# Hess’ Law Extra Practice Problems. 1.From the following enthalpy changes, S (s) + 3 / 2 O 2 (g)  SO 3 (g)  H  =  395.2 kJ 2 SO 2 (g) + O 2 (g)  2.

## Presentation on theme: "Hess’ Law Extra Practice Problems. 1.From the following enthalpy changes, S (s) + 3 / 2 O 2 (g)  SO 3 (g)  H  =  395.2 kJ 2 SO 2 (g) + O 2 (g)  2."— Presentation transcript:

Hess’ Law Extra Practice Problems

1.From the following enthalpy changes, S (s) + 3 / 2 O 2 (g)  SO 3 (g)  H  =  395.2 kJ 2 SO 2 (g) + O 2 (g)  2 SO 3 (g)  H  =  198.2 kJ calculate the value of  H  for the reaction S (s) + O 2 (g)  SO 2 (g) flip same S (s) + 3 / 2 O 2 (g)  SO 3 (g)  H  =  395.2 kJ 2 SO 3 (g) 2 S (s) 2 3  2 SO 2 (g) + O 2 (g)  H  = +198.2 kJ 2 2( ) 2 +2 O 2 (g)  2 SO 2 (g)  H  =  592.2 kJ 2 =  296.1 kJ

2.From the following enthalpy changes, H 2 (g) + ½ O 2 (g)  H 2 O (ℓ)  H  =  285.8 kJ N 2 O 5 (g) + H 2 O (ℓ)  2 HNO 3 (aq)  H  =  76.6 kJ ½ N 2 (g) + 3 / 2 O 2 (g) + ½ H 2 (g)  HNO 3 (aq)  H  =  174.1 kJ calculate the value of  H  for the reaction 2 N 2 (g) + 5 O 2 (g)  2 N 2 O 5 (g) flip same ½ N 2 (g)+ 3 / 2 O 2 (g)+ ½ H 2 (g)  HNO 3 (aq)  H  =  174.1 kJ 2 HNO 3 (aq)  N 2 O 5 (g) + H 2 O (ℓ)  H  = +285.8 kJ  H  = +76.6 kJ H 2 (g) + ½ O 2 (g)H 2 O (ℓ) flip  12.5 1 22( ) 3 N 2 (g) + 2.5 O 2 (g)  N 2 O 5 (g)  H  = 14.2 kJ 2 5 2 2( ) = 28.4 kJ

3.From the following enthalpy changes, C (s) + O 2 (g)  CO 2 (g)  H  =  393.5 kJ H 2 (g) + ½ O 2 (g)  H 2 O (l)  H  =  285.8 kJ C 5 H 12 (g) + 8 O 2 (g)  5 CO 2 (g) + 6 H 2 O (l)  H  =  3536 kJ calculate the value of  H  for the reaction 5 C (s) + 6 H 2 (g)  C 5 H 12 (g). same flip C (s) + O 2 (g)  CO 2 (g)  H  =  393.5 kJ  same H 2 (g) + ½ O 2 (g)  H 2 O (l)  H  =  285.8 kJ C 5 H 12 (g) + 8 O 2 (g)  H  = 5 CO 2 (g) + 6 H 2 O (l)+3536 kJ 5 5 5 5( ) 6 3 66( ) 5 C (s)+6 H 2 (g)  C 5 H 12 (g)  H  =  146.3 kJ

4.From the following enthalpy changes, CO (g) + SiO 2 (s)  SiO (g) + CO 2 (g)  H  = +520.9 kJ 3 SiO 2 (s) + 2 N 2 O (g) + 8 CO (g)  8 CO 2 (g) + Si 3 N 4 (s)  H  =  461.1 kJ calculate the value of  H  for the reaction 5 CO 2 (g) + Si 3 N 4 (s)  3 SiO (g) + 2 N 2 O (g) + 5 CO (g) flip same CO (g) + SiO 2 (s)  SiO (g) + CO 2 (g)  H  = +520.9 kJ 8 CO 2 (g) + Si 3 N 4 (s) 5 CO 2 (g) 33  3 SiO 2 (s) + 2 N 2 O (g) + 8 CO (g)  H  = +461.1 kJ 3 3( ) +Si 3 N 4 (s)  3 SiO (g)  H  = +2023.8 kJ 3 5 5 2 N 2 O (g) 5 CO (g) + +

5. From the following enthalpy changes, H 2 (g) + ½ O 2 (g)  H 2 O (l)  H  =  285.8 kJ SO 3 (g) + H 2 O (l)  H 2 SO 4 (l)  H  =  132.5 kJ H 2 SO 4 (l) + Ca (s)  CaSO 4 (s) + H 2 (g)  H  =  602.5 kJ Ca (s) + ½ O 2 (g)  CaO (s)  H  =  634.9 kJ calculate the value of  H  for the reaction CaO (s) + SO 3 (g)  CaSO 4 (s). same flip same SO 3 (g) + H 2 O (l)  H 2 SO 4 (l)  H  =  132.5 kJ H 2 (g) + ½ O 2 (g)  H 2 O (l)  H  =  285.8 kJ H 2 SO 4 (l) + Ca (s)  CaSO 4 (s) + H 2 (g)  H  =  602.5 kJ CaO (s)  H  =  Ca (s) + ½ O 2 (g) +634.9 kJ same SO 3 (g)+CaO (s)  CaSO 4 (s)  H  =  385.9 kJ

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