Question

Find the ΔH and ΔE for the reaction below:
C₂H₆(g) + 7/2 O₂(g) à 2 CO₂(g) + 3 H₂O (l)
given the following data:
ΔH_f C₂H₆ = -84.7 kJ/mol                  ΔH_f CO₂ = -393.5 kJ/mol
ΔH_f H₂O = -286 kJ/mol

Answer 1

Solution:

1) The balanced equation for the combustion of C₂H₆ (ethane) is:

2C₂H₆ + 7O₂ ---> 4CO₂ + 6H₂O

2) The enthalpy of the reaction is:

[sum of enthalpies of formation of products] minus [sum of enthalpies of formation of reactants]

[(2 moles CO₂)(-393.5 kJ/mole) + (6 moles H₂O)(-286 kJ/mole)] - [(2 moles C₂H₆)(-84.7 kJ/mole) + (7 moles O₂)(0 kJ/mole)]

-2503 kJ - (-169.4 kJ) = -2333.6 kJ

ΔH = -2333.6 kJ ------------------------------------------------1

H = E + PV, PV = RT,

H = E + n RT where n = no of moles of gaseous products - no of moles of gaseous reactants

E = H - n RT = -2333.6 kJ - (-5)( 8.314 X 10^-3 X 298) kJ

E = -2321.2121 kJ ------------------------------------------- 2