1. Enthalpy
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2. Enthalpy H
Enthalpy is the energy content of a substance ( potential energy).
We can calculate the enthalpy for a reaction by
∆H = Hp – H r
Exothermic reactions – energy is released to the surroundings.
In an exothermic reaction the products have less potential energy than the reactants.
Therefore ∆H = Hp – H r is negative.
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3. Endothermic reactions – energy is absorbed from the surroundings.
The products have more potential energy that the reactants.
Therefore ∆H = Hp – H r is positive.
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5. Examples Exothermic reactions
Combustions, neutralising acids with alkali’s.
Endothermic reactions
Dissolving some salts (NH4NO3)
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6. Enthalpy of Combustion
This is the enthalpy change when 1 mole of a substance is burned completely in oxygen.
We use the equation
Eh = c m ΔT
ΔT= change in Temperature
C = specific heat capacity of water
this is a constant – 4.18kJKg-1C-1
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7. Enthalpy of Combustion
m = mass of water that is absorbing the heat.
( e.g. 1 cm3 has a mass of 0.001kg,100cm3 = 0.1 kg etc)
Worked Example
Calculate the Enthalpy of combustion of Ethanol.
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8. Example Information Mass of burner at start = 80.63g
Mass of burner at end = 80.48g
T at start = 20.5 oC
T at end = 30.5 oC
Volume of water = 100 cm3
Mass of Ethanol = 0.15g
Rise in T = 10 oC
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9. Calculation Eh = c m ΔT 4.18 x 0.1 x 10 =4.18 kJ
0.15g of Ethanol produces 4.18kJ when burned.
Mass of 1 mole of Ethanol = 46g
So – 0.15g kJ
46g x 46/ =1282 kJ mol/l
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10. Enthalpy of Solution
This is the enthalpy change when 1 mole of a substance is dissolved in water.
Again we use: Eh = c m ΔT
Example
When 2g of NaOH is dissolved in 50 cm3 of water the temperature changes from 20 oC to 30 oC.
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11. Calculation c= 4.18 m = 0.05 ΔT = 10 Eh= c m ΔT 4.18 x 2 x 10
= 2.09 kJ mol/l
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12. The Answer ! Mass of 1 mole of NaOH = 40g 2g ----------2.09 kJ
40g /2 x 2.09
= 41.8 kJ mol/l
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13. Enthalpy of Neutralisation
This is the energy released when 1 mole of water is produced in a neutralisation reaction.
Example
Calculate the enthalpy of neutralisation when 50 cm3 of
Na OH ( 1 mol/l) neutralises 50 cm3 of H Cl. (1 mol/l)
The temperature rise is 6 oC.
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14. The calculation!
H Cl + Na OH —> Na Cl + H2O
Eh = c m ΔT
4.18 x 0.1 x 6 ( add mass of NaOH and HCl to get mass of water.)
2.52 kJ
Number of Moles of Water
From equation 1 mol H+ ions will give 1 mole H2O.
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15. More! n=cv 1 x 0.05 = 0.05 Heat released per mole of water
0.05 moles = 2.52 kJ
I mole = 1/0.05 x 2.52
= 50.4 kJ
Since heat is given out the reaction is exothermic –
ΔH neutralisation = Kj
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