Energy changes for AQA A-level Chemistry
This page covers the following topics:
1. Thermal energy
2. Enthalpy change
Thermal energy released in a chemical reaction can be obtained by heating a known mass of water in an isolated system and measuring the change in temperature. The energy released is then found by multiplying the mass of water, specific heat capacity of water and the change in temperature.
Standard enthalpy change is the amount of energy gained by a reaction mixture per one mole of a reactant at standard room temperature and pressure. Enthalpy of combustion refers to the combustion of material, enthalpy of neutralisation to neutralisation reactions, and enthalpy of formation is used when considering elements bonding together to make a compound of consideration.
Negative enthalpy change suggests that a reaction mixture is losing energy and therefore the reaction is exothermic. Positive enthalpy change means that a reaction mixture is gaining energy and thus the reaction is endothermic.
Define standard enthalpy of formation.
Standard enthalpy of formation is the amount of energy gained by elements bonding to make one mole of a compound at standard room temperature and pressure.
amount of energy gained by elements making one mole of a compound at standard conditions
Combustion of 4.00 g of carbon has increased the temperature of 500 g of water by 62.8 K. Find the enthalpy of combustion of carbon in kJ/mol. The specific heat capacity of water is 4180 J/(kgK).
500 g = 0.500 kg
E = mcΔT
E = 0.500 × 4180 × 62.8
E = 131 kJ (3 s. f.)
n = 4.00 ÷ 12 = 0.333 mol
ΔH = −E ÷ n
ΔH = −131 ÷ 0.333
ΔH = −394 kJ/mol
Delphine has mixed 0.25 mol of NaOH dissolved in 100 cm³ of water and 0.25 mol of HCl₃ dissolved in 100 cm³ of water. During the reaction the temperature of the total water within the solutions has increased from 10°C to 26°C. Find the enthalpy of neutralisation for this reaction in kJ/mol. The specific heat capacity of water is 4180 J/(kgK).
volume of water = 100 cm³ + 100 cm³ = 200 cm³
mass of water = 0.200 kg
E = mcΔT
E = 0.200 × 4180 × (26 − 10)
E = 13000 J = 13 kJ (2 s. f.)
ΔH = −E ÷ n
ΔH = −19 ÷ 0.25
ΔH = −54 kJ/mol (2 s. f.)
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