Question

Calculate the enthalpy of combustion of 1 mol decane, C10H22, (l), to form CO2 and H2O. ∆Hf0 for decane is —300.9 kJ/mol using the balanced chemical equation and Appendix Table B–14.

Answer 1

The enthalpy of combustion of 1 mol decane, C10H22, (l), to form CO2 and H2O. ∆Hf0 for decane is —300.9 kJ/mol using the balanced chemical is -6777.9 Kj/mole

Explanation:

Balanced chemical equation for combustion of 1 mole of decane:

2$C_{10}$$H_{22}$  + 31 $O_{2}$  ⇒  20 C$O_{2}$  + 22 $H_{2}$O

∆H◦ = −300.9 KJ/mole

enthalpy of combustion of 1 mole decane =?

for 1 mole ∆H◦ = -$\frac{300.9}{2}$

       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 C$O_{2}$ = -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

$\frac{-13555.8}{2}$

= -6777.9 kJ