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# Ch6 z5e thermo

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• Z5e 242 Section 6.1 - Nature of Energy
• Z5e 243
• Z5e Rf. Fig 6.2 p. 244
• Rf. Fig 6.3 p. 245
• Z5e 245
• Z5e 245
• Z5e 245
• Z5e 244
• d = distance
• Z5e Fig. 6.4 The piston, moving a distance delta h against pressure P, does work on the surroundings. Since V of cylinder = area of base x height, the delta V of gas = delta h x A
• w = -p(chg V) = -24 l-atm E = q + w = 2360 J + (-24 l-atm)(101.325 j/l-atm) = -71.8 J Want E = 0, so q + w = 0, so since w = -2430 J; q added must = + 2430 J
• Z5e 248 Section 6.2 Enthalpy &amp; Calorimetry
• Z5e 250
• Z5e 251
• Convert Kg to g; q = m  TC p = 2 880 000 J = 2880 kJ q gained = q lost; set both equations equal to each other. Ans. 0.203 J/g o C Be sure to discuss that we are looking at change in T, so since C and K degree increments are equal, do NOT convert change in T value to Kelvin. E.g., 10 Kelvin degree change should not be converted to a minus 263 Celsius change.
• Convert Kg to g; q = m  TC p = 2 880 000 J = 2880 kJ q gained = q lost; set both equations equal to each other. Ans. 0.203 J/g o C Be sure to discuss that we are looking at change in T, so since C and K degree increments are equal, do NOT convert change in T value to Kelvin. E.g., 10 Kelvin degree change should not be converted to a minus 263 Celsius change.
• Z5e 252
• Z5e 253 and see Fig. 6.6 p. 254
• Z5e 284 q. 50
• (2 * -394 + 3 * -286) - (1 *-278 - 3 * 0) = -1368 kJ/mol = exothermic Oxygen = O since element; liquid water value is not same as gaseous water
• Z5e 256 6.3 Hess’ Law
• Z5e 257 Fig. 6.7 - same change occurs whether in 1 or 2 steps. N 2 + 2O 2 --&gt; 2NO 2 ;  H 1 = 68kJ Or 2 step: N 2 + O 2 --&gt; 2NO ;  H 2 = 180kJ 2NO + O 2 --&gt; 2NO 2 l;  H 3 = -112kJ N 2 + 2O 2 --&gt; 2NO 2 ;  H 1 = 68kJ
• Z5e 259
• Hrw 520
• Hrw 520
• Hrw 520
• Hrw 520
• Reverse equation 1; multiply eq. 2 by 2 (include delta H) Ans. 226 kJ, so endothermic
• Reverse equation 1; multiply eq. 2 by 2 (include delta H) Ans. 226 kJ, so endothermic
• Cut all rxns in 1/2 Reverse #s 2 &amp; 3 Add up to = -426 kJ = exothermic
• Z5e Section 6.4 Standard Enthalpies of Formation
• Z53 261 a. Use 1/2 for coefficient of N 2 b. C(s) + 2H 2 (g) + 1/2 O 2 (g) = CH 3 OH
• Z53 261 a. Use 1/2 for coefficient of N 2 b. C(s) + 2H 2 (g) + 1/2 O 2 (g) = CH 3 OH
• Z5e 263
• (2 * -394 + 3 * -286) - (1 *-278 - 3 * 0) = -1368 kJ/mol = exothermic Oxygen = O since element; liquid water value is not same as gaseous water
• Z5e 264 SE 6.9
• Z5e 264 SE 6.9
• Z5e 264 SE 6.9
• Z5e 264 Fig. 6.10 used with SE 6.9.
