5.2 - Energy Changes from Temperature
Energy change from a temperature change
- When a reaction happens in a solution, the energy given out or taken in shows up as a temperature change in the solution.
- That temperature change can be used to calculate the energy transferred.
Formula for energy transferred
Where:
- = Energy transferred, in joules (J)
- = Mass of the water or solution being heated or cooled, in grams (g); the mass of a solid added to it is not included
- = Specific heat capacity of the solution, in joules per gram per degree Celsius (J/g/°C); for water, about 4.2 J/g/°C
- = Temperature change, in degrees Celsius (°C)
Enthalpy change per mole
- is the energy transferred by whatever mass happened to react.
- Dividing (in kJ) by the amount in moles that reacted gives the enthalpy change, ΔH, in kJ/mol.
The sign of ΔH
- Exothermic (temperature rose) → negative ΔH.
- Endothermic (temperature fell) → positive ΔH.
Worked example - Enthalpy change per mole
15.0 g of ammonium nitrate, NH4NO3, is dissolved in 85.0 g of water. The temperature of the solution falls from 23.7 °C to 12.3 °C. The specific heat capacity of the solution is 4.2 J/g/°C. Calculate the enthalpy change per mole of ammonium nitrate dissolved.
Step 1 - Find the temperature change
ΔT = 12.3 − 23.7 = −11.4 °C, so the temperature change is 11.4 °C
The temperature fell, so the reaction took the energy in.
Step 2 - Find the mass of the solution in grams
The mass heated is the mass of the water; the mass of the dissolved solid is not added to it:
m = 85.0 g
Step 3 - Multiply m × c × ΔT
Q = 85.0 × 4.2 × 11.4 = 4,069.8 J
In kilojoules, that is 4,069.8 ÷ 1,000 = 4.0698 kJ.
Step 4 - Find the amount in moles
Relative atomic masses: N = 14, H = 1, O = 16, so the relative formula mass of NH4NO3 = 14 + (4 × 1) + 14 + (3 × 16) = 80.
amount of NH4NO3 = 15.0 ÷ 80 = 0.1875 mol
Step 5 - Divide the energy by the amount in moles, and give the sign
4.0698 ÷ 0.1875 = 21.7056 kJ/mol
The temperature fell, so the reaction took energy in and the value is positive.
Step 6 - Round the answer
The masses and temperatures are all given to three significant figures, so:
ΔH = +21.7 kJ/mol