1. 6.7-6.9 Calorimetry, Hess’s Law, and Enthalpies of Formation

2. Constant Pressure Calorimetry
A simple calorimeter can be constructed by placing one Styrofoam cup inside of another, a lid placed on top, and two holes in the lid to tightly fit a thermometer and a stirrer.
This type of calorimeter occurs at constant pressure, and the amount of heat absorbed or released by the solution can be calculated.
qsoln = msoln X Cs, soln X ΔT

3. Let’s Try a Practice Problem
The addition of hydrochloric acid to a silver nitrate solution precipitates silver chloride according to the reaction: AgNO3(aq) + HCl(aq)  AgCl(s) + HNO3(aq) When 50.0 mL of M AgNO3 is combined with 50.0 mL of M HCl in a coffee cup calorimeter, the temperature changes from 23.40oC to 24.21oC. Calculate ΔHrxn for the reaction as written. Use 1.00 g/mL as the density of the solution and C = 4.18 J/goC as the specific heat capacity g 100. mLsoln X = 100. g solution 1.00 mL ΔT = – = 0.81oC qSoln = msoln X Cs soln X ΔT = (100. g) (4.18 J/goC) (0.81oC) = 339 J -qrxn -339 J Hrxn = = = -67,800 J / mol = kJ / mol mol AgNO3 (0.100 M X L)

4. Relationships involving ΔHrxn
If a chemical equation is multiplied by some factor, then ΔHrxn is multiplied by the same factor.
If a chemical reaction is reversed, the ΔHrxn changes sign
If a chemical equation can be expressed as the sum of a series of steps, then ΔHrxn for the overall equation is the sum of the heats of reactions for each step.
This is known as Hess’s Law.

5. An Example of how to Calculate Hrxn Given the Series of Steps of a Reaction
Find ΔHrxn for the reaction: N2O(g) + NO2(g)  3NO(g) Use these reactions with known ΔH’s. 2NO(g) + O2(g)  2NO2(g) ΔH = kJ N2(g) + O2(g)  2NO(g) ΔH = kJ 2N2O(g)  2N2(g) + O2(g) ΔH = kJ Step 1: Rearrange the series of reactions so that reactants and products match the reactants and products of the overall reaction (including # moles). Remember to flip sign on ΔH if the reaction is reversed. Reverse and halve reaction 1: NO2(g)  NO(g) + 1/2O2(g) ΔH = kJ Keep reaction 2 the way it is: N2(g) + O2(g)  2NO(g) ΔH = kJ Halve reaction 3: N2O(g)  N2(g) + 1/2O2(g) ΔH = kJ Cancel substances that are on both side of the reaction Problem continued on next slide 

6. Practice Problem Continued
Step 2: Calculate ΔHrxn ΔHrxn = ΔH1 + ΔH2 + ΔH3 ΔHrxn = ( kJ) + ( kJ)+ (-81.6 kJ) ΔHrxn = kJ Step 3: Write out the thermochemical equation for the overall reaction. N2O(g) + NO2(g)  3NO(g) ΔH = kJ

7. Let’s Try Another!
Find ΔHrxn for the reaction 3H2(g) + O3(g)  3H2O(g) Use these reactions with known ΔH’s: 2H2(g) + O2(g)  2H2O(g) ΔH = kJ 3O2(g)  2O3(g) ΔH = kJ Multiply reaction 1 be 1.5: 1.5 (2H2(g) + O2(g)  2H2O(g) ΔH = kJ) Reverse and halve reaction 2: .5 (2O3(g)  3O2(g) ΔH = kJ) 3H2(g) + 1 ½ O2(g)  3H2O(g) ΔH = kJ O3(g) 1 ½ O2(g) ΔH = kJ 3H2(g) + O3(g)  3H2O(g) ΔH = kJ

8. Determining Enthalpies of Reactions from Standard Enthalpies of Formation
Standard enthalpy of formation (ΔH o f ), also known as standard heat of formation, is the change in enthalpy when one mole of compound forms from its constituent elements in their standard states.
Here’s an example of how to write the heat of formation for a chemical reaction (these ΔH o f are found in appendix IIB in the back of your textbook)
6C(s, graphite) + 6H2(g) +3O2(g)  C6H12O6(s) ΔH o f = kJ/mol

9. Let’s Try One
Write equations for the formation of NaCl(s) from its elements in their standard states. 2Na(s) + Cl2(g)  2NaCl(s) Na(s) + ½ Cl2(g)  NaCl(s) ΔH o f = kJ/mol

10. Calculating Standard Enthalpy Change for a Reaction
If elements  Compound = ΔH o f Then, compounds to elements = - ΔH o f To calculate the heat ΔH o rxn subtract the enthalpies of formation of the reactants, multiplied by their stoichiometric coefficients from the enthalpies of formation of the products multiplied by their stoichiometric coefficients. ΔH o rxn = ΣnpΔH o f (products) - ΣnrΔH o f (reactants)

11. Let’s Try a Practice Problem!
The thermite reaction, in which powdered aluminum reacts with iron oxide, is highly exothermic. 2Al(s) + Fe2O3(s)  Al2O3(s) + 2Fe(s) Use standard enthalpies of formation to find ΔH o rxn Al(s) = 0 Fe2O3(s) = kJ/mol Al2O3(s) = kJ/mol Fe(s) = 0

12. ΔH o rxn = ΣnpΔH o f (products) - ΣnrΔH o f (reactants)
ΔH o rxn = (1(ΔH o f ,Al2O3(S)) + 2(ΔH o f ,Fe(S))) – (2(ΔH o f Al) + (1(ΔH o f ,Fe2O3(S)))
ΔH o rxn = (1( kJ/mol) + 2(0)) – (2(0) + (1( kJ/mol))
ΔH o rxn = kJ/mol

13. pgs. 289 #’s 80, 84, 86, 88, & 90 (a’s only)
Study for the Chapter 5-6 Exam