Question

What is the standard enthalpy change for the formation of $Li{F}_{\left(s\right)}$? Refer to the table given below.

Process	Reaction	$\Delta H^{\circ }\left(kJ/mol\right)$
Sublimation	$L{i}_{\left(s\right)}\to L{i}_{\left(g\right)}$	$155.2$
Dissociation	$\frac{1}{2}{F}_{2\left(g\right)}\to F_{\left(g\right)}$	$75.3$
Ionization	$L{i}_{\left(g\right)}\to L{i}_{\left(g\right)}^{+}+e^{-}$	$520$
Electron affinity	$F_{\left(g\right)}+e^{-}\to F_{\left(g\right)}^{-}$	$-328$
Formation of $LiF(s)	$L{i}_{\left(g\right)}^{+}+F_{\left(g\right)}^{-}\to Li{F}_{\left(s\right)}$	$-1017$

• A. $-594kJ/mol$
• B. $905kJ/mol$
• C. $1440kJ/mol$
• D. $2095kJ/mol$

Answer 1

The correct option is A $-594kJ/mol$

Enthalpy of formation = sublimation energy + dissociation energy + ionization energy + electron affinity + formation of LiF

Enthalpy of formation= 155.2 +75.2 +520-328-1017$\left(Kj/mol\right)$

Enthalpy of formation = $-594.5KJ/mol$