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 is 435 kJ mol-1 and 192 kJ mol-1 respectively. What is the bond energy (in kJ mol-1) of HBr?

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a

736

b

368

c

259

d

518

answer is B.

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Detailed Solution

H2(g)+Br2(g)2HBr(g);ΔH=109ΔH=(BE)R(BE)P=BEHH+BEBrBr2BEHBr109=(435)+(192)2BEHBrBEHBr=368kJmol1

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