QuestionApril 18, 2026

Bond Enthalpies in kJ/mol ABarrow A+B564 ACarrow A+C645 BCarrow B+C1,305 For the following reaction: A+BCarrow AC+B What is the change in enthalpy for the reaction? square

Bond Enthalpies in kJ/mol ABarrow A+B564 ACarrow A+C645 BCarrow B+C1,305 For the following reaction: A+BCarrow AC+B What is the change in enthalpy for the reaction? square
Bond Enthalpies in kJ/mol
ABarrow A+B564
ACarrow A+C645
BCarrow B+C1,305
For the following reaction:
A+BCarrow AC+B
What is the change in enthalpy for the reaction?
square

Solution
4.7(193 votes)

Answer

+660\ \text{kJ/mol} Explanation 1. Identify bonds broken and formed Bonds broken: BC (1,305 kJ/mol). Bonds formed: AC (645 kJ/mol). 2. Apply bond enthalpy formula **\Delta H = \text{Bonds broken} - \text{Bonds formed}** \Delta H = 1305 - 645 = 660\ \text{kJ/mol}

Explanation

1. Identify bonds broken and formed <br /> Bonds broken: $BC$ (1,305 kJ/mol). <br /> Bonds formed: $AC$ (645 kJ/mol). <br /><br />2. Apply bond enthalpy formula <br /> **$\Delta H = \text{Bonds broken} - \text{Bonds formed}$** <br /> $\Delta H = 1305 - 645 = 660\ \text{kJ/mol}$
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