7/27/2019 THERMOCHEMISTRY CHA4 FORM5
1/75
Thermochemistry
1.0 Energy Changes in Chemical Reaction
Definition:
Poem;
P.K.
EXOTHERMIC
ENDOTHERMIC
Very important
Bond breaking REQUIRES energy
Bond formation RELEASES energy
Thermochemistry is the study of changes in heat
energy which take place during chemical reaction
Heat energy, we shalllater studyIs involved in reactions ofthermochemistry,
A chemical reaction that GIVES OUT/
RELEASES heat to the surroundings
A chemical reaction that ABSORBS heat
from the surroundings
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
2/75
EXOTHERMIC REACTION
Energy Profile Diagram
Study the diagram carefully.
So that you will be able to draw the Energy Level Diagram.
What is activation energy?
Activation energy is the energy barrierthat must beovercome by the colliding particles of the reactants in order
for reaction to occur / tobecome the products
H = The change in the amount of heat in a chemical
reaction is called the heat of reaction.
Energy
Reaction path
Energy
releases
during bondformation
Energy
requires
during
bondbreaking
Reactants
Products
Energy Requires < Energy Releases
X
Y
Z
Activation Energy H = ve
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
3/75
What we can say about exothermic reaction?
The conclusion is;
- Bond breaking requires less energy than bond formation,
- So, the energy released to form the bond is higher /
greater than the energy absorbed to break the bonds.- Then, excess heat energy is released to the surroundings.
- Thus, during the reaction temperature of the mixture
inreases,
- The total energy of product is lower than the total energy
of reactant.
Energy Level Diagram for Exothermic Reaction
Simply mean like this;Project = rm100 000 [energy requires]Complete = rm150 000 [energy releases]
Loss = rm50 000 [ H, heat changes]Saving = less rm50 000 [negative]
Energy
Reactants
Products
H = (negative)
H = Hproducts
Hreactants
H = ve
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
4/75
ENDOTHERMIC REACTION
Energy Profile Diagram
Study the diagram carefully.
What we can say about endothermic reaction?
The conclusion is
- Bond breaking requires more energy than bond formation,
- So, the energy released to form the bond is lower / less
than the energy absorbed to break the bonds.
- Then, heat energy is absorbed from the surroundings.
- Thus, during the reaction temperature of the mixture
decreases,
- The total energy of product is higher than the total energy
of reactant.
Energy
Reaction path
Energy
releases
during bond
formation
Energy
requires
during
bondbreaking
Reactants
Products
Energy Required > Energy Released
Q
P
R
Activation Energy
H = +ve
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
5/75
Energy Level Diagram for Endothermic Reaction
Simply mean like this;Project = rm100 000 [energy requires]
Complete = rm50 000 [energy releases]
Profit = rm50 000 [energy heat changes]
Saving = up to rm50 000 [positive]
Example of exothermic reaction
Most of the chemical reaction is exothermic such as
- neutralization
- combustion- acid and metal
[Tip: better to memorize endothermic reaction, because not
many reaction is endothermic]
Energy
Reactants
Products
H = + (positive)
H = Hproduct
Hreactants
H = +ve
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
6/75
Example of endothermic reaction
1. Salt dissolves in water
NH4Cl (s) NH4+ (aq) + Cl- (aq)
KNO3 (s) K+ (aq) + NO3
- (aq)
2. Salt crystallization
CuSO4.5H2O (s) Cu2+ (aq) + SO4
2- (aq) + 5H2O (l)
3. Thermal decomposition
ZnCO3 (s) ZnO(s) + CO2 (g)
2Mg(NO3)2 (s) 2MgO
(s) + 4NO2 (g) + O2 (g)
4. Salt dissociation
NH4Cl (s) NH4+ (aq) + Cl- (aq)
CaCO3 (s) CaO(s) + CO2 (g)
5. Reaction between acid with sodium hydrogen carbonate
and potassium hydrogen carbonate;HCl(aq) + NaHCO3(aq) NaCl(aq) + H2O(l) + CO2(g)
HCl(aq) + KHCO3(aq) KCl(aq) + H2O(l) + CO2(g)
6. photosynthesis
6CO2 + 6H2O C6H12O6 + 6O2
7. process of melting, evaporation and boiling.
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
7/75
Example 1:
Mg + H2SO4 MgSO4 + H2
H = 467 kJ
Energy level diagram
Exercises
Based from the following equations, construct and explain
energy level diagram for the reaction.
(1) CaCO3 CaO + CO2 H = + 178 kJ
(2) 2H2 + O2 2H2O H = 572 kJ
(3) Zn + CuSO4 ZnSO4 + Cu H = 190 kJ
(4) H2 + I2 2HI H = + 53 kJ
Energy
Mg + H2SO
4
MgSO4 + H2
H = - 467 kJ
Explanation:
The reaction is an exothermic reaction
Temperature of mixture is increases
Total energy of 1 mole Mg and 1 mole H2SO
4is
higher than 1 mole of MgSO4 and 1 mole H2 by 467kJ
When 1 mole Mg reacts with 1 mole H2SO
4to form 1
mole of MgSO4
and 1 mole H2
, 467 kJ of heat is
released/produced.
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
8/75
Effective Practice pg 148 no. 1, 2, 3 & 4
Example 2:
2HgO 2Hg + O2
H = +182 kJEnergy level diagram;
Kamal Ariffin Bin Saaim
SMKDBL
http://www.kemhawk.webs.com/
Energy
2HgO
2Hg + O2
H = + 182
Explanation:
The reaction is an endothermic reaction
Temperature of mixture is decreases
Total energy of 2 mole HgO is lower than 2 mole ofHg and 1 mole O
2by 182 kJ
When 2 mole HgO decompose to form 2 mole of Hg
and 1 mole O2, 182 kJ of heat is absorbed
http://www.kemhawk.webs.com/http://www.kemhawk.webs.com/7/27/2019 THERMOCHEMISTRY CHA4 FORM5
9/75
exampleThe diagram shows an energy profile diagram.
Based on the above energy profile diagram, the amount of activation energy is..
A (Y X) kJ mol-1
B X kJ mol-1C (X Y) kJ mol-1
D Y kJ mol-1
25 The reaction between nitrogen and oxygen can be represented by the following equation:
N2 (g) + O2(g) 2NO(g) H = +181 kJ
Which of the following energy level diagrams represent the above reaction?
A B
C D
Energy
Reactants
X kJ mol-1
Y kJ mol-1
Products
2NO(g)
Energy
Energy
N2(g) + O
2(g)
2NO(g)
H = +181 kJ
N2(g) + O
2(g)
2NO(g)
H = +181 kJ
Energy
2NO(g)
N2(g) + O
2(g)
H = +181 kJ
Energy
N2(g) + O
2(g)
H = +181 kJ
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
10/75
6 A pupil carried out an experiment to determine the value of heat of neutralization.
Diagram 6 shows the set up of the apparatus used in the experiment.
The following data was obtained;
a) Why was a polystyrene cup used in this experiment?
..[1 mark
(b) Given that the specific heat capacity of the solution is 4.2 Jg-1oC-1 and the
density of the solution is 1.0 gcm-3.
(i) Calculate the change of heat in the experiment.
(ii) Calculate the heat of displacement in the experiment.
[3 marks
1
Initial temperature of hydrochloric acid = 28oCInitial temperature of sodium hydroxide solution = 28oC
Highest temperature of the mixture of product = 41oC
100 cm3of 2.0 mol dm-3
Sodium hydroxide solution
100 cm3of 2.0 moldm-3
hydrochloric acid
Thermometer
DIAGRAM 6
Polystyrene cup
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
11/75
(c) Draw the energy level diagram for the reaction.
[2 marks
(d) Based on the experiment, what is meant by theheat of neutralisation?
.......
