Question

1. a student uses a coffee - cup calorimeter to determine the enthalpy of reaction for hydrobromic acid and potassium hydroxide. the student mixes 100.0 ml of 0.50 mol/l hbr(aq) at 21.0°c with 100.0 ml of 0.50 mol/l koh(aq), also at 21.0°c. the highest temperature that is reached is 24.4°c. write a thermochemical equation for the reaction.

Explanation:

Step1: Calculate total volume of solution

Total volume = 100.0 mL + 100.0 mL = 200.0 mL
Assume density = 1 g/mL, so mass $m = 200$ g.

Step2: Calculate temperature change

$\Delta T = 24.4^\circ\text{C} - 21.0^\circ\text{C} = 3.4^\circ\text{C}$

Step3: Calculate heat absorbed by solution

$q_{\text{solution}} = mc\Delta T$, $c = 4.18$ J/(g·°C)
$q_{\text{solution}} = 200 \times 4.18 \times 3.4 = 2842.4$ J = 2.8424 kJ

Step4: Find heat released by reaction

$q_{\text{reaction}} = -q_{\text{solution}} = -2.8424$ kJ

Step5: Calculate moles of HBr/KOH

Moles = $0.50$ mol/L × $0.100$ L = $0.050$ mol

Step6: Calculate enthalpy change per mole

$\Delta H = \frac{q_{\text{reaction}}}{\text{moles}} = \frac{-2.8424}{0.050} \approx -57$ kJ/mol

Step7: Write balanced equation

$HBr(aq) + KOH(aq)
ightarrow KBr(aq) + H_2O(l)$

Answer:

$HBr(aq) + KOH(aq)
ightarrow KBr(aq) + H_2O(l) \quad \Delta H = -57\ \text{kJ/mol}$