Question

1. nitric acid is neutralized with potassium hydroxide in the following reaction.

hno₃(aq) + koh(aq) → kno₃(aq) + h₂o(l) δh = -53.4 kj/mol
55.0 ml of 1.30 mol/l solutions of both reactants, at 21.4°c, are mixed in a calorimeter. what is the final temperature of the mixture? assume that the density of both solutions is 1.00 g/ml. also assume that the specific heat capacity of both solutions is the same as the specific heat capacity of water. no heat is lost to the calorimeter itself.

Explanation:

Step1: Calculate moles of reactants

Volume = 55.0 mL = 0.055 L; Moles = $1.30\ \text{mol/L} \times 0.055\ \text{L} = 0.0715\ \text{mol}$

Step2: Calculate heat released

$\Delta H = -53.4\ \text{kJ/mol}$; Heat $Q = |0.0715\ \text{mol} \times (-53.4\ \text{kJ/mol})| = 3.8181\ \text{kJ} = 3818.1\ \text{J}$

Step3: Calculate total mass of solution

Total volume = $55.0 + 55.0 = 110.0\ \text{mL}$; Mass $m = 110.0\ \text{mL} \times 1.00\ \text{g/mL} = 110.0\ \text{g}$

Step4: Relate heat to temperature change

$Q = mc\Delta T$; $c = 4.184\ \text{J/(g·°C)}$; $\Delta T = \frac{Q}{mc} = \frac{3818.1}{110.0 \times 4.184} ≈ 8.25\ °C$

Step5: Find final temperature

Initial temp $T_i = 21.4°C$; $T_f = 21.4 + 8.25 ≈ 29.65°C$

Answer:

29.7°C (rounded to three significant figures)