[1 mark
(e) The pupil repeats the experiment by replacing hydrochloric acid with
ethanoic acid. All the other conditions remain unchanged.
(i) Predict the value of the heat of neutralisation?
.........
[1 mark]
(ii) Explain why?
........
....
........
....................................................................................................................
[2 marks
1
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
12/75
Thermochemistry2.0 Heat of precipitation
- precipitate is unsoluble salt
- precipitate must be prepared through double bond
decomposition orprecipitation method
Do you still remember what is meant by double bond
decomposition?
[please refer to salts notes]
General equation double bond decomposition/precipitation;
Ionic equation for precipitation reaction.
1
The heat of precipitation is the heat changewhen one mole of a precipitate is formed from
their ions in aqueous solution
- two aqueous solution of a substances/soluble salt was mix
together.
- one of the solution contain cation, while another onecontain anion forinsoluble salt/precipitate that need
to be prepared.
- both of the aqueous solution exchanging theirions to
produce 2 substances, which is unsoluble salt (precipitate)
and one soluble salt
MX (aq) + NY (aq) MY (s) + NX (aq)
M+ (aq) + Y- (aq) MY (s)
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
13/75
Salt Solubility in waterLi+, Na+, K+, NH4
+ All salt dissolve in water
Nitrate, NO3- All nitrate salt dissolve in water
ChlorideAll chloride salt dissolve dissolve in water except;PbCl2 - lead(II) chloride (dissolve in hot water)
AgCl - argentums/silver chloride
HgCl - hydroargentum chloride, mercury chloride
SulphateAll sulphate salt dissolve in water except;
PbSO4 , BaSO4 , CaSO4
CarbonateAll carbonate salt not dissolvein water except;
Li2CO3 Na2CO3 , K2CO3 , (NH4)2CO3
Oxide All oxide not dissolve in water except;Na2O , K2O , CaO
HydroxideAll hydroxide not dissolve in water except;
NaOH, KOH, Ca(OH)2 , Ba(OH)2
Formula to determine the heat change;
Heat released/absorbed, H = mc [unit = J or kJ]
Symbo
lDescription Unit
m mass of solution 1cm3 = 1 g
c specific heat capacity of solution 4.2 J g-1oC-1
temperature change oC
1
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
14/75
To determine precipitation heat of silver chloride, AgCl
1
Method to determine the heat of precipitation
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
15/75
In this experiment you must have the following data;
Data tabulation
Thermometer
Polystyrene
cup
Procedure
25 cm3 sodium chloride solution 0.5 mol dm-3 is measured with
measuring cylinder50ml, and poured intopolystyrene cup, recordthe temperature with termometer (0-110)oC.
25 cm3 silver nitrate solution 0.5 mol dm-3 is measured with
measuring cylinder 50ml, and poured into another polystyrene
cup, record the temperature with termometer (0-110)oC.
Sodium chloride solution is added to silver nitrate solution
quickly.
The reacting mixture is stirred using thermometer.Highest temperature obtained is recorded.
Repeat all the step by using different substance.
Precaution steps;
Use polystyrene cup. (polystyrene cup is insulator, to avoid loss of
heat)
Stir the mixture.
25 cm3 sodium
chloride solution
0.5 mol dm-3
Thermometer
Polystyrene
cup
25 cm3 silver
nitrate solution
0.5 mol dm-3
1
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
16/75
Initial temperature of sodium chloride
NaCl /oCx oC
Initial temperature of silver nitrate,
AgNO3 /oC
y oC
Average initial temperature for bothsolution
Highest temperature for the solution z oC
Temperature changez (x + y) oC = oC
2
Chemical equation for the reaction;
AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq)
Ionic equation for the reaction;
Ag+ (aq) + Cl- (aq) AgCl (s)Calculation of heat of precipitation for AgCl;
1. Calculate the number of mole of precipitate formed
No. of mol NaCl = = = 0.0125 mol
No. of mol AgNO3 = = = 0.0125 mol
FBCE;
No. of mol AgCl = 0.0125 mol
1
(x + y) oC
2
MV
1000
0.5 X 25
1000
MV
1000
0.5 X 25
1000
Formula of mole
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
17/75
2. Calculate the heat released/given out
[From the experiment]
Total volume of the mixture = 25 cm3 AgNO3 + 25 cm3 NaCl
= 50 cm3
Mass of solution = 50 g
Temperature change = oC
Heat given out, H =
= 50 4.2 J
= kJ
Therefore, heat given out during the experiment is kJ
3. Calculate the heat of precipitation
0.0125 mol of AgCl produces kJ
Therefore;
1 mol of AgCl produces = kJ mol-1
= kJ mol-1
H = kJ mol-1
Thus;
The heat of precipitation of silver chloride, AgCl;
1
50 4.2 1000
1 .
0.0125
m = mass of solution( 1cm3 = 1 g)
c = specific heat
capacity of solution
(4.2 J g-1oC-1)
= temperature change
oC
Heat given out
No. of mole
50 4.2
1000
50 4.2 12.5
50 4.2
1000
mc
1st
formula
2nd formula
50 4.2
1000
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
18/75
H = kJ mol-1
Example 1: Precipitation forlead(II) sulphate
Chemical equation;
Pb(NO3)2 + K2SO4 PbSO4 + 2KNO3
H = -50 kJmol-1
Ionic equation;
Pb2+ + SO42- PbSO4 H = -50 kJmol
-1
50 kJ heat released when 1 mol of lead(II) ions react with 1 mol of
sulphate ions to form 1 mol precipitate of lead(II) sulphate.
The heat of precipitation for PbSO4 = 50 kJmol-1
Example 2: Precipitation forsilver chloride
Energy
Pb2+ + SO4
2-
PbSO4
H = - 50 kJmol-1
Energy level diagram
1
50 4.2
12.5
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
19/75
Chemical equation ;
AgNO3 + KCl AgCl + KNO3
H = 65.5 kJmol-1
Ionic equation;
Ag+ + Cl- AgCl H = 65.5 kJmol-1
65.5 kJ heat released when 1 mol of silver ions react with 1 mol of
chloride ions to form 1 mol precipitate of silver chloride.
The heat of precipitation for AgCl = 65.5 kJmol-1
Example 3: Precipitation forcopper(II) hydroxide
Energy
Ag+ + Cl-
AgCl
H = 65.5 kJmol-1
Energy level diagram
1
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
20/75
Chemical equation;
CuSO4 + 2NaOH Cu(OH)2 + Na2SO4
H = -60 kJmol-1Ionic equation;
Cu2+ + 2OH- Cu(OH)2 H = -60 kJmol-1
60 kJ heat released when 1 mol of copper(II) ion react with 2 mol of
hydroxide ion to form 1 mol precipitate of copper(II) hydroxide.
The heat of precipitation for Cu(OH)2 = 60 kJmol-1
Calculation for heat of precipitate
Example 1
Energy
Cu2+
+ 2OH-
Cu(OH)2
H = 60 kJmol-1
Energy level diagram
2
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
21/75
When 50 cm3 calcium nitrate solution, Ca(NO3)2 2 mol dm-3
is added to 50 cm3 sodium carbonate solution, Na2CO3
2.0 mol dm-3,precipitate of calcium carbonate, CaCO3 is
produce.Temperature of the mixture solution decrease 3.0 oC.
Calculate the heat of precipitation of calcium carbonate.
[Specific heat capacity of solution: 4.2 J g-1oC-1.
Density of solution: 1 g cm-3]
Chemical equation;
Ca(NO3)2 + Na2CO3 CaCO3 + 2NaNO3
Ionic equation:
Ca2+ + CO3 CaCO3
Step 1: Calculate the number of mole of precipitate formed
No. of mol Ca(NO3)2 = = =
No. of mol Na2CO3 = = =
FBCE;
1 mol Ca2+ react with 1 mol CO32- , produce 1 mol calcium carbonat
Therefore;
0.1 mol Ca
2+
react with 0.1 mol CO32-
, produce 0.1 mol CaCO3Thus;
No. of mol CaCO3 = 0.1 mol
Step 2 : Calculate the total heat absorb in exp.
