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Question

If the bond energies of H–H, Br–Br, and H–Br are 433, 192, and 364 \( \text{kJ/mol} \) respectively, then \( \Delta H^\circ \) for the reaction: \( H_2(g) + Br_2(g) \rightarrow 2HBr(g) \) is

\( -261 \text{ kJ/mol} \)

\( +103 \text{ kJ/mol} \)

\( +261 \text{ kJ/mol} \)

\( -103 \text{ kJ/mol} \)

Solution

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