Question

You place 40.3 ml of 0.366 M NaOH in a coffee- cup calorimeter at 25.00°C and add 64.8 ml of 0.366 M HCl, also at 25.00°C. After stirring, the final temperature is 27.93°C. [Assume the total volume is the sum of the individual volumes and that the final solution has the same density (1.00 g/ml) and specific heat capacity (4.184 J/gK)]. Calculate the change in enthalpy (ΔH) of the reaction in kJ/mol of water formed. Enter to 1 decimal place. Hint: What does an increase in temperature mean? Enter the correct sign.

Answer 1

moles of NaOH = 40.3 x 0.366 /1000 = 0.0148

moles of HCl = 64.8 x 0.366 / 1000 = 0.0237

limiting reagent is NaOH

NaOH + HCl ------------------> NaCl + H2O

total volume = 40.3 + 64.8 = 105.1 mL

mass = volume x density

          = 105.1 x 1

           = 105.1 g

Q = m Cp dT

Q = 105.1 x 4.184 x (27.93 - 25.00)

Q = 1288.4 J

ΔH = - Q / n

       = - 1.288 kJ / 0.0148

       = -87.1 kJ / mol

ΔH = - 87.1 kJ / mol