ARTICLES
Understanding and Calculating the Heat Capacity of Solutions (continued)
100 g × 4.18 J.K –1 .g –1 = 418 J/K
K + (aq): 0.100 mol × 22 J.K –1 .mol –1 = 2.2 J/K
NO 3– (aq): 0.100 mol × –87 J.K –1 .mol– 1 = –8.7 J/K
4.22
4.21
Water:
TOTAL = 418 + 2.2 − 8.7 = 411.5 J/K, \actual temperature fall =
3600 J /(411.5 J/K) = 8.75 K
Dissolution of ammonium nitrate
We use about 100.0 g of water and make a solution that is close
to 1.00 molal by adding 8.00 gram of ammonium nitrate.
4.17
NH 4 NO 3 (s) → NH 4+ (aq) + NO 3– (aq) Δ H = +26 kJ
−366
−133
0
20
40
60
Temperature (°C)
80
100
Figure 1: The heat capacity of liquid water
−207
Hence again, by the usual method the heat absorbed dissolving
is computed to be 0.100 mol = 2.6 kJ
The heat capacity of liquid water is not simple. The 11 data points
were fitted using Excel’s built-in polynomial routines. Actually, a
quintic polynomial was required to get a visually acceptable fit.
Treating the heat capacity of the solution as the heat capacity of
the water,
Cp = –4.711538 × 10 –11 T 5 + 1.517920 × 10 –8 T 4
–1.861029 × 10 –6 T 3 + 1.161767 × 10 –4 T 2
–3.491806×10 –3 T + 4.217489
R² = 0.9997547
Heat capacity = m.Cp = 418 J/K
Temperature fall = 2600 J /(418 J/K) = 6.22 K
By once again including the amounts of dissolved ions:
Water: 100 g × 4.18 J.K –1 .g –1 = 418 J/K
NH 4+ (aq): 0.100 mol × 80 J.K –1 .g –1 = 8.0 J/K
NO 3– (aq):
4.19
4.18
The amount of NH 4 NO 3 dissolving is 0.100 mol. The amount of
heat released is calculated using standard heats of formation
and the equation.
4.20
Conclusion
Usually the heat capacity of the solution is treated as being
the heat capacity of the water. Whilst this is a good first
approximation since the great majority of the molecules present
are water, nevertheless, multiplying the mass of the solution by
4.18 J.K –1 .g –1 is not appropriate. In the examples above, the
heat capacity was less than the heat capacity of the water alone.
This is very common. The correction of adding the heat capacities
of the ions is computationally simple and should give more valid
and accurate results.
0.100 mol × –87 J.K –1 .mol –1 = –8.7 J/K
TOTAL = 418 + 8.0 – 8.7 = 417.3 J/K, ∴actual temperature fall
= 2600 J /(417.3 J/K) = 6.23 K
The temperature drops much less than with KNO 3 .
The models above are linear. In Figure 1, the data for water
are taken from the CRC Handbook of Chemistry and Physics.
In the temperature range that is commonly used for school
thermochemistry experiments, using the constant value
4.18 J.K –1 .mol –1 is an excellent approximation, so a linear model
works well. However, over the wider but limited temperature
range from 0 to 100˚C for liquid water at atmospheric pressure,
a polynomial fit is far better, so integration can be used to
calculate enthalpy differences. However, over the much broader
temperature range for gases, the asymptotic behaviour of
the model needs to be correct. The NIST chemical database
is searchable online, and provides excellent data for many
substances.
Reference
L G Hepler and J K Hovey, Can. J. Chem. 1996 74:639-
649. Available at https://www.nrcresearchpress.com/doi/
pdfplus/10.1139/v96-069
Note of thanks
I thank all my chemistry colleagues at NBHS for raising this matter
as a professional discussion, provoking me to think this through,
calculate, and share the results more broadly.
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SCIENCE EDUCATIONAL NEWS VOL 68 NO 3