QuestionAugust 6, 2025

What is the enthalpy change (Delta HDelta H) for the reaction H_(2)(g)+Cl_(2)(g)arrow 2HCl(g) if the bond energies are: H-H=436kJ/mol Cl-Cl=243kJ/mol H-Cl=431kJ/mol +183kJ/mol -183kJ/mol -862kJ/mol +862kJ/mol

What is the enthalpy change (Delta HDelta H) for the reaction H_(2)(g)+Cl_(2)(g)arrow 2HCl(g) if the bond energies are: H-H=436kJ/mol Cl-Cl=243kJ/mol H-Cl=431kJ/mol +183kJ/mol -183kJ/mol -862kJ/mol +862kJ/mol
What is the enthalpy change (Delta HDelta H) for the reaction H_(2)(g)+Cl_(2)(g)arrow 2HCl(g) if the bond energies are: H-H=436kJ/mol Cl-Cl=243kJ/mol H-Cl=431kJ/mol +183kJ/mol -183kJ/mol -862kJ/mol +862kJ/mol

Solution
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Answer

-183 \text{ kJ/mol} Explanation 1. Calculate energy required to break bonds Break 1 mole of H-H and 1 mole of Cl-Cl: 436 + 243 = 679 \text{ kJ/mol}. 2. Calculate energy released by forming bonds Form 2 moles of H-Cl: 2 \times 431 = 862 \text{ kJ/mol}. 3. Calculate enthalpy change \Delta H = \text{Energy required} - \text{Energy released} = 679 - 862 = -183 \text{ kJ/mol}.

Explanation

1. Calculate energy required to break bonds<br /> Break 1 mole of $H-H$ and 1 mole of $Cl-Cl$: $436 + 243 = 679 \text{ kJ/mol}$.<br />2. Calculate energy released by forming bonds<br /> Form 2 moles of $H-Cl$: $2 \times 431 = 862 \text{ kJ/mol}$.<br />3. Calculate enthalpy change<br /> $\Delta H = \text{Energy required} - \text{Energy released} = 679 - 862 = -183 \text{ kJ/mol}$.
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