Question
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
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
Answers
GPT 3.5
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.
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.