Transcript 1. From the following enthalpy changes, S (s) + 3/2 O2 (g) * SO3 (g

Hess’ Law
Extra Practice
Problems
same
flip
1.
From the following enthalpy changes,
S (s) + 3/2 O2 (g)  SO3 (g)
H = 395.2 kJ
2 SO2 (g) + O2 (g)  2 SO3 (g)
H = 198.2 kJ
calculate the value of H for the reaction
S (s) + O2 (g)  SO2 (g)
2
33
2 S (s) + /2 O2 (g) 2 SO3 (g)
H=2( 395.2 kJ)
2 SO3 (g)  2 SO2 (g) + O2 (g) H = +198.2 kJ
2 S (s) + 2 O2 (g)  2 SO2 (g)
H = 592.2 kJ
2
= 296.1 kJ
2.
From the following enthalpy changes,
flip H (g) + ½ O (g)  H O (ℓ)
H = 285.8 kJ
2
2
2
flip N O (g) + H O (ℓ)  2 HNO (aq)
H = 76.6 kJ
2 5
2
3
same ½ N2 (g) + 3/2 O2 (g) + ½ H2 (g)  HNO3 (aq) H = 174.1 kJ
calculate the value of H for the reaction
2 N2 (g) + 5 O2 (g)  2 N2O5 (g)
1
3 2.5
1
3
½ N2 (g)+ /2 O2 (g)+ ½ H2 (g) 2 HNO3 (aq)
H =2(174.1 kJ )
2 HNO3 (aq)  N2O5 (g) + H2O (ℓ) H = +76.6 kJ
H2O (ℓ)
2 N2 (g)
+
5
2.5 O2 (g)

H2 (g) + ½ O2 (g) H = +285.8 kJ

2 N2O5 (g)
H = 2( 14.2 kJ )
= 28.4 kJ
3.
From the following enthalpy changes,
same C (s) + O2 (g)  CO2 (g)
H = 393.5 kJ
same H2 (g) + ½ O2 (g)  H2O (l)
H = 285.8 kJ
flip C5H12 (g) + 8 O2 (g)  5 CO2 (g) + 6 H2O (l)
H = 3536 kJ
calculate the value of H for the reaction
5 C (s) + 6 H2 (g)  C5H12 (g).
5 C (s) + 5 O2 (g)  5 CO2 (g)
3
6 H2 (g) + ½ O2 (g)  6 H2O (l)
5 CO2 (g) + 6 H2O (l)  C5H12 (g) + 8 O2 (g)
5 C (s) +
6 H2 (g)  C5H12 (g)
H = 5(393.5 kJ )
H = 6(285.8 kJ )
H = +3536 kJ
H = 146.3 kJ
4.
From the following enthalpy changes,
same
CO (g) + SiO2 (s)  SiO (g) + CO2 (g)
H = +520.9 kJ
flip3 SiO2 (s) + 2 N2O (g) + 8 CO (g)  8 CO2 (g) + Si3N4 (s) H = 461.1 kJ
calculate the value of H for the reaction
5 CO2 (g) + Si3N4 (s)  3 SiO (g) + 2 N2O (g) + 5 CO (g)
3CO (g) + 3SiO2 (s)  3 SiO (g) + 3CO2 (g)
H= 3(+520.9 kJ )
5
5
8 CO2 (g) + Si3N4 (s)  3 SiO2 (s) + 2 N2O (g) + 8 CO (g) H = +461.1 kJ
5 CO2 (g) + Si3N4 (s)  3 SiO (g) + 2 N2O (g) + 5 CO (g)
H =
+2023.8 kJ
5. From the following enthalpy changes,
same H (g) + ½ O (g)  H O (l)
H = 285.8 kJ
2
2
2
same SO3 (g) + H2O (l)  H2SO4 (l)
H = 132.5 kJ
same H2SO4 (l) + Ca (s)  CaSO4 (s) + H2 (g)
H = 602.5 kJ
flip Ca (s) + ½ O2 (g)  CaO (s)
H = 634.9 kJ
calculate the value of H for the reaction
CaO (s) + SO3 (g)  CaSO4 (s).
SO3 (g) + H2O (l)  H2SO4 (l)
H2SO4 (l) + Ca (s)  CaSO4 (s) + H2 (g)
CaO (s)  Ca (s) + ½ O2 (g)
H2 (g) + ½ O2 (g)  H2O (l)
SO3 (g) +
CaO (s) 
CaSO4 (s)
H = 132.5 kJ
H = 602.5 kJ
H = +634.9 kJ
H = 285.8 kJ
H = 385.9 kJ