[From the experiment]2
MV
1000
2 X 50
10000.1 mol
MV
1000
2 X 50
10000.1 mol
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
22/75
Total volume of the mixture = 50 cm3Ca(NO3)2 + 50 cm3Na2CO3
= 100 cm3
Mass of solution = 100 g
Temperature change = 3o
C
Heat absorbs, H =
= 100 4.2 3 J
= 1260 J
Therefore, heat given out during the experiment is 1.26 kJ
Step 3 : Calculate the heat of precipitstion
0.1 mol CaCO3absorb at 1260 J of heat.
Therefore;
1 mol CaCO3absorb heat = J mol-1
= 12600 J mol-1
= 12.6 kJ mol-1Thus
The heat of precipitation CaCO3, H = + 12.6 kJ mol-1
Step 4 : Draw energy level diagram
Example 2:
Energy
Ca2+ + CO3
2-
CaCO3
H = +12.6 kJ mol-1
2
1 kJ = 1000 J
1260
0.1
si n (+) must write
m = mass of solution
( 1cm3 = 1 g)
c = specific heat
capacity of solution(4.2 J g-1oC-1)
= temperature change
oC
mc
1st formula
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
23/75
Thermochemical equation for precipitate of magnesium carbonate,
Mg(NO3)2.
Mg(NO3)2 (ak) + Na2CO3 MgCO3 (p) + 2NaNO3 (ak)
H = +25 kJmol-1
Calculate the changes of temperature when 50 cm3 magnesium nitrate
solution , Mg(NO3)2 2.0 mol dm-3 is added to 50 cm3 sodium carbonate
solution, Na2CO3 2.0 mol dm-3/
Chemical equation has been given the question.
Ionic equation: Mg2+ + CO32- MgCO3
Step 1: Calculate the number of mole of precipitate formed
No. of mol Mg(NO3)2 = = =
No. of molNa2CO3 = = =
FBCE;
1 mol Mg2+ react with 1 mol CO32-
, to produce 0.1 mol MgCO3
Therefore;
0.1 mol Mg2+
react with 0.1 mol CO32-
, to produce 0.1 mol MgCO3Thus;
No of mole MgCO3 = 0.1 mol
Step 2 : Calculate the total heat absorb in exp.2
MV
1000
2.0 X 50
10000.1 mol
MV
1000
2.0 X 50
10000.1 mol
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
24/75
Heat absorb = mc
(endothermic) = (50 + 50) x 4.2 x J
= 100 x 4.2 x J
= 420 J
[from chemical equation, (H = +25 kJ)]
The heat of precipitation for magnesium carbonate is +25 kJ,
Therefore;
1 mol precipitate of MgCO3 absorbs heat 25 kJ,
Thus;
0.1 mol precipitate MgCO3 absorb heat,
= 0.1 x 25 kJ,
= 2.5 kJ
= 2.5 x 1000 J
= 2500 J
Thus;
The heat absorbs in experiment is 2500 J
Step 3 : determine the value of , temperature change in exp.
In experiment,
Total heat absorb = mc
420 J = 2500 J
=
= 5.95 oC
The reaction is endothermic, temperature decrease 5.95 oC
2
m = solution mass
( 1cm3 = 1 g)
c = specific heat
capacity
(4.2 J g-1
o
C-1
) = tem erature
1 kJ = 1000 J
Change to J, we want
substitute it into formula
2500 J420 J
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
25/75
Energy level diagram of precipitate ofMgCO3
3. thermochemical equation for precipitate of copper(II)hydroxide,Cu(OH)2 given below,
Cu2+ (ak) + 2OH- (ak) Cu(OH)2 (p) H = -60 kJ
How many its volume solution for copper(II)sulphate, CuSO4 1.0
mol dm-3 that need to mixture with 50 cm-3 sodium hydroxide
solution, NaOH 2.0 mol dm-3 to increase temperature of the solution
mixture to 6.3oC?
Step 1: write the chemical equation for this reaction
Chemical equation has been state at the question.
Step 2 : calculate no. of mol for the substance
No. of mol CuSO4/Cu2+
= = =
No. of molNaOH /OH- = = =
Energy
Mg2+ + CO3
2-
MgCO3
H = +25 kJ mol-1
2
MV
1000
1.0 X V
1000
0.001 V mol
MV
1000
2.0 X 50
10000.1 mol
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
26/75
Step 3 : ratio for no. of mol
FBCE;
2 mol NaOH /OH- react with 1 mol CuSO4/Cu2+ , to produce1 mol
Cu(OH)2, copper(II) hydroxide,
so ;
0.1 molNaOH /OH- react with mol CuSO4/Cu2+, to produce
molCu(OH)2.
So ;
0.1 molNaOH /OH- react with 0.05 mol CuSO4/Cu2+, to produce 0.05
mol Cu(OH)2.
* carefull to calculate value of V, we must include the value of
temperature changes, (6.3 oC) from the question that give above.
Step 4 : calculate the total heat release/absorb in this exp.
Total heat released = mc(t/b exothermic) = (50 + V) x 4.2 x 6.3 J
= (50 + V) x 26.46 J
= (50 + V) 26.46 J
(V not yet known)
[from chemical equation], (H = -60 kJmol
-1
)]
Heat of precipitate for copper(II)hydroxide is -60 kJ, (from question)
1 mol precipitate of Cu(OH)2 releaseing heat 60 kJ,
So ;
0.05 mol precipitate of Cu(OH)2 releasing heat,2
= 6.3 oC
(substitute in
equation)
0.1
20.1
2
Dont calculate the volume of
CuSO4 from this value
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
27/75
= 0.05 x 60 kJ,
= 3 kJ
= 3 x 1000 J
= 3000 J
So ;
Total heat released by 0.05 mol precipitate of Cu(OH)2 in experiment
is 3000 J
Step 5 : determine the value ofV, solution volume of CuSO4, in exp
In experiment,
Total heat released = mc J
(50 + V) 26.46 J = 3000 J
(50 + V) =
(50 + V) = 113.38 cm3
V = (113.38 50) cm3
V = 63.38 cm3
Energy level diagram for precipitate ofCu(OH)2
energy
Cu
2+
(ak) + 2OH
-
(ak)
Cu(OH)2
(p)
H = -60 kJ mol-1
2
3000 J
26.46 J
1 kJ = 1000 J
Change to J, we want
substitute it into equation
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
28/75
4. when solution of 500 cm3 M2+ 2.0 mol dm-3 is mixture with
500 cm3 solution of ion Cl- , heat of solution increase to oC.
Calculate the heat of precipitate for the reacted ion M2+ and ion Cl-
to precipitate the formula substance MCl2.
Step 1: write the chemical equation
Ionic equation
M2+ (ak) + 2Cl- (ak) MCl2 (p)
Step 2 : calculate no. of mol for the substance
No. of mol M2+ = = =
No. of mol Cl- = = =
Step 3 : ratio for no. of mol
DPKYS;1 mol M2+ react with 2 mol Cl- to produce
1 mol precipitate MCl2
so;
1 mol M2+ reacted with 2 mol Cl- to produce
1 mol precipitate MCl2
Step 4 : calculate the total heat release/absorb in this exp.
