Chemistry - Thermochemistry Question with Solution | TestHub

To calculate the heat of formation of KOH(s), we use Hess's Law. The target equation is:

K(s) + 1/2 O₂(g) + 1/2 H₂(g) -> KOH(s)

Given equations:

(i) K(s) + H₂O(l) + aq -> KOH(aq) + 1/2 H₂(g) $\Delta H=-48.4$ kcal

(ii) H₂(g) + 1/2 O₂(g) -> H₂O(l) $\Delta H=-68.4$ kcal

(iii) KOH(s) + aq -> KOH(aq) $\Delta H=-14.0$ kcal

We need to manipulate these equations to get the target equation.

Reverse equation (iii):

(iii') KOH(aq) -> KOH(s) + aq $\Delta H=+14.0$ kcal

Now, add equations (i), (ii), and (iii'):

(i) K(s) + H₂O(l) + aq -> KOH(aq) + 1/2 H₂(g)

(ii) H₂(g) + 1/2 O₂(g) -> H₂O(l)

(iii') KOH(aq) -> KOH(s) + aq

----------------------------------------------------------------------------------

K(s) + H₂O(l) + aq + H₂(g) + 1/2 O₂(g) + KOH(aq) -> KOH(aq) + 1/2 H₂(g) + H₂O(l) + KOH(s) + aq

Cancel out common terms (H₂O(l), aq, KOH(aq), 1/2 H₂(g)):

K(s) + 1/2 H₂(g) + 1/2 O₂(g) -> KOH(s)

The enthalpy change for the formation of KOH(s) is the sum of the enthalpy changes of the manipulated equations:

$\Delta H_f$ = $\Delta H_{(i)}$ + $\Delta H_{(ii)}$ + $\Delta H_{(ii{i}^{\prime })}$

$\Delta H_f$ = -48.4 kcal + (-68.4 kcal) + 14.0 kcal

$\Delta H_f$ = -116.8 kcal + 14.0 kcal

$\Delta H_f$ = -102.8 kcal