Question

# Use bond energies to estimate the enthalpy of formation of HBr(g). BE(H-H) = 436 kJ/mol BE(Br-Br) = 192 kJ/mol BE(H-Br) = 366 kJ/mol

A

+152 kJ/mol

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B

−104 kJ/mol

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C

+262 kJ/mol

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D

−52 kJ/mol

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Solution

## Step 1:Enthalpy of reaction is the difference in energy absorbed in breaking bonds in reactants and energy released when new bonds are formed in products. ΔHo=∑bondenergy(reactants)−∑bondenergy(products) Step 2: Since two moles of HBr are formed in the reaction, for 1 mole 52 kJ/mol energy is required. Since energy is released during bond formation the value would be negative. ΔHo= 436 + 192 - 2× 366 = -104 KJ/mol Hence, option (B) is correct.

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