Energy Change in Chemical Rx'

1
Energy Change in Chemical Rx.
Webpages Study
(1)
Definition of Exothermic, Endothermic
Thermoneutral Rx. http//www.ilpi.com/msds/ref/e
xothermic.html
Examples a) Combustion of methane is an
exothermic rx. http//www.ilpi.com/msds/ref/exoth
ermic.html
2
Energy Change in Chemical Rx.
Suggested solution for Worked Example 1
(a)
In both expt. (I) (II), no. of moles of
acid/base in 50cm3 of 2M solution
Total vol. of solution 50 50 100cm3
Mass of the final solution 100g
In expt.(I),
Heat released by neutralization between 0.1mole
of NaOH(aq) 0.1mole of HCl(aq)
mc?T 100 x 4.2 x 13 5460J
Enthalpy of neutralization (?H1)
In expt.(II),
Heat released by neutralization between 0.1mole
of NaOH(aq) 0.1mole of RCOOH(aq)
mc?T 100 x 4.2 x 10.3 4410J
Enthalpy of neutralization (?H2)
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3
Energy Change in Chemical Rx.
Suggested solution for Worked Example 1 (continue)
According to the expt.
The enthalpy of neutralization of a strong acid
(HCl) a strong base (NaOH) is -54.6kJmol-1. The
enthalpy of neutralization of a weak acid (RCOOH)
a strong base (NaOH) is -44.1kJmol-1.
This is because RCOOH is a weak acid is only
slightly ionized to provide free hydrogen ions,
whereas hydrochloric acid is completely ionized.
The ionization of ethanoic acid to form H(aq)
absorbs some energy from the neutralization
reaction.
The difference in the values of enthalpy of
neutralization is due to the enthalpy of
dissociation of RCOOH.
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4
Energy Change in Chemical Rx.
Suggested solution for Worked Example 1 (continue)
(b)
Standard enthalpy of dissoication of RCOOH
10.5kJ of energy are required for the complete
dissociation of 1 mole of RCOOH into RCOO-(aq)
H(aq).
5
Energy Change in Chemical Rx.
Suggested solution for Worked Example 2
(a)
Energy released when 8g of anhydr. CuSO4 are
added to 100cm3 H2O
3399.99 J 3.4 kJ
When 1 mole of anhydr. CuSO4 is added to H2O,
energy released
67787.50J -67.79kJ
(b)
In calculating the enthalpy of solution, the
following assumptions are made
1. The thermal capacities of polystyrene cup
thermometer are negligible. 2. The specific heat
capacity of copper(II) sulphate soltuion is
similar to that of water. 3. Anhydrous
copper(II) sulphate dissolves quickly such that
heat losses are neglible. 4. Further dilution of
the solution would cause no further heat
change. 5. No volume change when copper(II)
sulphate is dissolved.
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6
Energy Change in Chemical Rx.
Suggested solution for Worked Example 2 (continue)
Sources of errors the methods of minimizing
errors
(2)
1. Heat may be lost to or gained from the
surrounding. Insulate thepolystyrene cup or
use a vacuum flask calorimeter. 2. The molar heat
capacity of copper(II) sulphate soltuion is not
the same as that of water. Determine the
acutal molar heat capacity of copper(II) sulphate
solution. 3. Anhydrous copper(II) sulphate may
have absorbed moisture from air. Gently heat the
sample in an oven for several hours keep it in
a desiccator. 4. Heat may be absorbed by the
polystyrene cup, i.e. its heat capacity may not
be negligible. Determine the value of the heat
capacity of the cup, or use a container of known
heat capacity take it into account in
calculation. 5. The copper(II) sulphate solutiion
may not be sufficiently dilute to eliminate
further heat change on addition of water. Use
smaller quantities of copper(II) sulphate in
powder form.
7
Energy Change in Chemical Rx.
Suggested solution for Worked Example 3
(a)
Cu is a good conductor of heat such that the heat
produced from combustion of ethanol is
transferred to the water bath.
Heat produced when 0.7g of ethanol undergoes
complete combustion
(b)
mc?T 250 x 4.18 x 20 20900J 20.9kJ
No. of mole of ethanol in 0.7g
Heat produced when 1 mole of ethanol undergoes
complete combustion
The standard enthalpy of combustion of ethanol is
-1373kJmol-1
(c)
Sources of error
1. Error in temperature reading. 2. Error in
weighing ethanol. 3. Some ethanol may be
incompletely burnt.
8
Energy Change in Chemical Rx.
Suggested solution for Worked Example 4
Molar mass of benzoic acid 7(12) 4 2(16)
120 g
Energy produced when 1 g of benzoic acid is
completely burnt 7800 x 3.39 26442J
26.442kJ
When one mole of benzoic acid, the energy
produced 26.442 x 120 3173.04kJ The enthalpy
of combustion of benzoic acid is -3173.04kJmol-1
9
Energy Change in Chemical Rx.
Suggested solution for Worked Example 5
1
(a)
Ca(s) C(s) 1---- O2(g) ? CaCO3(s)
2
Experimental difficulties in the direct
determination of the enthalpy of formation
1. The extent of reaction cannot be controlled
2. The heat evolved cannot be separated into
appropriate terms
3. Direct combustion of calcium can be violent
4. Side reactions may arise, e.g.
2Ca(s) O2(g) ? 2CaO(s)
C(s) O2(g) ? CO2(g)
(b)
CaCO3(s) 2HCl(aq) ? CaCl2(aq) CO2(g) H2O(l)
Ca(s) 2HCl(aq) ? CaCl2(aq) H2(g)
(c)
Apparatus Calorimeter plastic cup/insulated
beaker/electrical heater/temperature
bath, Thermometer, Measuring cylinder/pipete/buret
te, Pan balance/electronic balance.
10
Energy Change in Chemical Rx.
Suggested solution for Worked Example 6
Standard enthalpy of formation of a substance is
the enthalpy change when 1 mole of the substance
is made from its constituent elements under
standard conditions (298K 1 atm pressure).
3
(1) Fe2O3(s) ? 2Fe(s) ---O2(g) 822.2kJ
2
3
(2) 3CO(g) ? 3C(s) ---O2(g) 110.5 x 3
331.5kJ
2
(3) 3C(s) 3O2(g) ? 3CO2(g) -393.5 x 3
-1180.5kJ
Eqt. (1) (2) (3)
Fe2O3(s) 3CO(g) ? 2Fe(s) 3CO2(g)
822.2 331.5 1180.5 -26.8kJmol-1
11
Energy Change in Chemical Rx.
Suggested solution for Worked Example 7
Born-Haber Cycle
aq
aq
From Hesss law,


