Question

From the standard enthalpies of formation, calculate ΔH°rxn for the reaction

C₆H₁₂(l) + 9O₂(g) → 6CO₂(g) + 6H₂O(l)

For C₆H₁₂(l), ΔH°f = –151.9 kJ/mol
Substance ∆H°f , kJ/mol
C₆H₁₂(I) -151.9
O₂(g) 0
H₂O(I) -285.8

Answer 1

The standard enthalpy change ΔH°rxn for the given reaction is -3923.9 kJ per mole of C₆H₁₂ reacted, calculated using the provided standard enthalpies of formation and the formula ΔH°rxn = ∑ΔH°f(products) - ∑ΔH°f(reactants).

Step-by-step explanation:

To calculate the standard enthalpy change for the reaction C₆H₁₂(l) + 9O₂(g) → 6CO₂(g) + 6H₂O(l), we use the given standard enthalpies of formation (ΔH°f) and apply Hess's Law. The formula used is ΔH°rxn = ∑ΔH°f(products) - ∑ΔH°f(reactants).

The values given for the standard enthalpies of formation are ΔH°f for C₆H₁₂(l) which is -151.9 kJ/mol, for O₂(g) it is 0 kJ/mol (as it is the standard state of the element), for CO₂(g), it is -393.5 kJ/mol, and for H₂O(l) it is -285.8 kJ/mol. Inserting these values yields:

ΔH°rxn = {6(-393.5 kJ/mol) + 6(-285.8 kJ/mol)} - {(-151.9 kJ/mol) + 9(0 kJ/mol)}

= {6(-393.5) + 6(-285.8)} - {(-151.9)}

= {-2361 kJ - 1714.8 kJ + 151.9 kJ}

= -3923.9 kJ per mole of C₆H₁₂ reacted.