Answer to Question #114567 in Chemistry for Nikki

Question

Using CO2 = -393.5 kj/mol and H2O = -285.8 kj/mol determine delta H for C4H10

Answer 1

The combustion of butane can be shown as following:

2C₄H₁₀ + 13O₂ --> 8CO₂ + 10H₂O

According to the equation:

ΔH_reaction = ΔH_products - ΔH_reactants

From here:

ΔH_reactants = ΔH_products - ΔH_reaction

As ΔH_reaction = -2877.5 kJ/mol (butane combustion):

2ΔH_butane = ΔH_products - ΔH_reaction= (8ΔH_{carbon dioxide} + 10ΔH_water) - ΔH_reaction = 8 × (-393.5 kJ/mol) + 10 × (-285.8 kJ/mol) - (-2877.5 kJ/mol) = -251 kJ/mol

From here, the enthalpy of formation of butane equals:

ΔH_butane = -125.5 kJ/mol

Answer: -125.5 kJ/mol