☾ Thermochemical Equations ☽ Enthalpy Changes Measuring and Expressing ∆H ☾ Thermochemical Equations ☽

Thermochemical Equations In a thermochemical reaction, ∆H for the reaction can be written either as a reactant or product.

Thermochemical Equations In a thermochemical reaction, ∆H for the reaction can be written either as a reactant or product. CaO(s) + H2O(l) → Ca(OH)2(s) + 65.2 kJ

Thermochemical Equations In a thermochemical reaction, ∆H for the reaction can be written either as a reactant or product. CaO(s) + H2O(l) → Ca(OH)2(s) + 65.2 kJ When 1 mol of CaO reacts with 1 mol of H2O, 65.2 kJ of heat are produced at standard temperature, 298 K, and standard pressure, 1 atm.

Thermochemical Equations In a thermochemical reaction, ∆H for the reaction can be written either as a reactant or product. CaO(s) + H2O(l) → Ca(OH)2(s) + 65.2 kJ When 1 mol of CaO reacts with 1 mol of H2O, 65.2 kJ of heat are produced at standard temperature, 298 K, and standard pressure, 1 atm. More often we write the equation like this ...

Thermochemical Equations In a thermochemical reaction, ∆H for the reaction can be written either as a reactant or product. CaO(s) + H2O(l) → Ca(OH)2(s) + 65.2 kJ When 1 mol of CaO reacts with 1 mol of H2O, 65.2 kJ of heat are produced at standard temperature, 298 K, and standard pressure, 1 atm. More often we write the equation like this ... CaO(s) + H2O(l) → Ca(OH)2(s) ∆H = −65.2 kJ

Thermochemical Equations CaO(s) + H2O(l) → Ca(OH)2(s) ∆H = −65.2 kJ

Thermochemical Equations CaO(s) + H2O(l) → Ca(OH)2(s) ∆H = −65.2 kJ CaO(s) + H2O(l)

Thermochemical Equations CaO(s) + H2O(l) → Ca(OH)2(s) ∆H = −65.2 kJ CaO(s) + H2O(l) The reactants lose 65.2 kJ of heat to the surroundings.

Thermochemical Equations CaO(s) + H2O(l) → Ca(OH)2(s) ∆H = −65.2 kJ CaO(s) + H2O(l) The reactants lose 65.2 kJ of heat to the surroundings. ∆H = −65.2 kJ

Thermochemical Equations CaO(s) + H2O(l) → Ca(OH)2(s) ∆H = −65.2 kJ CaO(s) + H2O(l) ∆H = −65.2 kJ Ca(OH)2(s)

Thermochemical Equations CaO(s) + H2O(l) → Ca(OH)2(s) ∆H = −65.2 kJ CaO(s) + H2O(l) The reaction is exothermic. ∆H = −65.2 kJ Ca(OH)2(s)

Thermochemical Equations 2 NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) ∆H = +129 kJ

Thermochemical Equations 2 NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) ∆H = +129 kJ NaHCO3(s)

Thermochemical Equations 2 NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) ∆H = +129 kJ The reactant gains 129 kJ of heat from the surroundings. NaHCO3(s)

Thermochemical Equations 2 NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) ∆H = +129 kJ The reactant gains 129 kJ of heat from the surroundings. ∆H = +129 kJ NaHCO3(s)

Thermochemical Equations 2 NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) ∆H = +129 kJ Na2CO3(s) + H2O(g) + CO2(g) ∆H = +129 kJ NaHCO3(s)

Thermochemical Equations 2 NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) ∆H = +129 kJ Na2CO3(s) + H2O(g) + CO2(g) The reaction is endothermic. ∆H = +129 kJ NaHCO3(s)

Thermochemical Equations The ∆H of reactions is an extensive variable.

Thermochemical Equations The ∆H of reactions is an extensive variable. The value of ∆H depends on the amount of material.

Thermochemical Equations The ∆H of reactions is an extensive variable. The value of ∆H depends on the amount of material. Twice the amount of reactant produces twice the ∆H.

Thermochemical Equations The ∆H of reactions is an extensive variable. The value of ∆H depends on the amount of material. Twice the amount of reactant produces twice the ∆H. CaO(s) + H2O(l) → Ca(OH)2(s) ∆H = −65.2 kJ

Thermochemical Equations The ∆H of reactions is an extensive variable. The value of ∆H depends on the amount of material. Twice the amount of reactant produces twice the ∆H. CaO(s) + H2O(l) → Ca(OH)2(s) ∆H = −65.2 kJ 2 CaO(s) + 2 H2O(l) → 2 Ca(OH)2(s) ∆H = −130.4 kJ

Thermochemical Equations The ∆H of reactions is an extensive variable. The value of ∆H depends on the amount of material. Twice the amount of reactant produces twice the ∆H.

Thermochemical Equations The ∆H of reactions is an extensive variable.

