\text{O (g)} + \text{Cl}_2\text{O (g)} \rightleftharpoons 2\text{HOCl (g)} \]

Given data:
- \( K_{298} = 0.090 \)
- For \(\text{Cl}_2\text{O(g)}\), \( \Delta G_f = 97.9 \, \text{kJ/mol} \)
- \( \Delta H = 80.3 \, \text{kJ/mol} \)
- \( S = 266.1 \, \text{J/K} \cdot \text{mol} \)

(b) Use bond energy values (refer to Table 8.4) to estimate \( \Delta H^\circ \) for the reaction.

To estimate ΔHo for the reaction, we can use the bond energy values. The reaction involves breaking the bonds in H2O and Cl2O and forming the bonds in HOCl. The bond energy values for H-O, Cl-O, and Cl-Cl are 463 kJ/mol, 243 kJ/mol, and 240 kJ/mol, respectively.

We can calculate the enthalpy change using the formula: ΔHo = sum(bonds broken) - sum(bonds formed).

Bonds broken: 2(H-O) + 1(Cl-Cl) = 2(463 kJ/mol) + 1(240 kJ/mol) = 1166 kJ/mol
Bonds formed: 2(Cl-O) + 2(H-O) = 2(243 kJ/mol) + 2(463 kJ/mol) = 1412 kJ/mol

ΔHo = sum(bonds broken) - sum(bonds formed) = 1166 kJ/mol - 1412 kJ/mol = -246 kJ/mol

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