• Z5e 264 SE 6.9
• (2 * -394 + 3 * -286) - (1 *-278 - 3 * 0) = -1368 kJ/mol = exothermic Oxygen = O since element; liquid water value is not same as gaseous water
• Hrw 523
• (2 * -394 + 3 * -286) - (1 *-278 - 3 * 0) = -1368 kJ/mol = exothermic Oxygen = O since element; liquid water value is not same as gaseous water
• ### Ch6 z5e thermo

1. 1. Chapter 6 Energy Thermodynamics pp
2. 2. 6.1 The Nature of Energy <ul><li>Energy is . . . </li></ul><ul><li>The ability to do work. </li></ul><ul><li>Conserved (It’s the Law). </li></ul><ul><li>Made of heat and work . </li></ul><ul><li>A state function (unlike work and heat). </li></ul><ul><li>I.e. , it’s independent of the path, or how you get from point A to B. </li></ul>
3. 3. Enthalpy is a STATE FUNCTION . State functions are PATH-INDEPENDENT
4. 4. Work and Heat <ul><li>Work is a force acting over a distance. </li></ul><ul><li>Heat is energy transferred between objects because of temperature difference. </li></ul>
5. 5. The universe <ul><li>Is divided into two halves. </li></ul><ul><li>The system and the surroundings . </li></ul><ul><li>The system is the part you are concerned with. </li></ul><ul><li>The surroundings are the rest. </li></ul><ul><li>Exothermic reactions release energy to the surroundings. </li></ul><ul><li>Endothermic reactions absorb energy from the surroundings. </li></ul>
6. 6. Potential energy Heat
7. 7. Potential energy Heat
8. 8. Direction <ul><li>Every energy measurement has three parts. </li></ul><ul><li>Unit ( Joules or calories). </li></ul><ul><li>Number - how many. </li></ul><ul><li>Sign - to tell direction. </li></ul><ul><li>Negative - exo thermic </li></ul><ul><li>Positive - endo thermic </li></ul>
9. 9. System Surroundings Energy  E < 0
10. 10. System Surroundings Energy  E >0
11. 11. Same rules for heat and work pp <ul><li>Heat given off is negative . </li></ul><ul><li>Heat absorbed is positive . </li></ul><ul><li>Work done by system on surroundings is negative . </li></ul><ul><li>Work done on system by surroundings is positive . </li></ul><ul><li>Thermodynamics - The study of energy and the changes it undergoes. </li></ul>
12. 12. First Law of Thermodynamics <ul><li>The energy of the universe is constant . </li></ul><ul><li>Law of conservation of energy . </li></ul><ul><li>q = heat </li></ul><ul><li>w = work </li></ul><ul><li> E = q + w </li></ul><ul><li>Take the system’s point of view to decide signs. </li></ul>
13. 13. What is work? <ul><li>Work is a force acting over a distance. </li></ul><ul><li>w= F x  d </li></ul><ul><li>P = F/area </li></ul><ul><li>d = V/area </li></ul><ul><li>w = (P x area) x  (V/area) = P  V </li></ul><ul><li>Work can be calculated by multiplying pressure by the change in volume at constant pressure. </li></ul><ul><li>units of work: liter-atm or L-atm </li></ul>
14. 14. Work needs a sign <ul><li>If the volume of a gas increases , the system has done work on the surroundings (memorize this!). </li></ul><ul><li>And . . . work is negative : w = (-) </li></ul><ul><li>w = - P  V (don’t forget negative sign) </li></ul><ul><li>Expanding a gas: Work is negative . </li></ul><ul><li>Contracting : Surroundings do work on the system and w = (+). </li></ul><ul><li>1 L-atm = 101.325 J (need for conversions) </li></ul>