Total heat released = mc
(t/b exothermic) = (500 + 500) x 4.2 x J
= 1000 x 4.2 x J
= 4.2 x kJ
2
MV1000
2.0 X 5001000 1 mol
MV
1000
M X 500
10000.5 M mol
1 kJ = 1000 J
Dont use this no. of mol
because the actual no. of
mol we dont know yet
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
29/75
So ;
Total heat released by 1 mol precipitate of MCl2 in experiment is 4.2 x
kJ,
Step 5 determine the heat of precipitate of MCl2
Because the total heat released by 1 mol precipitate of
MCl2 in this experiment is 2 x kJ ,
So ;
Heat of precipitate for MCl2 is -4.2 x kJ mol-1
H= -4.2 x kJ mol-1
energy level diagram for heat of precipitate ofMCl2
5. In one experiment to determine the heat of precipitate between ion
M2+ and ion SO42- , found that when 250 cm3
of M2+ 2 mol dm3 solution is mix with 250 cm3 solution of
ion SO42- , temperature of the solution increase at 30 oC.
Calculate heat of precipitation for the reaction between
ion M2+ and ion SO42-
Energy
M2+ (ak) + Cl- (ak)
MCl2
(p)
H = -4.2 x kJ mol-1
2
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
30/75
Step 1: write the chemical equation for this reaction
Ionic equation
M2+
(ak) + SO42-
(ak) MSO4 (p)
Step 2 : calculate no. of mol for the substance
No. of mol M2+ = = =
No. of mol SO42-= = =
Step 3 : ratio for no. of mol
DPKYS;
1 mol ofM2+ react with 1 mol SO42- to produce
1 mol precipitate ofMSO4
So ;
0.5 mol M2+ react with 0.5 mol SO42- to produce
0.5 mol precipitate ofMSO4
Step 4 : calculate the total heat release/absorb in this exp.
Total heat release = mc
(t/b exothermic) = (250 + 250) x 4.2 x 30 J= 63000 J
= 63 kJ
So ;
Total heat release by 0.5 mol precipitate of MSO4 in experiment is
63 kJ3
MV
1000
2.0 X 250
10000.5 mol
MV
1000
M X 250
1000
0.25 M mol
1 kJ = 1000 J
Dont use this no. of mol
because the actual no. of
mol we dont know yet
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
31/75
Step 5 determine the heat of precipitate of MSO4In experiment;
0.5 mol precipitate of MSO4 release heat 63 kJ
So ;
1 mol precipitate of MSO4 releasing heat ;
=
= 126 kJ mol-1
So ;
Heat of precipitation of MSO4 is -126 kJ mol-1
H = -126 kJ mol-1
Energy level diagram for heat of precipitation ofMSO4
Energy
M2+ (ak) + SO4
2- (ak)
MSO4
(p)
H = -126 kJ mol-1
3
63 kJ
0.5
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
32/75
6. mixture between 75 cm3 hydrochloric acid solution 0.15 mol dm-3
and 75 cm3 silver nitrate solution 0.15 mol dm-3 has increase the
temperature at 1.9 oC.
i. how much the quantity of heat of energy that release in this
experiment?ii. heat of precipitate for this reaction
iii. draw energy level diagram for this reaction.
Step 1: write the chemical equation for this reaction
Chemical equation;
AgNO3 (ak) + HCl (ak) AgCl (p) + HNO3 (ak)
Ionic equation;
Ag+ (ak) + Cl- (ak) AgCl (p)
Step 2 : calculate no. of mol for the substance
No. of mol HCl = = =
No. of mol AgNO3 = = =
Step 3 : ratio for no. of mol
DPKYS;
1 mol HCl react with 1 mol AgNO3 to produce
1 mol precipitate of AgCl
So ;
0.01125 mol HCl react with 0.01125 mol AgNO3 to produce
0.01125 mol precipitate ofAgCl
3
MV
1000
0.15 X 75
10000.01125 mol
MV
1000
0.15 X 75
1000
0.01125 mol
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
33/75
Step 4 : calculate the total heat release/absorb in this exp.
Total heat release = mc
(t/b exothermic) = (75 + 75) x 4.2 x 1.9 J
= 1197 J
= 1.197 kJSo ;
Total heat release by 0.01125 mol precipitate of AgCl in experiment is
1.197 kJ
Step 5 determine the heat of precipitate of MSO4In experiment;
0.01125 mol precipitate of releasing heat at1.197 kJ
So ;
1 mol precipitate of AgCl releasing heat;
=
= 106.4 kJ mol-1
So ;Heat of precipitate of AgCl is -106.4 kJ mol-1
H = -106.4 kJ mol-1
Energy level diagram for heat of precipitation ofAgCl
Kamal Ariffin B Saaim
Energy
Ag+ (ak) + Cl- (ak)
AgCl (p)
H = -106.4 kJ mol-1
3
1 kJ = 1000 J
1.197 kJ mol-1
0.01125
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
34/75
Thermochemistry
3.0 Heat of Displacement
What is meant by displacement reaction?
Example 1;
Zn + Cu2+ Zn2+ + Cu H= 210 kJmol-1
Energy
Zn + Cu2+
Zn2+
+ Cu
When 1 mole Cu is displaced by Zn, 210 kJ heat energy
is released/given out.
H= 210 kJmol-1
3
The Heat of Diplacement is the heat change
when one mole of metal is diplaced from its solution
by a more electropositive metal.
Metal that more electropositive will displace
metal that is less electropositive from its salt solution.
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
35/75
Example 2;
Zn + Pb2+ Zn2+ + Pb H= 112 kJmol-1
Example 3;
Mg + Fe2+
Mg2+
+ Fe H= 80 kJmol-1
Energy
Zn + Pb2+
Zn2+ + Pb
When 1 mole Pb is displaced by Zn, 112 kJ heat energy
is released/given out.
H= 112 kJmol-1
Energy
Mg + Fe2+
Mg2+ + Fe
When 1 mole Fe is displaced by Mg, 80 kJ heat energy
is released/given out.
H= 80 kJmol-1
3
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
36/75
To determine the heat of diplacement of copper by zinc
Procedure;
50 cm3 copper(II) sulphate solution of 0.1 mol dm-3 is
measured with measuring cylinder 50ml and poured into
polystyrene cup, record the temperature withtermometer (0-110)oC.
1.0 g metal powder is weighed by using electronic balance and
quickly added into the polystyrene cup that contain copper(II)
sulphate solution.
The mixture is stirred using the thermometer.
The highest/maximum temperature of heat is recorded.
Repeat the step by using different substance. [if necessary]
Thermometer
Polystyrene
cup
50.0 cm3 copper(II)
sulphate solution
0.1 mol dm-3
Beaker that
contain
1 g zinc powder
(excess)
3
Method to determine heat o dis lacement
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
37/75
Data tabulation
Initial temperature of copper(II) sulphate ,
CuSO4 /oC
y
Highest/maximum temperature for the
solution /o
C
z
Temperature change /oC (z y) =
[Note : mass of zinc is used in excess]
Chemical equation for the reaction
Zn (s) + CuSO4 (aq) Cu (s) + ZnSO4 (aq)
Ionic EquationZn + Cu2+ Cu + Zn2+
Calculation the heat of displacement of copper by zinc;
1. Calculate the number of mole of Cu formed
No of moles of Zn = . mass . = 1.0 = 0.015 mol
molar mass 65
No. of moles CuSO4 = = = 0.005 mol
(No. of moles of CuSO4) 0.005 < 0.015 (No of moles of Zinc)
[Important notes: calculation MUST based on CuSO4 solution
because quantity of zinc used in excess]
FBCE;1 mole of CuSO4 produces 1 mole of Cu
Therefore;
0.005 mole of CuSO4 produces 0.005 mole of Cu
Thus;
No. of mole of Cu formed = 0.005 mol
3
M V
1000
0.1 X 50
1000
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
38/75
2. Calculate the heat released/given out
Heat release/given out = mc
(in exp.)= 50 x 4.2 x J
= kJ
3. Calculate the heat of diplacement for 1 mole copper by zinc
0.005 mole of copper diplaced produced kJ
Therefore;
When 1 mole of copper is diplaced by zinc, the heat released is
= kJ mol-1
= X kJ mol-1
= kJ mol-1
Thus;
The heat of diplacement of copper by zinc
H = kJ mol-1
3
50 x 4.2 x 1000
50 x 4.2 x
10001 .
