Question

write down the formation of born Haber cycle of cacl2
clearly please
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Answer 1

Answer:

The Born Haber cycle for the formation of calcium chloride from its constituent elements involves following steps.Ca (s) → Ca (g) ΔHa° (Ca) =178 kj/mol.Ca (g) → Ca+ (g) + e-Δ HIE°= 590 kj/mol. ...½ Cl2 (g) → Cl (g) ;Δ Ha°= ½Δ HCl-Cl°=121 kj/mol.Cl (g) + e- → Cl- (g) ;Δ Hea°= -364 kj/mol.Ca2+ (g) + 2Cl- (g) → CaCl2 (s)

Answer 2

The formation of $\sf{CaCl_2}$ from molecules is given below.

$\sf{Ca_{(s)}+Cl_2_{(g)}\longrightarrow CaCl_2_{(s)},\quad \Delta H_f\,\!^o}$

The enthalpy change here is the standard enthalpy of formation.

And that of $\sf{CaCl_2}$ from ions is given below.

$\sf{Ca^{2+}\,\!\!\!_{(g)}+2Cl^-\,\!\!\!_{(g)}\longrightarrow CaCl_2_{(s)},\quad\Delta H_{l}}$

The enthalpy change here is lattice enthalpy.

Here $\sf{Ca_{(s)}}$ is converted to $\sf{Ca^{2+}\,\!\!\!_{(g)}}$ during the formation from ions. This conversion is done in two successive steps.

1. Conversion of Calcium from solid state to gaseous state.

$\sf{Ca_{(s)}\longrightarrow Ca_{(g)},\quad \Delta H_s}$

The enthalpy change here is enthalpy of sublimation.

2. Conversion of Calcium into ions.

$\sf{Ca_{(g)}\longrightarrow Ca^{2+}\,\!\!\!_{(g)},\quad\Delta H_{i}}$

The enthalpy change here is the sum of first and second ionisation enthalpies, since two electrons are liberated.

And here $\sf{Cl_2_{(g)}}$ is converted to $\sf{2Cl^-\,\1\!\!_{(g)}}$ during the formation from ions. This is also done in two successive steps.

1. Conversion of Chlorine molecule into Chlorine atom.

$\sf{Cl_2_{(g)}\longrightarrow 2Cl_{(g)},\quad\Delta H_b}$

The enthalpy change here is the bond dissociation enthalpy.

2. Conversion of Chlorine into ions.

$\sf{Cl_{(g)}\longrightarrow Cl^-\,\!\!\!_{(g)},\quad \Delta H_{e}}$

The enthalpy change here is the electron gain enthalpy.

Since there are two chlorine atoms the electron gain enthalpy should be twice its standard value.

$\sf{2Cl_{(g)}\longrightarrow 2Cl^-\,\!\!\!_{(g)},\quad 2\Delta H_{e}}$

Hence the Born Haber Cycle for the formation of $\sf{CaCl_2}$ is,

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\put(0,0){ \sf{Ca_{(s)}} }\multiput(15,0)(0,-57){2}{+}\put(26,0){ \sf{Cl_2_{(g)}} }\put(45,1){\line(1,0){15}}\put(55,0){ \longrightarrow }\put(49.4,3){\footnotesize\sf{\Delta H_f\,\!^o}}\put(75,0){ \sf{CaCl_2_{(g)}} }\multiput(0,0)(0,-29){2}{\multiput(0,0)(27,0){2}{\put(3,-5){\line(0,-1){15}}\put(2,-21){ \downarrow }}}\put(0,-29){ \sf{Ca_{(g)}} }\put(26,-29){ \sf{2Cl_{(g)}} }\put(-1,-57){ \sf{Ca^{2+}\,\!\!\!_{(g)}} }\put(26,-57){ \sf{2Cl^-\,\!_{(g)}} }\put(45,-57){\line(1,0){37}}\put(82,-57){\line(0,1){50}}\put(81,-8){ \uparrow }\put(5,-13){\footnotesize\sf{\Delta H_s}}\put(5,-43){\footnotesize\sf{\Delta H_i}}\put(32,-13){\footnotesize\sf{\Delta H_b}}\put(32,-43){\footnotesize\sf{2\Delta H_e}}\put(85,-29){\footnotesize\sf{\Delta H_l}}\end{picture}$

From this we get,

$\longrightarrow\sf{\Delta H_f\,\!^o=\Delta H_s+\Delta H_i+\Delta H_b+2\Delta H_e+\Delta H_l}$

Thus we can find out the lattice enthalpy from this if others are known.

$\longrightarrow\sf{\Delta H_l=\Delta H_f\,\!\!^o-(\Delta H_s+\Delta H_i+\Delta H_b+2\Delta H_e)}$