Question

Given that the enthalpy of $\left( {i.} \right)$Sublimation of Ca $ = 121KJ\,mo{l^{ - 1\,}}\,\left( {ii} \right)\,$Dissociation of $C{l_{2\,}}\,to\,Cl\, = 242\,KJ\,mo{l^{ - 1\,}}\left( {iii} \right)\,$Ionization of Ca to $C{a^{2 + }}$$ = \,2422\,KJ\,mo{l^{ - 1\,}}\,\left( {iv} \right)$Electron gain enthalpy of Cl $ = \, - 355\,KJ\,mo{l^{ - 1}}\,\left( v \right)\,{\Delta _f}H$over all $ = \, - 795\,KJ\,mo{l^{ - 1}}$
If the lattice enthalpy of $CaC{l_2}$is $\left( {2900 - 10x} \right)KJ\,mo{l^{ - 1}}$ then what is x $?$

Answer 1

Hint:‌Born Haber cycle applies Hess’s law to calculate the lattice enthalpy of the compound by comparing the standard enthalpy change of formation of the ionic compound (from the elements) to the enthalpy required to make gaseous ions from the elements by considering enthalpy of sublimation, electron gain enthalpy, dissociation enthalpy etc.


Complete step by step solution:
Writing all the values given to us-
${\Delta _{sub}}{H_{Ca}} = 121\,KJ\,mo{l^{ - 1}}$${\Delta _l}{H_{CaC{l_2}}} = $$\left( {2900 - 10x} \right)KJ\,mo{l^{ - 1}}$
$\Delta {H_{Cl - Cl}} = 242\,KJ\,mo{l^{ - 1\,}}\,{\Delta _f}{H_{C{a^{2 + }}}} = 2422\,KJ\,mo{l^{ - 1}}\,{\Delta _{eg}}{H_{Cl}} = - 355\,KJ\,mo{l^{ - 1}}\,{\Delta _f}{H_{CaC{l_2}}} = - 795\,KJ\,mo{l^{ - 1}}$
Substituting the values in the formula, we got
 ${\Delta _f}{H_{CaC{l_2}}} = {\Delta _{sub}}{H_{Ca}}\, + \Delta {H_{Cl - Cl\,}} + {\Delta _f}{H_{C{a^{2 + }}}} + 2 \times {\Delta _{eg}}{H_{Cl\,}}\, + {\Delta _l}{H_{CaC{l_2}}}$
$ - 795 = 121 + 242 + 2422 + 2( - 355) + (2900 - 10x)$
$ - 795 - 121 - 242 - 2422 + 2(355) = 2900 - 10x$
$ - 2870 = 2900 - 10x$
$10x = 5770$
$x = 577$$K\,J\,mo{l^{ - 1}}$

Additional Information:Lattice energy can be defined in two ways:
1) It is the energy required to break apart an ionic solid and covert its component atoms into gaseous ions
2) It is the energy released when gaseous ions bind to form an ionic solid.

Note: The lattice enthalpy is the enthalpy change involved in the formation of an ionic compound from gaseous ions (an exothermic process)In Born Haber cycle, formation enthalpy is written as a sum of sublimation enthalpy (heat change to convert elements to gaseous atoms), bond dissociation enthalpy, ionization enthalpy, electron gain enthalpy and lattice enthalpy.
Lattice energy= heat of formation -heat of atomization- dissociation energy – ionization energy – electron gain energy.