20) In an experiment, 50 cm3 of 1 mol dm-3 NaCl is added to 50 cm3 of
1 mol dm-3 AgNO3 solution . The following data is obtained
Initial temp of NaCl
Initial temp of AgNO3 solution
28.0 o C
29.0 o C
Highest temperature of mixture
35.5 o C
Given Density of solution : 1 g cm-3 , specific heat of solution is 4.2 J g-1 o C -1
a) What is the name of this reaction?
Precipitation reaction
mcO = 100 x 4.2 x 7 = 2940 J
b) Calculate heat change
c) Calculate heat of reaction ∆H = mcO/mol = 2940/1000x 0.05 = - 58.8 kJ mol-1
d) If NaCl solution is replaced with HCl solution, predict the heat of reaction .
Justify your answer
Heat of rex is the SAME ,because the precipitate is still the same, that is AgCl
e) If NaCl solution is replaced with Na2CO3 solution, predict the heat of
reaction . Justify your answer
Heat of rex is the DIFF , because the precipitate is NOT the same, that is Ag2CO3
21) In an experiment, 100 cm3 of 1.0 mol dm-3 CaCl2 solution is added
to 100 cm3 of 1.0 mol dm-3 Na2CO3 solution . The following data is
obtained
Ca 2+ + CO3 2-  CaCO3
∆H = + 12.6 kJ mol-1
Given Density of solution : 1 g cm-3 , specific heat of solution is 4.2 J g-1 o C -1
Calculate the temperature change of the mixture
∆H
= mcO / mol
therefore O = ∆H x mol / mc
= 12.6 x 1000 x 0.1 / 200 x 4.2
= 1.5 oC
22) In an experiment, 1 g of zinc powder is added to 50 cm3 of
0.2 mol dm-3 CuSO4 solution . The following data is obtained
28.0 o C
33.0 o C
Initial temp of CuSO4 solution
Highest temperature of mixture
Given Density of solution : 1 g cm-3 , specific heat of solution is 4.2 J g-1 o C -1
a) What is the name of this reaction?
b) Calculate heat change
c) Calculate heat of reaction
Displacement reaction
mcO = 50 x 4.2 x 5 = 1050 J
∆H = mcO / mol
= 1050 / 1000x 0.01
= - 105 kJ mol-1
23) In an experiment, 100 cm3 of 2.0 mol dm-3 HCl is added to 100 cm3
of 2.0 mol dm-3 NaOH solution . The following data is obtained
o
Initial temp of HCl solution
Initial temp of NaOH solution
29.5
29.5
C
oC
Highest temperature of mixture
41.5 o C
Given Density of solution : 1 g cm-3 , specific heat of solution is 4.2 J g-1 o C -1
a) What is the name of this reaction?
Neutralisation reaction
b) Calculate heat change
mcO = 200 x 4.2 x 12 = 10080 J
c) Calculate heat of reaction
∆H = mcO / mol = 10080 / 1000x 0.2
= - 50.4 kJ mol-1
21) The thermochemical equation for the reaction between ethanoic
acid and sodium hydroxide solution is given below
CH3COOH + NaOH  CH3COONa + H2O ∆H = -55 kJ mol-1
Given Density of solution : 1 g cm-3 , specific heat of solution is 4.2 J g-1 o C -1
Calculate the heat given out when 200 cm 3 of ethanoic acid is
added to 100 cm3 of sodium hydroxide solution . Both solutions
have a concentration of 0.5 mol dm-3
Mol of ethanoic acid = MV/1000 = 0.5 x 200/1000 = 0.1 mol
Mol of sodium hydroxide = MV/1000 = 0.5 x 100/1000 = 0.05 mol
∆H = mcO/mol
Therefore mcO =∆H x mol = 55x1000 x 0.05
= 2750 J
25) In an experiment, the following data is obtained
Volume of water used
100 cm3
Initial temp of of water
29.0 o C
Highest temperature of water reached
51.0 o C
Mass of spirit lamp and methanol before combustion 156.55 g
Mass of spirit lamp and methanol after combustion 156.05 g
[Given Density of water : 1 g cm-3 , specific heat capacity of water is 4.2 J g-1 o C -1
RAM : H,1; C,12; O,16 ]
a) What is the name of this reaction?
b) Calculate heat change
c) Calculate heat of reaction
Combustion
mcO = 100x 4.2 x 22 = 9240 J
∆H = mcO/mol = 9249 /1000x 0.0156
= - 592.3 kJ mol-1
25) The heat of combustion of ethanol is 1376 kJ mol-1 What is its fuel value?
Fuel value = heat of combustion/ Molar mass
Fuel value = 1376/ 46 = - 29.9 kJ g -1
C3H7OH +
9/2
O2 3 CO2 +4 H2O
Heat released = mcO
=200 x 4.2 x 31
=26040 Joule
Mole = mass/molar mass
= 0.84/60 = 0.014
0.014 mole of propanol released heat 26040 joule
So 1 mol propanol  1x 26040/0.014
=1860 kJ
Heat of combustion = -1860 kJ mol -1
= -1860 kJ mol -1
Heat change
……………………………………….
when 1 mol of
Magnesium carbonate…..
