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BSAC Homework 4
Thermodynamics: Heat Capacities and Hess’ Law
Hand in on Thursday!
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Q1 : The molar heat of combustion for octane (C
\
= -1368 kJmol -1 .
8
H
18
, a major component of gasoline, density = 0.70 g/mL) is
∆
H
For ethanol (0.78 g/mL),
∆
H c a) Calculate the energy released when 1L of gasoline is burnt.
c
\
= -5471 kJmol b) Using 1L gasoline fuel, how many litres of water (c v
= 4.18 Jg -1 K -1 ) could be heated from room temperature (298 K) to boiling?
c) How many megajoules (MJ) of energy are released when 1L of ethanol is burnt?
-1 . d) Which of the two fuels is more efficient in terms of energy production per unit volume?
e) When 1L of ethanol froze, 736.5 kJ of heat was transferred. Was the process endo - or exo thermic? Calculate the liquid’s f) Why is the magnitude of
∆
H fus
\ significantly greater than for other organic liquids e.g. benzene (
∆
H fus
\
= 31 kJmol -1
∆
H fus
\
(latent heat of fusion) in kJmol -1
) and acetone (
∆
H fus
\
= 29 kJmol -1 )?
.
Q2: Q4:
Q3 : a) An electrical heater (power P = 200 Watts) supplied heat to a 25 g ice cube at -33 ºC in an isolated system. Given that after 81.7 sec it had become water at 333 K, calculate
∆
H fus
\ the latent heat of fusion (melting) of water (kJ/mol to 3 s.f.).
,
∆
H v
\ (H
2
O) = 40.7 kJ/mol heat capacity (c p
) of water :
= 2.1 Jg -1 K -1 (as vapour)
= 4.18 Jg -1 K -1 (as liquid)
= 2.0 Jg -1 K -1 (as solid)
Q5: a)
Q5 : Use Hess’ Law to indirectly calculate the heat of formation of aluminium chloride from its elements from the following data: b) How much more (extra) time would be required to heat the water at 333K until it became steam at 111 ºC? b) When 12 L sulphur dioxide at STP was reacted with 15 g oxygen how much heat was generated?
Q7 : a) Which of the statements below is true for a reaction that absorbs heat and produces products with smaller volume than the reactants?
i) the reaction does work and is endothermic.
ii)
∆
U is negative the reaction has work done on it and is exothermic.
∆ iii) the reaction has work done on it and is endothermic.
U is positive.
∆
U is positive iv) the reaction does work and is exothermic. v) the reaction does work and is exothermic.
∆
∆
U is negative.
U is positive b) Which of the above statements is true for a reaction that gives out heat and produces a large volume of gas from solid starting materials
When 1 mole of glucose (C
6
H
12
O
6
) burns in excess oxygen, the products are steam and carbon dioxide.
c) write a balanced equation for the combustion d) use the first law of thermodynamics to calculate how much work is done in expansion as the products are formed (assume vapours are at STP)
1

Q1 : The molar heat of combustion for octane (C
8
H component of gasoline, density = 0.70 g/mL) is
∆
H kJmol -1 . For ethanol (0.78 g/mL),
∆
H c
\
18 c
, a major
\
= -1368 kJmol
= -5471
-1 .
a) Calculate the energy released when 1L of gasoline is burnt.
1L = 700 g = 6.1283 mol heat = 6.1283 mol x 5471 x 10 3 = 3.353 x 10 7 J b) Using 1L gasoline fuel, how many litres of water ( c v
= 4.18 Jg
1 ) could be heated from room temperature (298 K) to boiling?
-1 K q = mc v m = q / c
∆
T v
∆
T
= 3.353 x 10 7 J / 4.18 x (100 - 25) = 1.0695 x 10 5 g ( 107 L )
Q1 : The molar hear of combustion for octane (C
8
H component of gasoline, density = 0.70 g/mL) is
∆
H kJmol -1 . For ethanol (0.78 g/mL),
∆
H c
\
18 c
\
, a major
= -1368 kJmol
= -5471
-1 .
c) How many megajoules (MJ) of energy are released when 1L of ethanol is burnt?
1L = 780 g = 16.9315 mol heat = 16.9315 mol x 1368 x 10 3 = 2.316 x 10 7 J ( 23.2 MJ ) d) Which of the two fuels is more efficient in terms of energy production per unit volume? Octane ( 33.5 MJ > 23.2 MJ )
2