15. 15. Fig 6.4 p. 246, Volume of a Cylinder <ul><li>The piston, moving a distance delta h against pressure P, does work on the surroundings. </li></ul><ul><li>b. Since V of cylinder = area of base x height, the delta V of gas = delta h x A </li></ul>
16. 16. Examples pp <ul><li>What amount of work is done when 15.00 L of gas is expanded to 25.00 L at 2.40 atm pressure? Steps . . . </li></ul><ul><li>w = -P∆V = -2.40 atm • 10.00L = -24 L-atm </li></ul><ul><li>If 2.36 kJ of heat are absorbed what is the change in energy? Steps . . . </li></ul><ul><li>∆ E = q + w = 2360 J + (-24 L-atm)(101.325 J /L-atm) = -71.8 J </li></ul><ul><li>How much heat to change the gas without changing the internal energy of the gas? </li></ul><ul><li>Want ∆E = 0, so q + w = 0 </li></ul><ul><li>since w = -2430 J; q added must = +2430 J </li></ul>
17. 17. 6.2 Enthalpy & Calorimetry <ul><li>Enthalpy is abbreviated as H </li></ul><ul><li>H = E + PV (that’s the definition of heat) </li></ul><ul><li>At constant pressure  H =  E + P  V </li></ul><ul><li>The heat at constant pressure q p can be calculated from . . . </li></ul><ul><li> E = q p + w = q p - P  V (w = - P  V) </li></ul><ul><li>q p =  E + P  V =  H </li></ul>
18. 18. Calorimetry <ul><li>Measuring heat -- use a calorimeter. </li></ul><ul><li>Two kinds: Constant P or constant V </li></ul><ul><li>Constant pressure calorimeter (called a coffee cup calorimeter, open to outside) </li></ul><ul><li>Heat capacity for a material, C p is calculated. </li></ul><ul><li>C p = heat absorbed/  T =  H/  T </li></ul><ul><li>Specific heat capacity = Cp/mass </li></ul>
19. 19. Calorimetry <ul><li>Specific heat capacity = Cp/mass </li></ul><ul><li>Heat = specific heat x m x  T </li></ul><ul><li>q = m•∆T•C p </li></ul><ul><li>Sometimes q is shown as ∆H ∆H = m•∆T•C p (online HW) </li></ul><ul><li>Molar heat capacity = C p / moles </li></ul><ul><li>Heat = molar heat x moles x  T </li></ul><ul><li>Make the units work and you’ve done the problem right (get dates in college). </li></ul><ul><li>Be careful to note if C p uses grams or moles . </li></ul>
20. 20. Calorimetry <ul><li>A coffee cup calorimeter measures  H. </li></ul><ul><li>It’s an insulated cup, full of water. </li></ul>
21. 21. Figure 6.5 A Coffee-Cup Calorimeter Made of Two Styrofoam Cups
22. 22. Calorimetry <ul><li>Specific heat of H 2 O = 1 cal/gºC = 4.184 J/g o C </li></ul><ul><li>Heat of reaction=  H = sh x mass x  T (see this form on online HW) </li></ul><ul><li>Also seen as: q = m  T C p </li></ul><ul><li>See, Table 6.1 p. 237 (next page) for C p of some common substances. Note the units! (J/ºC•g) </li></ul>
23. 23. Z5e 237 Table 6.1
24. 24. Examples <ul><li>C p of graphite is 0.710 J/gºC. Calculate q needed to raise the temperature of 75.0 kg of it from 294 K to 348 K. . . </li></ul><ul><li>Convert Kg to g; q = m  TC p . Ans . . . </li></ul><ul><li>= 2 880 000 J = 2880 kJ </li></ul>
