Quiz8KEY - Chem 6A 2011(Sailor Name Student ID Number...

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Unformatted text preview: Chem 6A 2011 (Sailor) Name: Student ID Number: Section Number: QUIZ #8 VERSION A KEY Some useful constants and relationships: Specific heat capacities (in J/g.K): H2O(l) = 4.184; Al(s) = 0.900; Cu(s) = 0.387; Steel(s) = 0.45 1 atm = 760 Torr 1J = 1kg.m2/s2 1 eV = 1.6022 x 10-19J 101.325 J = 1 L.atm . . -1 . -1 . -1 . -1 R = Ideal gas constant: 0.08206 L atm mol K = 8.31451 J mol K Avogadro constant: 6.022 x 1023 mole-1 Planck's constant = h = 6.6261 x 10-34 J.s 8 RH = 1.097 x 10-2 nm-1 c = speed of light: 3.00 x 10 m/s -2 . C2 = second radiation constant = 1.44 x 10 K m q = C pm Δ T 1 Emitted power (W) Tλmax = C2 = (cons tan t )T 4 e = mc 2 c = λν 5 Surface area (m2 ) ⎛ྎ 1 ⎛ྎ Z 2 ⎞ྏ 1 ⎞ྏ hc 1 = R H ⎜ྎ 2 − 2 ⎟ྏ E= E (in Joules) = −2.18 × 10−18 ⎜ྎ 2 ⎟ྏ E = hν λ λ ⎝ྎ n ⎠ྏ ⎝ྎ n1 n 2 ⎠ྏ € Chem 6A 2011 (Sailor) QUIZ #8 1. Acetylene gas burns in oxygen according to the equation shown below. Using the following table of bond energies, calculate ΔH°rxn for the reaction. Set up but do not solve; circle your final answer. (25 pts) Average Bond Energies, kJ/mol Bond H -H C ≡O C -H C -C Energy Bond 432 C=O (for CO2) 1070 O -O 413 C=C 347 C ≡C O=O 498 O -H This was a variant of homework problem 9.57 in the text Energy 799 204 614 839 467 ΔH°rxn = (all bond energies in reactants) – (all bond energies in products) = 2(C-H) + 1(C≡C) + 5/2(O=O) - 4(C=O) - 2(O-H) = 2(413) + 1(839) + 5/2(498) - 4(799) - 2(467) 10 points if they indicate the answer = (all bond energies in reactants) – (all bond energies in products) 5 points if they have all the right bonds in there even if the coefficients are incorrect (i.e., no 5/2 in front of O=O, etc) 10 points if they have all the right bonds in there AND if the coefficients are all correct (i.e., 5/2 in front of O=O, etc); thus the answer would correctly calculate to -1220 kJ/mol. Most common mistake will be to use 2(C=O) instead of 4(C=O). 2. Calculate the final temperature when a steel bolt, initially at 100.0 °C and weighing 50.0 g, is placed in 37 g of liquid water, initially at 11.0 °C. The relevant thermochemical data are given on the cover page of this quiz. Set up but do not solve; circle your answer. (25 pts) this was a variation of problems 6.26 and 6.28 in the text. Repeat problem from quiz 5. Chem 6A 2011 (Sailor) QUIZ #8 heat lost by bolt = - (heat gained by water), Tfinal is the same for both applying the expression q = C p mΔT to both, Cbolt mbolt ΔT = −Cwater mwater ΔT Cbolt mbolt (Tfinal − Tbolt ) = −Cwater mwater (Tfinal − Twater ) multiplying through, Cbolt mbolt Tfinal − Cbolt mbolt Tbolt = −Cwater mwaterTfinal + Cwater mwaterTwater collect terms, simplify, Cbolt mbolt Tfinal + Cwater mwaterTfinal = Cwater mwaterTwater + Cbolt mbolt Tbolt Tfinal (Cbolt mbolt + Cwater mwater ) = Cwater mwaterTwater + Cbolt mbolt Tbolt solve for Tfinal + Cbolt mbolt Tbolt (4.184 × 37 × 11) + (0.45 × 50 × 100) CmT Tfinal = water water water = (0.45 × 50 + 4.184 × 37) Cbolt mbolt + Cwater mwater € € Numerically this works out to 22.3 °C. 25 pts for correct answer. Partials: Give them 5 pts if they recognize that heat lost = - heat gained but solved it incorrectly; Give them 5 more points (total 10) if they get: Cbolt mbolt (Tfinal − Tbolt ) = −Cwater mwater (Tfinal − Twater ) But can’t solve it algebraically. Chem 6A 2011 (Sailor) Name: Student ID Number: Section Number: QUIZ #8 VERSION B KEY Some useful constants and relationships: Specific heat capacities (in J/g.K): H2O(l) = 4.184; Al(s) = 0.900; Cu(s) = 0.387; Steel(s) = 0.45 1 atm = 760 Torr 1J = 1kg.m2/s2 1 eV = 1.6022 x 10-19J 101.325 J = 1 L.atm . . -1 . -1 . -1 . -1 R = Ideal gas constant: 0.08206 L atm mol K = 8.31451 J mol K Avogadro constant: 6.022 x 1023 mole-1 Planck's constant = h = 6.6261 x 10-34 J.s 8 RH = 1.097 x 10-2 nm-1 c = speed of light: 3.00 x 10 m/s -2 . C2 = second radiation constant = 1.44 x 10 K m q = C pm Δ T 1 Emitted power (W) Tλmax = C2 = (cons tan t )T 4 e = mc 2 c = λν 5 Surface area (m2 ) ⎛ྎ 1 ⎛ྎ Z 2 ⎞ྏ 1 ⎞ྏ hc 1 = R H ⎜ྎ 2 − 2 ⎟ྏ E= E (in Joules) = −2.18 × 10−18 ⎜ྎ 2 ⎟ྏ E = hν λ λ ⎝ྎ n ⎠ྏ ⎝ྎ n1 n 2 ⎠ྏ € Chem 6A 2011 (Sailor) QUIZ #8 1. Acetylene gas burns in oxygen according to the equation shown below. Using the following table of bond energies, calculate ΔH°rxn for the reaction. Set up but do not solve; circle your final answer. (25 pts) Average Bond Energies, kJ/mol Bond H -H C ≡O C -H C -C Energy Bond 433 C=O (for CO2) 1071 O -O 414 C=C 348 C ≡C O=O 499 O -H This was a variant of homework problem 9.57 in the text Energy 800 205 615 840 468 ΔH°rxn = (all bond energies in reactants) – (all bond energies in products) = 2(C-H) + 1(C≡C) + 5/2(O=O) - 4(C=O) - 2(O-H) = 2(414) + 1(840) + 5/2(499) - 4(800) - 2(468) 10 points if they indicate the answer = (all bond energies in reactants) – (all bond energies in products) 5 points if they have all the right bonds in there even if the coefficients are incorrect (i.e., no 5/2 in front of O=O, etc) 10 points if they have all the right bonds in there AND if the coefficients are all correct (i.e., 5/2 in front of O=O, etc); thus the answer would correctly calculate to -1220.5 kJ/mol. Most common mistake will be to use 2(C=O) instead of 4(C=O). 2. Calculate the final temperature when a steel bolt, initially at 90.0 °C and weighing 40.0 g, is placed in 47.0 g of liquid water, initially at 14.0 °C. The relevant thermochemical data are given on the cover page of this quiz. (25 pts)this was a variation of problems 6.26 and 6.28 in the text Chem 6A 2011 (Sailor) QUIZ #8 heat lost by bolt = - (heat gained by water), Tfinal is the same for both applying the expression q = C p mΔT to both, Cbolt mbolt ΔT = −Cwater mwater ΔT Cbolt mbolt (Tfinal − Tbolt ) = −Cwater mwater (Tfinal − Twater ) multiplying through, Cbolt mbolt Tfinal − Cbolt mbolt Tbolt = −Cwater mwaterTfinal + Cwater mwaterTwater collect terms, simplify, Cbolt mbolt Tfinal + Cwater mwaterTfinal = Cwater mwaterTwater + Cbolt mbolt Tbolt Tfinal (Cbolt mbolt + Cwater mwater ) = Cwater mwaterTwater + Cbolt mbolt Tbolt solve for Tfinal + Cbolt mbolt Tbolt (4.184 × 47 × 14) + (0.45 × 40 × 90) CmT Tfinal = water water water = (0.45 × 40 + 4.184 × 47) Cbolt mbolt + Cwater mwater € € 25 pts for correct answer. Give them 5 pts if they recognize that heat lost = - heat gained but solved it incorrectly; Give them 5 more point (total 10) if they get: Cbolt mbolt (Tfinal − Tbolt ) = −Cwater mwater (Tfinal − Twater ) But can’t solve it algebraically. Chem 6A 2011 (Sailor) Name: Student ID Number: Section Number: QUIZ #8 VERSION C KEY Some