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Question

Given that bond energies of
HH and ClCl are 430 kJ mol1 and 240 kJ mol1 respectively and ΔfH for HCl is -90 kJ mol1. The bond enthalpy of HCl is:


A

245 KJ mol1

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B

2909 KJ mol1

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C

380 KJ mol1

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D

425 KJ mol1

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Solution

The correct option is D

425 KJ mol1


H2+Cl22HCl

One each of HH and ClCl bonds are broken whereas 2 HCl bonds are formed. Bond breaking requires energy and bond formation is always exothermic.

[BEHH+BEClCl][2BEHCl]=ΔHR

[430+240][2BEHCl]=180

or 6702(BE)HCl=850

BEHCl=425 kJ mol1


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