4. Use the standard molar entropies below to calculate the standard molar entropy change for dissolving KNO₃ in water.

Given $S^\circ$ values:
- KNO₃: 133.1 J/mol-K
- $K^+_{(aq)}$: 101.2 J/mol-K
- $NO_3^-_{(aq)}$: 146.70 J/mol-K

Calculate $\Delta S^\circ$:
\[
\Delta S^\circ = S^\circ_{\text{products}} - S^\circ_{\text{reactants}}
\]
\[
\Delta S^\circ = (146.70 + 101.2) - 133.1 = 114.8 \, \text{J/mol-K}
\]

5. Compare your calculated value for $\Delta S^\circ$ from part 4 with your experimentally obtained value of 106.8 J/mol-K. Comment on any difference in the values based on the experimental procedure.

The standard molar entropy change for dissolving KNO3 in water is calculated as 114.8 J/mol-K, which slightly differs from the experimentally obtained value of 106.8 J/mol-K. Differences can result from experimental inaccuracies or deviations from ideal behavior.

Explanation:

The student is asked to calculate the standard molar entropy change for dissolving potassium nitrate (KNO3) in water. Using the given standard molar entropies, the change is calculated as:

Sproducts - Sreactants = (SK+(aq) + SNO3-(aq)) - SKNO3(s)
= (101.2 J/mol-K + 146.7 J/mol-K) - 133.1 J/mol-K
= 114.8 J/mol-K

The student's experimentally obtained value for the entropy change is 106.8 J/mol-K. Differences between the calculated and experimental values can arise from many factors in the experimental procedure, such as measurement precision, assumptions made in the calculation (like ideal behavior), or errors in controlling external conditions (temperature, pressure, etc.).