The bond dissociation energies of gaseous H2, CI2 and HCI are 104, 58 and 103 kcal/mole respectively. Calculate the enthyalpy of formation of HCI gas.

Bond H – H CI – CI H – CI ∆H disso. 104 kcal 58 kcal 103 kcal Formation of hydrogen chloride can be represented as H – H + CI – CI → 2H – CI Thus the reaction involves Cleavage of one H – H bond, ∆H = 104 kcal Cleavage of one CI – CI bond, ∆H = 58 kcal Formation of two H – CI bonds, ∆H = 2 * - 103 kcal &there4; ∆H of the reaction = (104 + 58) – 2(103) = 162 – 206 = - 44 kcal Now since the enthalpy of formation of a compound is the change in heat content accompanied in the formation of one mole of the compound, the enthalpy of formation of HCI gas = - 44/2 = - 22 kcal

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Grade upto college level Physical Chemistry

The bond dissociation energies of gaseous H₂, CI₂ and HCI are 104, 58 and 103 kcal/mole respectively. Calculate the enthyalpy of formation of HCI gas.

Shane Macguire

12 Years agoGrade upto college level

1 Answer

Deepak Patra

12 Years ago

Bond H – H CI – CI H – CI

∆H disso. 104 kcal 58 kcal 103 kcal

Formation of hydrogen chloride can be represented as

H – H + CI – CI → 2H – CI

Thus the reaction involves

Cleavage of one H – H bond, ∆H = 104 kcal

Cleavage of one CI – CI bond, ∆H = 58 kcal

Formation of two H – CI bonds, ∆H = 2 * - 103 kcal

∴ ∆H of the reaction = (104 + 58) – 2(103) = 162 – 206 = - 44 kcal

Now since the enthalpy of formation of a compound is the change in heat content accompanied in the formation of one mole of the compound, the enthalpy of formation of

HCI gas = - 44/2 = - 22 kcal