Download C2H5OH + 3 O2 → 2 CO2 + 3 H2O + HEAT Q = mc ∆T

UNIT 1: THERMODYNAMICS : Energy Changes in Chemical Reactions (Answers)
1:1. Write a balanced equation for the combustion of ethanol ( C2H5OH ). What would be a third product of this reaction ?
C2H5OH + 3 O2
→ 2 CO2
+ 3 H2O
+
HEAT
Calorimetry
1:2 a. Attempt question # 6.40 on page 269 in your text
Q
= m c ∆T
= 0.175 kg * 4.184 kJ/kgK * 10 K
= 7.3 kJ
b. Attempt question # 6.41 on page 269 in your text
Q
c.
= m c ∆T
= 1.0 kg * 4.184 kJ/kgK * 74 K
= 3.1 x 102 kJ
Attempt question # 6.51 on page 269 in your text
Q
= m c ∆T
= 2.46 kg * 4.184 kJ/kgK * 2.07 K
= 21.3 kJ
1:3. If a concrete wall of mass 4.5 x 107 g reached a temperature of 38°C during the day, how much energy would it
release if its temp dropped to 18°C that night ? (c= 3.340 kJ.kg-1K-1 for concrete)
Q
= m c ∆T
= 4.5 x 104 kg * 3.340 kJ/kgK * 20 K
= 3.0 x 106 kJ
1:4. Attempt question # 6.46 on page 269 in your text.
Q
= 1.0 lb * 454 g/lb * 9.0 kCal/g * 0.85
= 3473 kCal released per lb of fat
# miles to run = 3476 kCal / 500 * 8
= 55.6 miles
Enthalpy Changes; ( 6.6 pg 253)
1:5. a. Attempt question # 6.14 on page 268 in your text.
Any property that does not depend on a systems history or future is called
a state function
EX: a systems temperature & enthalpy are state functions
b. Attempt question # 6.32 on page 268 in your text.
The 2 conditions necessary are: you must use only elements in the
standard states and you must produce only 1 mole of product = ∆H°f
1:6. Given * 3 C (s) + 2 Fe2O3 (s) + 462 kJ → 4 Fe (s) + 3 CO2 (g)
Rewrite this equation using 1 mole of carbon and ∆H notation
1C
+ 2/3 Fe2O3
→ 4/3 Fe
+ 1 CO2
1:7. Attempt questions # 6.56 – 6.59 on page 270 in your text.
56) given: 3 CO (g) + 3/2 O2 (g) → 3 CO2 (g)
a)
2 CO (g) + 2/2 O2 (g) → 2 CO2 (g)
b)
1 CO (g) + 1/2 O2 (g) → 1 CO2 (g)
∆ H = + 154 kJ
∆H = -849 kJ
∆H = -566 kJ
∆H = -283 kJ
57) 1 NH3 + 7/4 O2 → NO2 + 3/2 H2O
∆H = -283 kJ
2/3 NH3 + 7/6 O2 → 2/3 NO2 + 1H2O
58) Al
1/2 Fe2O3
+
4 Al + 2 Fe2O3
∆H = -188.7 kJ
→ 1/2 Al2O3 + Fe
→ 2 Al2O3
∆ H = - 426.9 kJ / mol of Al
+ 4 Fe
59) 1 C6H6 + 7.5 O2 → 6 CO2 + 3 H2O
3 C6H6 + 22.5 O2 → 18 CO2 + 9 H2O
∆ H = 4 (- 426.9 ) kJ
= - 1 707 kJ
∆H = –3271 kJ
∆H = –9813 kJ
1:8. Calcium is burnt according to the following eq’n Ca (s) + 1/2 O2 (g) → CaO (s) ∆H= –636 kJ/mol Ca
How much heat would be produced by burning 12 g of calcium ?