0.005
50 x 4.2 x
5
50 x 4.2 x
1000 .
0.005
50 x 4.2 x
1000
50 x 4.2 x
5
Volume of solution
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
39/75
Calculation heat of displacement
Question 1
Excess iron powder is added into 50 cm3 copper(II) chloride solution,
CuCl2 1.0 mol dm-3
, brown solid is formed and blue solution change togreen.
Iron has displace copper from its salt solution. The following data is
get from above experiment.
Initial temperature for copper(II)chloride solution = 28.0 oC
Highest temperature for mixture solution = 57.0 oC
Calculate heat changes when 1 mol of copper is displace by iron.
Solution
chemical equation;
Fe (s) + CuCl2 (aq) FeCl2 (aq) + Cu (s)
Ionic equation;Fe (s) + Cu2+ (aq) Cu (s) + Fe+2 (aq)
Step 1 : Calculate the number of mole of Cu formed
No. of mol CuCl2 = = =
No. of mol Fe = (no need to calculate because is in excess)
FBCE;
No. of mole of Cu formed = 0.05 mol
Step 2 : Calculate the heat released/given out3
MV
1000
1 X 50
10000.05 mol
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
40/75
Heat released/given out = mc
(exothermic) = 50 x 4.2 x J
Temperature change, = (highest temperature initial temperature)
= (57.0 28.0)o
C= 29.0 oC
Total heat release = mc
= 6090 J
= 6.09 kJ
Step 3 : determine the heat of displacement of Cu by Fe
Displacement of0.05 mol Cu releasing 6.09 kJ of heat.
Therefore;
When 1 mole of copper is diplaced by zinc, the heat released is
= kJ mol-1
= 121.8 kJ mol-1Thus;
The heat of displacement of Cuby iron;
H = 121.8 kJmol-1
Draw energy level diagram
Question2
Energy
Fe + Cu2+
Fe2+ + Cu
H = -121.8 kJ mol-1
4
(determine the changes
of temperature)
1 kJ = 1000 J
6.09
0.05
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
41/75
Study the following equation;
Fe (s) + CuSO4 (aq) Cu (s) + FeSO4 (aq)
H = -250 kJ mol-1
If excessive iron powder is add into 100 cm3
copper(II) sulphatesolution 0.25 mol dm-3, calculate
i. Heat released
ii. Temperature rise
iii. Mass of copper that is displace
iv. Mass of salt that formed if it crystalize
v. Draw energy level diagram
If magnesium powder is use to replace iron powder, is it the energy
that release is more higher, same or lower.
[Ar = Cu, 64; Fe, 56; S, 32; O, 16]
Solution
Chemical equation;
Fe (p) + CuSO4 (ak) FeSO4 (ak) + Cu (p)
Ionic equation;
Fe (p) + Cu2+ (ak) Fe+2 (ak) + Cu (p)
i : calculate no. of mole of Cu formed
No. of mol CuSO4 = = =
No. of mol Fe = (no need to calculate because excessive)
4
MV
1000
0.25 X 100
1000
0.025 mol
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
42/75
FBCE;
1 mol CuSO4 produced 1 mol Cu
Therefore;
0.025 mol CuSO4 produce 0.025 mol Cu
Thus;No. of mole of Cu formed = 0.025 mol
calculate the heat released by 0.025 mol of Cu in the exp.
[from question: H = -250 kJ mol-1]
Thats mean;
When 1 mole Cu is displaced by Fe, 250 kJ heat energy is
released/given out.
Therefore;
0.025 mol Cu is diplaced by Fe, heat released is;
= 0.025 250 kJ
= 6.25 kJ
= 6250 J
Thus;
Heat released/given out = 6.25 kJ or 6250 J
ii: determine the temperature changes during the reaction
[Heat released/given out = 6250 J]
Heat released/given out = mc6250 = 100 x 4.2 x
= 14.9 oC
Therefore;
The increases in temperature = 14.9 oC
iii: determine the mass of copper that is diplace4
Change to unit of kJ, because we
want to find
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
43/75
[from the previous information above]
No. of mole of Cu formed = 0.025 mol
No. of mol Cu = mass of Cu
ArCu
Mass of Cu = 0.025 64
= 1.6 g
iv: determine the mass of salt formed
FBCE;
1 mol CuSO4 produced 1 mol FeSO4Therefore;
0.025 mol CuSO4 produce 0.025 mol FeSO4Thus;
No. of mole of FeSO4 formed = 0.025 mol
No. of mol FeSO4 = mass of FeSO4MrFeSO4
Mass of FeSO4 = 0.025 [56 + 32 + 4(16)]
= 3.8 g
v : draw energy level diagram
Question 3
Energy
Fe + Cu2+
Fe2+ + Cu
H = -250 kJ mol-1
4
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
44/75
100 cm3copper(II) nitrate solution 0.2 mol dm-3 is poured into plastic
container. Temperature is recorded. Then excess magnesium powder
is added to the solution. The mixture is stirred and the temperature
rises is recorded. The temperature shows an increases of 5 oC.
(a) What is the colour of the solution in the plastic container;
i. before magnesium powder is place?
ii. after magnesium powder is place?
(b) Write the total ionic equation of the reaction.
(c) How many mole of copper(II) nitrate reacts?
(d) Calculate the heat releases in this experiment.
(e) Calculate the heat energy release when one mol of copper is
formed.
(f) What is the heat of displacement of copper?
(g) Draw energy level diagram for this experiment.
(h) Why magnesium used is in form of fine powder not granulated?
(i) i. if potassium hydroxide solution is mix with the solution in
plastic beaker in the end of the experiment, what can you
observe?
ii. write the chemical equation for the reaction in (i)(i).iii. write the ionic equation for the reaction in (i)(i).
SOLUTION4
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
45/75
(a) i. blue
ii. colourless
(b) Mg (aq) + Cu2+ (aq) Mg2+ (aq) + Cu (s)
(c) no. of mol Cu(NO3)2 = =
=
(d) Heat release = mc
(during reaction) = 100 x 4.2 x 5.0 J
= 2100 J
= 2.1 kJ
(e) FBCE;
1 mol CuSO4 produced 1 mol Cu
Therefore;
0.02 mol CuSO4 produce 0.02 mol Cu
Thus;
No. of mole of Cu formed = 0.02 mol
When 0.02 mol of Cu diplaced by Mg, 2.1 kJ of heat released.
Therefore;
1 mol of Cu diplaced by Mg will releases heat;
=
= 105 kJ mol
-1
(f) The heat of displacement of copper = -105 kJ mol-1
Heat change, H = -105 kJ mol-1
(g) Energy level diagram4
MV
1000
0.2 x 100
1000
0.02 mol
2.1 kJ mol-1
0.02
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
46/75
(h) To increase the total surface area per volume for magnesium,
thus it will increase the rate of reaction.
(i) i. white precipitate formed
ii. Mg(NO3)2 (aq) + 2KOH (aq) Mg(OH)2 (s) + 2KNO3 (aq)
iii. Mg2+ (aq) + 2OH- (aq) Mg(OH)2 (s)
Learning task: pg 158 no. 1 & 2
Effective Practise: pg. 158 no. 3
Kamal Ariffin B Saaim
SMKDBL
http://kemhawk.webs.com/
Energy
Mg (s) + Cu2+ (aq)
Mg2+ (aq) + Cu (s)
H = - 105 kJ mol-1
4
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
47/75
Thermochemistry
4.0 Heat of neutralisation
What is meant by neutralization?
