For the reaction shown below complete the following calculations.

H2(g) + C2H4(g) --> C2H6(g)

(a) Estimate the enthalpy of reaction using the bond energy values in Table 9.4.

(b) Calculate the enthalpy of reaction, using standard enthalpies of formation. (ΔH°f for H2, C2H4, and C2H6 are 0, 52.3 kJ/mol, and -84.7 kJ/mol, respectively.)

I'm not sure how to calculate this. I added the bond energies together from the left side, but wasn't sure if I even did that right

BE reactants - BE products

delta Hf products - delta Hf reactants.

(a) Bonds on the left side break by absorbing (+) energy and bonds on the right side form releasing (-) energy. Look up the bond energies and set up:

∆Hrxn = E(H-H) + E(C=C) + 4E(C-H) - E(C-C) - 6(C-H).
Look up and combine six positive energies on the left with seven negative energies on the right.

(b) set up
∆Hrxn = ∆Hf(C2H6) - {∆Hf(H2) + ∆Hf(C2H4)}
Look up the energies of formation and substitute in the above equation. When you combine them be careful with algebraic signs.

To estimate the enthalpy of reaction using bond energy values, you need to calculate the energy required to break the bonds in the reactants and the energy released in forming the bonds in the products.

To calculate the bond energy of a molecule, you can use the bond energy values listed in Table 9.4.

The bond energy for H-H is 432 kJ/mol and for the C=C bond is 612 kJ/mol.

(a) To estimate the enthalpy of reaction using bond energy values, you need to determine the total energy required to break the bonds in the reactants and the total energy released in forming the bonds in the products.

For H2: 1 mol H-H bond = 432 kJ/mol
For C2H4: 1 mol C=C bond = 612 kJ/mol

Therefore, in the reactants:
Total energy required to break H2 and C2H4 bonds = (1 mol H-H bond × 432 kJ/mol) + (1 mol C=C bond × 612 kJ/mol)

In the products:
For C2H6: 1 mol C-C bond = +347 kJ/mol, 1 mol C-H bond = +413 kJ/mol
Total energy released in forming C2H6 bonds = (1 mol C-C bond × +347 kJ/mol) + (6 mol C-H bond × +413 kJ/mol)

Now, calculate the change in enthalpy (∆H) of the reaction by subtracting the total energy required to break the reactant bonds from the total energy released in forming the product bonds:

∆H = (Total energy released in forming C2H6 bonds) - (Total energy required to break H2 and C2H4 bonds)

(b) To calculate the enthalpy of reaction using standard enthalpies of formation, you need to use the given standard enthalpies of formation of the reactants and products.

∆H° = Σn × ∆H°f(products) - Σm × ∆H°f(reactants)

Where:
n = stoichiometric coefficient of product
m = stoichiometric coefficient of reactant
∆H°f = standard enthalpy of formation

For the reaction:
H2(g) + C2H4(g) --> C2H6(g)

∆H° = (1 × ∆H°f(C2H6) ) - (1 × ∆H°f(H2) + 1 × ∆H°f(C2H4) )

∆H° = (1 × -84.7 kJ/mol) - (1 × 0 kJ/mol + 1 × 52.3 kJ/mol)

To estimate the enthalpy of reaction using bond energy values, you need to know the bond energies for the bonds broken and formed in the reaction. Bond energy is the energy required to break one mole of a particular bond in a gaseous molecule.

(a) To estimate the enthalpy of reaction using bond energies, you need to calculate the bond energies for the bonds broken and formed. In this case, you must consider the following bonds:

Bonds broken:
- 1 mole of H–H bonds (H2) with a bond energy of 436 kJ/mol.
- 1 mole of C=C bonds (C2H4) with a bond energy of 614 kJ/mol.

Bonds formed:
- 1 mole of C–H bonds (C2H6) with a bond energy of 413 kJ/mol.

To calculate the estimated enthalpy of reaction:
ΔH = (bonds broken) - (bonds formed)
ΔH = (1 mole of H–H bonds) + (1 mole of C=C bonds) - (1 mole of C–H bonds)

ΔH = (436 kJ/mol) + (614 kJ/mol) - (413 kJ/mol)
ΔH ≈ 637 kJ/mol

Therefore, the estimated enthalpy of reaction using bond energy values is approximately 637 kJ/mol.

(b) To calculate the enthalpy of reaction using standard enthalpies of formation, you need to consider the standard enthalpies of formation for the compounds involved.

In this case:
ΔH°f for H2 = 0 kJ/mol (given)
ΔH°f for C2H4 = 52.3 kJ/mol (given)
ΔH°f for C2H6 = -84.7 kJ/mol (given)

To calculate the enthalpy of reaction:
ΔH = (ΔH°f products) - (ΔH°f reactants)
ΔH = (ΔH°f for C2H6) - (ΔH°f for H2 + ΔH°f for C2H4)

ΔH = (-84.7 kJ/mol) - (0 kJ/mol + 52.3 kJ/mol)
ΔH ≈ -137.0 kJ/mol

Therefore, the enthalpy of reaction using standard enthalpies of formation is approximately -137.0 kJ/mol.

Note: The negative sign indicates an exothermic reaction (releases energy).