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Causes of Change

Causes of Change. Changes in Enthalpy During Chemical Reactions Using Hess’s Law in Calculations . On Separate Sheet of Paper: Turn in when finished, 10 pts. Define, in your own words, Hess’s Law What is the difference between bomb calorimetry and adiabatic calorimetry?

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Causes of Change

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  1. Causes of Change Changes in Enthalpy During Chemical Reactions Using Hess’s Law in Calculations

  2. On Separate Sheet of Paper:Turn in when finished, 10 pts • Define, in your own words, Hess’s Law • What is the difference between bomb calorimetry and adiabatic calorimetry? • What is standard thermodynamic temperature • What type of calorimetry do nutritionists use? • What is change in enthalpy? What does this represent? Hint: think of the definition of enthalpy.

  3. Standard Enthalpies of Formation

  4. Using Hess’s Law and Algebra • We can manipulate chemical equations using algebra just like we do with number equations • When equations are added or subtracted, enthalpy changes must be add or subtracted. • If the equations are multiplied by a constant, the ________________ must also be multiplies by that constant.

  5. Adding Equations • The enthalpy of the formation of CO, when CO2 and solid carbon are reactants, is found using the following equations: 2C(s) + O2(g) 2CO(g)ΔH = -221 kJ C(s) + O2(g) CO2(g) ΔH= -393 kJ • But, CO2 is not a reactant in these equations • What can we do?

  6. Adding Equations 2C(s) + O2(g) 2CO(g)ΔH = -221 kJ C(s) + O2(g)  CO2(g) ΔH = -393 kJ • If we reverse the second equation, we get: CO2(g) C(s) + O2(g) ΔH= ________ kJ • If we do this, we must also reverse the change in enthalpy for this equation. –(–393 kJ) = +393 kJ

  7. Adding Equations • So, reversing the equation causes the enthalpy of the new reaction to be the opposite of the original reaction. • Now we must add the two equations to get the equation for the formation of CO by using CO2 and C. • After we add the equations, we must cancel all items that appear on both sides, just like in algebra.

  8. Adding Equations 2C(s) + O2(g) 2CO(g)ΔH = -221 kJ CO2(g)  C(s) + O2(g) ΔH = 393 kJ 2C(s) + O2(g) + CO2(g)  2CO(g) + C(s) + O2(g) ΔH = 172 kJ • Now, cancel out all like terms: 2C(s)+ O2(g) + CO2(g)  2CO(g) + C(s) + O2(g) ΔH = 172 kJ • You are now left with: C(s)+ CO2(g)  2CO(g)ΔH = 172 kJ

  9. Standard Enthalpies of Formation • The enthalpy change in forming 1 mol of a substance from elements in their standard states is called the standard enthalpy of formation of the substance • This is denoted ΔHf° • Many ΔHf°values are listed on page 355 • ΔHf° values for elements are zero

  10. Standard Enthalpies of Formation • From their elements! • This means that the formation of NO would be represented by: N2 + O2 2NO • However, this is for 2 mol of NO, we want 1 mol so: ½ N2+ ½ O2 NO

  11. Using ΔHf°Values • From a list of standard enthalpies of formation, the enthalpy change for any reaction with available data can be calculated. • Consider the following reaction: SO2(g) + NO2(g) SO3(g) + NO(g)ΔH = ?? • There are _____ different compounds so there must be _____ different ΔHf°equations. • Look at Table 2 to find these four equations.

  12. Calculating a Standard Enthalpy of Formation • Step 1: Write out the equation we are to work on • Step 2: Plan your work. Determine which equations must be reversed, which equations must be multiplied, etc. • Step 3: Line up you work and do the calculations. • Step 4: Verify your results. The unnecessary reactants and products cancel to give you the correct equation.

