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Question: Given that the enthalpy of \(\left( {i.} \right)\)Sublimation of Ca \( = 121KJ\,mo{l^{ - 1\,}}\,\lef...

Given that the enthalpy of (i.)\left( {i.} \right)Sublimation of Ca =121KJ mol−1  (ii)  = 121KJ\,mo{l^{ - 1\,}}\,\left( {ii} \right)\,Dissociation of Cl2  to Cl =242 KJ mol−1 (iii) C{l_{2\,}}\,to\,Cl\, = 242\,KJ\,mo{l^{ - 1\,}}\left( {iii} \right)\,Ionization of Ca to Ca2+C{a^{2 + }} = 2422 KJ mol−1  (iv) = \,2422\,KJ\,mo{l^{ - 1\,}}\,\left( {iv} \right)Electron gain enthalpy of Cl = −355 KJ mol−1 (v) ΔfH = \, - 355\,KJ\,mo{l^{ - 1}}\,\left( v \right)\,{\Delta _f}Hover all = −795 KJ mol−1 = \, - 795\,KJ\,mo{l^{ - 1}}
If the lattice enthalpy of CaCl2CaC{l_2}is (2900−10x)KJ mol−1\left( {2900 - 10x} \right)KJ\,mo{l^{ - 1}} then what is x ??

Explanation

Solution

‌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-
ΔsubHCa=121 KJ mol−1{\Delta _{sub}}{H_{Ca}} = 121\,KJ\,mo{l^{ - 1}} ΔlHCaCl2={\Delta _l}{H_{CaC{l_2}}} = (2900−10x)KJ mol−1\left( {2900 - 10x} \right)KJ\,mo{l^{ - 1}}
ΔHCl−Cl=242 KJ mol−1  ΔfHCa2+=2422 KJ mol−1 ΔegHCl=−355 KJ mol−1 ΔfHCaCl2=−795 KJ mol−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
ΔfHCaCl2=ΔsubHCa +ΔHCl−Cl +ΔfHCa2++2×ΔegHCl  +ΔlHCaCl2{\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)
−795−121−242−2422+2(355)=2900−10x- 795 - 121 - 242 - 2422 + 2(355) = 2900 - 10x
−2870=2900−10x- 2870 = 2900 - 10x
10x=577010x = 5770
x=577x = 577 K J mol−1K\,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.