AP CHEMISTRY WKST: THERMOCHEMISTRY REVIEW
1)
The black ball has more PE than the white ball in the beginning. At the end, the white ball has more PE than the
black ball, but it doesn’t have as much PE as the black ball did in the beginning. The reason the white ball has more
PE than the black ball at the end is due to PE being transferred from the black to the white ball in the form of work
when the black ball collided with the white one and made it move to a new position. The white ball ended up with
less PE than the black ball initially had because both balls lost PE in the form of heat due to friction as they rolled
along the ground.
2)
Since both balls have the same PE in both scenarios, ΔE for both are the same. The black ball must have lost more
PE in ending #2 to friction than in ending #1. Consequently, it didn’t do any work on the white ball in ending #2 since
the white ball didn’t move. To summarize:
-Ending #1 had more work done than ending #2.
-Since ΔE is the same for both scenarios, more heat is lost to friction in ending #2 than ending #1 since ΔE = q + w
and w is greater in ending #1.
3)
ΔE = q + w
q = +120.0 J
w = −145 J
ΔE = +120.0 J – 145 J = −25 J
4)
ΔE = q + w
q = − 525 J (q = ΔH at constant P)
w = −PΔV
P = 2225 torr (
1 atm
) = 2. 2 atm
0 torr
w = −(2. 2 atm)(0. 50 L – 5.00 L) = −(2. 2 atm)(−4.15 L) = +12.1512 L∙atm
101. J
w = 12.1512 L∙atm (
) = +12 0. 1 J = +12 0 J
1 L∙atm
ΔE = q + w = −525 J + 12 0 J = + 05 J = +710 J
5)
ΔE = q + w
q = mCΔT
q = (30.0 mol)(20.8
J
mol∙ C
)(65.0°C – 25.0°C) = (30.0 mol)(20.8
J
mol∙ C
)(40.0°C) = +24960 J = +25.0 kJ
w = −PΔV
w = −(1.250 atm)(1 00. L – 1040. L) = −(1.250 atm)(
w=
101. J
50. L∙atm (
) =
1 L∙atm
2 5J=
0. L) = − 50. L∙atm
.2 kJ
ΔE = +25.0 kJ – 96.2 kJ = −71.2 kJ
1
AP CHEMISTRY WKST: THERMOCHEMISTRY REVIEW
6)
ΔE = +19.2 kJ
q = mCΔT
q = (40.2 mol)( 20.8
J
mol∙ C
)(62.5°C – 25.0°C) = (40.2 mol)( 20.8
J
mol∙ C
)(37.5°C) = +31356 J = +31.4 kJ
ΔE = q + w  w = ΔE – q
w = +19.2 kJ – 1.4 kJ = −12.2 kJ = −12200 J
w = −PΔV

w = 12200 J (
P=
w = −P(Vf – Vi)
1 L∙atm
) = 120. L∙atm
101. J
00. torr (
1 atm
) = 1.05 atm
0 torr
−120. L∙atm = −(1.05 atm)(1125 L – Vi)
114 L = 1125 L – Vi
Vi = 1011 L
7)
−mmCmΔTm = mwCwΔTw + CcalΔTw
mwCwΔTw + CcalΔTw
Cm = [
]
mmΔTm
(100.0 g)(4.1 4 gJC)(4 .2 C 22.5 C) + (22.2 JC)(4 .2 C 22.5 C)
Cm = [
]
( 5.00 g)(4 .2 C 5.0 C)
(100.0 g)(4.1 4 gJC)(20. C) + (22.2
Cm = [
( 5.00 g)( 51. C)
Cm =
8)
0.
q = mCΔT
J + 45 .54 J
=
440 g C

J
)(20.
C
C)
]
120.42 J
=
440 g C
q = mC(Tf – Ti)
+900.0 J = (250.0 g)(4.184 gJC)(Tf – 23.0°C)
0.8604°C = Tf – 23.0°C
Tf = 23.8604°C = 23.9°C
2
AP CHEMISTRY WKST: THERMOCHEMISTRY REVIEW
9)
−mHCHΔTH = mCCCΔTC
Since CH = CC, these can be dropped from the above equation.
mH (TfH TiH ) = mC (TfC TiC )
Since the Tf is the same for both water samples, the resulting equation is as follows.
−(500.0 g)(Tf – 82.00°C) = (400.0 g)(Tf – 23.50°C)
−500.0 Tf + 41000 = 400.0 Tf – 9400
50400 = 900.0 Tf
Tf = 56.00°C
10)
q = mCΔT
q = (25.0 g)(0.755 gJC)(30.0°C) = 566.25 J = 566 J
11)
Answer: Al
The higher the specific heat of a substance, the more heat it can absorb per degree. Thus, if Al and Fe both absorb
the same amount of heat, the Al wouldn’t get as hot as Fe would.
