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Bond dissociation enthalpy of H_{2}, Cl_{2} and HCl  are 434, 242, and 431 kJ mol^{-1} respectively. Enthalpy formation of HCl is:

 

Option: 1

93 kJmol^{-1}


Option: 2

-245kJmol^{-1}


Option: 3

-93kJmol^{-1}


Option: 4

245kJmol^{-1}


Answers (1)

best_answer

For the reaction,

H_2+Cl_2\rightarrow 2HCl

\Delta H_{reaction}=\sum B.E_{reactants}-\sum B.E_{products}\\ \Delta H_{reaction}=B.E_{H_2}+B.E_{Cl_2}-2B.E_{HCl}\\ =434+242-(2\times431)=-186kJmol^{-1}

Now,\Delta H_{formation} is the enthalpy change for the formation of 1 mole of molecules, while \Delta H_{reaction} is the enthalpy change for the reaction in which 2 moles of HCl are formed. Hence, for 1 mole of HCl

\Delta H_{formation}=\frac{\Delta H_{reaction}}{2}=\frac{-186kJmol^{-1}}{2}=-93kJmol^-^1

The Correct answer is option 3.

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