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Question6) Calculate the Gibbs free energy for the process of dissolving ammonium nitrate and for the process of dissolving sodium hydroxide. (Use the standard temperature, 298.15 K .) Do the results match your observations? Explain. ΔGstins 0=ΔH−1ΔS=22303.12−(218.15)(108.8)=−10135.6ΔG=−10135.6+(8.314)(248.15)=−7656.78=−7.7uy/mol⋅k\begin{array}{l} \Delta G_{\text {stins }}^{0}=\Delta H-1 \Delta S \\ =22303.12-(218.15)(108.8) \\ =-10135.6 \\ \begin{aligned} \Delta G & =-10135.6+(8.314)(248.15) \\ & =-7656.78 \\ & =-7.7 \mathrm{uy} / \mathrm{mol} \cdot \mathrm{k} \end{aligned} \end{array} ΔGG∘=Δ17−1aS=−35581.04−298.15(−16.16)=−30762.93ΔG=−30762.93+(8.314)(298.15)=−28284.12=−28.3 kJ/mol\begin{aligned} \Delta G G^{\circ} & =\Delta 17-1 a S \\ & =-35581.04-298.15(-16.16) \\ & =-30762.93 \\ \Delta G & =-30762.93+(8.314)(298.15) \\ & =-28284.12 \\ & =-28.3 \mathrm{~kJ} / \mathrm{mol} \end{aligned}

Studdy Solution
Compare the results with observations:
1. The negative ΔG\Delta G for both processes indicates that both dissolving ammonium nitrate and sodium hydroxide are spontaneous processes at 298.15 K298.15 \, \text{K}.
2. The magnitude of ΔG\Delta G for sodium hydroxide is larger (more negative) than for ammonium nitrate, suggesting that dissolving sodium hydroxide is more energetically favorable.
3. These results should match observations where sodium hydroxide dissolves exothermically and ammonium nitrate dissolves endothermically.
The calculated Gibbs free energies are −10.1388 kJ/mol-10.1388 \, \text{kJ/mol} for ammonium nitrate and −30.764696 kJ/mol-30.764696 \, \text{kJ/mol} for sodium hydroxide. These results align with the expected observations of the dissolution processes.

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