Calculating Heats of Reaction

1. Calculating Heats of Reaction Prentice-Hall Chapter 17.4 Dr. Yager

2. Objectives • State Hess’s Law of heat summation and describe how it is used in chemistry. • Solve for enthalpy changes by using Hess’s law or standard heats of formation.

3. Hess’s Law • Hess’s law allows you to determine the heat of reaction indirectly. • If you add two or more thermochemical equations to give a final equation, then you can also add the heats of reaction to give the final heat of reaction.

4. How much energy is required to convert diamond to graphite?

5. C(s,diamond)+ O2(g) CO2(g) ∆H = - 395.4 kJ C(s,graphite) + O2(g) CO2(g) ∆H = - 393.5 kJ Reverse the second reaction: C(s,diamond)+ O2(g) CO2(g) ∆H = - 395.4 kJ CO2(g)C(s,graphite) + O2(g)∆H = 393.5 kJ ∆H = -1.9 kJ Add the reactions: C(s,dmd)+ O2(g)+CO2(g) C(s,gph)+O2(g)+ CO2(g) ∆H=-1.9kJ Cancel same terms on each side: C(s,diamond) C(s,graphite) ∆H= -1.9 kJ

6. C(s)graphite + ½O2(g) CO(g) ∆H = ?

7. C(s,graphite) + ½O2(g) CO(g) ∆H = ? Actually doing this reaction in the lab is impossible. C(s,graphite)+ O2(g) CO2(g) ∆H = -393.5 kJ CO(g) + ½O2(g) CO2(g) ∆H = -283.0 kJ Reverse the second equation then add: C(s,graphite) + O2(g) CO2(g) ∆H = - 393.5 kJ CO2(g) CO(g) + ½O2(g)∆H = 283.0 kJ C(s,graphite) + ½O2(g) CO(g) ∆H = -110.5 kJ

8. Standard Heats of Formation • For a reaction that occurs at standard conditions, you can calculate the heat of reaction by using standard heats of formation. • The standard heat of formation(∆Hf0) of a compound is the change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states at 25°C.

10. The Standard Heat of Formation of Water Reactants in standard form at 25OC and 101.3 kPa

12. 2CO(g) + O2(g) 2CO2(g)

14. Calculate ∆Ho for the following reactions. • Br2(g) Br2(l) • CaCO3(s) CaO(s) + CO2(g) • 2NO(g) + O2(g) 2NO2 (g) • ∆Ho = (0.0 – 30.91) kJ/mol = -30.91 kJ/mol • ∆Ho= (-635.1 - 393.5) kJ/mol - (-1207.0) kJ/mol • = 178.4 kJ/mol • ∆Ho = (2 * 33.85) kJ/mol - (2 * 90.37 + 0.0) kJ/mol • = (67.70 - 180.74) kJ/mol = -113.04 kJ/mol

15. 1. According to Hess’s law, it is possible to calculate an unknown heat of reaction by using • heats of fusion for each of the compounds in the reaction. • two other reactions with known heats of reaction. • specific heat capacities for each compound in the reaction. • density for each compound in the reaction.

16. 1. According to Hess’s law, it is possible to calculate an unknown heat of reaction by using • heats of fusion for each of the compounds in the reaction. • two other reactions with known heats of reaction. • specific heat capacities for each compound in the reaction. • density for each compound in the reaction.

17. 2. The heat of formation of Cl2(g) at 25°C is • the same as that of H2O at 25°C. • larger than that of Fe(s) at 25°C. • undefined. • zero.

18. 2. The heat of formation of Cl2(g) at 25°C is • the same as that of H2O at 25°C. • larger than that of Fe(s) at 25°C. • undefined. • zero.

19. 3. Calculate HO for NH3(g) + HCl(g)  NH4Cl(s) Standard heats of formation:NH3(g) = 45.9 kJ/mol, HCl(g) = 92.3 kJ/mol, NH4Cl(s) = 314.4 kJ/mol • 176.2 kJ • 360.8 kJ • 176.2 kJ • 268 kJ

20. 3. Calculate HO for NH3(g) + HCl(g)  NH4Cl(s) Standard heats of formation:NH3(g) = 45.9 kJ/mol, HCl(g) = 92.3 kJ/mol, NH4Cl(s) = 314.4 kJ/mol • 176.2 kJ • 360.8 kJ • 176.2 kJ • 268 kJ