The values for the enthalpies of a reaction that are calculated from bond enthal

Question

The values for the enthalpies of a reaction that are calculated from bond enthalpies do not always match those calculated from standard enthalpies of formation. The reason for this is that the numbersin the tables are averages derived from data on a number of different compounds. Values calculated from different compounds do vary, sometimes widely. To illustrate this, lets look at the following reaction:

PF3 (g) + F2 (g) ---> PF5 (g)

The enthalpy offormation of PF5 (g) is -1593 kJ/mol. Use this and the delta-f-H-degree of PF3 (g) to determine the enthalpy change for this reaction. Then, use that value of delta-f-H-degree, along with the bond dissociation enthalpy of F2 (g) (159 kJ/mol), to calculate the enthalpy change for the formation of a P-F bond in this reaction.

Explanation / Answer

standard enthalpy change for the reaction

dHforxn = dHfoproducts) - dHfo(reactants)

             = -1593 -(- 945)

             = -648 kJ/mol

dH = using bond energies

     = dH(reactants) - dH(products)

     = (3 x 490 + 159) - (5 x 490)

     = -821 kJ/mol

So we have two different values for the same reaction If calculated in different ways.

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