Question

What is the standard enthalpy change for the formation of $LiF(s)$? Refer the table given below

Process	Reaction	$\mathrm{\Delta }{H}^0(KJmo{l}^{-1})$ $\mathrm{\Delta }{H}^0(KJmo{l}^{-1})$
Sublimation	$$Li(s)\to Li(g)$$	155.2
Dissociation	$${\displaystyle \frac{1}{2}}{F}_2(g)\to F(g)$$	75.3
Ionization	$$Li(g)\to L{i}^{+}(g)+e^{-}$$	520
Electron affinity	$$F(g)+e^{-}\to F^{-}(g)$$	-328
Formation of LiF	$$L{i}^{+}(g)+F^{-}(g)\to LiF(s)$$	-1017

a.) $-594KJmo{l}^{-1}$
b.) $905KJmo{l}^{-1}$
c.) $1440KJmo{l}^{-1}$
d.) $2095KJmo{l}^{-1}$

Answer 1

Hint: Hess law of constant heat summation will be used to solve this question. It states that regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes. This law is a manifestation that enthalpy is a state function.

Complete step by step solution:
- Sublimation: when a solid substance is directly vaporized without having any intermediate liquid phase then such process is known as sublimation. It is an endothermic process and energy will be absorbed.
Here sublimation of Lithium atom is taking place
$$Li(s)\to Li(g)$$ $\mathrm{\Delta }{H}^0=155.2KJmo{l}^{-1}$

- Dissociation: Atomization is the process of breaking a molecule into its atoms. This is also an endothermic process and energy will be absorbed.
$${\displaystyle \frac{1}{2}}{F}_2(g)\to F(g)$$ $\mathrm{\Delta }{H}^0=75.3KJmo{l}^{-1}$

- Ionization: ionization is the process of converting stable species into its ion. Here ionization of Lithium is taking place where it is losing one electron. Energy required for this process will be called ionization energy. Lithium absorbs energy to get converted into its ion, so this is also an endothermic process.
$$Li(g)\to L{i}^{+}(g)+e^{-}$$ $$$$ $\mathrm{\Delta }{H}^0=520KJmo{l}^{-1}$

- Electron affinity: It is the energy released absorbed or released when an electron is added into an atom. Because fluorine has a very high tendency to gain electrons, that’s why while accepting electrons it releases the energy and this process is an exothermic process.
$$F(g)+e^{-}\to F^{-}(g)$$ $\mathrm{\Delta }{H}^0=-328KJmo{l}^{-1}$

- Formation of molecules: Lithium ion and fluoride having opposite charges will attract towards each-other and get bound together by strong electrostatic forces. Energy will be released and this exothermic process.
$$L{i}^{+}(g)+F^{-}(g)\to LiF(s)$$ $\mathrm{\Delta }{H}^0=-1017KJmo{l}^{-1}$

- Total enthalpy change for the formation of 1 mole Lithium fluoride will be equal to the addition of the energy of all the steps above.
$\mathrm{\Delta }{H}_{rxn}=\mathrm{\Delta }{H}_{sub}+\mathrm{\Delta }{H}_{diss}+\mathrm{\Delta }H{}_{ion}+\mathrm{\Delta }{H}_{EA}+\mathrm{\Delta }{H}_{For}$
$$\mathrm{\Delta }{H}_{rxn}=155.2+75.3+520+(-328)+(-1017)$$
$$\mathrm{\Delta }{H}_{rxn}=750.5-328-1017$$
$$\mathrm{\Delta }{H}_{rxn}=750.5-1345$$
$$\mathrm{\Delta }{H}_{rxn}=-594.5KJmo{l}^{-1}$$
So enthalpy of formation of 1 mole of Lithium fluoride will be 594.5 KJ
So, the correct answer is “Option A”.

Additional Information:
Energy released or absorbed during the formation of one mole crystal of a compound is called lattice energy. So lattice energy of $LiF$ is 594.5 KJ.

Note: The change in enthalpy that accompanies the formation of one mole a compound from its elements, with all the substances are in their standard state. Standard states for atomic elements are given in terms of their most stable allotrope for each element. For example, white tin is the most stable allotrope of tin, so this will be the standard state of tin.