Combustion of a hydrocarbon, X, is given by the equation: 2 X + 5 O2 (g) --> 4 CO2 (g) + 6 H2O (g) ∆H = -75.0 kJ Use this information and the standard enthalpies of formation listed in the text or elsewhere, to determine the standard enthalpy of formation of the hydrocarbon, X.

Hof(X) = - 1474.9 KJ/mol

Explanation:

The reaction equation given to us is:

2 X + 5 O2 (g) --> 4 CO2 (g) + 6 H2O (g) ∆H = -75.0 kJ

Now, the equation for the standard enthalpy of formation of the reaction is written as follows:

ΔH = 4*Hof(CO2(g)) + 6*Hof(H2O(g)) - 2*Hof(X) - 5*Hof(O2(g))

where,

ΔH = - 75 KJ

Hof(CO2(g)) = Standard Heat of Formation of CO2(g) = -393.5 KJ/mol

Hof(H2O(g)) = Standard Heat of Formation of H2O(g) = -241.8 KJ/mol

Hof(X) = Standard Heat of Formation of X = ?

Hof(O2(g)) = Standard Heat of Formation of O2(g) = 0 KJ/mol

Therefore,

-75 KJ = (4)(-393.5 KJ/mol)+ (6)(-241.8 KJ/mol) - (2)(Hof (X)) - (5)(0 KJ/mol)

- 75 KJ + 3024.8 KJ = - (2)(Hof(X))

Hof(X) = -2949.8 KJ/2 mol

Hof(X) = - 1474.9 KJ/mol