ThermochemistryHard
Question
The enthalpy of formation of KCl(s) from the following data is
(i) KOH(aq) + HCl(aq) → KCl(aq) + H2O(l): ΔH = −13.7 kcal
(ii) H2(g) + ½ O2(g) → H2O(l): ΔH = −68.4 kcal
(iii) ½ H2(g) + ½ Cl2(g) + aq → HCl(aq): ΔH = −39.3 kcal
(iv) K(s) + ½ O2(g) + ½ H2(g) + aq → KOH(aq): ΔH = −116.5 kcal
(v) KCl(s) + aq → KCl(aq): ΔH = +4.4 kcal
Options
A.+105.5 kcal/mol
B.−105.5 kcal/mol
C.−13.7 kcal/mol
D.−18.1 kcal/mol
Solution
The required thermochemical equation is
$K(s) + \frac{1}{2}Cl_{2}(g) \rightarrow KCl(s);\Delta H = ?$
From (iv) + (iii) – (v) + (i) – (ii): we get,
$\Delta H = ( - 116.5) + ( - 39.3) - (4.4) + ( - 13.7) - ( - 68.4) = - 105.5\text{ Kcal}$
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