Ch. 16: Energy and Chemical Change

1. Ch. 16: Energy andChemical Change Sec. 16.3: Thermochemical Equations

2. Objectives • Write thermochemical equation for chemical reactions and other processes. • Describe how energy is lost or gained during changes of state. • Calculate the heat absorbed or released in a chemical reaction.

3. Review • The change in energy is an important part of chemical reactions so chemists include ΔH as part of the chemical equation.

4. Thermochemical Equations • A thermochemical equation is a balanced chemical equation that includes the physical states of all reactants and products and the energy change. • The energy change is usually expressed as the change in enthalpy, ΔH.

5. Thermochemical Equations • A subscript of ΔH will often give you information about the type of reaction or process taking place. • For example, ΔHcombis the change in enthalpy for a combustion reaction or, simply, the enthalpy (heat) of combustion.

6. Thermochemical Equations C6H12O6(s) + 6O2(g)→ 6CO2(g) + 6H2O(l) ΔHcomb = -2808 kJ/mol • This is the thermochemical equation for the combustion of glucose. • The enthalpy of combustion is -2808 kJ/mol.

7. Thermochemical Equations • The enthalpy of combustion (ΔHcomb) of a substance is defined as the enthalpy change for the complete burning of one mole of the substance. • That means, 2808 kJ of heat are released for every mole of glucose that is oxidized (or combusts). Two moles would release 5616 kJ: 2 mol glucose x -2808 kJ = -5616 kJ 1 mol

8. Standard Enthalpies • Standard enthalpy changes have the symbol ΔHo. • Standard conditions in thermochemistry are 1 atm pressure and 25 0C. **DO NOT CONFUSE THESE WITH THE STP CONDITIONS OF THE GAS LAWS.

9. Calculations • How much heat is evolved when 54.0 g of glucose (C6H12O6) is burned according to this equation? C6H12O6(s) + 6O2(g)→ 6CO2(g) + 6H2O(l) ΔHcomb = -2808 kJ/mol • Since the ΔHcomb given is for 1 mole of glucose, you must first determine the # of moles you have. 54.0 g x 1 mole = 0.300 moles glucose 180 g

10. Calculations (cont.) • You can now use the ΔHcomb value as a conversion factor: 0.300 mol glucose x -2808 kJ = -842 kJ 1 mole

11. Practice Problems • How much heat will be released when 6.44 g of sulfur reacts with O2 according to this equation: 2S + 3O2  2SO3ΔH0 = -395.7 kJ/mol • How much heat will be released when 11.8 g of iron react withO2 according to the equation: 3Fe + 2O2 Fe3O4 ΔH0 = -373.49 kJ/mol

12. Changes of State • The heat required to vaporize one mole of a liquid is called its molar enthalpy (heat) of vaporization (ΔHvap). • The heat required to melt one mole of a solid substance is called its molar enthalpy (heat) of fusion (ΔHfus). • Both phase changes are endothermic & ΔH has a positive value.

13. Changes of State • The vaporization of water and the melting of ice can be described by the following equations: • H2O(l) → H2O(g)ΔHvap = 40.7 kJ/mol • One mole of water requires 40.7 kJ to vaporize. • H2O(s) → H2O(l)ΔHfus = 6.01 kJ/mol • One mole of ice requires 6.01 kJ to melt.

14. Changes of State • The same amounts of energy are released in the reverse processes (condensation and solidification (freezing)) as are absorbed in the processes of vaporization and melting. • Therefore, they have the same numerical values but are opposite in sign. ΔHcond = - ΔHvap = -40.7 kJ/mol ΔHsolid = - ΔHfus = -6.01 kJ/mol

15. Practice Problems • How much heat is released when 275 g of ammonia gas condenses at its boiling point? (ΔHvap = 23.3 kJ/mol) • If water at 00 C releases 52.9 J as it freezes, what is the mass of the water? (ΔHfus= 6.01 kJ/mol) • How much heat is required to melt 25 g of ice at its MP? (ΔHfus= 6.01 kJ/mol)

16. Combination Problems • At times, you will need to calculate the amount of heat that is absorbed or released when a temperature change AND a phase change occur in sequence. • Recall, the heat involved in a temperature change is calculated by using: ΔH = mCΔT. NOTE: In this expression, ΔH is found in joules or cal. • Heat involved in a phase change is calculated using dimensional analysis. NOTE: ΔH will be in kJ.

17. Example • How much heat is released when 37.5 g of water that is at 20.0 0C freezes? (ΔHfus= 6.01 kJ/mol) • Water cannot freeze at 20.0 0C. It must be at 0 0C. Therefore, we must first determine how much energy is released when it is cooled from 20.0 0C to 0 0C. ΔH = mCΔT = (37.5 g)(4.184 J/g 0C)(- 20.0 0C) = - 3138 J = -3.14 kJ

18. Example • How much heat is released when 37.5 g of water that is at 20.0 0C freezes? (ΔHfus= 6.01 kJ/mol) • -3.14 kJ of heat are released when the sample is cooled to 0 0C. • Now we must calculate the energy released when the entire sample freezes. Recall that ΔHsolid= -ΔHfus= -6.01kJ/mol 37.5 g x 1 mole x -6.01 kJ = -12.5 kJ 18 g 1 mole

19. Example • How much heat is released when 37.5 g of water that is at 20.0 0C freezes? (ΔHfus= 6.01 kJ/mol) • -3.14 kJ of heat are released when the sample is cooled to 0 0C. • -12.5 kJ of heat are released when the sample freezes. • The total heat released is -3.14 + -12.5 or -15.6 kJ.

20. Practice Problems Use Cw = 4.184 J/g0C; ΔHfus = 6.01 kJ/mol; ΔHvap = 40.7 kJ/mol. • How much heat is needed to melt 8 g of ice at 0 0C to water at 15 0C? • How much heat is needed to change 28.0 g of water at 60.0 0C to steam?