gt
-92.3 75.1 -167.4kJmol-1
12
Energy Change in Chemical Rx.
Suggested solution for Worked Example 8
Given
?? -393.1kJmol-1
?? -285.5kJmol-1
?? -1367kJmol-1
The Born Haber Cycle with the equation for the
formation of ethanol is
3O2(g)
3O2(g)
-1367kJmol-1
2(-393.1kJmol-1) 3(-285.5kJmol-1)
By Hesss law,
(-1367kJmol-1)
2(-393.1kJmol-1) 3(-285.5kJmol-1)
gt
- 276.6kJmol-1
13
Energy Change in Chemical Rx.
Suggested solution for Worked Example 9
Born-Haber Cycle
2(-394) kJmol-1
490kJmol-1 2( )
2(-282) kJmol-1
2(964) kJmol-1
From Hesss law,



490 2( )
gt
2(-394)
2(-282)
2(964)

gt
607 kJmol-1
14
Energy Change in Chemical Rx.
Suggested solution for Worked Example 10
(a)
Enthalpy change of a reaction
sum of enthalpies of formation of products
sum of enthalpies of formation of reactants.
-

-393.5 2(-285.9) (-75.0) -890.3kJmol-1
?H-890.3kJmol-1 ---- eqt (1)
(b)
?H-801.7kJmol-1 ---- eqt (2)
(1) (2) gives
Ethanalpy change 88.6 kJ
Ethanalpy change to vaporize 2 moles of water
liquie 88.6 kJ
? Standard enthalpy of vaporization of water
15
Energy Change in Chemical Rx.
Suggested solution for Worked Example 11
/ kJmol-1
(1) KOH(s) aq --gt K(aq) OH-(aq)
-55
(2) K(s) H2O(l) aq --gt K(aq) OH-(aq)
H2(g)
-195
-286
(3) H2(g) O2(g)--gt H2O(l)
(2) (3) - (1) gives
K(s) H2(g) O2(g) --gt KOH(s)
Standard enthalpy of formation of potassium
hydroxide
-286 - 195 -(-55) -426 kJmol-1
16
Energy Change in Chemical Rx.
Suggested solution for Worked Example 12
-773 5(-286) kJmol-1
aq
aq
-66 kJmol-1
8 kJmol-1
From Hesss law,
(8) 5(-286) (-773) (-66),
-2277 kJmol-1
17
Energy Change in Chemical Rx.
Suggested solution for Worked Example 13
(a)
(1)
(2)
(b)
aq
aq
Free Hesss law,
-

-67.2 - 8.08 -75.3 kJmol-1
18
Energy Change in Chemical Rx.
Suggested solution for Worked Example 14
The Born-Haber cycle of the reactions are
(NaCl(s))(-285.9) kJmol-1
-92.3 kJmol-1
-3.9 kJmol-1
-425.6 (-71.9) kJmol-1
-57.3 kJmol-1
-425.6 - 57.3 285.9 - 92.3 - 71.9 - 3.9
-365.1 kJmol-1
(NaCl(s))
19
Energy Change in Chemical Rx.
Suggested solution for Worked Example 15
(Using technique similar to that in Q14) Answer
-1430kJmol-1
20
Energy Change in Chemical Rx.
Suggested solution for Worked Example 16
(a)
Max. temp. reached 37.1oC
21
Energy Change in Chemical Rx.
Suggested solution for Worked Example 16
(continue)
(b)
From the graph, temperature rise 37.1 - 28.6
8.5oC
Assuming the density of solution is the same as
that of water mass of solution 50 g
Enthalpy change
Enthalpy change
22
Energy Change in Chemical Rx.
Suggested solution for Worked Example 16
(continue)
(c)
Hence, by Hesss law,
-433.8 -285.8 ( -393.5) (-43.5) kJmol-1
-1069.6 kJmol-1
23
Energy Change in Chemical Rx.
Suggested solution for Worked Example 16
(continue)
(d)
24
Energy Change in Chemical Rx.
Suggested solution for Worked Example 1 7
Combustion of petrol is highly exothermic because
when burnt, it gives out a lot of energy. Octane
is energetically unstable relative to the
combustion products, carbon dioxide water. The
activation energy for the combustion of octane is
quite high. Therefore, octane is kinetically
stable relative to its combustion products. For
this reaction, motorists can fill up their petrol
tanks without the fear of an explosion.
25
Energy Change in Chemical Rx.
Suggested solution for Worked Example 1 8
Graphite is the relatively more stable allotropic
form. (According to the enthalpy of combustion,
diamond has a greater tendency to be
oxidized.) Conversion from diamond to graphite
does not occur at room temperature because this
involves the rearrangement of atoms in a giant
covalent network. A large activation energy is
required. Diamond is already very stable
kinetically.