Calculate the enthalpy of formation (in kcal/mol) of gaseous HCl using following data:<br>Substance: — Thermodynamics and Thermochemistry Chemistry Question
Question
Calculate the enthalpy of formation (in kcal/mol) of gaseous HCl using following data:<br>Substance: $\text{NH}_3\text{(g)}$ (-11), HCl(g) (X), $\text{NH}_4\text{Cl(s)}$ (-75) [Heat of formation]<br>Heat of solution: -8.5, -17.5, +3.9 kcal<br>and, $\text{NH}_3\text{(aq)} + \text{HCl(aq)} \to \text{NH}_4\text{Cl(aq)}; \Delta H = -12\text{ kcal}$
Answer: B
💡 Solution & Explanation
$\Delta H_f(\text{aq}) = \Delta H_f(\text{g/s}) + \Delta H_{\text{sol}}$. $\Delta H_f(\text{NH}_4\text{Cl, aq}) = -75 + 3.9 = -71.1$. $\Delta H_f(\text{NH}_3\text{, aq}) = -11 - 8.5 = -19.5$. $\Delta H_f(\text{HCl, aq}) = X - 17.5$. For the reaction: $-71.1 - [-19.5 + (X - 17.5)] = -12 \implies -71.1 + 37 - X = -12 \implies -34.1 - X = -12 \implies X = -22.1\text{ kcal/mol}$.
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