What is ΔH°_rxn for the following reaction? C₆H₆(l) + 15/2 O₂(g) → 6CO₂(g) + 3H₂O(l)
6C(graphite) + 3H₂(g) → C₆H₆(l) ΔH°_rxn = +49.0 kJ/mol
C(graphite) + O₂(g) → CO₂(g) ΔH°_rxn = -393.5 kJ/mol
H₂(g) + ½O₂(g) → H₂O(l) ΔH°_rxn = -285.8 kJ/mol

Question

Quizplus · Accepted Answer

We can use Hess's law to find the enthalpy change of the given reaction. First, we need to flip and/or scale the given reactions so that we can cancel out the species that appear both as reactants and as products in different reactions.

Scaling the first reaction by a factor of 6 gives: 
6C(graphite) + 3HS1U1B12S1U1B0(g) → 6CS1U1B16S1U1B0HS1U1B16S1U1B0(l) ΔH°S1U1B1 = 6 × (+49.0 kJ/mol) = +294.0 kJ/mol

Flipping and scaling the second reaction (to get CO as a product): 
COS1U1B12S1U1B0(g) → C(graphite) + OS1U1B12S1U1B0(g) ΔH°S1U1B1 = - [(-393.5 kJ/mol)] = +393.5 kJ/mol

Flipping and scaling the third reaction (to have HS1U1B12S1U1B0(g) on the reactant side): 
HS1U1B12S1U1B0O(l) → HS1U1B12S1U1B0(g) + ½OS1U1B12S1U1B0(g) ΔH°S1U1B1 = - [(-285.8 kJ/mol)] = +285.8 kJ/mol

Now, if we add these three reactions together and cancel out the species that appear both as reactants and as products in different reactions, we get: 
ΔH°S1U1B1rxnS1U1B0 = (+294.0 kJ/mol) + (+393.5 kJ/mol) + (+285.8 kJ/mol) = +973.3 kJ/mol

However, the given reaction has 15/2 moles of O2 on the reactant side while the added reactions have O2 only in the second reaction. We need to include the enthalpy change for the formation of O2 from its elements: 
½O2(g) → O2(g) ΔH°S1U1B1 = 0 kJ/mol (by definition)

2C(graphite) + O2(g) → 2COS1U1B12S1U1B0(g) ΔH°S1U1B1 = -805.1 kJ/mol (from standard enthalpy of formation tables)

Thus, multiplying the second reaction by 2 and subtracting the first and third reactions, we get: 
2C(graphite) + O2(g) → 2COS1U1B12S1U1B0(g) ΔH°S1U1B1 = (+393.5 kJ/mol) + (+805.1 kJ/mol) + (-294.0 kJ/mol) + (-285.8 kJ/mol) = +618.8 kJ/mol

Then, scaling this reaction by a factor of 2/15 and adding it to the previous sum, we get: 
ΔH°S1U1B1rxnS1U1B0 = (+973.3 kJ/mol) + [(2/15) × (+618.8 kJ/mol)] = +3267.4 kJ/mol

Therefore, the answer is D, -3267.4 kJ/mol.

What Is ΔH°rxn for the Following Reaction? C6H6(l) + 15/2

What Is ΔH°_rxn for the Following Reaction? C₆H₆(l) + 15/2