C2H4(g)+ 3O2(g) --> 2 CO2 (g) + 2H2O (l)
The complete combustion of 11.22g of ethylene, C2H4 (g), via the given equation, produces 564,4 KJ of heat. Calculate the standard enthalpy of formation for C2H4(g), using the standard enthalpies of formation given in the table.
In the table,CO2(g) -393.5 kJ/moL, H2O (l) -285.8 kJ/mol
dHrxn = (n*dHo products) - (n*dHo reactants)
dHrxn = (2*dHo CO2 + 2*dHo H2O) - (1*dHo C2H4 + 3*dHo O2)
dHrxn = 564.4 kJ
You have dHo for all but O2 and that is zero. I would recognize that the 564.4 kJ is for 11.22 grams C2H4 so for a mole of C2H4 that will be
564.4 kJ x (28/11.22) = ? kJ/mol. Make that substitution for dHrxn and solve for dHo C2H4. Post your work if you get stuck.
To calculate the standard enthalpy of formation (ΔHf) for C2H4(g), we need to use the equation:
ΔHrxn = ∑nΔHf(products) - ∑nΔHf(reactants)
Given that the reaction produces 564.4 kJ of heat, we can write:
564.4 kJ = (2 mol CO2)(-393.5 kJ/mol) + (2 mol H2O)(-285.8 kJ/mol) - ΔHf(C2H4)
Now, let's substitute the known values:
564.4 kJ = -787 kJ + (-571.6 kJ) - ΔHf(C2H4)
Rearrange the equation to solve for ΔHf(C2H4):
ΔHf(C2H4) = -787 kJ + (-571.6 kJ) - 564.4 kJ
ΔHf(C2H4) = -787 kJ - 571.6 kJ - 564.4 kJ
ΔHf(C2H4) = -1923 kJ
Therefore, the standard enthalpy of formation for C2H4(g) is -1923 kJ/mol.
To calculate the standard enthalpy of formation (ΔHf) for C2H4(g), we'll use the equation:
ΔHrxn = Σ(nΔHf products) - Σ(nΔHf reactants)
Where:
- ΔHrxn is the heat produced in the reaction (-564.4 kJ in this case)
- Σ(nΔHf products) is the sum of the products' standard enthalpies of formation multiplied by their respective coefficients
- Σ(nΔHf reactants) is the sum of the reactants' standard enthalpies of formation multiplied by their respective coefficients
Let's plug in the values:
ΔHrxn = (2 * ΔHf of CO2) + (2 * ΔHf of H2O) - (1 * ΔHf of C2H4)
We are given the standard enthalpies of formation for CO2 and H2O. However, we need to find the standard enthalpy of formation for C2H4. Rearranging the formula, we can solve for ΔHf of C2H4:
ΔHf of C2H4 = (2 * ΔHf of CO2) + (2 * ΔHf of H2O) - ΔHrxn
Plugging in the known values:
ΔHf of C2H4 = (2 * -393.5 kJ/mol) + (2 * -285.8 kJ/mol) - (-564.4 kJ)
ΔHf of C2H4 = -787 kJ/mol + -571.6 kJ/mol + 564.4 kJ
ΔHf of C2H4 = -794.2 kJ/mol + 564.4 kJ
ΔHf of C2H4 = -229.8 kJ/mol
Therefore, the standard enthalpy of formation for C2H4(g) is -229.8 kJ/mol.