25. 25. Examples pp <ul><li>46.2 g of copper heated to 95.4ºC & placed in calorimeter with 75.0 g water at 19.6ºC. Final temp. of both is 21.8ºC. What is C p of Cu? . . . </li></ul><ul><li>q gained by Cu = q lost by H 2 O; set both equations equal to each other, solve . . . </li></ul><ul><li>q Cu = m cu ∆T Cu Cp Cu = m wat ∆T wat Cp wat = q wat Cp wat = 4.18 J/ºC•g (memorize). Use absolute value for ∆T; plug-and-chug to get answer of . . . </li></ul><ul><li>0.203 J/gºC for Cu </li></ul><ul><li>Don’t need to convert ºC to K since both have the same degree increments & we are only interested in the change . </li></ul><ul><li>Need to know this for online HW05 Q8 </li></ul>
26. 26. Calorimetry <ul><li>Constant volume calorimeter is called a bomb calorimeter. </li></ul><ul><li>Material is put in a container with pure oxygen. Wires are used to start the combustion. The container is put into a container of water. </li></ul><ul><li>The heat capacity of the calorimeter is known and tested. </li></ul><ul><li>Since  V = 0, P  V = 0,  E = q </li></ul>
27. 27. Bomb Calorimeter <ul><li>thermometer </li></ul><ul><li>stirrer </li></ul><ul><li>full of water </li></ul><ul><li>ignition wire </li></ul><ul><li>Steel bomb </li></ul><ul><li>sample </li></ul>
28. 28. Bomb Calorimetry Example pp <ul><li>T increases by 3.2ºC when 0.1964 g of quinone (C 6 H 4 O 2 ) is burned in a bomb calorimeter with heat capacity of 1.56 kJ/ºC. What is the combustion energy of quinone per gram and also per mole ? </li></ul><ul><li>Heat gained by calorimeter = 1.56kJ/ºC x 3.2 ºC = 5.0 kJ = heat loss by quinone </li></ul><ul><li>∆ E comb = -5.0kJ/0.1964 g = -25kJ/ g </li></ul><ul><li>∆ E comb = -25kJ/g x 108.09g/mol = -2700kJ/ mol </li></ul><ul><li>Text problem #56 is like this (HW) </li></ul><ul><li>(q = m∆TC p + heat gained by calorimeter) </li></ul>
29. 29. Properties <ul><li>Intensive properties: Not related to the amount of substance. </li></ul><ul><li>E.g. , density, specific heat, temperature. </li></ul><ul><li>Extensive property - does depend on the amount of stuff. </li></ul><ul><li>E.g., heat capacity, mass, heat from a reaction. </li></ul>
30. 30. Note on Hydrates pp <ul><li>A hydrate means a salt (formula unit) that has some water hooked onto it . </li></ul><ul><li>Copper(II) chloride di hydrate is . . . </li></ul><ul><li>CuCl 2 •2H 2 O </li></ul><ul><li>The M of the hydrate is 170.49 g/mol </li></ul><ul><li>The molar mass of the anhydride is 134.45 g/mol (170.49 less 2 waters) </li></ul><ul><li>You need to know this for the online HW </li></ul>
31. 31. More online HW notes pp <ul><li>HW05 Q4 - answer in kJ/mol of Z (not J/g). Find the q, then ÷ by # of moles. </li></ul><ul><li>HW05 Q2 & Q8 - use the examples just done today. Use internet for C p of Pb. </li></ul><ul><li>HW05 Q11 - look at the example problems in your assigned reading. </li></ul>
32. 32. Note on online HW pp <ul><li>When a question asks, “how much heat is liberated ,” your answer will be positive because there is no “negative” heat. </li></ul><ul><li>When a question asks, “what is the change in heat” then you have to indicate the change by a (+) or (-) sign. </li></ul><ul><li>When you use energy or heat in a mathematical equation (e.g., q = m∆TC p then you also have to show the sign. </li></ul>
