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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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