1 mole of calcium → (produces) 636 kJ of heat ( from the rx eq'n )
so 40 g(1 mole) of Ca → 636 kJ of heat
&
12 g of Ca
→ x kJ
x
= 636 kJ x 12 g / 40 g
= 190.8 kJ would be produced from 12 g of Ca
1:9. a. Attempt question # 6.60 on page 270 in your text.
1 mole of Mg
so 24.3 g of Mg
& 6.54 g of Mg
x
→ 601.5 kJ of heat ( from the rx eq'n )
→ 601.5 kJ of heat
→ x kJ
= 161.9 kJ would be produced from 6.54 g of Mg
b. Attempt question # 6.61 on page 270 in your text.
1 mole of CH3OH → 599.5 kJ of heat ( from the rx eq'n )
so 32 g of CH3OH → 599.5 kJ of heat
& 46.0 g of CH3OH → x kJ
x = 861.8 kJ would be produced from 6.54 g of CH3OH
1:10. Write the equation for the formation of the following, include ∆H°f .
a) HC2H3O2 b) H2SO4 c) NaNO3 d) Ba(ClO3)2
a) 2 C (s) + 2 H2 (g) + O2 (g) →
HC2H3O2 (l)
∆H°f = - 487 kJ
H2SO4
∆H°f = - 811 kJ
b)
S (s) + H2 (g) + 2 O2 (g) →
c)
Na (s) + 1/2 N2 (g) + 3/2 O2 (g) → NaNO3 (s)
∆H°f = - 425 kJ
d) Ba (s) + Cl2 (g) + 3 O2 (g) → Ba (ClO3)2 (s)
∆H°f = - 760 kJ
1: 11. Attempt question # 6.76 & 6.77 on page 271 in your text.
76) * only (b) – you need elements & 1 mole product
77) * only (c) – you need elements & 1 mole product
1:12. Calculate ∆Hrx for :
a) 2 Fe (s) + O2 (g) →2 FeO (s)
a) 2 Fe (s) + O2 (g) → 2 FeO (s)
∆Hrx = 2 (-267 ) kJ = -534 kJ
→ N2 (g) + 3 H2 (g)
b) 2 NH3 (g)
b) 2 NH3 (g) → N2 (g) + 3 H2 (g)
∆Hrx = - 2 ( -46 ) kJ = + 92 kJ
Hess's Law of Heat Summation ( 6.7 pg 254)
1:13.
Attempt question # 6.65 & 6.66 on page 270 in your text.
65)
66)
2 KCl + 2 H2O → 2 KOH + 2 HCl
∆H= + 2(203.6) kJ
H2SO4 + 2 KOH → K2SO4 + 2 H2O
∆H = –342.4 kJ
2 KCl + H2SO4 → 2 HCl + K2SO4
∆H = + 64.8 kJ
NaNO2 → 1/2 Na2O + 1/2 NO2 + 1/2 NO
∆ H = 213.6 kJ
HCl + 1/2 Na2O → 1/2 H2O + NaCl
∆H = -253.7 kJ
1/2 N2O + 1/2 O2 + 1/2 H2O → HNO2
∆H = -17.2 kJ
1/2 NO + 1/2 NO2 → 1/2 N2 O + 1/2 O2
∆H = -21 .3 kJ
NaNO2
1:14.
69)
+ HCl → HNO2 +
CaO
+ 2 HCl → CaCl2
+ H2 O
∆H = -186 kJ
→ H2O + CaO
∆H = + 62.3 kJ
Ca(OH)2 (aq) → Ca(OH)2 (s)
∆H = + 12.6 kJ
Ca(OH)2 (aq) + 2 HCl → CaCl2 + 2 H2O
∆H = - 111.1 kJ
CaO + Cl2 → CaOCl2
∆H = –110.9 kJ
H2O + CaOCl2 + 2 NaBr → 2 NaCl + Ca(OH)2 + Br2
∆H = –60.2 kJ
Ca(OH)2
71)
∆H = -79.2 kJ
Attempt question # 6.69, 6.70, & 6.71 on page 271 in your text.