33. 33. 6.3 Hess’s Law <ul><li>Enthalpy is a state function. </li></ul><ul><li>So, it is independent of the path. </li></ul><ul><li>We can add the pathway equations to come up with the desired final product, and add the pathway  H’s. </li></ul><ul><li>Two rules: </li></ul><ul><li>(1) If the reaction is reversed the sign of  H is changed . </li></ul><ul><li>(2) If the reaction is multiplied, so is  H </li></ul>
34. 34. N 2 2O 2 O 2 2NO 68 kJ 2NO 2 180 kJ -112 kJ H (kJ)
35. 35. Hints for using Hess’ Law pp <ul><li>When using Hess’s Law, work backwards from the required reaction to decide how to manipulate each reaction step so you can add them up to make it look like the answer. </li></ul><ul><li>Reverse any reactions as needed (change  H sign) & multiply the coefficients so the extra parts cancel . </li></ul><ul><li>Note: H 2 O( l ) will not cancel H 2 O( g )! </li></ul>
36. 36. Example of Hess’ Law pp <ul><li>Calculate ∆Hº f for C (s) + 2H 2(g)  CH 4(g) given: C (s) + O 2(g)  CO 2(g) ∆Hº c = -393.5 kJ/mol H 2(g) + 1/2 O 2(g)  H 2 O (l) ∆Hº c = -285.8 kJ/mol CH 4(g) + 2O 2(g)  CO 2(g) + 2H 2 O (l) ∆Hº c = - 890.8 kJ/mol </li></ul><ul><li>Need CH 4 as a product, so reverse the 3rd equation & change its ∆H sign. </li></ul>
37. 37. Example of Hess’ Law pp <ul><li>Calculate ∆Hº f for C (s) + 2H 2(g)  CH 4(g) given: C (s) + O 2(g)  CO 2(g) ∆Hº c = -393.5 kJ/mol H 2(g) + 1/2 O 2(g)  H 2 O (l) ∆Hº c = -285.8 kJ/mol CO 2(g) + 2H 2 O (l)  CH 4(g) + 2O 2(g) ∆Hº f = + 890.8 kJ/mol </li></ul><ul><li>H 2 O is not in the original equation so need to cancel it out. </li></ul><ul><li>But, Eq. 2 has only 1 water as a product while Eq. 3 has 2 waters as a reactant . </li></ul><ul><li>Multiply Eq. 2 by 2 ( including ∆Hº c ) </li></ul>
38. 38. Example of Hess’ Law pp <ul><li>Calculate ∆Hº f for C (s) + 2H 2(g)  CH 4(g) given: C (s) + O 2(g)  CO 2(g) ∆Hº c = -393.5 kJ/mol 2H 2 (g) + O 2(g)  2 H 2 O (l) ∆Hº c = ( 2 )(-285.8 kJ/mol) CO 2(g) + 2H 2 O (l)  CH 4(g) + 2O 2(g) ∆Hº f = +890.8 kJ/mol </li></ul><ul><li>Add the 3 equations and cancel like terms on both sides of the reaction. </li></ul>
39. 39. Example of Hess’ Law pp <ul><li>Calculate ∆Hº f for C (s) + 2H 2(g)  CH 4(g) given : C (s) + O 2(g)  CO 2(g) ∆Hº c = -393.5 kJ/mol 2H 2 (g) + O 2(g)  2 H 2 O (l) ∆Hº c = ( 2 )(-285.8 kJ/mol) CO 2(g) + 2H 2 O (l)  CH 4(g) + 2O 2(g) ∆Hº f = +890.8 kJ/mol </li></ul><ul><li>C (s) + 2H 2(g)  CH 4(g) ∆Hº f = -74.3 kJ </li></ul><ul><li>Since ∆H is (-) the reaction is exo thermic. </li></ul>
40. 40. Standard Enthalpy <ul><li>The enthalpy change for a reaction at standard conditions (25ºC, 1 atm , 1 M solutions). (Std T = 0 ºC when working with the gas laws!) </li></ul><ul><li>Symbol  Hº </li></ul>
41. 41. Example <ul><li>Given calculate  Hº for this reaction </li></ul> Hº= -394 kJ  Hº= -286 kJ  Hº= -1300. kJ Calculate  Hº for this reaction
42. 42. Example <ul><li>Given </li></ul> Hº= -394 kJ  Hº= -286 kJ  Hº= -1300. kJ Calculate  Hº for this reaction Ans. . . Reverse equation 1; multiply eq. 2 by 2 (include delta H) Ans. 226 kJ , so endo thermic
43. 43. Example Given Calculate  Hº for this reaction. Ans. . .  Hº= +77.9kJ  Hº= +495 kJ  Hº= +435.9kJ Cut all reactions in 1/2 Reverse #s 2 & 3 Add up to = -426 kJ = exo thermic (not -426.5 d/t SF)
