Answer
-691.6 kJ/mol
Work Step by Step
Dissociation enthalpy: $1/2\ Cl_2(g)\rightarrow Cl(g),\ \Delta H_{1A}=+121.3\ kJ/mol$
Enthalpy of formation, gaseous Rb: $Rb(s)\rightarrow Rb(g),\ \Delta H_{2A}=+80.9\ kJ/mol$
Electron attachment energy, Cl: $Cl(g)+e^-\rightarrow Cl^-(g),\ \Delta H_{1B}=-349.0\ kJ/mol$
Ionization energy, Rb: $Rb(g)\rightarrow Rb^+(g)+e^-,\ \Delta H_{2B}=+403.0\ kJ/mol$
Lattice energy: $Rb^+(g)+Cl^-(g)\rightarrow RbCl(s),\ \Delta H_l$
Enthalpy of formation, RbCl: $Rb(s)+1/2\ Cl_2(g)\rightarrow RbCl(s),\ \Delta H_{f}=-435.4\ kJ/mol$
Combining equations we get:
$\Delta H_{1A}+\Delta H_{2A}+\Delta H_{1B}+\Delta H_{2B}+\Delta H_{l}=\Delta H_{f}$, so:
$\Delta H_{l}=-691.6\ kJ/mol$
