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General Chemistry I
Homework 4: Thermodynamics
(Heat Capacities and
Hess’ Law)
Answer Key
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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 (C8H18, a major component of gasoline, density = 0.70 g/mL) is ∆Hc\ = -5471 kJmol-1.
For ethanol (0.78 g/mL), ∆Hc\ = -1368 kJmol-1.
a) Calculate the energy released when 1L of gasoline is burnt.
b) Using 1L gasoline fuel, how many litres of water (cv = 4.18 Jg-1K-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?
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 exothermic? Calculate the liquid’s ∆Hfus\ (latent heat of fusion) in kJmol-1.
f) Why is the magnitude of ∆Hfus\ significantly greater than for other organic liquids e.g. benzene (∆Hfus\ = 31 kJmol-1) and acetone (∆Hfus\ = 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 ∆Hfus\ ,
the latent heat of fusion (melting) of water (kJ/mol to 3 s.f.).
Q5: Use Hess’ Law to indirectly calculate the
heat of formation of aluminium chloride from
its elements from the following data:
∆H v\ (H2O) = 40.7 kJ/mol
heat capacity (cp) of water :
= 2.1 Jg-1K-1 (as vapour)
= 4.18 Jg-1K-1 (as liquid)
= 2.0 Jg-1K-1 (as solid)
b) How much more (extra)
time would be required to
heat the water at 333K until
it became steam at 111 ºC?
Q5: a)
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. ∆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
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
b) When 12 L sulphur dioxide at STP was reacted with 15 g
oxygen how much heat was generated?
When 1 mole of glucose (C6H12O6) 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 (C8H18, a major
component of gasoline, density = 0.70 g/mL) is ∆Hc\ = -5471
kJmol-1. For ethanol (0.78 g/mL), ∆Hc\ = -1368 kJmol-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 103 = 3.353 x 107 J
b) Using 1L gasoline fuel, how many litres of water (cv = 4.18 Jg-1K1) could be heated from room temperature (298 K) to boiling?
q = mcv∆T
m = q / cv∆T
= 3.353 x 107 J / 4.18 x (100 - 25) = 1.0695 x 105 g (107 L)
Q1: The molar hear of combustion for octane (C8H18, a major
component of gasoline, density = 0.70 g/mL) is ∆Hc\ = -5471
kJmol-1. For ethanol (0.78 g/mL), ∆Hc\ = -1368 kJmol-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 103 = 2.316 x 107 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 (C8H18, a major
component of gasoline, density = 0.70 g/mL) is ∆Hc\ = -5471 kJmol-1.
For ethanol (0.78 g/mL), ∆Hc\ = -1368 kJmol-1.
e) When 1L of ethanol froze, 736.5 kJ of heat was transferred. Was the
process endo- or exothermic? Calculate the liquid’s ∆Hfus\ (latent heat
of fusion) in kJmol-1.
freezing is exothermic (molecular motion reduces so hydrogen bonds
strengthen)
1L = 780 g = 16.9315 mol
∆Hfus\ = -736.5 x 103 J / 16.9315 mol = -4.35 x 104 J/mol
(-43.5 kJmol-1)
negative = exothermic
f) Why is the magnitude of ∆Hfus\ significantly greater than for other
organic liquids e.g. benzene (∆Hfus\ = 31 kJmol-1) and acetone (∆Hfus\
= 29 kJmol-1)?
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 103) = 1.033 x 107 J or 10.3 MJ
b) q = mc∆T
therefore: m = q / c∆T
7
q = 1.033 x 10 J
c = 4.18 Jg-1K-1
∆T = 55
therefore m = 4.49 x 104 g = 44.9 kg
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 ∆Hfus\, 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 104 J
E = energy to heat ice + energy to melt ice + energy to heat water
n = no. moles H2O
E = (mc∆T)ice + (mc∆T)water + n∆H f\
1.634 x 104 J = (25 x 2.1 x 33) + (25 x 4.18 x 60) + (25/18.016)∆Hfus\
∆Hfus\ = 6008.3 J/mol
∆Hfus\ = 6.01 kJmol-1
∆H v\ (H2O) = 40.7 kJ/mol
heat capacity (cp) of water :
= 2.1 Jg-1K-1 (as vapour)
= 4.18 Jg-1K-1 (as liquid)
= 2.0 Jg-1K-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 ∆Hfus\, 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∆Hvap\
E = (25 x 4.18 x 40) + (25 x 2.0 x 11) + [(25/18.016) x 40.7 x 103]
E = 6.1208 x 104 J
t = E/P = 6.1208 x 104 J / 200 W = 306 s
4
Q4:
a) ∆T = 2.7 K
q = 2.7 x 525 = 1417.5 J
Na+
b) HNO3 + NaOH
net ionic reaction:
H+ +
OH-
H2O
+
NO3-
+
H2O
(1 : 1 acid: base stoichiometry)
moles HNO3 being neutralised = 50/1000 x 0.5 = 0.025 mol
molar energy = 1417.5 J / 0.025 = 5.67 x 104 J/mol
(57 kJmol-1)
Q5:
eqn.1
eqn.2
2 x eqn. 1 in reverse gives 2S on RHS:
2SO2
2S + 2O2
now add eqn. 2
2S + 3Ο2
2SO3
∆H\ = +593.66 kJmol-1
∆H\ = -791.44 kJmol-1
2SO2 + 2S + Ο2
2S + 2SO3
∆H\ = -197.78 kJmol-1
2SO2 + Ο2
2SO3
∆H\ = -197.78 kJmol-1
∆H\ = -198 kJmol-1
(3 marks)
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 SO2 = 12 /22.4 = 0.5357 mol
moles O2 = 12/32.00 = 0.375 mol
because SO2 : O2 stoichiometry is 2 : 1
⇒
∆H = 0.5357 mol/2 x 198 = -53.0 kJ
SO2 is limiting reagent
O2 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)
2AlCl3(aq) + 3H2 + 6HCl(aq)
-1049 + (6 x -73.5) = -1490
2Al (s) + 6HCl(g)
2AlCl3(aq) + 3H2
∆H = -1490
now add 3 x eqn 3 to eliminate H2 and HCl(g) introduce Cl2
3H2 + 3Cl2
6HCl(g)
-555
2Al (s) + 6HCl(g) + 3H2 + 3Cl2
6HCl(g) + 2AlCl3(aq) + 3H2
-1490 -555 = -2045
2AlCl3(aq)
∆H = -2045
2Al (s) + 3Cl2
Eqn 1
Eqn 2
Eqn 3
Eqn 4
Finally adjust state of AlCl3 form aqueous to solid by reversing
equation 4, doubling it and adding:
2Al (s) + 3Cl2
2AlCl3(aq.)
2AlCl3(aq)
2AlCl3(s)
2Al (s) + 3Cl2 + AlCl3(aq.)
2Al (s) + 3Cl2
∆H = -2045
∆H = +646
AlCl3(s) + 2AlCl3(aq)
-2045 + 646 = -1399
2AlCl3(s)
∆H = -1399
heats of formation must be expressed per mole of AlCl3 being formed:
Al (s) + 1.5Cl2
AlCl3(s)
∆Hf = -699.5 kJ/mol
= (AlCl3)
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 (C6H12O6) burns in excess oxygen the products
are steam and carbon dioxide
c) write a balanced equation for the combustion
C6H12O6 (g) + 6O2 (g)
6CO2 (g) + 6H2O (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 m3
w = -p∆V = 101325 Pa x 0.1344 m3 = -13,618 J = -13.6 kJ
8