Question

Use the data given below to construct a Born-Haber cycle to determine the heat of formation of KCl. Δ H°(kJ) K(s) → K(g) 89 K(g) → K (g) e- 418 Cl2(g) → 2 Cl(g) 244 Cl(g) e- → Cl-(g) -349 KCl(s) → K (g) Cl-(g) 717 Use the data given below to construct a Born-Haber cycle to determine the heat of formation of KCl. H°(kJ) K(s) → K(g) 89 K(g) → K (g) e- 418 Cl2(g) → 2 Cl(g) 244 Cl(g) e- → Cl-(g) -349 KCl(s) → K (g) Cl-(g) 717 158 kJ -1119 kJ -997 kJ 631 kJ -437 kJ

Answer 1

Explanation:

The net equation will be as follows.

          $K(s) + Cl_{2}(g) \rightarrow KCl(s)$

So, we are required to find $\Delta H_{formation}$ for this reaction.

Therefore, steps involved for the above process are as follows.

Step 1:  Convert K from solid state to gaseous state

          $K(s) \rightarrow K(g)$,    $\Delta H_{1}$ = 89 kJ

Step 2:  Ionization of gaseous K

           $K(g) \rightarrow K^{+}(g) + e^{-}$,    $H_{2}$ = 418 KJ

Step 3:  Dissociation of $Cl_{2}$ gas into chlorine atom .

            $\frac{1}{2} Cl_{2}(g) \rightarrow Cl(g)$,   $\Delta H_{3} = \frac{244}{2}$ = 122 KJ

Step 4: Iozination of chlorine atom.

              $Cl(g) + e^{-} \rightarro Cl^{-}(g)$,      $H_{4}$ = -349 KJ

Step 5:  Add $K^{+}$ ion and $Cl^{-}$ ion formed above to get KCl .

              $K^{+}(g) + Cl^{-}(g) \rightarrow KCl(s)$,   $H_{5}$ = -717 KJ

Now, using Born-Haber cycle, value of enthalpy of the formation is calculated as follows.

      $\Delta H_{f} = \DeltaH_{1} + \Delta H_{2} + \Delta H_{3} + \Delta H_{4} + \Delta H_{5}$

                  = 89 + 418 + 122 - 349 - 717

                  = - 437 KJ/mol

Thus, we can conclude that the heat of formation of KCl is - 437 KJ/mol.