Question

2. When a neutralization reaction was carried out using 100.0mL of 0.7890M NH3 water and 100.0mL of 0.7940M acetic acid, ΔT was found to be 4.76 degrees Celsius. The specific heat of the reaction mixture was 4.104 J g^-1 K^-1, and its density was 1.03 g mL-1. The calorimeter constant was 3.36 J K^-1.

a. Calculate the ΔHneutzn for the reaction of NH3 and acetic acid.

b. At the end of the experiment, it was discovered that the thermometer had not been calibrated. When it was calibrated, it was found that the thermometer read 0.50-degree Celsius low. What effect would this thermometer reading have on the reported ΔHneutzn calculated in a.?

c. When the temperature-time data graph was reviewed, it was found that an error had been made in determining ΔT. Instead of 4.76-degrees Celsius, ΔT was actually 4.70-degrees Celsius. Based on this change only, calculate the correct ΔHneutzn for the reaction of NH3 and acetic acid.

Answer 1

a)

mass of solution = (100 + 100 ) x 1.03 = 206 g

temperature change = 4.70

specific heat = 4.104 J / g oC

Q = m Cp dT + Cp dT

   = 206 x 4.104 x 4.76 + 3.36 x 4.76

Q = 4040.211 J

moles of NH3 = 100 x 0.7890 / 1000 = 0.0789

moles of CH3COOH = 100 x 0.7940 / 1000 = 0.0794

here limting reagent is NH3.

delta H = - Q / n = - 4040.211 x 10^-3 / 0.0789

           = - - 51.207 kJ /mol

heat of neutralization = - 51.21 kJ/mol

b)

Q = m Cp dT

if dT is low. then Q is also low.

then Q = n x delta H

if Q is low , then delta H neutralization is also low.

if temperture is low, then heat of neutralization is also low.

c)

mass of solution = (100 + 100 ) x 1.03 = 206 g

temperature change = 4.70

specific heat = 4.104 J / g oC

Q = m Cp dT + Cp dT

   = 206 x 4.104 x 4.70 + 3.36 x 4.70

Q = 3989.3 J

moles of NH3 = 100 x 0.7890 / 1000 = 0.0789

moles of CH3COOH = 100 x 0.7940 / 1000 = 0.0794

here limting reagent is NH3.

delta H = - Q / n = - 3989.3 x 10^-3 / 0.0789

           = - 50.56 kJ /mol

heat of neutralization = - 50.56 kJ/mol