Q1 : The molar hear of combustion for octane (C
8
H component of gasoline, density = 0.70 g/mL) is
∆
H
For ethanol (0.78 g/mL),
∆
H c
\
= -1368 kJmol -1 .
18 c
\
, a major
= -5471 kJmol -1 . e) When 1L of ethanol froze, 736.5 kJ of heat was transferred. Was the process endo - or exo thermic? Calculate the liquid’s
∆
H fus
\ (latent heat of fusion) in kJmol -1 .
freezing is exothermic (molecular motion reduces so hydrogen bonds strengthen)
1L = 780 g = 16.9315 mol
∆
H fus
\
= -736.5 x 10 3 J / 16.9315 mol = -4.35 x 10 4 J/mol
( -43.5 kJmol -1 ) negative = exothermic f) Why is the magnitude of
∆
H organic liquids e.g. benzene (
∆ fus
\
H fus
\
= 29 kJmol -1 )? significantly greater than for other
= 31 kJmol -1 ) and acetone (
∆
H fus
\ must overcome hydrogen bonds to vapourise
Q2 : a) mass = volume x density = 7 x 6 x 5 x 1.5 = 315 g molar mass = 12.01 g/mol no. moles = 315 g / 12.01 g/mol = 26.228 mol heat = 26.228 x (-394 x 10 3 ) = 1.033 x 10 7 J or 10.3 MJ b) q = mc
∆
T q = 1.033 x 10 7 J therefore: m = q / c c = 4.18 Jg -1 K -1
∆
T = 55 therefore m = 4.49 x 10 4 g = 44.9 kg
∆
T
3

Q3 : a) An electrical heater (power P = 200 Watts) supplied heat to a 25 g ice cube at -33 ºC in an isolated system. Given that after 81.7 sec it had become water at 333 K, calculate
∆
H fus
\
, the latent heat of fusion
(melting) of water (kJ/mol to 3 s.f.).
Heat change E = power x time = 200 x 81.7 = 1.634 x 10 4 J
E = energy to heat ice + energy to melt ice + energy to heat water
E = (mc
∆
T) ice
+ (mc
∆
T) water
+ n
∆
H f
\ n = no. moles H
1.634 x 10 4 J = (25 x 2.1 x 33) + (25 x 4.18 x 60) + (25/18.016)
∆
H
2
O fus
\
∆
H
∆
H fus
\ fus
\
= 6008.3 J/mol
= 6.01 kJmol -1
∆
H v
\ (H
2
O) = 40.7 kJ/mol heat capacity (c p
) of water :
= 2.1 Jg -1 K -1 (as vapour)
= 4.18 Jg -1 K -1 (as liquid)
= 2.0 Jg -1 K -1 (as solid)
Q3 : a) An electrical heater (power P = 200 Watts) supplied heat to a 25 g ice cube at -33 ºC in an isolated system. Given that after 81.7 sec it had become water at 333 K, calculate
∆
H fus
\
, the latent heat of fusion
(melting) of water (kJ/mol to 3 s.f.).
b) How much more (extra) time would be required to heat the water at
333 K until it became steam at 111 ºC?
E = (mc
∆
T) water
+ (mc
∆
T) steam
+ n
∆
H vap
\
E = (25 x 4.18 x 40) + (25 x 2.0 x 11) + [(25/18.016) x 40.7 x 10 3 ]
E = 6.1208 x 10 4 J t = E/P = 6.1208 x 10 4 J / 200 W = 306 s
4

