Standard+Enthalpies+of+Formation+(5.5)

**Page 332 #1**
Write formation equations for the compounds:

a) Benzene (C6H6), used as a solvent 3H2(g) + 6C(g) → C6H6(l)

b) Potassium bromate, used in commercial bread dough 3/2 O2 + ½ Br2 + K → KBrO3

c) Glucose (C6H12O6), found in soft drinks 6C + 6H2 + 3O2 → C6H12O6

d) Magnesium hydroxide, found in antacids H2 + O2 + Mg → Mg(OH)2

#2
a) C5H12 + 8O2 → 5CO2(g) + 6H2O(l)

ΔHr = ΣHfproducts - ΣHfreactants = [5(-393.5KJ/mol) + 6(-285.8 KJ/mol)] – [ 0 + 1(-173.5 KJ/mol)] = -3682KJ/mol + 173.5KJ/mol = -3508 KJ/mol

b) Fe2O3(s) + 3CO(g) → 2Fe +3CO2

ΔHr = ΣHfproducts - ΣHfreactants = [2(0KJ/mol) + 3(-393.5KJ/mol)] - [1(-824.2KJ/mol) + 3(-110.5KJ/mol)] = -1180.5KJ/mol + 1155.7 KJ/mol = -24.8 KJ/mol


 * 3

C6H12(l) + 9O2(g) → 6CO2 + 6H2O

ΔHr = ΣHfproducts - ΣHfreactants -3824KJ/mol = [6(-393.5KJ/mol) + 6(-285.8KJ/mol)] - [9(0) + Hfcyclohexane] -3824KJ/mol = -4075.8KJ/mol - Hfcyclohexane Hfcyclohexane = -251.8 KJ/mol


 * 4

a) CH4(g) + H2O(g) → CO2(g) + 3H2(g)

ΔHr = ΣHfproducts - ΣHfreactants = [1(-110.5KJ/mol) + 3(0KJ/mol)] - [1(-74.4KJ/mol) + 1(-241.8KJ/mol)] = -110.5KJ/mol + 316.2KJ/mol = 210.7 kJ/mol

b) CO2(g) + H2O(g) → CO2(g) + H2(g)

ΔHr = ΣHfproducts - ΣHfreactants = [1(-393.5KJ/mol) + 1(0KJ/mol)] - [ 1(-110.5KJ/mol) + 1(-241.8KJ/mol)] = -393.5KJ/mol + 352.3KJ/mol = -41.2 KJ/mol

c) N2(g) +3H2(g) → 2NH3(g)

ΔHr = ΣHfproducts - ΣHfreactants = [2(-45.9KJ/mol)] - [1(0KJ/mol) + 3(0KJ/mol)] = -91.8KJ/mol


 * 5


 * a) 4NH3(g)+ 5O2(g) → 4NO(g) + 6H2O(g)

ΔHr = ΣHfproducts - ΣHfreactants = [4(+90.2KJ/mol) + 6(-241.8KJ/mol)] - [4(-45.9KJ/mol) + 6(0KJ/mol] = -1090KJ/mol + 183.6KJ/mol = 906.4KJ/mol

b) 2NO(g) + O2(g) → 2NO2(g)

ΔHr = ΣHfproducts - ΣHfreactants = [2(33.2KJ/mol] - [2(90.2KJ/mol) + 1(0KJ/mol)] = 66.4KJ/mol - 180.4KJ/mol = -114KJ/mol

c) 3NO2(g) + H2O(l) → 2HNO3(l) + NO(g)

ΔHr = ΣHfproducts - ΣHfreactants = [2(-174.1KJ/mol) + 1(90.2KJ/mol)] - [ 3(33.2KJ/mol) + 1(-285.8KJ/mol)] = -258KJ/mol + 186.2KJ/mol = -71.8KJ/mol**