A volume of 100 ml of 0.5 N- solution is neutralized with 200 ml of 0.2 M- in a constant pressure ca — Thermodynamics and Thermochemistry Chemistry Question
Question
A volume of 100 ml of 0.5 N-$\text{H}_2\text{SO}_4$ solution is neutralized with 200 ml of 0.2 M-$\text{NH}_4\text{OH}$ in a constant pressure calorimeter which resulted $1.4^\circ\text{C}$ rise in temperature. The heat capacity of the calorimeter system is $1.5\text{ kJ/}^\circ\text{C}$. Some useful thermochemical equations are:<br>$\text{HCl} + \text{NaOH} \to \text{NaCl} + \text{H}_2\text{O} + 57\text{ kJ}$<br>$\text{CH}_3\text{COOH} + \text{NH}_4\text{OH} \to \text{CH}_3\text{COONH}_4 + \text{H}_2\text{O} + 48.1\text{ kJ}$<br>Which of the following statements are correct?
💡 Solution & Explanation
Moles $\text{H}^+ = 0.05$. Moles $\text{NH}_4\text{OH} = 0.04$ (limiting). Heat $Q = 1.5 \times 1.4 = 2.1\text{ kJ}$. $\Delta H_{\text{neut}}(\text{Strong Acid} + \text{NH}_4\text{OH}) = -2.1/0.04 = -52.5\text{ kJ/mol}$. Dissociation of $\text{NH}_4\text{OH} = 57 - 52.5 = 4.5\text{ kJ/mol}$. Dissociation of $\text{CH}_3\text{COOH} = 57 - 48.1 - 4.5 = 4.4\text{ kJ/mol}$ (c is incorrect). For 2 water molecules, $\Delta H = 2 \times 57 = 114\text{ kJ}$.