Let's solve each part step by step using the provided thermodynamic data.

Reaction setup

The balanced chemical equation for the formation of SO₃(g) from SO₂(g) and O₂(g) is:

$2 \, \text{SO}_2(g) + \text{O}_2(g) \rightarrow 2 \, \text{SO}_3(g)$

Given Data

• ΔHf° (Standard enthalpy of formation, in kJ/mol)

  • SO₂(g): -297 kJ/mol
  • O₂(g): 0 kJ/mol
  • SO₃(g): -395 kJ/mol

• ΔS° (Standard entropy, in J/mol-K)

  • SO₂(g): 249 J/mol-K
  • O₂(g): 205 J/mol-K
  • SO₃(g): 256 J/mol-K

Calculations

1. Enthalpy of the reaction (ΔH°)

The enthalpy change of the reaction, ΔH°, is calculated using the formula:

$\Delta H^\circ = \sum \Delta H_f^\circ \text{(products)} - \sum \Delta H_f^\circ \text{(reactants)}$

Substituting in values:

[ \Delta H^\circ = \left[2 \times (-395)\right] - \left[2 \times (-297) + 1 \times 0\right] ]

$= (-790) - (-594) = -196 \, \text{kJ/mol}$

So, the enthalpy of the reaction, ΔH°, is -196 kJ/mol.

2. Entropy change of the reaction (ΔS°)

The entropy change of the reaction, ΔS°, is calculated using the formula:

$\Delta S^\circ = \sum S^\circ \text{(products)} - \sum S^\circ \text{(reactants)}$

Substituting in values:

[ \Delta S^\circ = \left[2 \times 256\right] - \left[2 \times 249 + 1 \times 205\right] ]

$= 512 - 703 = -191 \, \text{J/mol-K}$

So, the entropy change of the reaction, ΔS°, is -191 J/mol-K.

3. Entropy change of the surroundings (ΔS° surroundings)

The entropy change of the surroundings is related to the enthalpy change of the reaction and the temperature at which it occurs. Assuming standard conditions (298 K), the formula is:

$\Delta S_{\text{surroundings}} = -\frac{\Delta H^\circ}{T}$

Converting ΔH° to J/mol (since 1 kJ = 1000 J):

$\Delta H^\circ = -196 \, \text{kJ/mol} = -196000 \, \text{J/mol}$

Now,

$\Delta S_{\text{surroundings}} = -\frac{-196000}{298} = 657.72 \, \text{J/mol-K}$

So, the entropy change of the surroundings is 657.72 J/mol-K.

4. Is this reaction spontaneous?

To determine spontaneity, we evaluate the Gibbs free energy change (ΔG°) using the formula:

$\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ$

Substitute the values (using T = 298 K):

$\Delta G^\circ = -196000 - (298 \times -191)$

$= -196000 + 56818 = -139182 \, \text{J/mol} \quad \text{or} \quad -139.18 \, \text{kJ/mol}$

Since ΔG° is negative, the reaction is spontaneous.

Summary of Answers

1. Enthalpy of the reaction, ΔH°: -196 kJ/mol
2. Entropy change of the reaction, ΔS°: -191 J/mol-K
3. Entropy change of the surroundings, ΔS° surroundings: 657.72 J/mol-K
4. Is this reaction spontaneous? Yes

Would you like further clarification on any of these calculations or concepts?

Additional Questions:

1. How does the value of ΔS° influence the spontaneity of a reaction?
2. What effect does temperature have on the entropy change of the surroundings?
3. Why is the enthalpy of formation for O₂(g) set to zero?
4. Can a reaction with a negative entropy change still be spontaneous?
5. How would these values change if the reaction were conducted at a different temperature?

Tip: Always keep track of units when performing thermodynamic calculations to ensure consistency and accuracy.