1. Enthalpy and Hess’s Law

2. From the homework, you may have realized that  H can have a negative number. It relates to the fact that energy as heat has either entered or left the system. It relates to the fact that energy as heat has either entered or left the system. If it was positive, it meant that the heating of the sample required energy. If it was positive, it meant that the heating of the sample required energy. Energy in means endothermic Energy in means endothermic So a positive  H means that the process was endothermic. So a positive  H means that the process was endothermic. The opposite is true if  H was negative. The opposite is true if  H was negative. A negative  H means the process was exothermic (energy left the system) A negative  H means the process was exothermic (energy left the system)

3. Thermodynamics The branch of science that is concerned with the energy changes that accompany chemical and physical changes The branch of science that is concerned with the energy changes that accompany chemical and physical changes To standardize the enthalpies of reactions, scientists have agreed upon a standard thermodynamic temperature of 25.00  C (298.15 K). To standardize the enthalpies of reactions, scientists have agreed upon a standard thermodynamic temperature of 25.00  C (298.15 K).

4. Equations and Enthalpy As a result, we can incorporate more information, like  H values and temperatures, into chemical equations. As a result, we can incorporate more information, like  H values and temperatures, into chemical equations. For example: For example: Fe (s, 300 K)  Fe (s, 1100K)  H = 20.1 kJ/mol Fe (s, 300 K)  Fe (s, 1100K)  H = 20.1 kJ/mol H 2 (g, 298 K) + Br 2 (l, 298 K)  2 HBr (g, 298K)  H = -72.8 kJ/mol H 2 (g, 298 K) + Br 2 (l, 298 K)  2 HBr (g, 298K)  H = -72.8 kJ/mol

5. How do they get  H values? To get  H values, scientists use calorimetry. To get  H values, scientists use calorimetry. It is defined as the measurement of heat- related constants (specific heat, latent heat) It is defined as the measurement of heat- related constants (specific heat, latent heat) It is an experimental measurement of an enthalpy change. It is an experimental measurement of an enthalpy change. A calorimeter is utilized – a device used to measure the heat absorbed or released in a chemical or physical change. A calorimeter is utilized – a device used to measure the heat absorbed or released in a chemical or physical change.

6. Hess’s Law It is stated that the amount of heat released or absorbed in a chemical reaction does not depend on the number of steps in the reaction. It is stated that the amount of heat released or absorbed in a chemical reaction does not depend on the number of steps in the reaction. Example: Phosphorous pentachloride can be made two ways. Example: Phosphorous pentachloride can be made two ways. Way #1 – 1 Step Way #1 – 1 Step P 4 (s) + 10 Cl 2  4 PCl 5 (g)  H = -1596 kJ P 4 (s) + 10 Cl 2  4 PCl 5 (g)  H = -1596 kJ

7. Hess’s Law Way #2 takes two steps Way #2 takes two steps P 4 (s) + 6 Cl 2 (g)  4 PCl 3 (g)  H =-1224 kJ P 4 (s) + 6 Cl 2 (g)  4 PCl 3 (g)  H =-1224 kJ PCl 3 (g) + Cl 2 (g)  PCl 5 (g)  H = -93 kJ PCl 3 (g) + Cl 2 (g)  PCl 5 (g)  H = -93 kJ Notice that we need to do the second step 4 times to use all the PCl 3 from step 1. Notice that we need to do the second step 4 times to use all the PCl 3 from step 1. So, -1224 +4(-93) = -1596 kJ So, -1224 +4(-93) = -1596 kJ Comparing the 1 step way from the last slide to the 2 step way from this slide: Comparing the 1 step way from the last slide to the 2 step way from this slide: -1596 kJ = -1596 kJ, Hess’s Law works -1596 kJ = -1596 kJ, Hess’s Law works

8. So, what does Hess’s Law mean?  H reaction =  H products -  H reactants  H reaction =  H products -  H reactants So we need to use Standard enthalpy of formation – the enthalpy change in forming one mole of a substance from elements in their standard states. These are listed on Pg 355 and 833. So we need to use Standard enthalpy of formation – the enthalpy change in forming one mole of a substance from elements in their standard states. These are listed on Pg 355 and 833. What these tables mean is when a compound is formed from the elements that make it up,  H of formation will be the value listed. What these tables mean is when a compound is formed from the elements that make it up,  H of formation will be the value listed. That means that we can find each compound on these lists and use that number in the equation above. That means that we can find each compound on these lists and use that number in the equation above.

9. Things to remember When compounds are multiplied by a constant (coefficient), the enthalpy change must be multiplied by the same constant. So equations will have to be balanced. When compounds are multiplied by a constant (coefficient), the enthalpy change must be multiplied by the same constant. So equations will have to be balanced.  H of elements is zero, so those equations may not be in the book.  H of elements is zero, so those equations may not be in the book. Remember the diatomics and that S by itself is S 8. This may help you find their values. Remember the diatomics and that S by itself is S 8. This may help you find their values.

10. Now we can look at an example We want to look at the enthalpy change when carbon and carbon dioxide are reacted together to make carbon monoxide. (C + CO 2  2 CO) We want to look at the enthalpy change when carbon and carbon dioxide are reacted together to make carbon monoxide. (C + CO 2  2 CO) So, we need to look at the formation of CO and CO 2 to complete this problem. So, we need to look at the formation of CO and CO 2 to complete this problem. We see that CO,  H = -110.5 kJ/mol and We see that CO,  H = -110.5 kJ/mol and CO 2,  H = -393.5 kJ/mol (C,  H = 0)  H reaction =  H products -  H reactants = 2(-110.5)-[(0)+(-393.5)] = 2(-110.5)-[(0)+(-393.5)] = 172.5 kJ/mol

11. Examples Calculate the enthalpy change for the reaction: Calculate the enthalpy change for the reaction: 2H 2 (g) + 2CO 2 (g)  2H 2 O (g) + 2CO (g) What enthalpy change accompanies the reaction: What enthalpy change accompanies the reaction: 2Al(s) + 3H 2 O (l)  Al 2 O 3 (s) + 3H 2 (g) Calculate  H for the decomposition of calcium carbonate into calcium oxide and carbon dioxide. Calculate  H for the decomposition of calcium carbonate into calcium oxide and carbon dioxide.

12. Homework Page 371: 26, 27, 28, 29 Page 371: 26, 27, 28, 29