Question

The following temperature-time data where recorded for the reaction between 50.00ml of 1.14M HA (a weak acid) and 50.00ml of 0.98M NH4OH, ammonium hydro (a weak base, also known as aqueous ammonia). The solutions are mixed after 60s of approximately constant temperature readings. The Delta T is 7 degrees Celsius (31.24 degrees Celsius at 165s minus 24.25 degrees Celsius at 0 seconds).

1)Given that the specific heat of the final salt solution is 3.96J/g - degrees Celcius and the density of the weak acid is 1.028g/ml, the density of NH4OH (aq) is 0.992g/ml and 1.0 mol/L at 298 degrees K. Calculate the heat capacity of the system in J/degrees Celsius.

2)Using this information calculate the qCAL for the reaction in kJ

3)Given the Volumes, concentration and that the reactants has a 1:1 stoichiometric ratio between the reactants calculate the moles of limiting reactant in the reaction

4)Using the answer to question 2&3, calculate the molar heat of reaction (delta H) in kJ/mol. Remember qCAL and delta H have opposite signs.

Answer 1

1)

mass of weak acid = 50 x 1.028 = 51.4 g

mass of NH4OH = 50 x 0.992 = 49.6 g

total mass of solution = 51.4 + 49.6 = 101 g

specific heat of the final salt solution = 3.96J/g oC

Delta T = 7 oC

Q = m Cp dT

    = 101 x 3.96 x 7

     = 2800 J

Q = 2800 J

heat capacity of the system = Q / dT = 2800 / 7

heat capacity of the system = 400 J/ oC

2)

q cal = Cp x dT

         = 400 x 7

         = 2800 J

q cal = 2.8 kJ

3) moles of HA = 50 x 1.14 = 57

moles of NH4OH = 50 x 0.98 = 49

HA + NH4OH --------------> products

1          1

57       49

here limiting reagent is NH4OH

4)

delta H = - qcal / n

            = - 2800 / 49

delta H = - 57.14 kJ/mol