Question

. In a coffee-cup calorimeter, 100.0 mL of 1.0 M NaOH and 100.0 mL of 1.0 M HCl are mixed. Both solutions were originally at 24.6 oC. After the reaction, the temperature is 31.3 oC. What is the enthalpy change for the neutralization of HCl by NaOH?

Answer 1

HCl + NaOH ---> NaCl + H2O

Total volume = 100 ml + 100 ml = 200 ml

density of water = 1 g/ml

total mass = 200 ml x 1 g/ml = 200 g

Cp of water = 4.18 J/g.oC

change in temperature (dT) = 31.3 - 24.6 = 6.7 oC

heat change = q = mCpdT

                     = 200 x 4.18 x 6.7 = 5601.2 J

moles of HCl = moles of NaOH = 1 M x 0.1 L = 0.1 mol

Enthalpy change (dH) for neutralization = 5601.2 J/0.1 mol x 1000

                                                                = 56.012 kJ/mol