Ionic equation:
Heat of neutralization can be divide into two types;
i. Reaction between strong acid with strong alkali.
ii. Reaction between weak acid with weak alkali.
4
The heat of neutralization is
the heat change when one mole of water is formed
from the reaction between an acid and an alkali.
H+ (aq) + OH- (aq) H2O (l)
Acid reacts with alkali/oxide base to produce salt and water
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
48/75
Heat Of Neutralization Between Strong Acid With
Strong Alkali
Example of reaction;
Neutralization reactionNeutralization heat,
H (kJ mol-1)
HCl + NaOH NaCl + H2O -57.3
HCl + KOH KCl + H2O -57.3
HNO3 + KOH KNO3 + H2O -57.3
HNO3 + NaOH NaNO3 + H2O -57.3
Therefore;
Energy level diagram,
57.3 kJ heat is released when 1 mol of water is produced from
a reaction between 1 mol of hydrogen ion, H+and1 mol of
hydroxide ion, OH- .
Energy
H+ (aq) + OH- (aq)
H2O (l)
H = - 57.3 kJmol-1
4
In neutralization reaction between strong acid and
strong alkali, the heat of neutralization is -57.3 kJmol
-1
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
49/75
1. Study this reaction
What is the value of heat of neutralization for the above
reaction?
Ionic equation:
H+ (aq) + OH- (aq) H2O (l) H = -57.3 kJ
- 57.3 kJ heat is released when 1 mole of water is produced
from a reaction between 1 mole of hydrogen ion, H+
and1 mole of hydroxide ion, OH- .
- Thus,the heat of neutralizationis -57.3 kJmol-1
Remember: HCl is monoprotic acid
[Monoprotic acid: When 1 mole of HCl acid dissolves in water,
it produces 1 mol of hydrogen ion]
Energy level diagram,
Energy
H+ (aq) + OH- (aq)
H2O (l)
H = - 57.3 kJmol-1
4
HCl (aq) + NaOH (aq) NaCl (aq) + H2O (l)
H = -57.3 kJ
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
50/75
2. Study this reaction
What is the value of heat of neutralization for the above
reaction?
- H2SO4is diprotic acid. It produces 2 moles of hydrogen
ions when it dissolves in water.
H2SO4 (aq) 2H+ (aq) + SO4
2- (aq)
- 2 moles of H+ ions produce 2 moles of water when reacted
with alkali.
Ionic equation:
2H+ (aq) + 2OH- (aq) 2H2O (l) H = -114.6 kJ
- 114 kJ heat is released when 2 moles of water is produced
from a reaction between 2 moles of hydrogen ion, H+ and
2 mol of hydroxide ion, OH- .
- Thus,the heat of neutralizationis still -57.3 kJmol-1
5
H2SO4 (ak) + 2NaOH (ak) Na2SO4 (ak) + 2H2O (ce)
H = -114 kJ
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
51/75
Energy level diagram,
Energy
H+ (aq) + OH- (aq)
H2O (l)
H = - 12 kJmol-1
5
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
52/75
Heat of neutralization between weak acid with strong
alkali
Example of reaction;
Neutralization reaction
Heat of
neutralization,
H (kJ mol-1)
CH3COOH + NaOH CH3COONa + H2O -55
HCN + KOH KCN + H2O -12
Why ?
Study the ionization equation for this weak acid;
- Weak acid only ionize partially in water. Only small amount of
hydrogen ion is produce.
- Most of the molecules of weak acid still exist as a molecule.- Therefore, some of the heat energy is use to overcome the
molecular bond in acid, so that in can be ionize in water.
- Thus, this process cause the value of heat of neutralization,
less than 57 kJmol-1
5
In neutralization between weak acid and strong alkali,
heat of neutralization is less than -57.3 kJmol-1
CH3COOH (aq) CH3COO- (aq) + H+ (aq)
Reversible rocess
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
53/75
To determine the heat of neutralizationbetween HCl and NaOH
Thermometer
Polystyrene
cup
Procedure ;
50 cm3 hydrochloric acid solution 2.0 mol dm-3 is measured
with measuring cylinder 50ml and pour into polystyrene cup and
the temperature is recorded using thermometer (0-110)oC.50 cm3 potassium hydroxide solution 2.0 mol dm-3 is measured
and pour into anotherpolystyrene cup and the temperature is
recorded.
Quicklythe hydrochloric acid solution is added to the
potassium hydroxide solution,the mixture is stirred using
thermometer.
highest/maximum temperature of the mixture is recorded.Repeatthe step by using different substance.
50 cm3
hidrochloric acid
2.0 mol dm-3
Thermometer
Polystyrene
cup
50 cm3 potassium
hydroxide solution
2.0 mol dm-3
5
Method to determie the heat of neutralization
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
54/75
Data tabulation
Initial temperature for hydrochloric acid,
HCl /oCx
Initial temperature for potassium hydroxide,NaOH /oC
y
Average temperature for the both
solution /oC
Highest/maximum temperature for the
solution /oCz
Temperature changes /
o
C z - =
Calculation heat of neutralization HCl and NaOH
Chemical equation;
HCl (aq) + NaOH (aq) NaCl (aq) + H2O (l)
Ionic equation;
H+ (aq) + OH (aq) H2O (l)
Calculation the heat of neutralization;
1. Calculate the number of mole of water formed
No. of mol hydrogen ions, H+ / HCl ;
= = = 0.1 mol
No. of mol of hydroxide ions, OH- / NaOH
= = = 0.1 mol
5
(x + y)
2
(x + y)2
MV
1000
2.0 X 50
1000
MV
1000
2.0 X 50
1000
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
55/75
FBCE;
1 mole of H+ reacts with 1 mole of OH- to form 1 mole of H2O
Therefore;
0.1 mole of H+ reacts with 0.1 mole of OH- to form 0.1 mole of H2O
Thus;No. of mole of H2O formed = 0.1 mol
2. Calculate the heat given out/releases during reaction;
[From the experiment]
Total volume of the mixture = 50 cm3 HCl + 50 cm3 NaOH
= 100 cm3
Mass of solution = 100 g
Temperature change = oC
Heat given out, H =
= 100 4.2 J
= kJ
Therefore, heat given out during the experiment is;
= kJ
5
100 4.2
1000
mc
100 4.2
1000
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
56/75
3. Calculate the heat of neutralization
0.01 mol of H2O produces kJ
Therefore;1 mol of H2O produces = kJ mol
-1
= kJ mol-1
= kJ mol-1
Thus;
The heat of neutralization;
H = kJ mol-1
H = - 4.2 x kJ mol-1
Energy level diagram,
Energy
H+ (aq) + OH- (aq)
H2O (l)
H = - 4.2 x kJ mol-1
5
1 .
0.1
Heat given outNo. of mole
100 4.2
1000
100 4.2 100
100 4.2
1000
100 4.2
100
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
57/75
Question 1
When 50 cm3 potassium hydroxide solution, NaOH 1.0 mol dm-3 is
added with 50 cm3 hydrochloric acid HCl 1.0 mol dm-3, the temperature
of mixture increase 6.2 oC.
Calculate the heat of neutralization for this reaction.
