Chapter 6, Problem 6.91CP

Interpretation Introduction

Interpretation:

The reaction of solid sodium with gaseous chlorine produces solid sodium chloride ( $N{a}^{+}C{l}^{-}$ ). Calculate the energy change for the other reaction of solid sodium with gaseous chlorine that produces chlorine sodide ( $C{l}^{+}N{a}^{-}$ ). Also, explain why sodium chloride formation is preferred


     $2Na\left(s\right)\text{ + }C{l}_2\left(g\right)\to \text{ 2}C{l}^{+}N{a}^{-}\left(s\right)$

The data is given as follows:

$E_{ea}forNa(g)=-52.9kJ/mol$

$E_{i1}forCl(g)=\text{ +1251 }kJ/mol$

The lattice energy of $C{l}^{+}N{a}^{-}$ is the same as that of the lattice energy of $N{a}^{+}C{l}^{-}$

Concept Introduction:
Calculate the net energy change for the formation of the compound $C{l}^{+}N{a}^{-}$. Then compare the energy change with $N{a}^{+}C{l}^{-}$. The compound with more negative energy change is more likely to be formed in the reaction.