Determine the Enthalpy of combustion of ethyne (C2H2) using the following enthalpies of formation;
Data:
DHf°(H2O)= -285.9kj/mol
DHf°(2H2)= +226.7Kj/mol
DHf°(CO2)= -393.5Kj/mol
Equation; C2H2--------->CO2 + H2O
Balance the equation.
2C2H2 + 5O2 ==>4CO2 + 2H2O
dHfo rxn = (n*dHfo products)-(n*dHfo reactanta). That will give you dHfo for the rxn which is for 2 mols C2H2. Divide by 2 for kJ/mol.
To determine the enthalpy of combustion of ethyne (C2H2), you can use the concept of Hess's Law. Hess's Law states that the enthalpy change of a reaction is the same regardless of the pathway taken.
The equation for the combustion of ethyne (C2H2) is:
C2H2 + O2 -> CO2 + H2O
To determine the enthalpy of this combustion reaction, you need to express it in terms of the given enthalpies of formation. The enthalpy change of a reaction is calculated by subtracting the sum of the enthalpies of formation of the reactants from the sum of the enthalpies of formation of the products.
First, let's balance the equation:
C2H2 + 5/2 O2 -> 2 CO2 + H2O
Now, let's determine the enthalpy change:
ΔH = Σ(ΔHf° of products) - Σ(ΔHf° of reactants)
Using the given enthalpies of formation:
ΔH = [2 × ΔHf°(CO2) + 1 × ΔHf°(H2O)] - [1 × ΔHf°(C2H2) + 5/2 × ΔHf°(O2)]
Substituting the values:
ΔH = [2 × (-393.5 kj/mol) + 1 × (-285.9 kj/mol)] - [1 × ΔHf°(C2H2) + 5/2 × 0 kj/mol]
Since ΔHf°(O2) is zero, we exclude it from the equation.
Now, substitute the given value of ΔHf°(C2H2):
ΔH = [2 × (-393.5 kj/mol) + 1 × (-285.9 kj/mol)] - [1 × (ΔHf°(C2H2))]
You need to know the value of ΔHf°(C2H2) to calculate the enthalpy of combustion of ethyne (C2H2).