Study CheMisTry with LKY: 2011

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, CaCl₂solution 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 0}C^-1 . Density of solution : 1 g cm^-3 ]

(ans: 1.26 kJ)

Heat of displacement (Experiment)

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 0}C^-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 0}C^-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^-3sodium 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^-3hydrochloric 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 0}C^-1 . Density of solution : 1 g cm^-3 ]

Calculate the heat of neutralization.

(Ans: - 54.6 kJ mol^-1)

Heat of combustion (experiment)

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