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Enthalpy Changes

Enthalpy Changes. Measuring and Expressing ∆H Thermochemical Equations. Objectives. When you complete this presentation, you will be able to Write and interpret a thermochemical equation

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Enthalpy Changes

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  1. Enthalpy Changes Measuring and Expressing ∆H Thermochemical Equations

  2. Objectives • When you complete this presentation, you will be able to • Write and interpret a thermochemical equation • Determine whether a chemical reaction is endothermic or exothermic on the basis of the change in enthalpy of the reaction • Use change in enthalpy as an extensive property in solving thermochemical problems

  3. 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

  4. Thermochemical Equations • Let’s look more closely at the previous equation: 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 The reaction is exothermic. The product, Ca(OH)2(s), has less energy than the reactants, CaO(s) + H2O(l). Ca(OH)2 (s)

  5. Thermochemical Equations • Let’s look at a different equation: 2 NaHCO3(s) → NaCO3(s) + H2O(g) + CO2(g) ∆H = +129 kJ NaCO3(s) + H2O(g) + CO2(g) 2 NaHCO3(s) The reactants gain 129 kJ of heat from the surroundings. ∆H = +129 kJ The reaction is endothermic. The products, NaCO3(s) + H2O(g) + CO2(g), have less energy than the reactant, NaHCO3(s).

  6. Thermochemical Equations • The ∆H of reactions is an extensivevariable. • The value of ∆H depends on the amountof material. • Twice the amount of reactant produces twice the value of ∆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

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

  8. 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.

  9. 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.

  10. 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.

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