Ca(OH)2 (s)
70)
NaCl
→ CaO + H2O
∆H = +65.1 kJ
Cl2 + 2 NaBr → 2 NaCl + Br2
∆H = –106 kJ
1/2 Cl2 + NaBr → NaCl + 1/2 Br2
∆H = –53 kJ
Cu + 1/2 S → 1/2 Cu2S
∆H = -39.75 kJ
1/2 SO2 → 1/2 S + 1/2 O2
∆H = + 148.5 kJ
1/2 Cu2S + O2 → CuO + 1/2 SO2
∆H = –263.75
Cu + 1/2 O2 →
CuO
∆H = - 155 kJ
Calculating ∆Hrx using the ∑ method ( 6.9 pg 262)
1:17.
Attempt question # 6.81 on page 272 in your text.
CH4 + Cl2 (g) → CH3Cl + HCl (g)
81a)
∆Hrx = ∑ prod - ∑ react
= (-92.3 kJ + -82 kJ) - ( -74.85 kJ + 0.0 kJ )
= – 99.5 kJ
2 NH3 + CO2 (g) → CO(NH2)3 + H2O (l)
81b)
∆Hrx = ∑ prod - ∑ react
= (-285.9 kJ + -333.2 kJ) - ( 2(-46.2) kJ + (-393.5 kJ ))
= – 133.2 kJ
1:18.
Attempt question # 6.85 on page 272 in your text.
2 C2H2 + 5 O2 (g) → 4 CO2 + 2 H2O (l)
85)
∆Hrx = –2599.3 kJ
∆Hrx = ∑ prod - ∑ react
1/2(–2599.3 kJ) = (-285.9 kJ + 2(-393.5 kJ)) – ( ∆H°f (C2H2) + 2.5(0.0 kJ ))
∆H°f (C2H2) = +226.8 kJ
1:19. The standard heat of combustion (∆Hrx) for eicosane (C20H42), a typical component of candle wax,
is 47.3 kJ/g when it burns in pure oxygen. The only products are CO2 (g) and H2O (g). Calculate
the standard heat of formation (∆H°f) of eicosane in kJ/mol.
The rx
C20H42
+ 30.5 O2
→
20 CO2
+
21 H2O
∆H (comb)
= 47.3 kJ/g = 47.3 kJ/g x 282 g/mol = – 13 338.6 kJ /mol
∆Hcomb
= ∑ prod – ∑ react
so – 13 338.6 kJ
– 13 338.6 kJ
= 21 ( -286 ) + 20 (-394 ) - ∆H°f ( eicosane )
= 13 886 kJ – ∆H°f ( eicosane )
∆H°f ( eicosane ) = – 547.4 kJ /mol
1:20. 3.0 g of hydrazine ( N2H4 ) are burned in a bomb calorimeter causing 3.9 L of water to increase
its temperature by 3.5°C.
Calculate the heat of combustion (∆Hcomb) for hydrazine in kJ/g and kJ/mol.
Q
= 3.9 kg x 4. 184 kJ kg–1 K–1 x 3.5 K
= 57.1 kJ / 3 g of hydrazine
so
∆Hcomb
= – 19.0 kJ / g
or
∆Hcomb
= - 19.0 kJ / g x 32 g/mol
= – 608 kJ / mol
1:21. a) Write the equation for the complete combustion of benzene ( C6H6 (l)).
b) 2.1 g of benzene is burned in a metal bomb calorimeter (mass = 80 g , c= 0.825 kJ/kgK)
containing 3800 mL of water The temperature rose from 22°C to 27.2°C. Calculate the
experimental heat of combustion (∆Hcomb) for benzene with this data.
c) Now use your data sheet to calculate the theoretical ∆H(combustion) for benzene
d) Calculate the % error in the experiment and do a short error analysis.
