We are tasked with finding the enthalpy change (∆H) for the reaction:

$\text{H}_2\text{SO}_4 (\text{l}) \rightarrow \text{SO}_3 (\text{g}) + \text{H}_2\text{O} (\text{g})$

using the given reactions and their corresponding ∆H values:

1. $\text{H}_2\text{S} (\text{g}) + 2 \text{O}_2 (\text{g}) \rightarrow \text{H}_2\text{SO}_4 (\text{l})$, ∆H = -235.5 kJ

2. $\text{H}_2\text{S} (\text{g}) + 2 \text{O}_2 (\text{g}) \rightarrow \text{SO}_3 (\text{g}) + \text{H}_2\text{O} (\text{l})$, ∆H = -207 kJ

3. $\text{H}_2\text{O} (\text{l}) \rightarrow \text{H}_2\text{O} (\text{g})$, ∆H = 44 kJ

Step-by-Step Solution:

We can use Hess's Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes of the steps that lead to the overall reaction.

We need to manipulate the given reactions to match the desired equation. The reaction we're interested in is:

$\text{H}_2\text{SO}_4 (\text{l}) \rightarrow \text{SO}_3 (\text{g}) + \text{H}_2\text{O} (\text{g})$

Analyzing the Reactions:

• Reaction 1 produces H₂SO₄ (l), so we can reverse this reaction to get H₂SO₄ (l) as a product.
• Reaction 2 produces SO₃ (g) and H₂O (l). We need to reverse this reaction and turn the H₂O (l) into H₂O (g).
• Reaction 3 gives us the transformation of liquid water to gas (H₂O (l) → H₂O (g)).

Reversing and combining:

1. Reverse Reaction 1:

$\text{H}_2\text{SO}_4 (\text{l}) \rightarrow \text{H}_2\text{S} (\text{g}) + 2 \text{O}_2 (\text{g}) \quad \Delta H = +235.5 \text{ kJ}$

2. Reverse Reaction 2 and also turn H₂O (l) into H₂O (g):

$\text{SO}_3 (\text{g}) + \text{H}_2\text{O} (\text{l}) \rightarrow \text{H}_2\text{S} (\text{g}) + 2 \text{O}_2 (\text{g}) \quad \Delta H = +207 \text{ kJ}$

3. Use Reaction 3 as is (it’s already in the correct direction):

$\text{H}_2\text{O} (\text{l}) \rightarrow \text{H}_2\text{O} (\text{g}) \quad \Delta H = 44 \text{ kJ}$

Now, combine all reactions:

• We need to eliminate H₂S (g) and 2O₂ (g) from the combined equation.

Adding the reversed Reaction 1 and Reaction 2:

The $\text{H}_2\text{S} (\text{g})$ terms cancel out, and we are left with:

$\text{H}_2\text{SO}_4 (\text{l}) \rightarrow \text{SO}_3 (\text{g}) + \text{H}_2\text{O} (\text{l}) \quad \Delta H = +235.5 \text{ kJ} + 207 \text{ kJ} = 442.5 \text{ kJ}$

Finally, we add Reaction 3 (which turns H₂O (l) into H₂O (g)):

$\text{SO}_3 (\text{g}) + \text{H}_2\text{O} (\text{l}) \rightarrow \text{SO}_3 (\text{g}) + \text{H}_2\text{O} (\text{g}) \quad \Delta H = 44 \text{ kJ}$

So, the overall reaction is:

$\text{H}_2\text{SO}_4 (\text{l}) \rightarrow \text{SO}_3 (\text{g}) + \text{H}_2\text{O} (\text{g})$

Final ∆H:

Summing all the enthalpy changes:

$\Delta H_{\text{total}} = 442.5 \, \text{kJ} + 44 \, \text{kJ} = 486.5 \, \text{kJ}$

Thus, the enthalpy change for the reaction is:

$\boxed{\Delta H = 486.5 \, \text{kJ}}$

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Would you like any further explanation or details about the steps taken? Here are some related questions:

1. How can Hess's Law be used to solve reactions that cannot be directly observed?
2. What is the significance of reversing reactions when using Hess’s Law?
3. How would you calculate the ∆H if there was a phase change involved, like from solid to liquid?
4. How does the enthalpy of formation relate to Hess's Law?
5. Can Hess's Law be used for reactions involving different states of matter, like gases and solids?

Tip: Always check that the elements on both sides of the reaction are balanced before adding the equations together!