To determine the heat of neutralization, a student mixes 50.0 mL of 1.00 M HCl with 50.0 mL of 1.00 M NaOH in a polystyrene calorimeter. Initial temperature of both solutions = 21.5 C. Maximum temperature after mixing = 28.2 C. Assume density 1.00 g/mL and specific heat 4.18 J/(g*C). Heat capacity of the calorimeter is negligible.
What is the molar enthalpy of neutralization (delta H) for HCl + NaOH?
- A
-112 kJ/mol
- Bcheck_circle
-56.0 kJ/mol
- C
-2.80 kJ/mol
- D
+56.0 kJ/mol
Explanation
q = m<em>c</em>deltaT = 100.0 * 4.18 * 6.7 = 2801 J released. Moles H2O formed = 0.0500 mol. delta H = -2801/0.0500 = -56,020 J/mol = -56.0 kJ/mol.