Heat of precipitation

In an experiment to determine the heat of precipitation of silver nitrate,
50 cm3 of 1.0 mol dm-3 hydrochloric acid is added to 50 cm3 of 1.0 mol dm-3 silver nitrate solution.
The following results are obtained.

Initial temperature of silver nitrate solution (oC)
28.0
Initial temperature of hydrochloric acid solution (oC)
29.0
Highest temperature of the mixture (oC)
35.5

Calculate the heat of precipitation of silver chloride, AgCl
[Specific heat capacity of water is 4.2 J g-1 oC-1 and density of water is 1 g cm-3]

(ans: - 58.8 kJ mol-1)

Heat of precipitation (Find the heat change)

Calculate the heat change when 200 cm3 of 0.5 mol dm-3 calcium chloride, CaCl2  solution  is added to 200 cm3 of 0.5 mol dm-3 sodium carbonate, Na2CO3 solution if the heat of precipitation of calcium carbonate, CaCO3 is  +12.6 kJ mol-1
[ Specific heat capacity of solution : 4.2 J g-1 0C-1 . Density of solution : 1 g cm-3 ]

(ans: 1.26 kJ)

Heat of displacement (Calculation)

To determine the heat of displacement of copper from a copper (ll) sulphate solution by zinc.

Procedure :
1. Measure 25 cm3 0.2 mol dm-3 copper(ll) sulphate solution and pour into a polystyrene cup.
2. Put the thermometer into the copper(ll) sulphate solution. Record the initial temperature,
3. Add half a spatula of zinc powder (in excess) quickly into copper(ll) sulphate solution.
5. Stir the mixture with the thermometer and record the highest temperature achieved.

Result :
Initial temperature of copper(II) sulphate solution        =  30.0 0C
Highest temperature of the mixture                              =  40.0 0C
Calculate the heat of displacement of copper from a copper (ll) sulphate solution by zinc.
[ Specific heat capacity of solution : 4.2 J g-1 0C-1 . Density of solution : 1 g cm-3 ]

(Ans: - 210 kJ mol-1)

Heat of displacement (Find the change of temperature)

In an experiment, excess magnesium powder is added to 50 cm3 of 0.25 mol dm-3 
iron(ll) sulphate solution. The thermochemical equation is shown below,

       Mg(s)   +   Fe2+ (aq)  à  Mg2+ (aq)  +   Fe (s)                  ΔH =  -80.6 kJ mol - 1  

What is the change in temperature?
[ Specific heat capacity of solution : 4.2 J g-1 0C-1 . Density of solution : 1 g cm-3 ]

(ans: 4.8 oC)

Heat of neutralisation (Experiment)

Heat of neutralisation (Calculation)

To determine the heat of neutralization between a strong acid (hydrochloric acid) and a strong alkali (sodium hydroxide)

Procedure :
1. Measure 50 cm3 2.0 mol dm-3 sodium hydroxide solution and pour it into the polystyrene cup,
2. Put the thermometer into the sodium hydroxide solution. Record the initial temperature,
3. Measure 50 cm3 2.0 mol dm-3 hydrochloric acid solution  and record the initial temperature,
4. Pour the hydrochloric acid solution quickly into the sodium hydroxide solution in the
    polystyrene cup.
5. Stir the mixture with the thermometer and record the highest temperature achieved.

Result :
Initial temperature of sodium hydroxide solution =  29.0 0C
Initial temperature of hydrochloric acid  solution =  29.0 0C
Highest temperature of the mixture                     =  42.0 0C
[ Specific heat capacity of solution : 4.2 J g-1 0C-1 . Density of solution : 1 g cm-3 ]

Calculate the heat of neutralization.
(Ans: - 54.6 kJ mol-1)

Electrolysis (Types of electrodes)

III) Types of electrodes used
Electrolysis of 0.2 mol dm-3 copper(II) sulphate solution
Using copper electrodes

