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Energy changes for AQA A-level Chemistry

Energy changes

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.

Thermal energy

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.

Enthalpy change

1

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

Define standard enthalpy of formation.

2

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 = mcฮ”T
E = 0.500 ร— 4180 ร— 62.8
E = 131 kJ (3 s. f.)

n = 4.00 รท 12 = 0.333 mol
ฮ”H = โˆ’E รท n
ฮ”H = โˆ’131 รท 0.333
ฮ”H = โˆ’394 kJ/mol

โˆ’394 kJ/mol

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).

3

Delphine has mixed 0.25 mol of NaOH dissolved in 100 cmยณ of water and 0.25 mol of HClโ‚ƒ dissolved in 100 cmยณ of water. During the reaction the temperature of the total water within the solutions has increased from 10ยฐC to 26ยฐC. 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 cmยณ + 100 cmยณ = 200 cmยณ
mass of water = 0.200 kg

E = mcฮ”T
E = 0.200 ร— 4180 ร— (26 โˆ’ 10)
E = 13000 J = 13 kJ (2 s. f.)

ฮ”H = โˆ’E รท n
ฮ”H = โˆ’19 รท 0.25
ฮ”H = โˆ’54 kJ/mol (2 s. f.)

โˆ’54 kJ/mol

Delphine has mixed 0.25 mol of NaOH dissolved in 100 cmยณ of water and 0.25 mol of HClโ‚ƒ dissolved in 100 cmยณ of water. During the reaction the temperature of the total water within the solutions has increased from 10ยฐC to 26ยฐC. Find the enthalpy of neutralisation for this reaction in kJ/mol. The specific heat capacity of water is 4180 J/(kgK).

4

Andrew has mixed 1.00 mol of KOH dissolved in 50 cmยณ of water and 1.00 mol of HNOโ‚ƒ dissolved in 100 cmยณ of water. During the reaction the temperature of the total water within the solutions has increased from 5.0ยฐC to 95.9ยฐC. Find the enthalpy of neutralisation for this reaction. The specific heat capacity of water is 4180 J/(kgK).

volume of water = 50 cmยณ + 100 cmยณ = 150 cmยณ
mass of water = 0.150 kg

E = mcฮ”T = 0.1500 ร— 4180 ร— (95.9 โˆ’ 5.0) = 57000 J (3 s. f.)
ฮ”H = โˆ’E รท n
ฮ”H = โˆ’57000 รท 1.00 = โˆ’57000 kJ/mol

โˆ’57000 J/mol

Andrew has mixed 1.00 mol of KOH dissolved in 50 cmยณ of water and 1.00 mol of HNOโ‚ƒ dissolved in 100 cmยณ of water. During the reaction the temperature of the total water within the solutions has increased from 5.0ยฐC to 95.9ยฐC. Find the enthalpy of neutralisation for this reaction. The specific heat capacity of water is 4180 J/(kgK).

5

Nikesh has carefully placed a small piece of sodium in 200 cmยณ of water. Calculate the temperature change of the water if the energy released during the process is 75 kJ. The specific heat capacity of water is 4180 J/(kgK).

200 cmยณ of water have a mass of 200 g, which is 0.200 kg.
75 kJ = 75000 J

E = mcฮ”T
75000 = 0.200 ร— 4180 ร— ฮ”T
75000 = 836 ร— ฮ”T
ฮ”T = 75000 รท 836 = 89.7 K (3 s. f.)

89.7 K

Nikesh has carefully placed a small piece of sodium in 200 cmยณ of water. Calculate the temperature change of the water if the energy released during the process is 75 kJ. The specific heat capacity of water is 4180 J/(kgK).

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