Termokimia-part I

salam...

Fokus pada kali ini adalah berkaitan tajuk termokimia.Daripada analisis yang telah cikgu buat topik ini amat popular dan antara topik yang wajib keluar dalam SPM.Beberapa penekanan perlu dibuat untuk menguasai topik ini dengan baik antaranya ialah

1. Tindakbalas exothermic
reactant released heat to surrounding >>>the container feels hot when touch>>>temperature of the surroundings increases
- pelajar seharusnya dapat menyenaraikan contoh2 tindakbalas exorthermic (e.g neutralisation, combustion, metal with acid)

2. Tindakbalas endothermic
reactant absorbs heat from surrounding >>>the container feels cool when touch>>>temperature of the surroundings drops
- contoh tindakbalas: decomposition

3.Energy level diagram
-pelajar mesti mengetahui perbezaan antara energy level diagram bagi kedua-dua jenis tindakbalas.

4.Heat change and Heat of Reaction
-ramai pelajar keliru antara heat change dan heat of reaction.

HEAT CHANGE 
- haba yang dibebaskan/diserap apabila berlakunya perubahan suhu dan rumus yang mesti digunakan ialah
                                                        Q=mcT        , T-perubahan suhu
contoh:

in the experiment, 25.0 cm³ of 0.2 mol dm^-3 copper(II) sulphate solution reacted with 0.5g of zinc. The result of the experiment is shown below :

            Initial temperature copper(II) sulphate : 29°C

            Final temperature for the mixture         : 38°C

Calculate the heat of change.

v  Chemical equation	CuSO4 + Zn → Cu + Zn SO4
v  Thermochemical reaction	Cu2+ + Zn → Cu + Zn2+
v  Number of mole for displacement metal	Mole of Cu2+     =  MV                             1000                          = (0.2)(25.0)                                  1000                          = 0.005 mole
v  Changes of heat in the reaction	Q = mcθ J             m = 25.0 cm3= 25 g              c = 4.2 Jg-1°C-1  θ = 38°C - 29°C     = 9°C             Q = mcθ J     = 25(4.2)(9) J     = 945 J     Therefore, 0.005 mole of Cu2+ released 945 J of heat.

HEAT OF REACTION 
-perubahan haba apabila 1 mol bahan tindakbalas bertindakbalas atau 1 mol hasil tindakbalas terbentuk
- boleh dikenalpasti daripada persamaan termokimia iaitu H= x kj/mol
- terbahagi kepada 4 tindakbalas ( PN. CD)
                                                     P - precipitation
                                                     N - neutralisation
                                                     C - combustion
                                                     D - displacement
- boleh dikira apabila kita mengetahui heat change untuk tindakbalas tersebut
contoh:

In the experiment, 25.0 cm³ of 0.2 mol dm^-3 copper(II) sulphate solution reacted with 0.5g of zinc. The result of the experiment is shown below :

            Initial temperature copper(II) sulphate : 29°C

            Final temperature for the mixture         : 38°C

Calculate the heat of displacement.

v  Chemical equation	CuSO4 + Zn → Cu + Zn SO4
v  Thermochemical reaction	Cu2+ + Zn → Cu + Zn2+
v  Number of mole for displacement metal	Mole of Cu2+     =  MV                             1000                          = (0.2)(25.0)                                  1000                          = 0.005 mole
v  Changes of heat in the reaction	Q = mcθ J             m = 25.0 cm3= 25 g              c = 4.2 Jg-1°C-1  θ = 38°C - 29°C     = 9°C             Q = mcθ J     = 25(4.2)(9) J     = 945 J     Therefore, 0.005 mole of Cu2+ released 945 J of heat.
v  Heat of displacement	0.005 mole of Cu2+ : 945 J     1.0 mole of Cu2+ : x J x = 1.0(945)    J          0.005    = 189000 J    = 189 kJ Therefore, heat of displacement of Cu2+ is -189 kJ mol-1

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