Answer to Question #246513 in General Chemistry for Aiko

Question

Determine the enthalpy of formation of decane (C10H22(l)) if the complete combustion of 1 mol of decane releases 6777.9 kJ when gaseous carbon dioxide and liquid water are the products.

Answer 1

Balanced chemical equation for combustion of 1 mole of decane

2C₁₀H₂₂+31O₃-->>20CO₂+22H₂O

H◦ = −300.9 KJ/mole

enthalpy of combustion of 1 mole decane =?

for 1 mole ∆H◦= -300.9/2

for 1 mole ∆H◦ = - 150.45 KJ

Formula used:

∆H◦reaction of combustion =m. ∆H◦ (products) - n. ∆H◦ (reactants)

m = number of moles of product

n = number of moles of reactant.

from the table it is seen that,

∆H° for CO₂= -393.5 kj/mole

∆H◦ for water = -285.8kj/mole

∆H◦ for oxygen = 0

∆H◦ for decane = -300.9 kj/mole

putting the values in the equation:

∆H◦ = 20(-393.5) + 22(-285.5) = 2(-300.9) +22(0)

= -14151 = -601.8

= -13555.8 kj

The value of enthalpy of reaction or combustion calculated is for 2 moles, hence for 1 mole

= -6777.9 kJ