The Δ H f of KF(s) is −563 kJ/mol. The ionisation energy of potassium K(g) is 419 kJ/mol and enthalpy of sublimation of potassium is 88 kJ/mol. The electron affinity of F(g) is −322 kJ/mol and F−F bond enthalpy is 158 kJ/mol. Calculate the absolute value of lattice energy (in kJ/mol) of KF(s)

Question

The       Δ        H    f   of KF(s) is −563 kJ/mol. The ionisation energy of potassium K(g) is 419 kJ/mol and enthalpy of sublimation of potassium is 88 kJ/mol. The electron affinity of F(g) is −322 kJ/mol and F−F bond enthalpy is 158 kJ/mol. Calculate the absolute value of lattice energy (in kJ/mol) of KF(s)

GrierveAlierskgc · Accepted Answer

For the formation of KF, the Born-Haber cycle is: Now, using Hess&#039;s law :      Δ        H    f    =      Δ        H    1    +      Δ        H    2    +      Δ        H    3    +      Δ        H    4    +      Δ        H    5    −  563  =  88  +  419  +  79  −  322  +      Δ        H    5        Δ        H    5    =  −  827  k  J      /    m  o  l∴ Lattice energy of KF (s) is −827 kJ/mol

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