for a particular isomer C8H18, the following reaction produces 5113.3 kj of heat per mole of C8H18 consumed under standard conditions
c8h18(g) + 25/2 O2 (g) ---> 8co2(g) + 9h2o(g)
what is the standard enthalpy of formation of this isomer of c8h18?
o2= 0
co2(g)= -393.5
h2o(g)= -241.8
delta h= -5113.3
Are all those kJ/mol?
is that -5112.3 kJ/mol for delta H rxn (combustion)?
-5113.3 = [(8*-393.5) +(9*-241.8)]-[dHf C8H18]
To find the standard enthalpy of formation of the isomer of C8H18, you need to use the equation:
ΔH°f = Σ(nΔH°f(products)) - Σ(nΔH°f(reactants))
First, let's calculate the moles of reactants and products based on the balanced equation:
Reactant:
C8H18(g): 1 mole
Products:
CO2(g): 8 moles
H2O(g): 9 moles
Next, we substitute the given enthalpy values into the equation:
ΔH°f = (8 × ΔH°f(CO2)) + (9 × ΔH°f(H2O)) - (1 × 0) - (25/2 × ΔH°f(O2))
ΔH°f = (8 × -393.5) + (9 × -241.8) - (1 × 0) - (25/2 × 0)
ΔH°f = -3148 + -2176.2
ΔH°f ≈ -5324.2 kJ/mol
Therefore, the standard enthalpy of formation of this isomer of C8H18 is approximately -5324.2 kJ/mol.