44. 44. Standard Enthalpies of Formation <ul><li>Hess’s Law is much more useful if you know lots of reactions. </li></ul><ul><li>The amount of heat needed to form 1 mole of a compound from its elements in their standard states; e.g., H 2 O(l) v. H 2 O(g) </li></ul><ul><li>Standard states are 1 atm, 1 M and 25ºC </li></ul><ul><li> (Std T = 0ºC for gas laws like PV = nRT) </li></ul><ul><li>For an element  H f º = 0 (don’t forget!!) </li></ul><ul><li>See Table 6.2 p. 247; also Appendix 4 (pg A19) </li></ul>
45. 45. 6.4 Standard Enthalpies of Formation <ul><li>Need to be able to write the equations. </li></ul><ul><li>What is equation for formation of NO 2 ? </li></ul><ul><li>½N 2 (g) + O 2 (g)  NO 2 (g) because . . . </li></ul><ul><li>Have to make one mole to meet the definition. So, what coeff. for N 2 ? </li></ul><ul><li>Use 1/2 for coefficient of N 2 </li></ul><ul><li>1/2 N 2 (g) + O 2 (g)  NO 2 (g) </li></ul><ul><li>Why not N 2 (g) + 2 O 2 (g)  2 NO 2 (g) ? </li></ul><ul><li>Because by definition only looking at one mole of NO 2 </li></ul>
46. 46. Standard Enthalpies of Formation cont. <ul><li>Write equation for formation of methanol, CH 3 OH from its elements . Answer . . . </li></ul><ul><li>C(s) + 2H 2 (g) + 1/2 O 2 (g)  CH 3 OH </li></ul><ul><li>Note: heat of combustion is the reverse of heat of formation (sign also changes). This will be on the test!! </li></ul><ul><li>The above reaction showing heat of combustion instead of formation is . . . </li></ul><ul><li>CH 3 OH  C(s) + 2H 2 (g) + 1/2 O 2 (g) </li></ul>
47. 47. Since we can manipulate the equations <ul><li>We can use heats of formation to figure out the heat of reaction. </li></ul><ul><li>Let’s look at the schematic (see fig. 6.9, p. 248, next slide) . . . </li></ul>
48. 48. Figure 6.9. A Schematic Diagram of the Energy Changes for the Reaction CH 4 (g) + 2O 2 (g )  CO 2 (g ) + 2H 2 O (l)
49. 49. Since we can manipulate the equations <ul><li>We can use heats of formation to figure out the heat of reaction. </li></ul><ul><li>Lets do it with this equation. </li></ul><ul><li>C 2 H 5 OH (l) +3O 2(g)  2CO 2(g) + 3H 2 O (l) ( 2 • -394 + 3 • -286 ) - ( 1• -278 + 3 • 0 ) = -1368 kJ = exothermic </li></ul><ul><li>Oxygen = O since element; liquid water value is not same as gaseous water </li></ul><ul><li>Which leads us to this rule: </li></ul>
50. 50. Example 1 pp <ul><li>Using the standard enthalpies of formation listed in Table 6.2, p. 247, calculate the enthalpy change for the following reaction . . . Steps </li></ul><ul><li>4NH 3(g) + 7O 2(g)  4NO 2(g) + 6H2O (l) </li></ul>
51. 51. Example 1 pp <ul><li>4NH 3(g) + 7O 2(g)  4NO 2(g) + 6H2O (l) </li></ul><ul><li>Step 1 : Decompose NH 3(g) into its elements. 4NH 3(g)  2N 2(g) + 6H 2(g) </li></ul><ul><li>The reverse of this is the formation reaction for NH 3(g) 1/2N 2(g) + 3/2H 2(g)  NH 3(g)  H f º = - 46 kJ/mol </li></ul><ul><li>We need 4 times the reverse of the formation reaction for our equation above. </li></ul>
52. 52. Example 1 pp <ul><li>4NH 3(g) + 7O 2(g)  4NO 2(g) + 6H2O (l) </li></ul><ul><li>Step 2 : Since O 2(g) is in standard state its  H f º = 0. </li></ul><ul><li>We now have the elements N 2(g) , H 2(g) , and O 2(g) , which can be combined to form the products of the overall reaction as follows: . . . </li></ul>
53. 53. Figure 6.10 A Pathway for the Combustion of Ammonia