+ 7 1/2 O2
→
a)
C6H6
6 CO2
+ 3 H2O
b)
Q(water)
= 3.8 kg x 4. 184 kJ kg–1 K–1 x 5.2 K
= 82.7 kJ / 2.1 g benzene
Q(calor)
= 0.080 kg x 0.825 kJ kg–1 K–1 x 5.2 K
= 0.343 kJ / 2.1 g
∆H (comb)
= - 83 kJ / ( 2.1 g /78 g/mol )
= – 3084.5 kJ / mol
( experimental answer )
∆H (Theor) = ∑ prod -∑ react
= (6 (-394) kJ + 3 ( - 242) kJ) - ( 49 kJ )
= – 3139 kJ
( theoretical answer )
% error
= 3139 kJ - 3084.5 / 3139 kJ x 100
= 1.7 %
Bond Energies and Heats of Reaction ( 18.10 pg 830 )
1:22. What energy would be required to completely disassemble a mole of butane gas ( C4H10 )
molecules into individual gaseous atoms. (= total bonding energy )
Butane ( C4H10 ) contains 3 C-C bonds and 10 C - H bonds
the rx.
C4H10
→
4 C + 10 H
∆H (TBE ) = 3 ( 348 kJ ) + 10 ( 412 kJ )
= 5.16 x 103 kJ
b) Attempt question # 18.97 on page 841 in your text.
acetone ( C3H6O) contains 2 C-C bonds , 6 C - H bonds and 1 C=O
the rx.
C3H6O →
3 C + 6 H +1O
∆H (TBE ) = 2 ( 348 kJ ) + 6 ( 412 kJ ) + 1 ( 743 kJ)
= 3.91 x 103 kJ released
1.23 Attempt questions # 18.99, 18.100, & 18.102 on page 841 in your text.
99)
2 C(graphite)
+
2 H2
(g)
→ C2H4
(g)
∆H °f = +52.284 kJ
• the elements C & H2 have to be vaporized to gaseous atoms this is endothermic
• then the bonds in C2H4 will be made – this is exothermic
∆H °f = +52.284 kJ = (2 ( 716.7 kJ ) + 4 ( 218 kJ )) – ( 4(412 kJ) + C=C )
+52.284 kJ = (2305.4 kJ) –1648 kJ – C=C )
C=C bond energy = 605 kJ/mol
`
100)
C(graphite)
2S
+
(s)
→ CS2
(g)
∆H °f = +115.3 kJ
∆H °f = +115.3 kJ = ( 716.7 kJ ) + 2 ( 277 kJ )) – ( 2( C=S ) )
+115.3 kJ = (1270.7 kJ) – ( 2( C=S ) )
C=S bond energy = 1155.4 /2 kJ/mol = 577.7 kJ/mol
102)
3 F2 (g)
+
S
(s)
→ SF6
(g)
∆H °f = –1096 kJ
∆H °f = –1096 kJ = ( 277 kJ ) + 6( 79.14 kJ )) – ( 6( S–F bonds ) )
–1096 kJ = (751.84 kJ) – (6( S–F bonds ))
S–F bond energy = 1847.84 /6 kJ/mol = 308 kJ/mol
Spontaneity / Entropy / Gibbs Free Energy ( 18.2 pg 797, 18.3 pg 800, 18.4 pg 805 )
1:24. Attempt questions # 18.57 (a) & (b) on page 838 in your text.
57 a)
2 C2H2 + 5 O2 (g) → 4 CO2 + 2 H2O (g)
∆Hrx = ∑ prod - ∑ react
∆Hrx = (2(-242 kJ) + 4(-394 kJ)) – ( 2 (227 kJ) + 5(0.0 kJ ))
∆Hrx = –2514 kJ
so rx favoured to proceed spontaneously
57 b)
C2H2 + 5 N2O (g) → 2 CO2 + H2O (g) + 5 N2
∆Hrx = ∑ prod - ∑ react
∆Hrx = (5(0.0) +(-242 kJ) + 2(-394 kJ))–( 5 (81.5 kJ)+(227 kJ ))
∆Hrx = –1665 kJ
so rx favoured to proceed spontaneously
1:25. Attempt questions # 18.60 & 18.61 on page 839 in your text.
