To answer these questions, let’s follow each calculation step by step based on the thermodynamic data provided.

Given Data

$\Delta H_f^\circ \text{ for SO}_2(g) = -297 \, \text{kJ/mol}$ $\Delta H_f^\circ \text{ for O}_2(g) = 0 \, \text{kJ/mol}$ $\Delta H_f^\circ \text{ for SO}_3(g) = -395 \, \text{kJ/mol}$ $\Delta S^\circ \text{ for SO}_2(g) = 249 \, \text{J/mol-K}$ $\Delta S^\circ \text{ for O}_2(g) = 205 \, \text{J/mol-K}$ $\Delta S^\circ \text{ for SO}_3(g) = 256 \, \text{J/mol-K}$

Let's assume the balanced reaction is: $2\text{SO}_2(g) + \text{O}_2(g) \rightarrow 2\text{SO}_3(g)$

(1) Enthalpy of the Reaction $(\Delta H_{\text{rxn}})$

The enthalpy of reaction is calculated using: $\Delta H_{\text{rxn}} = \sum \Delta H_f^\circ (\text{products}) - \sum \Delta H_f^\circ (\text{reactants})$

For the balanced reaction: [ \Delta H_{\text{rxn}} = [2 \times \Delta H_f^\circ (\text{SO}_3)] - [2 \times \Delta H_f^\circ (\text{SO}_2) + 1 \times \Delta H_f^\circ (\text{O}_2)] ]

Substituting the values: [ \Delta H_{\text{rxn}} = [2 \times (-395)] - [2 \times (-297) + 0] = -790 + 594 = -196 , \text{kJ/mol} ]

Answer for (1): $-196 \, \text{kJ/mol}$

(2) Entropy Change of the Reaction $(\Delta S_{\text{rxn}})$

The entropy change for the reaction is calculated using: $\Delta S_{\text{rxn}} = \sum \Delta S^\circ (\text{products}) - \sum \Delta S^\circ (\text{reactants})$

For the balanced reaction: [ \Delta S_{\text{rxn}} = [2 \times \Delta S^\circ (\text{SO}_3)] - [2 \times \Delta S^\circ (\text{SO}_2) + 1 \times \Delta S^\circ (\text{O}_2)] ]

Substituting the values: [ \Delta S_{\text{rxn}} = [2 \times 256] - [2 \times 249 + 205] = 512 - 703 = -191 , \text{J/mol-K} ]

Answer for (2): $-191 \, \text{J/mol-K}$

(3) Entropy Change of the Surroundings $(\Delta S_{\text{surroundings}})$

The entropy change of the surroundings at constant temperature can be found using: $\Delta S_{\text{surroundings}} = -\frac{\Delta H_{\text{rxn}}}{T}$

Assuming $T = 298 \, \text{K}$: $\Delta S_{\text{surroundings}} = -\frac{-196 \times 1000}{298} = 657.72 \, \text{J/mol-K}$

Answer for (3): $657.72 \, \text{J/mol-K}$

(4) Spontaneity of the Reaction

To determine if the reaction is spontaneous, we evaluate the Gibbs free energy change $(\Delta G_{\text{rxn}})$ using: $\Delta G_{\text{rxn}} = \Delta H_{\text{rxn}} - T \Delta S_{\text{rxn}}$

Using $T = 298 \, \text{K}$: $\Delta G_{\text{rxn}} = -196 \, \text{kJ/mol} - (298 \, \text{K} \times -0.191 \, \text{kJ/mol-K})$ $\Delta G_{\text{rxn}} = -196 + 56.918 = -139.082 \, \text{kJ/mol}$

Since $\Delta G_{\text{rxn}} < 0$, the reaction is spontaneous.

Answer for (4): Yes (the reaction is spontaneous)

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Let me know if you’d like more details on any part of this calculation or have any questions!

Related Questions:

1. How would a change in temperature affect the spontaneity of this reaction?
2. Can the reaction’s equilibrium constant be determined from the $\Delta G_{\text{rxn}}$ value?
3. What would happen to $\Delta S_{\text{surroundings}}$ if the temperature were increased?
4. How does the entropy of the system relate to the overall disorder of the molecules involved?
5. Could you explain why $\Delta H_{\text{rxn}}$ is negative in this case?

Tip:

In thermodynamic calculations, always keep track of units and conversions (e.g., converting kJ to J when needed) to ensure accuracy.