Chapter 6, Problem 6.83SP

Interpretation Introduction

Interpretation:

The compound which has lower heat of formation is more likely to be formed.

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:

i) The electron affinity of Cl(g) to $C{l}^{-}(g)$ is (-)348.6 kJ/mole.
ii) The 1^st Ionisation energy (I.E.1) of Ca(g) to $C{a}^{+}(g)$ is (+)589.8 kJ/mole.
iii) The 2^nd Ionisation energy (I.E.2) of Ca(g) to $C{a}^{2+}(g)$ is (+)1145kJ/mole.
iv) The heat of sublimation (S) of Ca(s) to Ca(g) is (+)178.2 kJ/mole.
v) The bond dissociation energy of $C{l}_2(g)$ to 2 $C{l}^{-}(g)$ is (+) 243 kJ/mole.
vi) Lattice energy of $CaC{l}_2(s)$ is (+) 2258kJ/mole
vii) Lattice energy of $CaCl(s)$ is (+) 717kJ/mole.

To determine:

The heat of formation of $CaC{l}_2(s)$ from its elements.
The compound which is more likely to be formed in between $CaC{l}_2(s)$ and CaCl(s).