Question

The bond dissociation enthalpy for chlorine is +242 kJ mol−1 and that for fluorine is +158 kJ mol−1. The standard enthalpy of formation of ClF(g) is −56 kJ mol−1. (i) Write an equation, including state symbols, for the reaction that has an enthalpy change equal to the standard enthalpy of formation of gaseous ClF

Answer 1

The enthalpy change for the reaction of 1 mole of H₂(g) with 1 mole of Cl₂(g) is 864 kJ.

The enthalpy change for the reaction of 1 mole of H₂(g) with 1 mole of Cl₂(g) can be calculated using the bond dissociation enthalpies and the standard enthalpy of formation of HCl(g).

1. To form two moles of HCl, one mole of H-H bonds and one mole of Cl-Cl bonds must be broken. The energy required to break these bonds is the sum of the bond energy of the H-H bond (436 kJ/mol) and the Cl-Cl bond (243 kJ/mol).
2. During the reaction, two moles of H-Cl bonds are formed, releasing 2 x 432 kJ/mol or 864 kJ. The reaction releases more energy than it consumes because the bonds in the products are stronger than those in the reactants.

Therefore, the enthalpy change for the reaction of 1 mole of H₂(g) with 1 mole of Cl₂(g) is 864 kJ.

Complete Question:

The bond dissociation enthalpy for chlorine is +242 kJ mol−1 and that for fluorine is +158 kJ mol−1. The standard enthalpy of formation of ClF(g) is −56 kJ mol−1.

(i) Write an equation, including state symbols, for the reaction that has an

enthalpy change equal to the standard enthalpy of formation of gaseous ClF. Calculate a value for the bond enthalpy of the Cl - F bond.