Energy changes
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I am doing an experiment about the determination of the concentrations of two acids, hydrochloric acid and ethanoic acid by thermometric titration, to calculate the enthalpy change for each titration. I would like to ask:
1. The enthalpy change of neutralization for a very dilute strong acid reacting with a very dilute strong base is constant at -57.6 kJ mol-1. Why is the value constant?
2. Why are the experimental results usually a little less negative than -57.6 kJ mol-1? Is it anything about the heat loss when performing the experiment?
3. Why the heats of neutralization for reactions involving weak acids and/or weak alkalis are always less negative than for strong acids and alkalis?
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Corrie writes .....
Strong acids and bases are, by definition completely ionised, to give H+ and OH- ions respectively, in dilute solutions. Here is part of an answer to a similar question which my colleague Jenkin gave a while ago:
1. As long as we are considering a strong acid reacting with a strong base, we are always dealing with the same reaction, that is:
H+ (aq) + OH– (aq) -> H2O (l)
This is because strong acids and bases are assumed to be completely ionised.
3. Notice that weak acids (such as CH3CO2H) and weak bases (such as ammonia) have different values, because some energy is needed to bring about ionisation.
2. Yes, there will be heat losses in this experiment - a polystyrene cup is a good but not perfect insulator, heat can be lost from the surface of the solution, and the cup and thermometer will absorb some heat themselves. Furthermore, in your calculation you will be assuming the solution has the specific heat capacity of pure water (4.18 J g-1 K-1) and that 100 cm3 of it has the same mass as that of pure water (density = 1 g cm-3). All these little losses or approximations add up. The greatest source of measurement error will be in the temperature rise, especially if only a few degrees.
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updated: 28 March 2008
