Question

1.) Describe the relationship between∆Esys and ∆Esurr,.

2.) Calculate ΔH° for the following reaction using standard enthalpies of formation.

2C2H6(g) + 7O2(g)—> 6H2O(g) + 4CO2(g)

Answer 1

1)

Since Energy is conserved,

∆Etotal = 0

use:

∆Etotal = ∆Esys + ∆Esurr

∆Esys + ∆Esurr = 0

∆Esys = - ∆Esurr

They have same magnitude but opposite sign

2)

we have:

Hof(C2H6(g)) = -84.68 KJ/mol

Hof(O2(g)) = 0.0 KJ/mol

Hof(H2O(g)) = -241.818 KJ/mol

Hof(CO2(g)) = -393.509 KJ/mol

we have the Balanced chemical equation as:

2 C2H6(g) + 7 O2(g) ---> 6 H2O(g) + 4 CO2(g)

deltaHo rxn = 6*Hof(H2O(g)) + 4*Hof(CO2(g)) - 2*Hof( C2H6(g)) - 7*Hof(O2(g))

deltaHo rxn = 6*(-241.818) + 4*(-393.509) - 2*(-84.68) - 7*(0.0)

deltaHo rxn = -2856 KJ/mol

Answer: -2856 KJ/mol