12)
2
KE = ½ mv
−1 2
KE = ½ (4500 kg)(1.79 ms ) = 7209.225 J = 7200 J
1 cal
200 J (
) = 1 20. 41 cal =
4.1 4 J
13)
q = mCΔT
q = (580.00 g)(4.184 gJC)(58.47°C – 23.00°C) = (580.00 g)(4.184
(
.0 kJ
40.00 g a H
)(
) =
0.00 g a H
1 mol a H
3
J
g C
)(35.47°C) = +86075.7584 J = +86.08 kJ
AP CHEMISTRY WKST: THERMOCHEMISTRY REVIEW
14)
Mg(s) + N2(g) + 3O2(g) → Mg(
3)2(s)
8Mg(s) + Mg(NO3)2(s) → Mg3N2(s) + 6MgO(s)
Mg3N2(s) → 3Mg(s) + N2(g)
2Mg (s) → 2Mg(s) + O2(g)
ΔH = −
4 kJ
ΔH = +4 kJ
ΔH = +120 kJ
Mg3N2(s) + 6MgO(s) → 8Mg(s) + Mg(NO3)2
3Mg(s) + N2(g) → Mg3N2(s)
6Mg(s) + 3O2(g) → 6MgO(s)
ΔH = +
4 kJ
ΔH = −463 kJ
ΔH = −3609 kJ
Mg(s) + N2(g) + 3O2(g) → Mg(NO3)2(s)
15)
2N2(g) + 5O2(g) → 2
ΔH = −188 kJ
2O5(g)
2H2(g) + O2(g) → 2H2O(l)
N2O5(g) + H2 (l) → 2HNO3(l)
1/2 N2(g) + 3/2 O2(g) + 1/2 H2(g) → HNO3(l)
ΔH = −5 1.5 kJ
ΔH = − . kJ
ΔH = −1 4 kJ
2N2(g) + 5 6O2(g) + 2H2(g) → 4HNO3(l)
4HNO3(l) → 2N2O5(g) + 2H2O(l)
2H2O(l) → 2H2(g) + O2(g)
2N2(g) + 5O2(g) → 2N2O5(g)
16)
17)
3Fe2O3(s) + C (g) → 2Fe3O4(s) + CO2(g)
a)
100.0 g Fe2
b)
500.0 kJ (
(
1 mol Fe2
15 . 0 g Fe2
2 mol Fe
1 kJ
CH4(g) + 2O2(g) → C
a)
4
2(g)
)(
)(
+1 kJ
mol Fe2
2 1.55 g Fe
1 mol Fe
+ 2H2 (l)
ΔH = +1
4
ΔH = −
kJ
ΔH = +15 .2 kJ
ΔH = +5 1.5 kJ
ΔH = +2 . kJ = +29 kJ
kJ
) =
) = 1
5.
4
g Fe
4
=
ΔH = − 0 kJ
heat absorbed by water: q = mCΔT
q = (500.0 g)(4.184 gJC)(99.8°C – 22.0°C) = (500.0 g)(4.184 gJC)(77.8°C) = +162757.6 J = +163 kJ
1
kJ (
1 mol CH4 1 .05 g CH4
)(
) =
0 kJ
1 mol CH4
b)
V=
L∙atm
( .2 g)(0.0 20 mol∙
)(2 .15 )
g T
=
=
g
P(MM)
(1 atm)(1 .05 mol
)
4
AP CHEMISTRY WKST: THERMOCHEMISTRY REVIEW
18)
q = mCΔT
C=
q
mΔT
C=
(45. g)(
20
J
(
gC
1
0.12
19)
.2 J
.2 J
=
=
. C 20.0 C)
(45. g)(1 . C)
. 2 g Pb
) =
mol Pb
q = CcalΔT
q = (2.561 kJC)(24.75°C – 23.00°C) = (2.561 kJC)(1.75°C) = 4.48 kJ
4.4 kJ
=
0. 2 5 g
5.42
20)
kJ
g
(
22.14 g C12 H22
1 mol C12 H22 1
ΔH = ∑ ΔHf n
1
ΔHf
kJ
−427mol
=
∑ ΔHf nreactants
roducts
2NaOH(s) + CO2(g) →
a)
kJ
mol
) = 1 4
kJ
.5 mol
−
a2CO3(s) + H2O(g)
kJ
−11 1 mol
kJ
−242 mol
kJ
kJ
ΔH = [1 mol(−11 1 mol
) + 1 mol (−242 mol
)] – [2 mol(−42
ΔH = (−11 1 kJ – 242 kJ) – (− 54 kJ – 393.5 kJ)
ΔH = −126 kJ
kJ
mol
) + 1 mol(−
kJ
.5 mol
)]
2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(s)
b)
ΔHf
0
kJ
mol
− 2
−1
kJ
mol
0
kJ
ΔH = [1 mol(−1
)] – [1 mol(− 2
mol
ΔH = −1
kJ + 2 kJ
ΔH = −850. kJ
10N2O(g) + C3H8(g) → 10
c)
ΔHf
kJ
+82 mol
kJ
−104 mol
2(g)
kJ
mol
)]
+ 3CO2(g) + 4H2O(g)
0
−
kJ
.5 mol
kJ
−242 mol
kJ
kJ
kJ
kJ
ΔH = [ mol(−
.5 mol
) + 4 mol(−242 mol
)] – [10 mol(82 mol
) + 1 mol(−104 mol
)]
ΔH = (−11 0.5 kJ – 968 kJ) – (820. kJ – 104 kJ)
ΔH = −2 4.5 kJ = −2864 kJ
3Cu(s) + 8HNO3(aq) → Cu(
d)
ΔHf
0
−20
kJ
mol
−
3)2(aq)
kJ
. 4 mol
+ 2NO(g) + 4H2O(l)
kJ
+90. mol
−2
kJ
kJ
ΔH = [ mol(− . 4 mol
) + 2 mol(90. mol
) + 4 mol(−2
ΔH = (−250. 2 kJ + 1 0. kJ – 1144 kJ) – (−1656 kJ)
ΔH = +441 kJ
5
kJ
mol
kJ
mol
)] – [ mol(−20
kJ
mol
)]