………………………………….
is formed from
magnesium ions and carbonate ions
…………………………………………
…..in solution
Mg
2+
+ CO3
2-
 MgCO3
mco = 50 x 4.2 x 5.5 = 1155J
Mole = MV/1000 = 25 x2 /1000 = 0.05 mol
No of mole of MgCO3 = no of mole of magnesium ion
= 0.05 mol
In thermochemistry,
do not use from equation
0.05 mol of MgCO3 release heat 1155 J
So 1 mole of MgCO3  1 x 1155/0.05 Joule = 23.1 kJ
Heat of precipitation = + 23.1 kJ mol -1
H= positive
Same precipitate is formed, that is magnesium carbonate
/ sodium ion or potassium ion does not take part in the reaction
Oxidised: iron (II) sulphate / Fe 2+
Reduced : Acidified potassium manganate(VII)
Oxidation:
Reduction:
K
Mn
O4 = 0
A)AIM OF EXPERIMENT
Able to give the aim of the experiment
correctly
Sample answers
To investigate the effect of Metal Mg and Cu
when it in contact with iron in the rusting of
iron//
To study the effect of Metal Cu can faster the
rusting of iron while metal Mg prevent the iron
from rusting
a. variable
Manipulated variable :
Metal X and metal Y
//two different metals(one metals is less
electropositive and one is more
electropositive than iron)
//pairs of metal X / iron and Y/iron
Responding variable :
Rusting of iron
//iron rust
//the formation of brown solid
//formation of blue spot
Constant variables:
Iron nail
//jelly solution
//temperature
c. hypothesis
Able to state the relationship between the
manipulated variable and the responding variable
with direction correctly.
Sample answer
When a more electropositive metal is in contact with
iron, the metal inhibits rusting.
// When a less electropositive metal is in contact
with iron, the metal speeds up rusting.
d. Materials/substances
Material & apparatus
Two Iron nails,
Magnesium/zinc/aluminium strip,
tin/copper/lead/silver strip
Potassium hexacyanoferrate(III) solution +
phenolphthalein
Apparatus Test tube/boiling tube, Sand paper
e. procedure
Sample answer:
•Clean the iron nails and metals strip with sand paper.
•Coil iron nails with magnesium ribbon and copper
strips.
•Put/place the coiled iron nail into different test tube.
•Pour/add/fill the hot jelly solution containing
potassium hexacyanoferrate(III)solution and
phenolphthalein into the test tube.
e. procedure
Sample answer:
•Leave the test tube in a test tube rack for few days.
•Record the observation.
•Steps 1 to 6 are repeated using different metal/Y with
iron(if steps 2 does not mention two different test
tube).
f. Tabulated data
Pair of metals
Mg/Fe
Cu/Fe
Observation
Revision redox
Redox reaction or non redox?
• Which one is REDOX? State your reason
a) AgNO3 + NaCl  AgCl + Na NO3
b) Cl2 +2 KI  2KCl + I2
Reaction a) is not a redox reaction because
oxidation number
of Ag does not change, that is from +1 to +1
Reaction b) is a redox reaction because
oxidation number
of Cl change, from 0 to -1
State oxidation number for the underlined
(+1 + Mn + 4O = 0 ) , (+1 + Mn -8 = 0 ),
• KMnO4
Mn
=
+7
•KI
(+1 + I = 0 ) , I = -1
• Fe Cl2
+2
• I2
0
• Cu
• CuSO4 0
• MnO4 – +2
(Mn + 4O = -1) , Mn = +7
• H3O +
(3H -2 = + 1) , H = +1
0
oxidation
+2
Mg + 2FeCl3  2FeCl2 + MgCl2
+3
reduction
+2
Which is oxidized ? reason
Mg because ON of Mg increase from 0 to +2
Which is reduced? Reason
FeCl3 because ON of Fe decrease from +3 to +2
Which is oxidizing agent, why
FeCl3 because it oxidise Mg to MgCl2
Which is reducing agent?why
Mg because it reduce FeCl3 to FeCl2
0
reduction
-1
Br2 + 2NaI  2NaBr + I2
-1
oxidation
0
Which is oxidized ? reason
Sodium iodide because ON of iodine increase from -1 to 0
Which is reduced? Reason
Bromine because ON of Br decrease from 0 to -1
Which is oxidizing agent, why
Bromine because it oxidise NaI to I2
Which is reducing agent? why
NaI because it reduce Br2 to NaBr
0
oxidation
+2
Zn + CuSO4  ZnSO4 + Cu
+2
reduction
0
Which is oxidized ? reason
Zinc because ON of zinc increase from 0 to +2
Which is reduced? Reason
CuSO4 because ON of copper decrease from +2 to 0
Which is oxidizing agent, why
CuSO4 because it oxidise Zn to ZnSO4
Which is reducing agent? why
Zinc because it reduce CuSO4 to Cu
Reaction A is NOT a redox reaction because Oxidation
number of Na does not change, that is from +1 to +1
Reaction B is a REDOX reaction because Oxidation
number of Mg change from 0 to +2
In P, +2
P is copper(II) oxide
Because ON of copper is +2
In Q, +1
Q is copper(I) oxide because
ON of copper is +1
+2
Oxidisin
g agent
reduction
0
Reducing
agent
0
oxidation
+1
+2
reductio
n
0
0
+1
oxidation
H2
Because ON of hydrogen
increased from 0 to +1
Because ON of copper
decreased from +2 to 0
Because it oxidise
CuO
hydrogen to hydrogen
oxide
H2
Because it reduce CuO to Cu
CuO
State the fuction of
i) Glass wool
Prevent metal from mixing with KMnO4 ( can cause explosion)
i) Potassium manganate (VII) powder
release oxygen when heated to react with metal
powder