Q4 : a)
∆
T = 2.7 K q = 2.7 x 525 = 1417.5 J b) HNO
3
+ NaOH net ionic reaction:
H + + OH -
Na + + NO
3
+ H
2
O
H
2
O (1 : 1 acid: base stoichiometry) moles HNO
3 being neutralised = 50/1000 x 0.5 = 0.025 mol molar energy = 1417.5 J / 0.025 = 5.67 x 10 4 J/mol ( 57 kJmol -1 )
Q5 : eqn.1
eqn.2
2 x eqn. 1 in reverse gives 2S on RHS:
2SO
2
2S + 2O
2 now add eqn. 2
2S
+ 3Ο
2
2SO
2
+ 2S
+ Ο
2
2SO
2
+ Ο
2
∆
H
\
= -198 kJmol -1
2SO
3
2S + 2SO
2SO
3
3
∆
H
\
= +593.66 kJmol -1
∆
H \ = -791.44 kJmol -1
∆
H \ = -197.78 kJmol -1
∆
H
\
= -197.78 kJmol
( 3 marks
-1
)
5

Q5 : eqn.1
eqn.2
b) When 12 L sulphur dioxide at STP was reacted with 15 g oxygen how much heat was generated?
Determine limiting reagent: moles SO
2 moles O
2
= 12 /22.4 = 0.5357 mol
= 12/32.00 = 0.375 mol because SO
2
: O
2 stoichiometry is 2 : 1
⇒
∆
H = 0.5357 mol/ 2 x 198 = -53.0 kJ
SO
2
O
2 is limiting reagent is in excess
( 3 marks )
Q5 : Use Hess’ Law to indirectly calculate the heat of formation of aluminium chloride from its elements from the following data:
6

Eqn 1
Eqn 2
Eqn 3
Eqn 4 reaction of interest is:
Add eqn 1 and 6 x eqn 2 to eliminate HCl(aq)
2Al (s) + 6HCl(aq) + 6HCl(g) 2AlCl
3
(aq) + 3H
2
-1049 + (6 x -73.5) = -1490
2Al (s) + 6HCl(g) 2AlCl
3
(aq) + 3H
2
+ 6HCl(aq)
∆
H = -1490 now add 3 x eqn 3 to eliminate H
2
3H
2
+ 3Cl
2 and HCl(g) introduce Cl
2
6HCl(g) -555
2Al (s) + 6HCl(g) + 3H
2
2Al (s) + 3Cl
2
+ 3Cl
2
6HCl(g) + 2AlCl
3
(aq) + 3H
2
-1490 -555 = -2045
2AlCl
3
(aq)
∆
H = -2045
Eqn 1
Eqn 2
Eqn 3
Eqn 4
Finally adjust state of AlCl
3 form aqueous to solid by reversing equation 4, doubling it and adding:
2Al (s) + 3Cl
2AlCl
3
(aq.)
2
2AlCl
2AlCl
3
3
(s)
(aq)
∆
H = -2045
∆
H = +646
2Al (s) + 3Cl
2
2Al (s) + 3Cl
2
+ AlCl
3
(aq.)
2AlCl
3
AlCl
3
(s) + 2AlCl
3
(aq)
(s)
-2045 + 646 = -1399
∆
H = -1399 heats of formation must be expressed per mole of AlCl
Al (s) + 1.5Cl
2
AlCl
3
(s)
∆
H f
=
3 being formed:
-699.5 kJ/mol
= (AlCl
3
)
7

Q7 : a) Which of the following statements is true for a reaction that absorbs heat and produces products with smaller volume than the reactants?
b) Which of the following statements is true for a reaction that gives out heat and produces a large volume of gas from solid starting materials i) the reaction does work and is endothermic.
∆
U is negative ii) the reaction has work done on it and is exothermic.
∆
U is positive.
iii) the reaction has work done on it and is endothermic.
∆
U is positive iv) the reaction does work and is exothermic.
∆
U is negative.
v) the reaction does work and is exothermic.
∆
U is positive
Q7 :
When 1 mole of glucose (C
6
H
12
O
6
) burns in excess oxygen the products are steam and carbon dioxide c) write a balanced equation for the combustion
C
6
H
12
O
6
(g) + 6O
2
(g) 6CO
2
(g) + 6H
2
O (g) d) use the first law of thermodynamics to calculate how much work is done in expansion as the products are formed (assume vapours are at
STP) w = -p
∆
V
∆
V = (12 - 6) mol = 6 mol
Avogadros’ Law:
6 mol = 134.4 L = 0.1344 m 3 w = -p
∆
V = 101325 Pa x 0.1344 m 3 = -13,618 J = -13.6 kJ
8