60 a)
60 b)
60 c)
60 d)
∆S = –ve
∆S = –ve
∆S = –ve
∆S = +ve
→ number of moles of gaseous molecules decreases
→ number of moles of gaseous molecules decreases
→ number of moles of gaseous molecules decreases
→ number of moles of gaseous molecules increases
61 a)
61 b)
61 c)
61 d)
∆S = +ve
∆S = –ve
∆S = –ve
∆S = –ve
→ number of moles of gaseous molecules increases
→ number of moles of gaseous molecules decreases
→ number of moles of gaseous molecules decreases
→ the relatively random liquid disappears→ solid
1:26. Attempt question # 18.62 (c) on page 839 in your text.
62 a)
∆S° = ∑ prod - ∑ react
= (2 ( 192.5 ) J) - ( 3( 130.6) + 191.5 J )
= 385 - 583.2 J
= – 198.2 J
rx not favoured to proceed spontaneously
b)
∆S° = ∑ prod - ∑ react
= (126.8 J) - ( 2( 130.6) + 197.9 J )
= 126.8 - 459.1 J
= –332.3 J
rx not favoured to proceed spontaneously
c)
∆S° = ∑ prod - ∑ react
= (4 ( 213.6 ) + 6 ( 188.7) J) - ( 7( 205) + 2( 229.5) J )
= 1986.6 - 1894.0 J
= + 92.6 J
rx favoured to proceed spontaneously
d)
∆S° = ∑ prod - ∑ react
= (2 ( 69.96 ) + 107 J) - ( 157 + 76.1 J )
= 246.92 - 233.10 J
= + 13.82 J
rx favoured to proceed spontaneously
e)
∆S° = ∑ prod - ∑ react
= (2 ( 191.5 ) + 248.5 J) - ( 2( 220) + 31.9 J )
= 631.5 - 471.9 J
= + 159.6 J
rx favoured to proceed spontaneously
1:27. Attempt questions # 18.66 & 18.67 on page 839 in your text.
3 NO2 + H2O (l) → 2 HNO3 + NO (g)
66)
∆S° = ∑ prod - ∑ react
= (2 ( 155.6 ) + 210.6 J) - ( 3( 240.5) + 69.96 J )
= 521.8 - 791.46 J
= - 269.7 J
rx not favoured to proceed spontaneously
C2H5OH (l) + O2 (g) → HC2H3O2
67)
(l)
+ H2O (l)
∆S° = ∑ prod - ∑ react
= ( ( 160 ) + 69.96 J) - ( 161 + 205 J )
= 229.96 - 366 J
= - 136 J
rx not favoured to proceed spontaneously
1.28 Attempt question # 18.68 on page 839 in your text.
68)
C (s) + 1/2 O2 (g)
+ Cl2 (g) → COCl2 (g)
a)
∆H°f = – 223 kJ
b)
∆S°f = ( 284) - ( 5.69 + 1/2(205) + 223 )
= – 47 J or –0.047 kJ
c)
∆G°f = ∆H°f - T∆S°f
= -223 kJ - (298) (–0.047) kJ
= – 209 kJ
∆G°f
1:29. Attempt questions # 18.82, 18.84, & 18.85 on page 840 in your text.
