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?

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?

  1. A

    259

  2. B

    368

  3. C

    736

  4. D

    518

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    Solution:

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

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