Question

set the bond strengths: a - a 300 kj/mol b - b 200 kj/mol a - b 380 kj/mol using these bond enthalpies (energies), calculate \\( \delta h_{rxn} \\) for the reaction \\( a_{2}+b_{2}\to 2ab \\)

Explanation:

Step1: Calculate energy to break bonds

To break 1 mol of \( A - A \) (300 kJ/mol) and 1 mol of \( B - B \) (200 kJ/mol), total energy input is \( 300 + 200 = 500 \) kJ/mol.

Step2: Calculate energy released from forming bonds

Forming 2 mol of \( A - B \) (each 380 kJ/mol), total energy released is \( 2\times380 = 760 \) kJ/mol.

Step3: Calculate \( \Delta H_{\text{rxn}} \)

\( \Delta H_{\text{rxn}} = \text{Energy released} - \text{Energy input} = 760 - 500 = 260 \)? Wait, no—wait, bond enthalpy formula: \( \Delta H_{\text{rxn}} = \sum \text{bond enthalpies of reactants broken} - \sum \text{bond enthalpies of products formed} \). Wait, correction: \( \Delta H_{\text{rxn}} = (\text{Energy to break reactant bonds}) - (\text{Energy released from product bonds}) \). Wait, reactants: \( A_2 \) (1 A - A bond) and \( B_2 \) (1 B - B bond). Products: 2 AB (2 A - B bonds). So \( \Delta H_{\text{rxn}} = (300 + 200) - (2\times380) = 500 - 760 = -260 \) kJ/mol.

Answer:

\(-260\) kJ/mol