54. 54. Example 1 pp <ul><li>4NH 3(g) + 7O 2(g)  4NO 2(g) + 6H 2 O (l) </li></ul><ul><li>So far, we’ve figured  H f º for the reactants . </li></ul><ul><li>Now we do the same for the products . </li></ul><ul><li>Step 3 : Synthesis of NO 2 . We need 4 times the formation reaction (from Table 6.2). </li></ul><ul><li>Step 4 : Need 6 times formation reaction for H2O (l) ( not H 2 O ( g ) ). </li></ul>
55. 55. Example 1 cont. pp <ul><li>4NH 3(g) + 7O 2(g)  4NO 2(g) + 6H2O (l) </li></ul><ul><li>Use: </li></ul><ul><li>[ 4x(34kJ) + 6x(-286 kJ) ] - [ 4x(-46kJ) ] = -1396 kJ . </li></ul>
56. 56. Calculating  H o reaction using standard enthalpies of formation - Values for  H o f are looked up in tables such as Appendix 4 in your book
57. 57. Calculate  H o rxn for CaCO 3(s)  CaO (s) + CO 2(g) compound  H o f (kJ/mol) CaCO 3(s) -1206.9 CaO (s) -635.1 CO 2(g) -393.5 [ -635.1 + -393.5 ] - [ -1206.9 ] = 178.3 = endothermic
58. 58. Calculation of  H o <ul><li> H o =  H f o products -  H f o reactants </li></ul>
59. 59. Example What is the value of  H rx for the reaction: 2 C 6 H 6(l) + 15 O 2(g)  12 CO 2(g) + 6 H 2 O (g) <ul><li>C 6 H 6(l)  H f o = + 49.0 kJ/mol </li></ul><ul><li>Use above plus Appendix 4, p. A19 to get . . . </li></ul><ul><li>O 2(g)  H f o = 0 kJ/mol </li></ul><ul><li>CO 2(g)  H f o = - 393.5 kJ/mol </li></ul><ul><li>H 2 O (g)  H f o = - 241.8 kJ/mol </li></ul><ul><li> H rx   H f o  product -   H f o  reactants </li></ul>
60. 60. Example What is the value of  H rx for the reaction: 2 C 6 H 6(l) + 15 O 2(g)  12 CO 2(g) + 6 H 2 O (g) from Appendix 4 p. A21ff C 6 H 6(l)  H f o = + 49.0 kJ/mol; O 2(g)  H f o = 0 CO 2(g)  H f o = - 393.5; H 2 O (g)  H f o = - 241.8  H rx   H f o  product -   H f o  reactants <ul><li> H rx  - 393.5) + 6(- 241.8)  product </li></ul><ul><li>-  2(+ 49.0 ) + 15(0)  reactants kJ </li></ul><ul><li>= - 6.27 x 10 3 kJ </li></ul>
61. 61. Nitroglycerine is a powerful explosive, giving four different gases when detonated C 3 H 5 (NO 3 ) 3(l)  3 N 2(g) + O 2(g) + 6 CO 2(g) + 5 H 2 O (g) Given that  H o f for nitroglycerine is -364 kJ/mol, and consulting a table for the enthalpies of formation of the other compounds, calculate the enthalpy change when 10.0 g of nitroglycerine is detonated.  H o rxn and Stoichiometry Find enthalpy for 1 mole of NTG (use previous method) Then convert 10.0 g NTG to moles and use mole ratio between that and 1 mole. 10g ÷ 227 g/mol = 0.044 mol
62. 62. C 3 H 5 (NO 3 ) 3(l)  3 N 2(g) + O 2(g) + 6 CO 2(g) + 5 H 2 O (g) calculate ∆H rxn when 10.0 g of NTG is detonated.  H o rxn and Stoichiometry [6•-394 + 5•-242] - [-364] = -3.21 x 10 3 kJ/mol This is for 1 mol. Have 10.0 g, convert to moles 10g ÷ 227 g/mol = 0.044 mol Then 0.044 mol • -3.21 x 10 3 kJ = -141 kJ exothermic
63. 63. Note on online HW pp <ul><li>When a question asks, “how much heat is liberated ,” your answer will be positive because there is no “negative” heat. </li></ul><ul><li>When a question asks, “what is the change in heat” then you have to indicate the change by a (+) or (-) sign. </li></ul><ul><li>When you use energy or heat in a mathematical equation (e.g., q = m∆TC p then you also have to show the sign. </li></ul>