C2H4 (g) + 2 HNO3 (l) → HC2H3O2
82)
i)
∆H
(l)
+ H2O (l) + NO (g) + NO2 (g)
= (–487 + (-286) + (90.4) + (34)) - (2(-174.1 ) + (51.9) ) kJ
= –648.6 + 296.3 kJ
= – 352.3 kJ
ii)
iii)
= ((160 + (70) + (210.6) +(240.5) ) - ( 2 (155.6) + (219.8) )
= 681.1 - 531
= + 150.1 J or 0.1501 kJ
∆G° = ∆H - T∆S
= –352.3 kJ – (298) (0.1501) kJ
= – 397 kJ
very spontaneous
C2H4 (g) + H2 (g) → C2H6 (g)
84)
i)
∆H
ii)
∆S°
iii)
85)
∆S°
=
=
=
=
(–84.5) - (( 51.9 ) + (0.0) ) kJ
– 136.4 kJ
(229.5) - (( 130.6 ) + (219.8) ) J
– 120.9 J or 0.1209 kJ
∆G° = ∆H - T∆S
= –136.4 kJ – (373) (0.1209) kJ
= – 91.3 kJ
very spontaneous
5 SO3 (g) + 2 NH3 (g) → 2 NO (g) + 5 SO2 (g)
i) ∆H
ii) ∆S°
iii) ∆G
=
=
=
=
=
=
+ 3 H2O (g)
(2 ( 90.4 ) +5(-297) + 3(-242) ) - (2(–46 ) + 5(–396) ) kJ
-2026 + 2067 kJ
+ 42 kJ
(2 (211) +5 (249) + 3 (189) ) - ( 2 (193) +5 (256) )
2234 - 1666
+ 564 J or 0.564 kJ
= ∆H - T∆S
= 42 kJ - (373) (0.562) kJ
= – 168 kJ
very spontaneous
1:30. Given: * Cu2O (s) + 0.5 O2 (g) → 2 CuO (s) Calculate ∆G° for this rx then discuss the 2 laws of Thermodynamics as they
apply to this reaction. Determine the temperature at which there will be no reaction (equilibrium) and discuss the spontaneity of the
reaction above and below this temperature ( ie the temperature range of this rx)
a)
∆H° = 2 ( -155 ) – (-167 + 0.0) kJ
= –143 kJ
Ist law states: PE spontaneously decreases. The negative ∆H indicates a drop in PE thus the 1st law
favours the products
b)
∆S° = 2 (44) - ( 1/2 (205) +101 )
= – 115 .5 J or 0.1155 kJ
2nd law states: Entropy spontaneously increases. The negative ∆S indicates a drop in entropy thus the
2nd law favours the reactants so any product formed would break this law
c)
∆G° = ∆H - T∆S
= -143 kJ - (298) (-0.1155) kJ
= - 108 . 6 kJ
spontaneous at this temp →( enthalpy drive greater than the entropy drive )
d) at equil'm ∆G = 0
= ∆H -T∆S
= -143 kJ - T (-0.1155 kJ )
T = 1238 K ≡ 965 °C
rx is spontaneous at any temperature below 965°C
Free Energy and equilibrium(ie ∆G = 0)
1:31. Attempt questions # 18.78, & 18.81 on page 840 in your text.
78)
∆H (vap) = 31.4 kJ
∆S = 94.2 J
**When a substance boils the liquid & gas are at equil'm so ∆G = 0
soooo T = ∆H / ∆S
= 31.4 kJ / 0.0942 kJ
= 333.3 K = 60.3 °C → ( BOILING POINT )
81)
∆H (vap) = 31.9 kJ
BP(acetone) = 56.2 °C ∆S = ?????? J
**When a substance boils the liquid & gas are at equil'm so ∆G = 0
soooo
T
= ∆H / ∆S
329.2 K = 31.9 kJ / ∆S
∆S = 0.0969 kJ = 96.9 J
*1:32. Microwaves are used to heat food. The microwave radiation is absorbed by moisture in the food.
This heats the water and thus the food. How many photons having a wavelength of 3.0 mm
would have to be absorbed by 1.0 g of water to raise its temperature by 1°C.
( HINT: Find Q (req'd) then E per microwave using
E(photon) = hc/λ where h (planck's constant) = 6.6 x 10–34 J.s & c = speed of light
Q(required)
E(photon)
= 1 x 10–3 kg x 4. 184 kJ kg–1 K–1 x 1 K
= 4.184 J
= 6.6 x 10–34 J.s x 3.0 x 108 m/s / 3 x 10–3 m
= 6.6 x 10–23 J/photon
# photon req’d = 4.184 J / 6.6 x 10–23 J/photon
= 6.3 x 1022 photons