Question

Given the following reactions 2S (s) + 3O2 (g) → 2SO3 (g) ΔH = -300 kJ S (s) + O2 (g) → SO2 (g) ΔH = -200 kJ the enthalpy of the reaction in which sulfur dioxide is oxidized to sulfur trioxide 2SO2 (g) + O2 (g) → 2SO3 (g) is ________ kJ

Answer 1

For the given reaction the enthalpy of the reaction in which sulfur dioxide is oxidized to sulfur trioxide 2SO2 (g) + O2 (g) → 2SO3 (g) is 100 kJ.

To determine the enthalpy change
`(\(\Delta H\))` for the reaction in which sulfur dioxide is oxidized to sulfur trioxide:

`\[2SO_2 (g) + O_2 (g) \rightarrow 2SO_3 (g)\]`

you can use Hess's Law. This law states that the overall enthalpy change for a reaction is the sum of the enthalpy changes of its individual steps.

Given reactions:

1.
`\(2S (s) + 3O_2 (g) \rightarrow 2SO_3 (g)\) \(\Delta H = -300 \, \text{kJ}\)`

2.
`\(S (s) + O_2 (g) \rightarrow SO_2 (g)\) \(\Delta H = -200 \, \text{kJ}\)`

To find the enthalpy change for the target reaction, subtract the enthalpy change of the second reaction from the first:

`\[\Delta H = (-300 \, \text{kJ}) - (-200 \, \text{kJ})\]`

`\[\Delta H = -300 \, \text{kJ} + 200 \, \text{kJ}\]`

`\[\Delta H = -100 \, \text{kJ}\]`

So, the enthalpy change for the oxidation of sulfur dioxide to sulfur trioxide is
`\(-100`,
`\text{kJ}\)`.