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Q.

At standard conditions, if the change in the enthalpy for the following reaction is –109 kJ mol-1.H2(g) + Br2g → 2HBr(g)Given that bond energy of H2 and Br2 is435 kJ mol-1 and 192 kJ mol-1, respectively,what is the bond energy (in kJ mol-1) of HBr?

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Detailed Solution

H2(g)+Br2(g)⟶2HBr(g);ΔH=−109ΔH=∑(BE)R−∑(BE)P=BEH−H+BEBr−Br−2BEH−Br−109=(435)+(192)−2BEH−BrBEH−Br=368kJmol−1
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