Summary iGCSE Chemistry Pearson Edexcel Topic 3 Physical Chemistry Complete Notes

IGCSE CHEMISTRY – TOPIC 3
PHYSICAL CHEMISTRY

a) ENERGETICS

• EXOTHERMIC – a reaction in which energy is GIVEN OUT to it’s
surroundings (temperature of environment increases) – the energy
changes is NEGATIVE (e.g. burning, neutralization, reaction between
water and calcium oxide)
• ENDOTHERMIC – reaction in which energy is TAKEN IN from
surroundings (temperature of environment decreases) – the energy
change is POSITIVE (e.g. electrolysis, thermal decomposition of copper
(II) carbonate, reaction between ethanoic acid and sodium carbonate)
HEAT ENERGY CHANGE CALCULATIONS:

• From the data we gain in a calorimetry experiment, we can calculate
the heat energy change of a reaction:

Q = m x c x △T
Q – energy transferred to water
m – mass of water (if water is used, replace with other
solutions)
c – the specific heat capacity – the value is 4.18
△T – the change in temperature of water
Example:

The energy from burning 0.5 g of propane was transferred to 100 cm3 of water to raise
its temperature by 20°C. Calculate the heat energy change (in kJ)

Mass of Water = 100 cm3, Heat capacity of Water = 4.2 j / g / C, Temperature
rise = 20°C

Energy transferred = 100 x 4.2 x 20 = 8400 J (1000 J = 1 kJ)

So 8400 J = 8.4 kJ
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, HEAT ENERGY CHANGE CALCULATIONS:

• ΔH is the symbol that represents the AMOUNT OF ENERGY LOST OR
GAINED in a reaction

• +ΔH is endothermic (because it gains heat from the surroundings)

• -ΔH is exothermic (because it loses heat to the surroundings)

• The unit of ∆H is kJ/moL

• TO WORK OUT THE MOLAR ENTHALPY CHANGE WE NEED TO WORK
OUT THE MOLES AND THEN USE THIS EQUATION:


ΔH = Q/moles
Q – energy transferred to water
ΔH – molar enthalpy change – in Kj/mol
Moles – mass/RFM

Example:

The energy from burning 0.5 g of propane was transferred to 100 cm3 of water to
raise its temperature by 20°C. Calculate the molar enthalpy change (in kJ / mol).

Mass of Water = 100 cm3, Heat capacity of Water = 4.2 j / g

Temperature rise = 20°C, Energy
transferred = 100 x 4.2 x 20 = 8400 J

1000 J = 1 kJ So 8400 J = 8.4 kJ

Mr of Propane = 44 - Moles of Propane burned = 0.5 ÷ 44 = 0.01136

Molar Enthalpy change = 8.4 ÷ 0.01136 = 739

This Reaction is Exothermic so Enthalpy Change Needs to be Negative

Molar Enthalpy change = – 739 kJ / mol
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