Question

For a particular isomer of C8H18, the following reaction produces 5108.7 kJ of heat per mole of C8H18(g) consumed, under standard conditions.

C8H18(g)+12.5O2(g)-->8CO2(g)+9H2O(g)

deltaHrxn=-5108.7

What is the standard enthalpy of formation of this isomer of C8H18(g)?

delta Hf= ? kJ/mol

Answer 1

C8H18(g)+12.5O2(g)-->8CO2(g)+9H2O(g)   H = -5108.7Kj/mole
rxn = sum of enthalpies of products - sum of enthalpies of reactaNTS
o2(g) = 0
Co2(g) = -393.5 kj
H2o(g) = -241.8kj
C8H18   = X

-5108.7 = 8*-393.5 + 9*-241.8-X
-5108.7 = -3148 -2176 -X
-5108.7 = -5324.2-X
X        = -5324.2+5108.7
         = -215.5 Kj/mole
threfore standared enthalpy of formation of C8H18 is -215.5 kj/mole