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From gas to a liquid phase, the disorder of a system decreases. In this reaction, two gas molecules are combining to form one liquid molecule. Considering fewer molecules and lower randomness, we can predict a negative \(\Delta S^{\circ}\).

To calculate the change in entropy, we will use the formula: \[\Delta S^{\circ} =\Delta S^{\circ}_{\text{products}} - \Delta S^{\circ}_{\text{reactants}}\] Let's find the entropy changes in the reactants and products: - For H鈧�(g): \(S^{\circ}_{\text{H鈧�(g)}} = 130.6 \frac{J}{mol K}\) - For O鈧�(g): \(S^{\circ}_{\text{O鈧�(g)}} = 205.1 \frac{J}{mol K}\) - For H鈧侽(l): \(S^{\circ}_{\text{H鈧侽(l)}} = 69.9 \frac{J}{mol K}\) Now, we can substitute these values in the formula and calculate the change in entropy: \[\Delta S^{\circ}= 69.9 - (130.6 + (0.5 \times 205.1))\] \[\Delta S^{\circ}= -164.2 \frac{J}{mol K}\] #b. \(2 \mathrm{CH}_{3} \mathrm{OH}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\) #

We have five moles of gas in the reactants and six moles of gas in the products. This means an increase in the number of molecules involved in the reaction which results in higher disorder. We can predict a positive \(\Delta S^{\circ}\).

Let's find the entropy changes in the reactants and products: - For CH鈧僌H(g): \(S^{\circ}_{\text{CH鈧僌H(g)}} = 239.7 \frac{J}{mol K}\) - For O鈧�(g): \(S^{\circ}_{\text{O鈧�(g)}} = 205.1 \frac{J}{mol K}\) - For CO鈧�(g): \(S^{\circ}_{\text{CO鈧�(g)}} = 213.8 \frac{J}{mol K}\) - For H鈧侽(g): \(S^{\circ}_{\text{H鈧侽(g)}} = 188.8 \frac{J}{mol K}\) Now we can substitute these values in the formula and calculate the change in entropy: \[\Delta S^{\circ}= (2 \times 213.8 + 4 \times 188.8) - (2 \times 239.7 + 3 \times 205.1)\] \[\Delta S^{\circ}= 243.6 \frac{J}{mol K}\] #c. \(\mathrm{HCl}(g) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{Cl}^{-}(a q)\) #

Gaseous HCl dissociates into aq H鈦� and Cl鈦� ions, resulting in a higher disorder due to the involvement of a significant amount of solvent molecules to solvate these ions. Consequently, we can predict a positive \(\Delta S^{\circ}\).

Let's find the entropy changes in the reactants and products: - For HCl(g): \(S^{\circ}_{\text{HCl(g)}} = 186.9 \frac{J}{mol K}\) - For H鈦�(aq): \(S^{\circ}_{\text{H鈦�(aq)}} = 0 \frac{J}{mol K}\) (by convention) - For Cl鈦�(aq): \(S^{\circ}_{\text{Cl鈦�(aq)}} = 56.5 \frac{J}{mol K}\) Now we can substitute these values in the formula and calculate the change in entropy: \[\Delta S^{\circ} = (0 + 56.5) - (186.9)\] \[\Delta S^{\circ} = -130.4 \frac{J}{mol K}\] To summarize the results: - For reaction (a), the sign of \(\Delta S^{\circ}\) is negative, and the calculated value is -164.2 \(\frac{J}{mol K}\). - For reaction (b), the sign of \(\Delta S^{\circ}\) is positive, and the calculated value is 243.6 \(\frac{J}{mol K}\). - For reaction (c), the sign of \(\Delta S^{\circ}\) is negative, and the calculated value is -130.4 \(\frac{J}{mol K}\).