Question

A student reacted 100.0 mL of 0.9800 M HCl with 100.0 mL of 0.9900 M NH3. The density of                          
the reaction mixture was 1.02 g/mL and the heat capacity was 4.016 J/g K.                          
Calculate the enthalpy of neutralization by plotting and using the data shown below.

Time(min) Temp(oC)
0.0 23.25
0.5 23.27
1.0 23.28
1.5 23.30
2.0 23.30
3.0 23.35
4.0 23.44
4.5 23.47
mix ---------
5.5 28.75
6.0 28.50
7.0 28.55
8.0 28.48
9.0 28.32
10.0 28.25
11.0 28.20
12.0 28.05
13.0 27.96
14.0 27.80
15.0 27.75

Using the data provided in the excel file, show all of your work for the following calculations:

a.) mean temperature of unmixed reagents (oC)

b.) δελταT from graph (oC)

c.) q absorbed by reaction mixture (J)

d.) q absorbed by calorimeter, stirrer, and thermometer (J)

e.) q total absorbed (J)

f.) q total released (J)

g.) calculation to show limiting reagent

h.) deltaH neutralization for the reaction (kJ/mole of acid)

Answer 1

A student reacted 100.0 mL of 0.9800 M HCl with 100.0 mL of 0.9900 M NH3. The density of                          
the reaction mixture was 1.02 g/mL and the heat capacity was 4.016 J/g K.          

Graph will be

Answers :

a)

a.) mean temperature of unmixed reagents (oC)

Mean temperature = 23.34 ⁰C

b.) ?????T from graph (oC) : symbols not clear

c.) q absorbed by reaction mixture (J)

Q = Specific heat X mass of solution X change in temperature

Q = Specific heat X volume of solution X density X change in temperature

Q = 4.016 J/g K X 200 X 1.02 X 4.89

Q = 4006.2 Joules

d.) q absorbed by calorimeter, stirrer, and thermometer (J)

We need heat capacity of calorimeter

e.) q total absorbed (J)

For q total we need the heat absorbed by calorimeter etc

f.) q total released (J)

For q total we need the heat absorbed by calorimeter etc

g.) calculation to show limiting reagent

The Moles of HCl present = MolarityX volume = 0.98 X 0.1 = 0.098

Moles of NH3 = 0.99 X 0.01 = 0.099

so the limiting reagent is HCl

h.) deltaH neutralization for the reaction (kJ/mole of acid)

Heat released by 0.098 moles of acid neutralization = Total heat absorbed

So heat released by 1 mole = Total heat released / 0.098

The delta H of neutralization = Total heat released / 0.098