Example What is the ∆H of the decomposition of 105.0g of sodium bicarbonate to sodium carbonate, carbon dioxide, and water. 2 NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) ∆H = +129 kJ

Example What is the ∆H of the decomposition of 105.0g of sodium bicarbonate to sodium carbonate, carbon dioxide, and water. 2 NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) ∆H = +129 kJ Knowns: mNaHCO3 = 105.0 g MNaHCO3 = 84.01 g/mol ∆H = +129 kJ/2 mol NaHCO3

Example What is the ∆H of the decomposition of 105.0g of sodium bicarbonate to sodium carbonate, carbon dioxide, and water. 2 NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) ∆H = +129 kJ Knowns: mNaHCO3 = 105.0 g MNaHCO3 = 84.01 g/mol ∆H = +129 kJ/2 mol NaHCO3 Calculation:

Example What is the ∆H of the decomposition of 105.0g of sodium bicarbonate to sodium carbonate, carbon dioxide, and water. 2 NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) ∆H = +129 kJ Knowns: mNaHCO3 = 105.0 g MNaHCO3 = 84.01 g/mol ∆H = +129 kJ/2 mol NaHCO3 Calculation: 105.0 g NaHCO3 1

Example What is the ∆H of the decomposition of 105.0g of sodium bicarbonate to sodium carbonate, carbon dioxide, and water. 2 NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) ∆H = +129 kJ Knowns: mNaHCO3 = 105.0 g MNaHCO3 = 84.01 g/mol ∆H = +129 kJ/2 mol NaHCO3 Calculation: 105.0 g NaHCO3 1 mol NaHCO3 × 1 84.01 g NaHCO3

Example What is the ∆H of the decomposition of 105.0g of sodium bicarbonate to sodium carbonate, carbon dioxide, and water. 2 NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) ∆H = +129 kJ Knowns: mNaHCO3 = 105.0 g MNaHCO3 = 84.01 g/mol ∆H = +129 kJ/2 mol NaHCO3 Calculation: 105.0 g NaHCO3 1 mol NaHCO3 × 1 84.01 g NaHCO3

Example What is the ∆H of the decomposition of 105.0g of sodium bicarbonate to sodium carbonate, carbon dioxide, and water. 2 NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) ∆H = +129 kJ Knowns: mNaHCO3 = 105.0 g MNaHCO3 = 84.01 g/mol ∆H = +129 kJ/2 mol NaHCO3 Calculation: (105.0)(1) mol NaHCO3 (1)(84.01)

Example What is the ∆H of the decomposition of 105.0g of sodium bicarbonate to sodium carbonate, carbon dioxide, and water. 2 NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) ∆H = +129 kJ Knowns: mNaHCO3 = 105.0 g MNaHCO3 = 84.01 g/mol ∆H = +129 kJ/2 mol NaHCO3 Calculation: (105.0)(1) mol NaHCO3 +129 kJ × (1)(84.01) 2 mol NaHCO3

Example What is the ∆H of the decomposition of 105.0g of sodium bicarbonate to sodium carbonate, carbon dioxide, and water. 2 NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) ∆H = +129 kJ Knowns: mNaHCO3 = 105.0 g MNaHCO3 = 84.01 g/mol ∆H = +129 kJ/2 mol NaHCO3 Calculation: (105.0)(1) mol NaHCO3 +129 kJ × (1)(84.01) 2 mol NaHCO3

Example What is the ∆H of the decomposition of 105.0g of sodium bicarbonate to sodium carbonate, carbon dioxide, and water. 2 NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) ∆H = +129 kJ Knowns: mNaHCO3 = 105.0 g MNaHCO3 = 84.01 g/mol ∆H = +129 kJ/2 mol NaHCO3 Calculation: (105.0)(1)(+129) kJ (1)(84.01)(2)

Example What is the ∆H of the decomposition of 105.0g of sodium bicarbonate to sodium carbonate, carbon dioxide, and water. 2 NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) ∆H = +129 kJ Knowns: mNaHCO3 = 105.0 g MNaHCO3 = 84.01 g/mol ∆H = +129 kJ/2 mol NaHCO3 Calculation: (105.0)(1)(+129) kJ = 80.6 kJ (1)(84.01)(2)

Example What is the ∆H of the decomposition of 105.0g of sodium bicarbonate to sodium carbonate, carbon dioxide, and water. 2 NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) ∆H = +129 kJ Knowns: mNaHCO3 = 105.0 g MNaHCO3 = 84.01 g/mol ∆H = +129 kJ/2 mol NaHCO3 Calculation: (105.0)(1)(+129) kJ = 80.6 kJ (1)(84.01)(2)

Thermochemical Equations Thermochemical equations also require that we give the physical states of the reactants.

Thermochemical Equations Thermochemical equations also require that we give the physical states of the reactants. H2O(l) → H2(g) + ½ O2(g) ∆H = 285.8 kJ

Thermochemical Equations Thermochemical equations also require that we give the physical states of the reactants. H2O(l) → H2(g) + ½ O2(g) ∆H = 285.8 kJ H2O(g) → H2(g) + ½ O2(g) ∆H = 241.8 kJ

Thermochemical Equations Thermochemical equations also require that we give the physical states of the reactants. H2O(l) → H2(g) + ½ O2(g) ∆H = 285.8 kJ H2O(g) → H2(g) + ½ O2(g) ∆H = 241.8 kJ Even though the stoichiometry is the same, the state of the reactants are different.

Thermochemical Equations Thermochemical equations also require that we give the physical states of the reactants. H2O(l) → H2(g) + ½ O2(g) ∆H = 285.8 kJ H2O(g) → H2(g) + ½ O2(g) ∆H = 241.8 kJ Even though the stoichiometry is the same, the state of the reactants are different. The difference in ∆H is 44.0 kJ.

Summary In a thermochemical reaction, ∆H for the reaction can be written either as a reactant or product. More often we write the equation like this ... CaO(s) + H2O(l) → Ca(OH)2(s) ∆H = −65.2 kJ Where ∆H is negative, the reaction is exothermic. Where ∆H is positive, the reaction is endothermic.

Summary The ∆H of reactions is an extensive variable and depends on the amount of material. Thermochemical equations also require that we give the physical states of the reactants.