Calculate ΔH° for the reaction S(s) + O2(g) -> SO2(g) from the following data: S(s) + (3/2)O2(g) -> SO3(g) ΔH° = -395.2 kJ 2SO2(g) + O2(g) -> 2SO3(g) ΔH° = -198.2 kJ

Steps to Solve

1. Write down the target reaction and the given reactions

Target reaction: $S(s) + O_2(g) \rightarrow SO_2(g)$

Given reactions: Reaction 1: $S(s) + \frac{3}{2}O_2(g) \rightarrow SO_3(g) \quad \Delta H_1^\circ = -395.2 , \text{kJ}$ Reaction 2: $2SO_2(g) + O_2(g) \rightarrow 2SO_3(g) \quad \Delta H_2^\circ = -198.2 , \text{kJ}$

2. Manipulate Reaction 2

We need $SO_2(g)$ on the product side, so we reverse Reaction 2 and divide it by 2:

$\frac{1}{2} \times$ [ Reversing Reaction 2: $2SO_3(g) \rightarrow 2SO_2(g) + O_2(g)$] $SO_3(g) \rightarrow SO_2(g) + \frac{1}{2}O_2(g) \quad \Delta H^\circ = \frac{198.2}{2} = 99.1 , \text{kJ}$

3. Add the manipulated reactions

Adding Reaction 1 and the modified Reaction 2: $S(s) + \frac{3}{2}O_2(g) \rightarrow SO_3(g) \quad \Delta H_1^\circ = -395.2 , \text{kJ}$ $SO_3(g) \rightarrow SO_2(g) + \frac{1}{2}O_2(g) \quad \Delta H^\circ = 99.1 , \text{kJ}$

Adding these equations together:

$S(s) + \frac{3}{2}O_2(g) + SO_3(g) \rightarrow SO_3(g) + SO_2(g) + \frac{1}{2}O_2(g)$

4. Simplify the equation

Cancel out any common species on both sides of the equation: $SO_3(g)$ appears on both sides, so we can remove it. Also, we have $\frac{3}{2}O_2(g)$ on the left and $\frac{1}{2}O_2(g)$ on the right. Subtracting $\frac{1}{2}O_2(g)$ from both sides gives us: $S(s) + (\frac{3}{2} - \frac{1}{2})O_2(g) \rightarrow SO_2(g)$ $S(s) + O_2(g) \rightarrow SO_2(g)$

This is our target reaction.

5. Calculate the overall enthalpy change

Add the enthalpy changes for the reactions we added: $\Delta H^\circ = -395.2 , \text{kJ} + 99.1 , \text{kJ} = -296.1 , \text{kJ}$