Two solutions, initially at , were mixed in an insulated bottle. One contained 200 ml of 0.4 M weak — Thermodynamics and Thermochemistry Chemistry Question
Question
Two solutions, initially at $25^\circ\text{C}$, were mixed in an insulated bottle. One contained 200 ml of 0.4 M weak monoprotic acid solution. The other contained 100 ml of a solution having 0.5 mole NaOH per litre. After mixing, the temperature rose to $26^\circ\text{C}$. Assume that the densities of both the solutions are 1.0 g/ml and that their specific heat capacities are all 1.0 cal/g-K. The amount of heat evolved (in kcal) in the neutralization of 1 mole of the acid is
💡 Solution & Explanation
Moles of acid = $0.2 \text{ L} \times 0.4 \text{ M} = 0.08$. Moles of NaOH = $0.1 \text{ L} \times 0.5 \text{ M} = 0.05$. NaOH is limiting, so 0.05 moles of water are formed. Total mass of solution = $200 + 100 = 300\text{ g}$. $\Delta T = 26 - 25 = 1\text{ K}$. Heat evolved $q = mc\Delta T = 300 \times 1.0 \times 1 = 300\text{ cal} = 0.3\text{ kcal}$. Heat per mole of acid neutralized (per mole of limiting reagent) = $0.3 / 0.05 = 6\text{ kcal/mol}$.