Solution
Write the chemical equation for the reaction
Chemical equation;
NaOH (aq) + HCl (aq) NaCl (aq) + H2O (l)
Ionic equation;
H+(aq) + OH- (aq) H2O (l)
Calculation the heat of neutralization;
1. Calculate the number of mole of water produced
No. of mol H+ / HCl = = =
No. of mol OH- / NaOH = = =
FBCE;
1 mole of H+ reacts with 1 mole of OH- to form 1 mole of H2O
Therefore;
0.05 mole of H+ reacts with 0.05 mole of OH- to form
0.05 mole of H2OThus;
No. of mole of H2O formed = 0.05 mol
5
MV
1000
1.0 X 50
10000.05 mol
MV
1000
1.0 X 50
10000.05 mol
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
58/75
2. Calculate the heat given out/releases during reaction;
m = (50 + 50) cm3 = 6.2 oC
= 100 g
Heat released/given out = mc(exothermic reaction) = (50+50) x 4.2 x 6.2 J
= 2604 J
= 2.604 kJ
3. Calculate the heat of neutralization
0.05 mol of H2O formed produces 2.640 kJ
Therefore;
1 mol of H2O produces = kJ mol-1
= kJ mol-1
= 52.08 kJ mol-1
Thus;
The heat of neutralization;
H = 52.08 kJ mol-1
Energy level diagram,
Energy
H++ OH-
H2O
H = -52.08 kJ mol-1
5
2.604
0.05
1 kJ = 1000 J
Heat given outNo. of mole
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
59/75
Question 2
When 50 cm3 potassium hydroxide solution, NaOH 1.0 mol dm-3 is
added into 50 cm3 sulphuric acid, H2SO4 0.5 mol dm-3, the
temperature of solution mixture has increase 6.2o
C?
(a) calculate the heat releases for the following reaction;
i. H2SO4 (ak) + NaOH (ak) Na2SO4 (ak) + H2O (ce)
ii. H2SO4 (ak) + 2NaOH (ak) Na2SO4 (ak) + 2H2O (ce)
SOLUTIONCalculate no. of mol of water formed
No. of mole H2SO4 = = =
Therefore;
No. of mole of H+ = 2 0.025 = 0.05 mol
No. of mole of OH- / NaOH = = =
FBCE;
1 mole of H2SO4 reacts with 2 mole of NaOH to form 2 mole of H2O
[Remember: H2SO4 is diprotic acid]
2 mole of H+ reacts with 2 mole of OH- to form 2 mole of H2O
5
MV
1000
0.5 X 50
10000.025 mol
MV
1000
1.0 X 50
10000.05 mol
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
60/75
Therefore;
(0.025 2) mole of H+ reacts with 0.05 mole of OH- to form
0.05 mole of H2O
Thus;
No. of mole of H2O formed = 0.05 mol
Calculate the total heat release during the reaction;
Heat released/given out = mc
= (50+50) x 4.2 x 6.2 J
= 2604 J
= 2.604 kJ
Calculate the heat of neutralization
0.05 mol of H2O formed produces 2.640 kJ
Therefore;
1 mol of H2O produces = kJ mol-1
= 52.08 kJ mol-1
Thus;
The heat of neutralization;H = 52.08 kJ mol-1
Experiment (i) : H2SO4 + NaOH Na2SO4 + H2O
Ionic equation: H+ + OH- H2O [1 mole of water]
Therefore;
Heat changes in this reaction is 52.08 kJ
Experiment (ii) H2SO4 + 2NaOH Na2SO4 + 2H2O
Ionic equation: 2H+ + 2OH- 2H2O [2 moles of water]
Therefore;
Heat changes in this reaction is 2 52.08 kJ = 104.16 kJ6
2.604
0.05
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
61/75
Question 3
HCl (ak) + NaOH (ak) NaCl (ak) + H2O (ce)
H = -57 kJmol-1
Based on the thermochemistry equation above, answer the followingequation;
a) Write the ionic equation for the reaction above
b) When 100 cm3 excessive potassium hydroxide solution is mix
with 100 cm3 hydrochloric acid, solution temperature increase
13.6oC. calculate the concentration of hydrochloric acid use.
Solution
Ionic equation : H+ (ak) + OH- (ak) H2O (ce)
Heat release during reaction = mc
= (100 + 100) x 4.2 x 13.6 J
= 11424 J
= 11.424 kJ
[Heat changes, H = -57 kJmol-1]57 kJ heat releases by 1 mol water
1kJ heat releases by mol water
Therefore;
11.424 kJ release x 11.424 mol air,
= 0.2 mol H2O
Thus;
No. of mol of water = 0.2 mol
6
1 .
57
1 .
57
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
62/75
FBCE;
1 mol water is produce by 1 mol HCl
Therefore;
0.2 mol water produce by 0.2 mol HCl
Calculate the concentration of HCl
No. of mole HCl = =
0.2 mol =
M =
Concentration of HCl = 2.0 mol dm-3
Learning Task 4.6 no. 1, 2, 3 pg. 164
Effective Practise no. 1, 2, 3
Kamal Ariffin Bin Saaim
SMKDBL
http://kemhawk.webs.com/
6
MV
1000M x 100
1000
M x 100
1000
0.2 x 1000
100
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
63/75
Thermochemistry
5.0 Heat of combustion
Heat of combustion of various alcohol
Name
Molecular
formula
(CnH2n+1OH)
No. of carbon
atom per alcohol
molecule
Mass of
molecula
r relative
Heat
combustion
H (kJ mol-1)
Methanol CH3OH 1 32 -725
Ethanol C2H5OH 2 46 -1376
Propan-1-ol C3H7OH 3 60 -2015
Butan-1-ol C4H9OH 4 74 -2676
What inferens can be made from the table above?
- If the number of carbon atom per molecule of alcohol is
higher, the heat of combustion also increases.
- The difference in heat of combustion between alcoholmember is almost the same because each alcohol member
difference is in one group of CH2
6
The heat of combustion is the heat change when
1 molof substance is completely burnt in oxygen
under standard conditions.
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
64/75
Example of reaction of heat combustion in substance ;
i. Methanol,
CH3OH (ce) + O2 (g) CO2 (g) + 2H2O (ce)
H = -725 kJmol-1
ii
Ethanol,
C2H5OH (ce) + 3O2 (g) 2CO2 (g) + 3H2O (ce)
H = -1376 kJmol-1
iii. Propan-1-ol,
C3H7OH (ce) + O2 (g) 3CO2 (g) + 4H2O (ce)H = -2015 kJmol-1
iv. Butan-1-ol,
C4H9OH (ce) + 6O2 (g) 4CO2 (g) + 5H2O (ce)
H = -2676 kJmol-1
6
3
2
9
2
When 1 mole of methanol burnt, 725 kJ of heat is releases.
Method to determine the heat o combustion o uel
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
65/75
To determine the heat of combustion of methanol
250 cm3
water
Thermometer
Asbestos
screen
Copper
container
Tripod stand
Spirit lampMethanol
Wooden block
Procedure;
250 cm3 of wateris measured with measuring cylinder 100ml and
poured into copper container, the temperature of water isrecorded with thermometer (0-110)oC.
A spirit lamp is filled with methanol until half full.
A spirit lamp and are weighed with electrical balanced. and the mass is
recorded.
The spirit lamp is placed under copper container, and the wick is lighted.
The water is stirred.
When the temperature of water is increases 30 oC, the spirit lamp is
distinguished.
Spirit lamp is weighed immediately, and the mass is recorded.
The experiment is repeatedusing different alcohol.
6
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
66/75
Precautions steps;
- Use copper container or any suitable metal.
(metal is condustor, all heat from combustion of fuel was absorb
by water)
- Spirit flame is placed on a wooden block, so the flame contactdirectly the copper container.
(A bigger area of the flame can be in contact with the copper
container)
- Wire gauze is not used.
(to prevent wire gauze absorb heat energy)
- Asbestos screen is placed around the copper container.
(to avoid heat loss to surrounding)
- The water must always be stirred.
(temperature changes is uniform)
In this experiment, you need to get the following data;
Data tabulation;
Initial temperature of water /oC x
Highest temperature of water /oC y
Temperature rise /oC 30
Initial mass of metanol + spirit lamp /g a
Final mass of methanol + spirit lamp /g b
Mass of methanol burnt /g (a b) g
Calculation of heat of combustion
Chemical equation
6
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
67/75
CH3OH + O2 CO2 + 2H2O
No ionic equation!
