### Calorimetry: Heat of Neutralization

```Calorimetry: Heat of Neutralization
Prepared by Dongling Fei
Manatee Community College
Purpose: Determine the calorimeter constant for a calorimeter. Determine the enthalpy
of neutralization for the reaction of a strong acid and a strong base.
Background: When a system absorbs heat (q) its temperature changes by ∆ T.
Experimental measurements demonstrate that the heat absorbed by a system and its
corresponding temperature change are directly proportional. q=C x ∆ T. C is called the
heat capacity, which is often defined as the quantity of heat required to change its
temperature by 1oC
C= q/∆ T=J/ oC
Heat capacity is an extensive property; it depends on the amount of matter being heated.
The measure of intrinsic capacity of a substance to absorb heat is called specific heat
capacity (Cs), the amount of heat required to raise the temperature of 1 gram of the
substance by 1 oC
Cs=q/(∆ T *m)
The Cs of H2O is 4.184J/g oC. For the case where all the heat is used to heat water, the
amount of heat is calculated by the relation
q=Cs x mwater x ∆ T
The enthalpy change for a chemical reaction, ∆Hrxn is also called the enthalpy of
reaction or heat of reaction, which is also an extensive property. For many aqueous
reacions, ∆Hrxn can be measured simply using a coffee-cup calorimeter
1
In part I of this experiment, you will determine the calorimeter constant of the styroforam
coffee cup calorimeter. Although heat is transferred essentially instantaneously at the
time of mixing, thermometers can not respond quickly enough to give an accurate
indication of the temperature at the time of mixing and shortly thereafter. From a plot of
the time-temperature data, you will determine the ΔT for both the hot and the cold water.
The number of Joules of heat energy gained (or lost) by water is given by the equation:
q=mwaterxΔTx4.184J/goC
Where q is the quantity of heater, m is the mass of water, ΔT is the temperature change
and 4.184J/goC is the heat capacity of water. In the experiment just completed, we would
expect the final temperature to be exactly halfway between in the initial temperature of
the hot and cold water, but any absorption of heat by the calorimeter will result in a final
temperature below that which is anticipated; the heat capacity of calorimeter (at constant
pressure) Cp is calculated as follows:
Cwatermhotwater(TH-TF)=Cwatermcoldwater(TF-Tc)+Ccal(TF-Tc)
In part II of this experiment, the enthalpy of neutralization for the reaction of a strong
acid with a base going to be determined. Upon pouring the acid into the base solution, the
temperature data will be collected for a period of time. Using ΔT determined from plotted
2
time-temperature data, the volume and molarities of the reactants and the density and
specific heat of your reaction mixture to calculate ΔHn for the acid.
Table 1. Heat capacities and densities of reaction mixtures.
solution
Sodium chloride
Sodium sulfate
water
heat capacity J/goC
3.89
3.76
4.184
concentration M
1.00
1.00
-----
Density g/mL
1.04
1.12
1.00
Materials
50-mL beaker
Thermometer X2
3 Styrofoam cups
20.0 mL 2M HCl
20.0 mL 2M NaOH
Phenolphthelein
Procedures
Part I. The Heat Capacity of the Calorimeter.
1. Place 20.0 mL (20.0g) of room-temperature DI water in a clean, dry Styrofoam cup,
determine the actual temperature(±0.1oC) every minute over a 4-minute period and record
on data table.
2. Trim the top cup just below the rim, pierce the top cup and use a cork borer to make a
tight-fitting hole to suspend thermometer.
3. Weigh exactly 20.0g of water in a 50 mL of beaker, which you then place in a hot
water bath (40-50oC) for at least 5 min; this is to allow the temperature of the water in the
beaker to equal that of the bath. Add the warm water all at once and record the temp of
water. (It is critical that there be no delay in transferring the hot water immediately to the
room temperature water in the calorimeter; hot water will cool rapidly in air.
4. Record the temperature ((±0.1oC) every 30 seconds over 4 minutes. Make a group of
temperature versus time and determine the final temperature and ΔT.
3
Part II. Heat of Reaction
1. Place 20.0mL of 2.0 M of HCl at room temperature to Styrofoam calorimeter. Add 2
drops of phenolphthalein to this solution. Suspend a thermometer like in part I and
then record the temperature of HCl over 4 mins at 30 seconds interval.
2. Add 20.0mL of 2.0 M of NaOH at room temperature to Styrofoam, after mixing the
solution should be pink indicating a small excess of base. If not pink, add a few more
drops of NaOH till it is pink. Then cap the calorimeter and record the temperature
over 4mins at 30 secons interval. Determine the final temperature and ΔT.
3. The waste can be disposed to the sink.
Data Table.
Part 1.
Mass of cold water : ____________________
Mass of warm water: ____________________
Temperature of warm water: ______________
Temperature of cold water
Temperature (±0.1oC)
Time
0 min
1min
2min
3min
4min
4
Temperature of water after mixing
Temperature (±0.1oC)
Time
0 min
0.5 min
1 min
1.5 min
2 min
2.5min
3 min
3.5min
4 min
Plot Temp vs time
ΔT=Thot-Tfinal=____________________
ΔT=Tfinal-Tcold=__________________
Cwatermhotwater(TH-TF)=Cwatermcoldwater(TF-Tc)+Ccal(TF-Tc)
Ccal=_________________________J/oC
5
Part II.
Concentration of HCl _____________ M
Concentration of NaOH ____________M
Volume of 2.0 M HCl _____________mL
Volume of 2.0 M NaOH ___________mL
Temperature of HCl before mixing
Temperature (±0.1oC)
Time
0 min
0.5 min
1 min
1.5 min
2 min
2.5min
3 min
3.5min
4 min
6
Temperature of reaction mixture after mixing
Temperature (±0.1oC)
Time
0 min
0.5 min
1 min
1.5 min
2 min
2.5min
3 min
3.5min
4 min
Plot Temp vs Time
ΔT=______________
Volume of reaction mixture_______________mL
Density of reaction mixture (NaCl, 1.0M)____________g/mL
Specific heat of reaction mixture (NaCl)________________ J/goC
Heat absorbed by the
ΔT=_________________J
calorimeter
=
calorimeter
constant
J/oC
x
Heat absorbed by the reaction mixture=(Volume of reaction mixture)x(density of
RM)xΔTx(Specific heat of RM)=__________________J
7
Total heat absorbed= (Heat absorbed by the calorimeter)+( Heat absorbed by the reaction
mixture)=___________________J
HCl(aq) + NaOH (aq) NaCl(aq) + H2O
The reaction is Exothermic
Endothermic (circle one)
HCl is the limiting reagent in this reaction.
The number of moles of HCl reaction = MxV=_____________mol
The molar enthalpy of neutralization of HCl _______________KJ/mol