  13. Example: SO2(g) + NO2(g) SO3(g) + NO(g)ΔH = ?? • How do you make SO2(g)? • How do you make NO2(g)? • How do you make SO3(g)? • How do you make NO(g)? S8(g) + O2(g) SO2(g) N2(g) + O2(g) NO2(g) S8(g) + O2(g) SO3(g) N2(g) + O2(g) NO(g)

  14. Example: • Now, balance the previous 4 equations S8(g) + O2(g) SO2(g) ⅛S8(g) + O2(g) SO2(g) N2(g) + O2(g) NO2(g) ½N2(g) + O2(g) NO2(g) S8(g) + O2(g) SO3(g) ⅛ S8(g) + 3/2 O2(g) SO3(g) N2(g) + O2(g) NO(g) ½N2(g) + ½O2(g) NO(g)

  15. Example: • Next, write them as reactants and products • Original eq: SO2(g) + NO2(g) SO3(g) + NO(g) • Reactants are __________ and __________ and must be reversed. Products are alright. SO2(g)  ⅛S8(g) + O2(g)ΔH = -ΔHf°= -(-296.8 kJ/mol) NO2(g) ½N2(g) + O2(g)ΔH = -ΔHf° = -(-33.1 kJ/mol)

  16. ΔH of the Reaction • When using standard enthalpies of formation to determine the enthalpy change of a chemical reaction, remember the following equation: ΔHreaction = ΣΔHproducts - ΣΔHreactants

  17. Enthalpy Change of Reaction • If you apply Hess’s Law, the enthalpies can be added as follows: SO2(g) + NO2(g) SO3(g) + NO(g) ΔH = (ΔHf°, SO3 + ΔHf°,NO ) - (ΔHf°,NO2 + ΔHf°,SO2 ) ΔH= (90.3 kJ/mol + -395.8 kJ/mol) - (33.1 kJ/mol+ -296.8 kJ/mol) = = -41.8 kJ/mol

  18. Practice  • Calculate the standard enthalpy of formation of pentane, C5H12, using the given information. (1) C(s) + O2(g) CO2(g) ΔHf° = -393.5 kJ/mol (2) H2(g) + ½ O2(g) H2O(l) ΔHf° = -285.8 kJ/mol (3) C5H12(g) + 8O2(g)  5CO2(g) + 6H2O(l)ΔH = -3535.6 kJ/mol

  19. Practice  • Step 1: Write out the equation we are to work on • How do you make C5H12(g)? 5C(s) + 6H2(g) C5H12(g)

  20. Practice  (1) C(s) + O2(g) CO2(g) ΔHf° = -393.5 kJ/mol (2) H2(g) + ½O2(g) H2O(l) ΔHf° = -285.8 kJ/mol (3) C5H12(g) + 8O2(g)  5CO2(g) + 6H2O(l)ΔH = -3535.6 kJ/mol 5C(s) + 6H2(g) C5H12(g) • Step 2: Plan your work. Determine which equations must be reversed, which equations must be multiplied, etc. Do we have to reverse any of the three reactions? Which one? Do we have to multiple by any coefficients? By what?

  21. Practice  (1) C(s) + O2(g) CO2(g) ΔHf° = -393.5 kJ/mol (2) H2(g) + ½O2(g) H2O(l) ΔHf° = -285.8 kJ/mol (3) C5H12(g) + 8O2(g)  5CO2(g) + 6H2O(l)ΔH = -3535.6 kJ/mol 5C(s) + 6H2(g) C5H12(g) • Step 3: Line up you work and do the calculations. (1) 5C(s) + 5O2(g) 5CO2(g) ΔH= 5(-393.5 kJ/mol) (2) 6H2(g) + 3O2(g) 6H2O(l) ΔH= 6(-285.8 kJ/mol) (3) 5CO2(g) + 6H2O(l) C5H12(g) + 8O2(g) ΔH = -(-3535.6 kJ/mol) 5C(s) + 6H2(g) C5H12(g)ΔHf°= -145.7 kJ/mol

  22. Practice  • Step 4: Verify your results. The unnecessary reactants and products cancel to give you the correct equation. Question: Calculate the standard enthalpy of formation of pentane, C5H12, using the given information. Answer: 5C(s) + 6H2(g) C5H12(g)ΔHf° = -145.7 kJ/mol • Is this reaction exothermic or endothermic? ____________

  23. Practice On Your Own: • Calculate the change in enthalpy for the reaction below by using data from Table 2. 2H2(g) + 2CO2(g) 2H2O(g) + 2CO(g) • Then State whether the reaction is exothermic or endothermic.

  24. HW: “Changes in Enthalpy During Chemical Reactions” Packet • Homework: Finish the last two problems from the packet from last week. Due tomorrow at the beginning of class. • Ask if you are having trouble!

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