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 = CpmΔT
1
Emitted power (W)
Tλmax = C2
= (constan 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)
C m T
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 = CpmΔT
1
Emitted power (W)
Tλmax = C2
= (constan 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)
C m T
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 = CpmΔT
1
Emitted power (W)
Tλmax = C2
= (constan 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
2 O
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)
C m T
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.