Sulfur trioxide reacts with water to form sulfuric acid, a major
contributor to acid rain. One origin of SO3 is the combustion of
sulfur, which is present in small quantities in coal, according to
the following equation.
S(s) + 3/2 O2(g) → SO3(g)
Given the thermochemical information below, determine the
heat of reaction for this reaction.
S(s) + O2(g) → SO2(g) ∆𝐻
𝑜 = -296.8 kJ
2SO2(g) + O2(g) → 2SO3(g) ∆𝐻
𝑜 = -197.0 kJ
To find the heat of reaction for the combustion of sulfur to form sulfur trioxide, we can use Hess's Law.
Hess's Law states that the overall enthalpy change in a chemical reaction is equal to the sum of the enthalpy changes of the individual steps of the reaction.
We are given the enthalpy changes for the following reactions:
1) S(s) + O2(g) → SO2(g) ∆𝐻 = -296.8 kJ
2) 2SO2(g) + O2(g) → 2SO3(g) ∆𝐻 = -197.0 kJ
To find the enthalpy change for the reaction S(s) + 3/2 O2(g) → SO3(g), we need to manipulate these equations so that the desired reaction is obtained.
From equation 1): S(s) + O2(g) → SO2(g)
multiply this by 2 to get: 2S(s) + 2O2(g) → 2SO2(g)
From equation 2): 2SO2(g) + O2(g) → 2SO3(g)
Now, we can cancel out the common compounds in the two equations:
2S(s) + 2O2(g) → 2SO2(g)
2SO2(g) + O2(g) → 2SO3(g)
Combine the two equations:
2S(s) + 2O2(g) + 2SO2(g) + O2(g) → 2SO2(g) + 2SO3(g)
S(s) + 3O2(g) → 2SO3(g)
The enthalpy change for this combined reaction can be calculated by summing up the enthalpy changes for the individual reactions:
∆𝐻(combined) = ∆𝐻(reaction 1) + ∆𝐻(reaction 2)
∆𝐻(combined) = (-296.8 kJ) + (-197.0 kJ)
∆𝐻(combined) = -493.8 kJ
Therefore, the heat of reaction for the reaction S(s) + 3/2 O2(g) → SO3(g) is -493.8 kJ.