Chapter 6, Problem 6.78SP

Interpretation Introduction

Interpretation:

The lattice energy of a solid material can be obtained from the Born-Haber cycle in presence of necessary data.

Concept introduction:

• Born-Haber Cycle: It is a thermodynamic cycle correlating the heat of formation of a compound with other thermochemical quantities; among these are included the energy terms involved in a various processes which may be carried on successively to form the same substance in a different path.

  The construction of Born-Haber cycle may be illustrated with the reference to formation of an alkali halide (MX) from the elemental state. We consider the stepwise formation as represented by the following processes.

  Process	Energy
  $M(s)=M(g)$	Heat of Sublimation (+)S
  $M(g)=M^{+}(g)$	Ionization Energy (+)I
  $\frac{1}{2}{X}_2(g)=X(g)$	Heat of dissociation (+) $\frac{1}{2}D$
  $X(g)=X^{-}(g)$	Electron Affinity (-)E.A.
  $M^{+}(g)+X^{-}(g)=MX(s)$	Lattice Energy (-)U
  $M(s)+\frac{1}{2}{X}_2(g)=MX(s)$	Heat of Formation (-) $\Delta H_f$

The algebraic sum of the energy terms should be equal to the heat of formation ( $\Delta H_f$ ) of MX. As heat is actually involved in the reaction, $\Delta H_f$ will be negative. So, $(-)\Delta H_f=(+)S+(+)I+(+)\frac{1}{2}D-E.A.-U$

The lattice energy can be calculated as-$U=S+I+\frac{1}{2}D-E.A.+\Delta H_f$.

Given:

Born-Heber cycle are called cycles because they form close loop. If any five of the six energy changes in the cycle are known, the value of the sixth can be calculated.
i) The electron affinity of H (g) to $H^{-}(g)$ is (-)72.8kJ/mole.
ii) Ionisation energy (I.E.) of Na (g) to $N{a}^{+}$ (g) is (+) 495.8 kJ/mole.
iii) Heat of sublimation (S) of Na(s) to Na (g) = (+) 107.3kJ/mole.
iv) Bond dissociation energy (D) of $H_2$ (g) = (+) 435.9 kJ/mole.
v) Net energy change for the formation of $NaH$ (s) from its elements, ( $\Delta H_f$ ) (-) 60kJ/mole.

To determine:

The lattice energy of NaH (s).