Challenge Your Exam Readiness With Our Free ACS General Chemistry Practice Questions

In constant-pressure (coffee-cup) calorimetry, q_solution = m c ΔT. Here, m = 103.0 g, ΔT = 31.0 − 25.0 = 6.0 °C, so q_solution ≈ 103.0 × 4.18 × 6.0 ≈ 2.58 × 10³ J = +2.58 kJ. The solution warmed, so the process released heat: q_rxn = −q_solution = −2.58 kJ for dissolving 3.00 g NaOH. Convert to molar enthalpy using n(NaOH) = 3.00 g ÷ 40.00 g·mol⁻¹ = 0.0750 mol. Thus ΔH ≈ (−2.58 kJ) ÷ 0.0750 mol ≈ −34.4 kJ·mol⁻¹ (choice B). ACS thermochemistry learning goals include sign conventions (exothermic negative ΔH), isolating the system vs surroundings, and appropriate significant figures. Typical errors include forgetting to include the mass of solute in the heat capacity calculation, using 100 g instead of 103 g, or assigning the wrong sign. Although we neglect the calorimeter constant here, some questions may require adding q_calorimeter; always read premises carefully. This method and reasoning apply broadly to solution processes and neutralization reactions in first-term thermochemistry.