What is the change in entropy when 6.03 mL of liquid benzene (C6H6, d = 0.879 g/mL) is combusted in the presence of 26.8 L of oxygen gas, measured at 298 K and 1 atm pressure? (R = 0.0821 L · atm/(K · mol))
2C6H6(l) + 15O2(g) → 12CO2(g) + 6H2O(l); ∆S° = –437.7 J/K at 298 K
dSrxn = (n*dSo products) - (n*dSo reactants)
Look the standard So values in tables in your text/notes/web.
To find the change in entropy (∆S) for the given reaction, we can use the equation:
∆S = Σn(products)S°(products) - Σn(reactants)S°(reactants)
where n represents the stoichiometric coefficient of each substance, S° represents the standard molar entropy of each substance, and the Σ symbol indicates the sum of all the terms.
First, let's calculate the moles of benzene (C6H6) and oxygen gas (O2) in the reaction:
Molar mass of C6H6 = 12.01 g/mol * 6 + 1.01 g/mol * 6 = 78.11 g/mol
Mass of C6H6 = 6.03 mL * 0.879 g/mL = 5.29 g
Moles of C6H6 = 5.29 g / 78.11 g/mol ≈ 0.068 mol
Volume of O2 = 26.8 L
Moles of O2 = (26.8 L * 1 mol) / (22.4 L/mol) ≈ 1.196 mol
Next, let's calculate the change in entropy (∆S):
∆S = (12 * S°(CO2) + 6 * S°(H2O)) - (2 * S°(C6H6) + 15 * S°(O2))
Using a data source, we can find the standard molar entropies (S°) of each substance:
S°(CO2) = 213.7 J/(K * mol)
S°(H2O) = 69.9 J/(K * mol)
S°(C6H6) = 173.4 J/(K * mol)
S°(O2) = 205.2 J/(K * mol)
Substituting the values into the equation:
∆S = (12 * 213.7 J/(K * mol) + 6 * 69.9 J/(K * mol)) - (2 * 173.4 J/(K * mol) + 15 * 205.2 J/(K * mol))
∆S = (2564.4 J/K) - (346.8 J/K)
∆S = 2217.6 J/K