Problem 46 - Chemistry An Atoms First Approach

Question

Predict the sign of \(\Delta S^{\circ}\) and then calculate \(\Delta S^{\circ}\) for each of the following reactions. a. \(\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\) b. \(2 \mathrm{CH}_{3} \mathrm{OH}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\) c. \(\mathrm{HCl}(g) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{Cl}^{-}(a q)\)

Step-by-Step Solution

Answer

The sign of \(\Delta S^{\circ}\) and the calculated values for the given reactions are: a) Negative; -164.2 \(\frac{J}{mol K}\) b) Positive; 243.6 \(\frac{J}{mol K}\) c) Negative; -130.4 \(\frac{J}{mol K}\)

Step 1: Predict the sign of \(\Delta S^{\circ}\)

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 \(\Delta S^{\circ}\).

Step 2: Calculate \(\Delta S^{\circ}\)

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 H₂(g): \(S^{\circ}_{\text{H₂(g)}} = 130.6 \frac{J}{mol K}\) - For O₂(g): \(S^{\circ}_{\text{O₂(g)}} = 205.1 \frac{J}{mol K}\) - For H₂O(l): \(S^{\circ}_{\text{H₂O(l)}} = 69.9 \frac{J}{mol K}\) 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. \(2 \mathrm{CH}_{3} \mathrm{OH}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\) #

Step 1: Predict the sign of \(\Delta S^{\circ}\)

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 \(\Delta S^{\circ}\).

Step 2: Calculate \(\Delta S^{\circ}\)

Let's find the entropy changes in the reactants and products: - For CH₃OH(g): \(S^{\circ}_{\text{CH₃OH(g)}} = 239.7 \frac{J}{mol K}\) - For O₂(g): \(S^{\circ}_{\text{O₂(g)}} = 205.1 \frac{J}{mol K}\) - For CO₂(g): \(S^{\circ}_{\text{CO₂(g)}} = 213.8 \frac{J}{mol K}\) - For H₂O(g): \(S^{\circ}_{\text{H₂O(g)}} = 188.8 \frac{J}{mol K}\) 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. \(\mathrm{HCl}(g) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{Cl}^{-}(a q)\) #

Step 1: Predict the sign of \(\Delta S^{\circ}\)

Gaseous HCl dissociates into aq H⁺ and Cl⁻ 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 \(\Delta S^{\circ}\).

Step 2: Calculate \(\Delta S^{\circ}\)

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⁺(aq): \(S^{\circ}_{\text{H⁺(aq)}} = 0 \frac{J}{mol K}\) (by convention) - For Cl⁻(aq): \(S^{\circ}_{\text{Cl⁻(aq)}} = 56.5 \frac{J}{mol K}\) 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 \(\frac{J}{mol K}\).