Question

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

Answer 1

BE reactants - BE products
delta Hf products - delta Hf reactants.

Answer 2

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