Question Video: Calculating the Enthalpy of Dilution of Sodium Hydroxide Using Enthalpy of Solution Values | Nagwa Question Video: Calculating the Enthalpy of Dilution of Sodium Hydroxide Using Enthalpy of Solution Values | Nagwa

Question Video: Calculating the Enthalpy of Dilution of Sodium Hydroxide Using Enthalpy of Solution Values Chemistry • First Year of Secondary School

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The enthalpy change of a solution for NaOH with differing amounts of water is shown. NaOH(s) + 3H₂O(l) ⟶ NaOH(aq), Δ𝐻_sol = −28.9 kJ/mol NaOH(s) + 300H₂O(l) ⟶ NaOH(aq), Δ𝐻_sol = −42.3 kJ/mol What is the enthalpy change of dilution, Δ𝐻_dil?

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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.

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