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!!! :(

-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

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answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.