Ions present:
Electrode
Anode (+)
Cathode (-)
Ions attracted


Ion discharged


Half Equation


Observation


Product formed
-

Colour of the electrolyte

Electrolysis of 1.0 mol dm-3 sodium chloride solution

II) The concentration of ions in the solution
Electrolysis of 1.0 mol dm-3 sodium chloride solution

Ions present:
Electrode
Anode (+)
Cathode (-)
Ions moving


Ion discharged


Reason


Half Equation


Observation


Product formed


Electrolysis of Aqueous solutions

1. Example, sodium chloride solution consists of 4 type of ions

Ions
Cation
Anion
Sodium chloride
Na+
Cl-
Water
H+
OH-


2. Three factors that affect the selective discharge of ions at the
    electrodes during electrolysis of an aqueous solution:
I    The position of ions in the electrochemical series.
II   The concentration of ions in the solution
III Types of electrodes used  



Observation
inference
Colourless gas released
H2 / O2
Greenish yellow gas
(Blue litmus red white)
Cl2
Brown gas
Br2
Grey solid deposited
Pb/Al
Brown solid deposited
Cu


Electrolysis (molten)

Electrolysis of molten lead(II) bromide

Ion present:
Electrode
Anode (+)
Cathode (-)
Ions moving


Half Equation


Observation


Product formed





Answer:

Reactivity of Group 1 elements



When going down Group 1, the alkali metals become more reactive when react with water

Carbon Compound





Properties of Covalent compound

(a) Low melting points and boiling points
Because:
-          Molecules held by weak intermolecular forces.
-          Needs less energy to overcome these forces.
-         
(b) Cannot Conduct electricity
Because:
-          consists of neutral molecules.
-          Do not have ions that can move freely.

(c) does not dissolve in water
(d) dissolve in organic solvents.

Properties of Ionic compound

(a) High melting points and boiling points.
Because:
-          Ions held by strong electrostatic forces.
-          Needs a lot of energy to overcome these forces.

(b) Conduct electricity when in molten or aqueous solution.
Because:
-          In solid, ions are fixed in positions and do not move freely.
-          In molten state or aqueous solution, the ions can move freely.

(c) dissolve in water
(d) does not dissolve in organic solvents.

Covalent Bond (non-metal + non-metal) Sharing of electron

[proton number: C:6, O:8]
- Electron arrangement of atom C is 2.4
- Atom C needs 4 more electron to achieve stable octet electron arrangement.
- Atom C contributes 4 electron for sharing
- Electron arrangement of atom O is 2.6
- Atom O needs 2 more electron to achieve stable octet electron arrangement
- Atom O contributes 2 electron for sharing.
- covalent compound, CO2 is formed

Ionic Bond (metal + non metal) Transfer of electron

[number of proton: Mg: 12, O: 8]
- Electron arrangement of magnesium atom is 2.8.2
- Magnesium atom donates 2 electron to achieve stable octet electron arrangement.
  Mg Mg2+ + 2e
- Electron arrangement of oxygen atom is 2.6
- Oxygen atom receives 2 electron to achieve stable octet electron arrangement.
   O + 2e O2-
- Mg2+ ion and O2‑ ion attract by strong electrostatic force to form ionic bond.
- Ionic compound, MgO is formed.

Period+Transition element

Quiz:

Answer:



Group 17

Quiz:

Answer:

Group 1

Quiz:

Answer:

Heat of combustion (calculation)


An experiment was carried out to determine the heat of combustion of ethanol.
The table below shows the experimental results
Volume of water in copper tin (cm3)
200
Mass of spirit lamp before heating (g)
189.55
Mass of spirit lamp after heating (g)
188.80
Initial temperature of water (C)
28.5
Maximum temperature of water (C)
53.5
a) Calculate the number of mole of ethanol used.
b) Calculate the heat absorbed by the water.
c) Calculate the heat of combustion of ethanol.
[Specific heat capacity of solution : 4.2 J g-1 0C-1 . Density of solution : 1 g cm-3 ] 
Relative atomic mass: H = 1; C = 12; O = 16]

Answer: a) 0.0163 mol  b) 21 kJ           c) – 1288.34 kJ mol-1