Question

I am so lost. Does anyone know how to do this??? Enthalpies of reaction stuff

C2H4(g)+3O2(g)→2CO2(g)+2H2O(l) ΔH∘1=?

The combustion of C2H4(g) is represented by the equation above.

(a) Use the enthalpies of formation in the table below to calculate the value of ΔH∘1 for the reaction.

ΔH∘f(kJ/mol)
C2H4(g) 52
CO2(g) −394
H2O(l) −286
O2(g) 0
Please help!!! :(

Answer 1

Answer:

-1464 kJ/mol

Explanation:

The balanced equation for the reaction is given as;

C2H4(g)+3O2(g)→2CO2(g)+2H2O(l) ΔH∘1=?

The enthalpy of the reaction is given by the equation;

Enthalpy of reaction = Enthalpy of products - Enthalpy of reactants

Products:

2CO2(g)+2H2O(l)

Enthalpy of Products = 2 (−394) + 2(−286)

Enthalpy of Products = -1360 kJ/mol

Reactants:

C2H4(g)+3O2(g)

Enthalpy of Reactants = 2 (52) + 3(0)

Enthalpy of Reactants = 104 kJ/mol

Enthalpy of Reaction =  -1360 - 104 = -1464 kJ/mol