Video Transcript
The enthalpy change of a solution
for NaOH with differing amounts of water is shown. NaOH solid plus three H2O liquid
react to form NaOH aqueous, Δ𝐻 sol equals negative 28.9 kilojoules per mole. NaOH solid plus 300 H2O liquid
react to form NaOH aqueous, Δ𝐻 sol equals negative 42.3 kilojoules per mole. What is the enthalpy change of
dilution, Δ𝐻 dil?
Chemists measure and calculate many
different kinds of enthalpy changes. In this question, we will be
working with the enthalpy of solution, which is abbreviated Δ𝐻 sol, and the
enthalpy of dilution, which is abbreviated Δ𝐻 dil.
The enthalpy of solution is the
enthalpy change when one mole of substance is diluted infinitely. It represents the point at which
adding more water no longer causes any more heat to be absorbed or released by the
solution. This is a theoretical value, and
scientists use large amounts of solvent to dissolve the solute to mimic this. However, this value doesn’t always
align with the enthalpy change values observed in laboratory experiments. This is where the enthalpy of
dilution comes in.
The enthalpy of dilution is the
enthalpy change when a solution of a substance is diluted infinitely per mole of
substance. This is a value that helps us
connect the theoretical enthalpy of solution with a more practical one.
When looking at the two enthalpy of
solution values that are provided, we see that they are different. We also see that the values are
negative. This means that dissolving sodium
hydroxide in water is an exothermic process overall and releases heat to the
surroundings.
The first value represents a
situation in which solid sodium hydroxide is dissolved in a few moles of water to
carry out an experiment in the laboratory. The second value represents
infinite dilution. We can see that there are a very
large number of moles of water used to dissolve the sodium hydroxide, which would
not be practical for an experiment. In other words, if we could
infinitely dilute sodium hydroxide in water, 42.3 kilojoules of energy would be
released per mole of NaOH. However, infinite dilution of NaOH
is not possible in the laboratory. So 28.9 kilojoules of energy could
be expected to be released per mole of NaOH under standard conditions.
To calculate the enthalpy of
dilution, we must find the extra amount of energy that would be released if we
diluted the smaller solution of sodium hydroxide infinitely by adding a large amount
of water. In this case, we’re adding 100
times more moles of water. So, we will need to subtract the
practical laboratory value of the enthalpy of solution from the enthalpy of solution
of the infinitely diluted solution. Let’s take negative 42.3 kilojoules
per mole and subtract negative 28.9 kilojoules per mole. We get the answer negative 13.4
kilojoules per mole.
In conclusion, the enthalpy change
of dilution for sodium hydroxide is negative 13.4 kilojoules per mole.
