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# Enthalpy Hess's Law

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### Enthalpy Hess's Law

1. 1. Enthalpy and Hess’s Law By Shawn P. Shields, Ph.D. This work is licensed by Shawn P. Shields-Maxwell under a Creative Commons Attribution-NonCommercial-ShareAlike
2. 2. Hess’s Law and the Conservation of Energy Since enthalpy (H) is a state function, the change in the enthalpy of reaction (∆Hrxn) does not depend on how reactants are converted to products (i.e., in one step, two steps, or more).
3. 3. Hess’s Law and the Conservation of Energy ∆Hrxn is only dependent on two factors: The mass of the substances that are reacting and their physical states (solid, liquid, gas, aqueous solution, etc.)
4. 4. Hess’s Law and ∆Hrxn Consider: N2O4(g) → 2NO2(g) ∆Hrxn = qp = +57.2 kJ/mol ½N2O4(g) → NO2(g) ∆Hrxn = qp = + = + 28.6 kJ/mol 2NO2(g) → N2O4(g) ∆Hrxn = qp = − 57.2 kJ/mol What do you notice? 
5. 5. Hess’s Law and ∆Hrxn Consider: N2O4(g) → 2NO2(g) ∆Hrxn = qp = +57.2 kJ/mol Let’s use the first as the original reaction so we can compare it to the others. ½N2O4(g) → NO2(g) ∆Hrxn = qp = + = + 28.6 kJ/mol If we cut the amount of N2O4 reacting in half, only half the heat is absorbed! 
6. 6. Hess’s Law and ∆Hrxn Consider: N2O4(g) → 2NO2(g) ∆Hrxn = qp = +57.2 kJ/mol Let’s use this as the original reaction. 2NO2(g) → N2O4(g) ∆Hrxn = qp = − 57.2 kJ/mol If we reverse the reaction so that 1 mol N2O4 is produced rather than consumed, the same amount of heat is released in the reaction as was absorbed in the original.
7. 7. Hess’s Law Hess’s Law states that when two or more reactions are added to give a new chemical reaction, addition of the corresponding enthalpies will provide the enthalpy of the new reaction. In other words, whether the reaction is done in one step, two steps, or more, the overall enthalpy change of the reaction will be the same.
8. 8. “Rules” for Hess’s Law Calculations Rule 1: If we add two chemical equations together, then we need to add the corresponding ∆Hrxn values. A + B → C ∆Hrxn = +20 kJ/mol C + D → F ∆Hrxn = +10 kJ/mol A + B + D → F ∆Hrxn = 20 + 10 = +30 kJ/mol
9. 9. “Rules” for Hess’s Law Calculations Rule 2: If we reverse the direction of a chemical reaction, then the sign of ∆Hrxn should be reversed. A + B → C ∆Hrxn = +20 kJ/mol C → A + B ∆Hrxn = − 20 kJ/mol
10. 10. “Rules” for Hess’s Law Calculations Rule 3: If we multiply a chemical reaction by a factor, then we need to multiply ∆Hrxn by the same factor. Original reaction C → A + B ∆Hrxn = − 20 kJ/mol 2(C → A + B) multiply each coefficient by 2 2C → 2A + 2B ∆Hrxn = 2(− 20) = − 40 kJ/mol
11. 11. Hess’s Law Example Calculation Suppose we need ∆Hrxn for the following “target” reaction N2(g) + 2O2(g) → N2O4(g) ∆Hrxn = ???? kJ/mol We have information for the following two reactions N2(g) + 2O2(g) → 2NO2(g) ∆Hrxn = 66.4 kJ/mol 2NO2(g) → N2O4(g) ∆Hrxn = − 57.2 kJ/mol Calculate ∆Hrxn for the target reaction using this information.
12. 12. Hess’s Law Example We need to add the two reactions so that they match our target reaction. N2(g) + 2O2(g) → N2O4(g) ∆Hrxn = ???? kJ/mol Look at the first reaction to see if reactants and/or products are on the correct side of the equation (matching the target): N2(g) + 2O2(g) → 2NO2(g) ∆Hrxn = 66.4 kJ/mol 2NO2(g) → N2O4(g) ∆Hrxn = − 57.2 kJ/mol
13. 13. Hess’s Law Example We need to add the two reactions so that they match our target reaction. N2(g) + 2O2(g) → N2O4(g) ∆Hrxn = ???? kJ/mol N2(g) + 2O2(g) → 2NO2(g) ∆Hrxn = 66.4 kJ/mol The reactants are on the correct side of the equation, so move on to the second equation: 2NO2(g) → N2O4(g) ∆Hrxn = − 57.2 kJ/mol
14. 14. Hess’s Law Example We need to add the two reactions so that they match our target reaction. N2(g) + O2(g) → N2O4(g) ∆Hrxn = ???? kJ/mol N2(g) + 2O2(g) → 2NO2(g) ∆Hrxn = 66.4 kJ/mol 2NO2(g) → N2O4(g) ∆Hrxn = − 57.2 kJ/mol The products are on the correct side of the equation, and the NO2 should cancel out. Add the two equations together.
15. 15. Hess’s Law Example We need to add the two reactions so that they match our target reaction. N2(g) + 2O2(g) → N2O4(g) ∆Hrxn = ???? kJ/mol N2(g) + 2O2(g) → 2NO2(g) ∆Hrxn = 66.4 kJ/mol 2NO2(g) → N2O4(g) ∆Hrxn = − 57.2 kJ/mol N2(g) + 2O2(g) → N2O4(g) ∆Hrxn = 66.4 + (− 57.2 ) ∆Hrxn = 9.20 kJ/mol
16. 16. Why Does Hess’s Law Work? ∆H is a state function. Only the initial and final states matter, not how we get there!  The initial state is the moles of each reactant.  The final state is the moles of each product.  We end up with the same reaction whether we run it directly or do it in two steps (i.e., add two reactions together.)
17. 17. Why Does Hess’s Law Work? ∆H is a state function. Only the initial and final states matter, not how we get there! ∆Hrxn H (kJ/mol) N2(g) + 2O2(g) 2NO2(g) N2O4(g) Target reaction Reaction 2 Reaction 1
18. 18. What You Should Be Able to Do (so far) Describe how Hess’s Law can be used to calculate the enthalpy of a reaction. Use Hess’s Law to calculate the enthalpy of reaction for a target reaction. (∆Hrxn) Be able to use the rules for manipulating chemical equations in Hess’s Law calculations. Determine whether the target reaction is exo- or endothermic from the sign of ∆H.