Enthalpy Changes

1. Enthalpy Changes • Enthalpy is a state function whose value for a particular system cannot be measured. • We can, however, measureenthalpy changes (DH). • Since DH = Hfinal – Hinitial, the enthalpy change for a chemical reaction is found by: DHrxn = Hproducts – Hreactants where DHrxn = enthalpy of reactionor theheat of reaction

2. P H R time Enthalpy Changes • Endothermic reaction: • DHrxn = positive • Heat must be added to the system • The decomposition of water is endothermic! 2 H2O (g)  2 H2 (g) + O2 (g) DH = + 483.6 kJ

3. R H P time Enthalpy Changes • Exothermic reaction: • DHrxn = negative • Heat is lost by the system • Combustion of methane is exothermic! CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (l) DH = -890. kJ

4. Enthalpy Changes • Some special enthalpy changes: • DHrxn = heat of reaction • DHsoln = heat of solution • The enthalpy change associated with dissolving a solute in a solvent • DHcomb = heat of combustion • Enthalpy change associated with burning a substance in excess oxygen • DHvap = heat of vaporization • Enthalpy change associated with l  g • DHfus = heat of fusion • Enthalpy change associated with s  l • DHf = heat of formation • Enthalpy change associated with preparing 1 mole of substance from its elements

5. Thermochemical Equations • The enthalpy change that occurs during a particular chemical reaction or process is often depicted using a thermochemical equation. • balanced chemical equatioins that show the associated enthalpy change • Thermochemical equations must include: • Balanced equation • Enthalpy change (DHrxn)

6. Thermochemical Equations • Examples of thermochemical equations: • The coefficients in the balanced equation show the number of moles of reactants and products that produce the associated DH. • Since DH is an extensive property, if the number of moles of reactant used or product formed changes, then DH will change as well. CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (l) DH = -890. kJ H2O (l)  H2O (g)DH = 44.01 kJ

7. Thermochemical Equations • The conversion factors below can be obtained from the following reaction: CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (l) DH = -890. kJ -890. kJ-890. kJ 1 mol CH4 2 mol O2 -890. kJ-890. kJ 1 mol CO2 2 mol H2O • These conversion factors can be used to calculate the amount of heat lost when the amount of methane burned is NOT exactly 1 mole.

8. Thermochemical Equations • Guidelines for Using Thermochemical Equations: • Since enthalpy is an extensive property,the magnitude of DH is directly proportional to the amount of reactant consumed or product formed. H2O (l)  H2O (g) DH = 44.01 kJ 2 H2O (l)  2 H2O (g) DH = 88.02 kJ

9. Thermochemical Equations • Guidelines for Using Thermochemical Equations (cont): • The enthalpy change for a reaction is equal in magnitude but opposite in sign to the DH for the reverse reaction. 2 H2O2 (l)  2 H2O (l) + O2 (g) DH = -196 kJ 2 H2O (l) + O2 (g)  2 H2O2 (l) DH = +196 kJ

10. Thermochemical Equations • Guidelines for Using Thermochemical Equations (cont): • The enthalpy change for a reaction depends on the physical state of the reactants and products. CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (l) DH = -890. kJ CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (g) DH = -802 kJ

11. Thermochemical Equations • Why does the magnitude of DHrxn depend on the physical state of the reactants and products? • Energy is either absorbed or released when chemicals change from one physical state to another. H2O (l)  H2O (g) DH = + 44 kJ

12. Thermochemical Equations Example:Calculate the enthalpy change that occurs when 10.0 g of butane, C4H10 (l) are burned at constant pressure? 2 C4H10 (l) + 13 O2 (g)  8 CO2 (g) + 10 H2O (g) DH = -5271 kJ

13. Thermochemical Equations Example: Calculate the enthalpy change that occurs when 17.6 g of nitrogen dioxide decomposes according to the following reaction. 2 NO (g) + O2 (g)  2 NO2 (g) DH =-114.6 kJ

14. Heat Capacity and Specific Heat • If you leave your keys and your chemistry book sitting in the sun on a hot summer day, which one is hotter? • Why is there a difference in temperature between the two?

15. Heat Capacity and Specific Heat • The temperature increase observed when an object absorbs heat depends on its mass and its heat capacity (C). • The amount of heat needed to raise the temperature of an object by 1oC (or by 1K) • Larger heat capacities indicate that more heat must be added to produce a specific temperature increase.

16. Heat Capacity and Specific Heat • For pure substances, heat capacity is usually given for a specified amount of the substance: • Molar heat capacity (Cm): • Amount of heat needed to raise the temperature of 1 mole of a substance by 1oC • Specific heat (Cs): • Amount of heat need to raise the temperature of 1 g of a substance by 1oC

17. Heat Capacity and Specific Heat • Cm = amount of heat transferred moles x temperature change = ___q___ mol x DT • Cs = amount of heat transferred mass x temperature change = ___q___ mass x DT Typical units = J/mol.K or cal/mol.oC Typical units = J/g.K or cal/g.oC

18. Heat Capacity and Specific Heat Example:Calculate the specific heat of copper if 1584 J of heat are needed to raise the temperature of 65.325 g of Cu from 32oC to 95oC.

19. Heat Capacity and Specific Heat Example:What is the molar heat capacity of aluminum if it takes 9.00 J to raise the temperature of 5.00 g of aluminum from 298.0 K to 300.0 K?

20. Heat Capacity and Specific Heat Example:If the specific heat of Al (s) is 0.90 J/g.K, how much heat is required to raise the temperature of 2.25 kg of Al from 25.0oCto its melting point, 660.3oC?

21. Heat Capacity and Specific Heat Example:If the specific heat of Fe (s) is 0.45 J/g.K, calculate the temperature observed when 2.21 kJ of heat is added to 112.223 g of Fe (s)?