Chemistry 11 HL
Unit 5 / IB Topic 5.3
Energetics 3
Practice Problems - Hess’s Law - SOLUTIONS
1.
see your notes
2.
a)
C(s) + O2(g)  CO2(g)
b)
Eq 1 x 1
C(s) + 1/2 O2(g) → CO(g)
∆H1º = - 110.5 kJ
Eq 2 x 1
CO(g) + 1/2 O2(g) → CO2(g)
∆H2º = - 280.3 kJ
––––––––––––––––––––––––––––
C(s) + O2(g)  CO2(g)
––––––––––––––––
∆H = -390.8 kJ
3.
a)
b)
Eq 1 x -1
2 Fe(s) + 3 CO2(g)  Fe2O3(s) + 3 CO(g)
∆H1 = +6.39 kcal
Eq 2 x 3
3CO(g) + 3/2 O2(g)  3CO2(g)
∆H2 = -202.89 kcal
––––––––––––––––––––––––––––––––––––
2 Fe(s) + 3/2 O2(g)  Fe2O3(s)
––––––––––––––––
∆H = -196.50 kcal
∆H = -196.50 kcal x 4.18 kJ/kcal = -821.37 kJ
c)
mol Fe = 2500 g x
1 mol
55.58 g
= 44.8 mol
821.37 kJ
x 44.8 mol Fe
2 mol Fe
= 1.84 x 10 4 kJ
heat released =
4.
5.
a)
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
b)
Eq 1 x 1
C(s) + O2(g) → CO2(g)
∆H1 = -393.5 kJ
Eq 2 x 2
2H2(g) + O2(g) → 2H2O(l)
∆H2 = -571.2 kJ
Eq 3 x -1
CH4(g)  C(s) + 2H2(g)
∆H3 = +74.8 kJ
–––––––––––––––––––––––––––––––––
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
––––––––––––––
∆Hc = -889.9 kJ
Eq 1 x -1/2
HCl(g) + ½ Na2O(s)  NaCl(s) + ½ H2O(l)
∆H1 = -253.65 kJ
Eq 2 x -1/2
NaNO2(s)  ½ NO(g) + ½ NO2(g) + ½ Na2O(s)
∆H2 = +213.57 kJ
Eq 3 x ½
½ NO(g) + ½ NO2(g)  ½ N2O(g) + ½ O2(g)
∆H3 = -21.34 kJ
Eq 4 x -1/2
½ N2O(g) + ½ O2(g) + ½ H2O(l)  HNO2(l)
∆H4 = -17.18 kJ
–––––––––––––––––––––––––––––––––––––––––
HCl(g) + NaNO2(s)  NaCl(s) + HNO2(g)
––––––––––––––––
∆H = -78.6 kJ
Chemistry 11 HL
Unit 5 / IB Topic 5.3
6.
∆H2 + 2∆H3 = ∆H1 + ∆H4
-393 + 2(-286) = ∆H1 + (-890)
∆H1 = -75 kJ
-1
Note: The units for this answer should be kJ mol because the question is asking for
an enthalpy of formation value. You will learn more about this value in Energetics 5.