SPM Chemistry Form 5 Notes

SPM Chemistry Form 5 Notes – Thermochemistry (Part 2)

by BERRYBERRYTEACHER on 07/01/2011

in BERRY REFERENCE (NOTES)

From Part 1 of Berry Berry Easy‘s SPM Form 5 Chemistry chapter of Thermochemistry, Berry Readers would have known that this topic is all about uncovering the simplified theory behind energy and heat coming from chemical reactions or physical transformations. If Part 1 was heavily inclined towards definitions f, this post (Part 2) would include a lot of reactions and some simple calculations. The focus of this post would be onenergy change during formation and breaking of bonds, application of exothermic and endothermic reactions in everyday life, heat of precipitation and heat of displacement.

Tips to learn: The key point for this post would be to get yourself familiarise with the simple calculation of relationship between energy change and the formation and breaking of bonds. You need to understand which is positive and which is negative for ‘bond forming’ and ‘heat of reaction’ if you wish to do well in the calculation part.

SPM Chemistry Form 5 – Terminology and Concepts: Thermochemistry (Part 2)

Energy Change during Formation and Breaking of Bonds

Bond breaking

• Usually chemical bonds of the reactant.
• Heat energy is absorbed.

Bond forming

• Usually new chemical bonds of the product.
• Heat energy is given out.

Relationship between energy change and the formation and breaking of bonds

• In a chemical reaction, if the heat energy absorbed in bond breaking is lower than the heat energy given out in bond forming, the reaction is an exothermic reaction.
• Example: ΔH (bond breaking) = +600 kJ, ΔH (bond forming) = -800 kJ, ΔH (heat of reaction) = [(+600) + (-800)] kJ= -200 kJ
• In a chemical reaction, if the heat energy absorbed in bond breaking is higher than the heat energy given out in bond forming, the reaction is an endothermic reaction.
• Example: ΔH (bond breaking) = +800 kJ, ΔH (bond forming) = -600 kJ, ΔH (heat of reaction) = [(+800) + (-600)] kJ= +200 kJ

Applications of Exothermic and Endothermic Reaction in Everyday Life

Hot pack

• Contains of anhydrous calcium chloride / anhydrous magnesium sulphate / wet iron powder and sodium chloride / calcium oxide.
• Uses: reduce swelling and mucles or joint sprain.

Cold pack or Ice pack

• Contains of ammonium nitrate / potassium nitrate / sodium thiosulphate.
• Uses: reduce swelling, muscles or joint sprain and reduce fever.

*Berry Important Notes

You must be able to

1. Calculate the number of moles of salt precipitated / metal displaced / water produced / fuel used;
2. Calculate the heat energy released (ΔH); and
3. Calculate the heat of precipitation / heat of displacement / heat of neutralization / heat of combustion.

Heat of Precipitation (Form 4, Chapter 8 Salts)

1. Heat of precipitation – the heat change when one mole of a precipitate is formed fromtheir ions in aqueous solution.
2. Precipitation reaction = double decomposition which is used to prepare insoluble salts.
3. Heat change of a solution = mcθ Joule [m = mass of the solution (g), c = specific heat capacity of the solution (J g^-1˚C^-1), θ = temperature change in the solution (˚C)]
4. Heat change in a reaction, mcθ = n x ΔH
5. Heat of reaction / Heat of precipitation, ΔH = mcθ / n

Example 1:
Chemical reaction: Pb(NO₃)₂(aq) + 2KI(aq) –> PbI₂(s) + 2KNO₃(aq)
Ionic reaction: Pb²⁺(aq) + 2I^-(aq) –> PbI₂(s)
Heat of precipitation of PbI₂ = – Heat change / Number of moles of PbI₂

Example 2:
Chemical reaction: BaCl₂(aq) + Na₂SO₄(aq) –> BaSO₄(s) + 2NaCl(aq)
Ionic reaction: Ba²⁺(aq) + SO₄^2-(aq) –> BaSO₄(s)
Heat of precipitation of BaSO₄ = – Heat change / Number of moles of BaSO₄

Heat of Displacement (Form 4, Chapter 6 Electrochemistry & Form 5, Chapter 3 Oxidation and Reduction)

1. Heat of displacement – the heat change when one mole of a metal is displaced from its salt solution by a more electropositive metal.
2. Heat change of the reaction mixture / Heat energy released / Heat given out in the reaction = mcθ Joule
3. Heat change in a reaction, mcθ = n x ΔH
4. Heat of reaction / Heat of displacement, ΔH = mcθ / n

Example 1:
Chemical equation: Mg(s) + FeCl₂(aq) –> MgCl₂(aq) + Fe(s)
Ionic equation: Mg(s) + Fe²⁺(aq) –> Mg²⁺(aq) + Fe(s)

Example 2:
Chemical equation: Zn(s) + CuSO₄(aq) –> ZnSO₄(aq) + Cu(s)
Ionic equation: Zn(s) + Cu²⁺(aq) –> Zn²⁺ (aq) + Cu(s)

The upcoming Part 3 in this series of notes on Thermochemistry would feature solely on the typically easily understood subtopic on Heat of Neutralisation

SUMBER:http://berryberryeasy.com/2011/01/spm-chemistry-form-5-notes-thermochemisty-part-2/