useful constants and relationships: Specific heat capacities (in J/g.K): H2O(l) = 4.184; Al(s) = 0.900; Cu(s) = 0.387; Steel(s) = 0.45 1 atm = 760 Torr 1J = 1kg.m2/s2 1 eV = 1.6022 x 10-19J 101.325 J = 1 L.atm . . -1 . -1 . -1 . -1 R = Ideal gas constant: 0.08206 L atm mol K = 8.31451 J mol K Avogadro constant: 6.022 x 1023 mole-1 Planck's constant = h = 6.6261 x 10-34 J.s 8 RH = 1.097 x 10-2 nm-1 c = speed of light: 3.00 x 10 m/s -2 . C2 = second radiation constant = 1.44 x 10 K m q = C pm Δ T 1 Emitted power (W) Tλmax = C2 = (cons tan t )T 4 e = mc 2 c = λν 5 Surface area (m2 ) ⎛ྎ 1 ⎛ྎ Z 2 ⎞ྏ 1 ⎞ྏ hc 1 = R H ⎜ྎ 2 − 2 ⎟ྏ E= E (in Joules) = −2.18 × 10−18 ⎜ྎ 2 ⎟ྏ E = hν λ λ ⎝ྎ n ⎠ྏ ⎝ྎ n1 n 2 ⎠ྏ € Chem 6A 2011 (Sailor) QUIZ #8 1. Acetylene gas burns in oxygen according to the equation shown below. Using the following table of bond energies, calculate ΔH°rxn for the reaction. Set up but do not solve; circle your final answer. (25 pts) H C C H + 5/2 O 2O O C O + O H H Average Bond Energies, kJ/mol Bond H -H C ≡O C -H C -C Energy Bond 434 C=O (for CO2) 1072 O -O 415 C=C 349 C ≡C O=O 500 O -H This was a variant of homework problem 9.57 in the text Energy 801 206 616 841 469 ΔH°rxn = (all bond energies in reactants) – (all bond energies in products) = 2(C-H) + 1(C≡C) + 5/2(O=O) - 4(C=O) - 2(O-H) = 2(415) + 1(841) + 5/2(500) - 4(801) - 2(469) 10 points if they indicate the answer = (all bond energies in reactants) – (all bond energies in products) 5 points if they have all the right bonds in there even if the coefficients are incorrect (i.e., no 5/2 in front of O=O, etc) 10 points if they have all the right bonds in there AND if the coefficients are all correct (i.e., 5/2 in front of O=O, etc); thus the answer would correctly calculate to -1221 kJ/mol. Most common mistake will be to use 2(C=O) instead of 4(C=O). 2. Calculate the final temperature when a steel bolt, initially at 80.0 °C and weighing 30.0 g, is placed in 27.0 g of liquid water, initially at 15.3 °C. The relevant thermochemical data are given on the cover page of this quiz. (25 pts)this was a variation of problems 6.26 and 6.28 in the text Chem 6A 2011 (Sailor) QUIZ #8 heat lost by bolt = - (heat gained by water), Tfinal is the same for both applying the expression q = C p mΔT to both, Cbolt mbolt ΔT = −Cwater mwater ΔT Cbolt mbolt (Tfinal − Tbolt ) = −Cwater mwater (Tfinal − Twater ) multiplying through, Cbolt mbolt Tfinal − Cbolt mbolt Tbolt = −Cwater mwaterTfinal + Cwater mwaterTwater collect terms, simplify, Cbolt mbolt Tfinal + Cwater mwaterTfinal = Cwater mwaterTwater + Cbolt mbolt Tbolt Tfinal (Cbolt mbolt + Cwater mwater ) = Cwater mwaterTwater + Cbolt mbolt Tbolt solve for Tfinal + Cbolt mbolt Tbolt (4.184 × 27 × 15.3) + (0.45 × 30 × 80) CmT Tfinal = water water water = (0.45 × 30 + 4.184 × 27) Cbolt mbolt + Cwater mwater € € 25 pts for correct answer. Give them 5 pts if they recognize that heat lost = - heat gained but solved it incorrectly; Give them 5 more points (total 10) if they get: Cbolt mbolt (Tfinal − Tbolt ) = −Cwater mwater (Tfinal − Twater ) But can’t solve it algebraically. ...
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This note was uploaded on 11/21/2011 for the course CHEM 6A 6A taught by Professor Sailor during the Fall '11 term at UCSD.

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