1. Calculate the number of mole of methanol (fuel used)
Mass of methanol = (a b) g
Molar mass of methanol = 12 + 3(1) + 16 + 1
= 32 g mol-1
No. of mole of methanol = = = c mol
2. Calculate the heat given out/releases during reaction;
[From the experiment]
Heat release during reaction = mc
= 200 x 4.2 x 30 J
= kJ
3. Calculate the heat of neutralization
c mol of methanol produces heat
Therefore, heat release by burnt of one mol of methanol,
= kJ mol-1
Thus;6
massMolar mass
(a - b) .32
100 x 4.2 x 30
1000
3
2
Volume of water
100 x 4.2 x 30
1000 .
c
100 x 4.2 x 30
1000
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
68/75
The heat of combustion of methanol;
H = - [ ] kJmol-1
Very easylah!
Fuel Value
Fuel value can be use to compare the energy cost..
Name
Relative
molecula
r mass
Heat of
combustion,
H (kJ mol-1)
Mass of 1
mol (g)
Burnt substance
(kJ g-1)
Metanol 32 -725 32 725/32 = 22.66
Etanol 46 -1376 46 1376/46 = 29.91
Propan-1-
ol60 -2015 60 2015/60 = 33.58
Butan-1-ol 74 -2676 74 2676/74 = 36.16
How much the price of 1 g for each burnt substance above?
Calculation for heat of combustion
Question 1
6
The amount of heat energy give out when 1 g of the fuel is
completely burnt in excess of oxygen.
100 x 4.2 x 30
1000 x c
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
69/75
Heat from combustion of 0.28 g oktane, C8H18 increasing temperature
of 200 cm3 water from 30 oC to 46 oC. Based on this information,
[ Ar: H, 1; C, 12; O, 16; 1 mole gas is occupy 24 dm3]
i. calculate the heat of combustion for octane
ii. value of fuel for octaneiii. volume of oxygen that need
iv. mass of water produce
Solution
Chemical equation;
C8H18 (ce) + O2 (g) 8CO2 (g) + 9H2O (ce)
Calculate no. of mol of fuel
No. of mol for octane = = = 0.0025 mol
Calculate the heat changes/total heat release in this experiment
Temperature changes = (highest temp initial temp) oC= ( 46 30) oC
= 16 oC
Heat release = mc
(exothermic react.) = 200 x 4.2 x 16 J
= 13440 J
= 13.44 kJ
Calculate the heat of combustion
Combustion 0.0025 of mol of octane releasing 13.44 kJ of heat.6
mass .
molar mass
0.28
114
m = volume of water
( 1cm3 = 1 g)
1 kJ = 1000 J
25
2
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
70/75
Therefore;
Combustion of 1 mol octane releases;
= kJ mol-1
= 5376 kJ mol-1
Thus;
Heat of combustion for octane = - 5376 kJ mol-1
H = - 5376 kJ mol-1
Caculate the fuel value of octane
Mr oktane, C8H18 = (8 x 12) + 18 = 114 gmol-1
Therefore;
1 mole of octane, C8H18 = 114 g
Thus;
Fuel value of octane =
= kJ g-1
= 47.15 kJ g-1
Calculate the volume of oxygen
FBCE;7
13.44
0.0025
Heat of combustion for octane
Mass of one mol of octane
5376
114Make sure the
unit is correct
25
2
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
71/75
1 mol octane need mol oxygen to react completely
Therefore;
0.0025 mol octane need x 0.0025 mol oxygen;
= 0.0312 mol oxygen
The number of O2 used in the reaction = 0.0312 mol oxygen
Thus;
The volume of oxygen needed = 0.0312 x 24 dm3
= 0.7488 dm3
= 748.8 cm3
Calculate the mass of water produce
FBCE;
1 mole of oktane react completely to produce 9 mole of water
Therefore;
0.0025 mol oktane produce 0.0025 x 9 mol water;
= 0.0225 mol water
The number of mole of water = 0.0225 mol
Thus;
No. of mol for water =
0.0225 mol =7
25
2
Mass of water
Mr of water
Mass of water
(2 x 1) + 16
Make sure the
unit is correct
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
72/75
Mass of water = 0.0225 x 18 g
= 0.405 g
Question 2
Following equation show heat of combustion for ethanol, C2H5OH
C2H5OH (ce) + 3O2 (g) 2CO2 (g) + 3H2O (ce)
H = -1400 kJmol-1
Heat combustion for 0.46 g etanol, is use to heat200 cm3 of water, calculate the temperature rise of water .
Solution
Step 1 : calculate no. of mol for ethanol that burnt
No. of mol etanol = = = 0.01 mol
Step 2 : calculate the heat changes/total heat release in this experimen[heat combustion for etanol = -1400 kJmol-1] (from ques.)
1 mol etanol burnt releasing 1400 kJ of heat;So ;
0.1 mol etanol burnt releasing;
= 0.01 x 1400 kJ
= 14 kJ= 14000 J of heat
Step 3 : calculate the temperature rise for water
Total heat release = mc
14000 J = 200 x 4.2 x
7
Make sure the
unit is correct
Mass etanol
JMR etanol
0.46
46
Change kJ to J, we wantsubstitute it into equation
mc
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
73/75
= oC
= 16.7 oC
Question 3
Complete heat combustion for 1 mol butanol, C4H9OH release 2600 kJ
of heat. Calculate the mass for butanol that need to burnt completely so
heat release can increase the temperature for 500 cm3 of water to 36 oC
[J.A.R: C, 12; O, 16; H, 1]
Solution
Step 1 : calculate the heat changes/total heat release in this experiment
total heat release = mc
= 500 x 4.2 x 36 J
= 75600 J
= 75.6 kJ
Step 2 : calculate no. of mol for butanol that burnt[heat combustion for butanol = -2600 kJmol-1] (from ques.)
2600 kJ heat release by 1 mol butanol that burnt;
so ;
75.6 kJ heat release by mol butanol that burnt;
= 0.03 mol
No. of mol of etanol that burnt = 0.03 mol
Step 3 : calculate the mass of butanol that burnt
No. of mol butanol =7
14000 .
200 X 4.2
mass butanol
JMR butanol
1 x 75.6 kJ
2600 kJ
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
74/75
0.03 mol =
Mass of butanol = 0.03 x 74
= 2.22 gQuestion 4
Energy level diagram for combuction reaction for methanol, CH3OH i
shown below.
Heat that release from the complete combustion 8.0 g of methanol, isuse to hot 1 dm3 of water. Calculate the temperature rise of the water.
Solution
Step 1 : calculate no. of mol for ethanol that is burn
Mol no for etanol = = = 0.25 mol
Step 2 : calculate the total heat released/absorb in exp.
[heat of combustion for ethanol = -720 kJmol-1] (from question)
Energy
CH3OH (ce) + O
2(g)
CO2
(g) + 2H2O (ce)
H = - 720 kJmol-1
3
2
7
mass butanol .
(4 x 12) + 9 + 16 + 1
Mass of metanol
JMR metanol
8.0 .
12+(3x1)+16+1
7/27/2019 THERMOCHEMISTRY CHA4 FORM5
75/75
1 mol metanol burn releasing 720 kJ of heat;So ;
0.25 mol methanol that burn releasing;
= 0.25 x 720 kJ
= 180 kJ= 180000 J of heat
Step 3 : calculate the rising in water temperature
Volume of water = 1 dm3
= 1000 cm3
Total heat released = mc
180 000 J = 1000 x 4.2 x
= oC
= 42.85 oC
Learning Task 4.8: Problem solving pg 168 no. 1, 2, 3Effective Practise: pg 173 no. 1, 2, 3
Review Questions: pg 176 to 179
Kamal Ariffin B Saaim
SMKDBL
180 000 .
1000 X 4.2
Change kJ to J, we want
substitute it into formula
mc