Chemistry Lab report 5
DCP1=1,DCP2=1,DCP3=0
CE1=1,CE2=1,CE3=0
Calorimetric Study of Heat of
Neutralization
Yumi Nakayama
2009 Mar 5
Subject Teacher: Helen Xu
1
Aim:
To estimate the heat of Neutralization of sodium hydroxide and hydrochloric acid using
calorimetric method.
Materials:
Apparatus:
1. Calorimeter
Chemicals:
1. 50ml NaOH (1.0M)
2. 50ml HCl (2.0M)
2. Thermometer (±0.1℃)
3. Stopwatch (±0.01sec)
4. Beaker
5. Graduated cylinder (±1ml)
Procedure: referred to the first page of instruction paper
Data collection:
Table 1:
Individual student raw data showing the relationship between time and the temperature of the
mixture after hydrochloric acid is added to sodium hydroxide. The temperature was recorded
every 30 seconds, and the experiment was conducted twice.
Time (sec)
Temperature (℃)
Temperature (℃)
Trial 1
Trial 2
0
22.0±0.1
22.5±0.1
30
27.0±0.1
26.1±0.1
60
27.0±0.1
26.2±0.1
90
27.3±0.1
26.2±0.1
120
27.5±0.1
26.0±0.1
150
27.6±0.1
26.0±0.1
180
27.5±0.1
25.9±0.1
210
27.5±0.1
25.6±0.1
240
27.2±0.1
25.4±0.1
270
27.2±0.1
25.4±0.1
300
27.2±0.1
25.1±0.1
330
27.1±0.1
25.1±0.1
360
27.0±0.1
25.0±0.1
390
27.0±0.1
25.0±0.1
2
Data processing:
Graph 1:
Graph showing the temperature change after sodium hydroxide and hydrochloric acid is mixed
together in the calorimeter. The graph is based on Table 1. Temperature is measured by
thermometer with uncertainty of 0.1℃, and time is measured by stopwatch with uncertainty
0.01 second. The initial temperatures of NaOH and HCl were both 19.0℃.
30.0
Temperature (℃)
25.0
20.0
Trial 1
Trial 2
15.0
10.0
5.0
0.0
0
100
200
300
Time (seconds)
400
500
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Calculation of heat of neutralization:
From Data Collection and Data Processing, the temperature change of the solution could be
calculated. If we assume that the solution has the same density and specific heat energy as
water, we can calculate the heat of neutralization per mole of water formed. Calculation is
shown below:
1) Specific heat capacity
Specific heat energy of water: 4.18 KJ/mol (assume there is no heat loss to the surroundings)
2) Calculation of mass
Equation of reaction: NaOH + HCl  NaCl + H2O
Volume of NaOH (0.1M) used is 50±1ml  0.05L±0.001L
Volume of HCl (0.2M) used is 50±1ml  0.05L±0.001L
When they are added,
0.05 + 0.05 = 0.1 L ±0.002L
Because it is assumed that solution has same density with water, 0.1L is equivalent to 0.1g.
3) Calculation of temperature change
Initial Temperature of HCl and NaOH: 19.0±0.1℃
Highest temperature of the mixture: Trial 1 = 27.6±0.1℃
△T1 = 27.6 – 19 = 8.6
△T2 =26.2 – 19 = 7.2
and Trial 2 = 26.2±0.1℃
with uncertainty: 8.6±0.2℃
with uncertainty: 7.2±0.2℃
4) Calculation of energy
The energy (Q) can be calculated by the equation:
Q= (mass) x (temperature difference) x (specific heat energy)
Trial 1 T = 27.6±0.1℃
Q = 0.1 x 8.6 x 4.18 = 3.59 KJ/mol
Trial 2 T = 26.2±0.1℃
with uncertainty: 3.59±0.2 KJ/mol
Q = 0.1 x 7.2 x 4.18 = 3.01 KJ/mol
with uncertainty: 3.01±0.1 KJ/mol
The mean of the Q for Trial 1 and Trial 2 is (3.59 + 3.01) / 2 = 3.3
with uncertainty: 3.3±
0.15 KJ/mol
However, because this value is calculated with 0.1mol, we need to adjust it to that of 1mol.
Therefore, we need to divide the value by 0.1.
3.3 / 0.1 = 33.0 KJ/mol
with uncertainty: 33.0±1.5 KJ/mol
4
Conclusion:
From the experiment, the heat energy of the neutralization between HCl and NaOH is
calculated to be about 33.0±1.5 KJ/ mol. However, according to the data booklet, the literature
value is about 56.6 KJ/mol. This means that the energy of the neutralization between
hydrochloric acid and sodium hydroxide calculated in this experiment is much smaller than the
literature value. This is mainly because the heat has escaped to the surroundings even though
the mixture of solution is insulated from the surroundings by calorimeter.
Analyzing Error:
Percentage discrepancy: ｜(33 – 56.6) / 56.6｜x 100 = 41.7%
The experimental value is much smaller than the literature value. Therefore, this error is related
to the precision error.
Evaluation:
Limitation of the experiment:
・Heat loss to the surroundings
 Even though the calorimeter is used, heat loss to the surroundings is inevitable. The
temperature change is smaller than the actual temperature change. This is the major factor of
precision error in the experiment.
・Assumption that the solution has same density and specific heat capacity with water
 Although the reaction between NaOH and HCl will produce water, it is not a pure water
because the solution contains NaCl. Also, because the molarity of HCl is 0.2M while that of
NaOH is 0.1M, HCl is in excess after 50ml of each solution is mixed.
Amount of NaOH is Mol = Concentration X Volume = 0.1 X 0.05 = 0.005 mol
Amount of HCl is
Mol = Concentration X Volume = 0.2 X 0.05 = 0.010 mol
Thus, the solution is a mixture of water, NaCl, and HCl even after the reaction has taken place.
It is not proper to assume that the solution has same density and specific heat capacity with
water.
Weakness of the experiment:
・Large uncertainty caused by equipment
 To measure 50ml of NaOH and HCl exactly, it was necessary to use graduated cylinder with
relatively large uncertainty of 1ml.
・Time taken before the first temperature is read
 After HCl is added to NaOH, it was necessary to immediately put the lid on the calorimeter,
but this takes time. Also, reading the temperature on thermometer at 0 second is impossible,
and therefore the data is not precise.
5
Improvement of the experiment:
・Use more precise instrument
Use more accurate thermometer and cylinder to minimize the uncertainty. Also, using
calorimeter with higher degree of insulation will lessen the heat loss to the surroundings.
・Use concentrated reagent
By using more concentrated NaOH and HCl, the temperature change will increase. The
increase in temperature will minimize the uncertainty ratio.
・Repeat experiment
Because the temperature varied greatly between the two trials, it is important to repeat the
experiment to get more accurate mean value.
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