Question

Butane (C4H10(9), AH = -125.6 kJ/mo) reacts with oxygen to produce carbon dioxide (CO2. AH+ = -393.5 kJ/mol) and

water (H20, AH+ = -241.82 kJ/mol) according to the equation below.
2C4H10(g) + 1302(g) → 8C02(g) + 10H2O(g)
What is the enthalpy of combustion (per mole) of C4H10(9)?
Use AH yen-(AHproducts) - ( He reactants)
-2,657.5 kJ/mol
0-5315.0 kJ/mol
--509.7 kJ/mol
O-254.8 kJ/mol

Answer 1

ΔH = -5315kJ/mol

Step-by-step explanation:

Based on Hess's law, ΔH of a reaction is defined as the sum of the ΔH of the products times their reaction coefficient - The sum of the ΔH of the reactants times their reaction coefficient.

For the reaction of the problem, ΔH is:

ΔH = 8ΔH CO2 + 10ΔH H2O - (2ΔH C4H10 - 13ΔH O2)

Where:

ΔH CO2 = -393.5kJ/mol

ΔH H2O = -241.82kJ/mol

ΔH C4H10 = -125.6kJ/mol

ΔH O2 = 0kJ/mol because O2(g) is the state of oxygen under STP.

Replacing:

ΔH = 8*(-393.5kJ/mol) + 10*(-241.82kJ/mol ) - (2*-125.6kJ/mol - 13*0kJ/mol)

ΔH = -5315kJ/mol