Question

Using values from the Appendix of Thermodynamic data, calculate the value of H° for each of the following reactions.
(a) 3 Fe(s) + 4 CO2(g) 4 CO(g) + Fe3O4(s)
H° = 13.6

Correct: Your answer is correct.
kJ
(b) CH4(g) + 4 Cl2(g) CCl4(l) + 4 HCl(g)
H° = 308.6

Incorrect: Your answer is incorrect.
kJ
(c) Fe2O3(s) + 3 CO(g) 2 Fe(s) + 3 CO2(g)
H° = 24.8

Incorrect: Your answer is incorrect.
kJ
(d) 4 NH3(g) + O2(g) 2 N2H4(g) + 2 H2O(l)
H° = -286

Incorrect: Your answer is incorrect.
kJ

Answer 1

a) the given reaction is

3Fe + 4C02 ---> 4C0 + Fe304

we know that

dHorxn = dHfo products - dHfo reactants

so

dHorxn = ( 4 x dHfo C0 + dHfo Fe304) - ( 3 x dHfo Fe) - ( 4 x dHfo C02)

dHo rxn = ( 4 x -110.525) - 1118 - ( 3 x 0) - ( 4 x -393.509)

dHo rxn = 13.936

so

the dHo rxn value is 13.936 kJ

b)

the given reaction is

CH4 + 4Cl2 ---> CCl4 + 4 HCl

so

dHo rxn = ( dHfo CH4 ) + ( 4 x dHfo Cl2) - dHfo CCl4 - ( 4 x dHfo HCl)

dHro rxn = -74.87 + ( 4 x 0) - (-135.4) - ( 4 x -92.3)

dHo rxn = 429.73

so

dHo rxn value is 429.73 kJ

c) Fe203 + 3CO ---> 2Fe + 3C02

dHo rxn = ( 2 x dHfo Fe) + ( 3 x dHfo C02) - dHfo Fe203 - ( 3 x dHfo C0)

dHo rxn= ( 2 x 0) + ( 3 x -393.509) - (-824.2) - ( 3 x -110.525)

dHo rxn = -24.752

so

dHo rxn value is -24.752 kJ

d)

4 NH3 + 02 --> 2 N2H4 + 2 H20

dHo rxn = ( 2 x dHfo N2H4) + ( 2 x dHfo H20) - dHfo 02 - ( 4 x dHfo NH3)

dHo rxn = ( 2 x 95) + ( 2 x -285.8) - 0 - ( 4 x -45.9)

dHo rxn = -198

so

dHo rxn value is -198 kJ