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serenityjohnson98765
11.03.2021 •
Chemistry
Benzene is a minor component of gasoline. The standard molar enthalpy of formation of benzene C7H16(l) is 48.95 kJ/mol. For the balanced reaction equation of benzene use:
C6H6(l) + 7.5O2 --> 6CO2 + 3H2O.
Benzene is a liquid, but all the other chemicals in this reaction are gases. What is the enthalpy change of this reaction in kJ/mol?
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Ответ:
-3135.47 kJ/mol
Explanation:
Step 1: Write the balanced equation
C₆H₆(l) + 7.5 O₂(g) ⇒ 6 CO₂(g) + 3 H₂O(g)
Step 2: Calculate the standard enthalpy change of the reaction (ΔH°r)
We will use the following expression.
ΔH°r = ∑np × ΔH°f(p) - ∑nr × ΔH°f(r)
where,
n: moles
ΔH°f: standard enthalpies of formation
p: products
r: reactants
ΔH°r = 6 mol × ΔH°f(CO₂(g)) + 3 mol × ΔH°f(H₂O(g)) - 1 mol × ΔH°f(C₆H₆(l)) - 7.5 mol × ΔH°f(O₂(g))
ΔH°r = 6 mol × (-393.51 kJ/mol) + 3 mol × (-241.82 kJ/mol) - 1 mol × (48.95 kJ/mol) - 7.5 mol × 0 kJ/mol
ΔH°r = -3135.47 kJ
Since this enthalpy change is for 1 mole of C₆H₆(l), we can express it as -3135.47 kJ/mol.
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