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) Average temperature of unmixed reagents is obtained by extrapolating temperature data of 0 to 4.5min at 5min, which is Tinitial = 23.5C

b) DT = Tfinal - Tinitial.

Tfinal data is also obtained by extrrapolation at 5min of temperature data from 5.5min to 15min. Tfinal = 29C

DT = 29-23.5 = 5.5C

c) q absorbed by reaction mixture, q = m*c*DT

Volume of mixture, V = 100+100=200ml. Mass of mixture, m = density*V=1.02*200=204gm

Specific heat, c = 4.016J/(g K).

q = 204*4.016*5.5=4506J

d) , e) and f) Cannt be determined as the heat capacity data for calorimeter system is not available

g) HCl is limiting reagent as its concentration is less than NH3 and both have same volume.