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Hess’ Law Extra Practice Problems

Hess’ Law Extra Practice Problems. same. flip.

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Hess’ Law Extra Practice Problems

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  1. Hess’ Law Extra Practice Problems

  2. same flip 1. From the following enthalpy changes,S (s) + 3/2 O2 (g)  SO3 (g) H = 395.2 kJ2 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 3 S (s) + 3/2 O2 (g)  SO3 (g) H= 395.2 kJ 2( ) 2 2  2 SO2 (g) + O2 (g) H = +198.2 kJ 2 SO3 (g) 2 S (s) + 2 O2 (g) 2 SO2 (g)  H = 592.2 kJ 2 = 296.1 kJ

  3. flip flip same 2. From the following enthalpy changes,H2 (g) + ½ O2 (g)  H2O (ℓ) H = 285.8 kJN2O5 (g) + H2O (ℓ)  2 HNO3 (aq) H = 76.6 kJ½ N2 (g) + 3/2 O2 (g) + ½ H2 (g)  HNO3 (aq) H = 174.1 kJcalculate the value of H for the reaction 2 N2 (g) + 5 O2 (g)  2 N2O5 (g) 1 2.5 3 1 ½ N2 (g)+ 3/2 O2 (g)+ ½ H2 (g)  HNO3 (aq) H = 174.1 kJ 2 2( ) N2O5 (g) + H2O (ℓ) H = +76.6 kJ 2 HNO3 (aq)  H2O (ℓ) H2 (g) + ½ O2 (g) H = +285.8 kJ  5 N2 (g) N2O5 (g) H = 2( ) 14.2 kJ 2 + 2.5 O2 (g)  2 = 28.4 kJ

  4. same same flip 3. From the following enthalpy changes,C (s) + O2 (g)  CO2 (g) H = 393.5 kJ H2(g) + ½ O2 (g)  H2O (l) H = 285.8 kJ C5H12(g) + 8 O2 (g)  5 CO2 (g) + 6 H2O (l) H = 3536 kJcalculate the value of H for the reaction 5 C (s) + 6 H2 (g)  C5H12 (g). C (s) + O2 (g)  CO2 (g) H = 393.5 kJ 5( ) 5 5 5 3 H2 (g) + ½ O2 (g)  H2O (l) H = 285.8 kJ 6 6 6( ) 5 CO2 (g) + 6 H2O (l) +3536 kJ H = C5H12 (g) + 8 O2 (g)  146.3 kJ H = C5H12 (g)  5 C (s) + 6 H2 (g)

  5. same flip 4. From the following enthalpy changes,CO (g) + SiO2 (s) SiO (g) + CO2 (g) H = +520.9 kJ3 SiO2 (s) + 2 N2O (g) + 8 CO (g)  8 CO2 (g) + Si3N4(s) H = 461.1 kJ calculate the value of H for the reaction5 CO2 (g) + Si3N4 (s)  3 SiO (g) + 2 N2O (g) + 5 CO (g) CO (g) + SiO2 (s) SiO (g) + CO2 (g)H= +520.9kJ 3 3( ) 3 3 3 5 5  3 SiO2 (s) + 2 N2O (g) + 8 CO (g) H = +461.1 kJ 8 CO2 (g) + Si3N4 (s) 5 CO2 (g) + 2 N2O (g) + Si3N4 (s) 3SiO(g) 5 CO (g)  + H = +2023.8 kJ

  6. same same same flip 5. From the following enthalpy changes, H2(g) + ½ O2 (g)  H2O (l) H = 285.8 kJ SO3(g) + H2O (l)  H2SO4 (l) H = 132.5 kJ H2SO4(l) + Ca (s)  CaSO4 (s) + H2 (g) H = 602.5 kJCa(s) + ½ O2 (g) CaO (s) H = 634.9 kJ calculate the value of H for the reactionCaO (s) + SO3 (g)  CaSO4 (s). SO3 (g) + H2O (l)  H2SO4 (l) H = 132.5 kJ H2SO4 (l) + Ca (s)  CaSO4 (s) + H2 (g) H = 602.5 kJ  H = +634.9 kJ CaO (s) Ca (s) + ½ O2 (g) H2 (g) + ½ O2 (g)  H2O (l) H = 285.8 kJ H = 385.9 kJ SO3 (g) + CaO (s)  CaSO4 (s)

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