When (0.5g) of propane is burned, the heat produced is used to raise the temperature of (100cm3) of water from (20°C to 40°C), calculate the enthalpy change for the reaction.
In this example, you are measuring the heat change indirectly. You measure the rise in temperature of the water. The temperature change in the water as it gets hotter will help you calculate the enthalpy change of the methane as it is burned. This method isn’t very accurate because some heat is lost to the environment (not all goes to the water) and the loss isn’t recorded. But for a small experiment held in the laboratory, this is an effective method.
q = m • c • Δt
Note: In case you’ve not come across this before, in America the • sign centered on the cap height is used interchangeably with the x sign to indicate multiplication.
Step 1: m = mass of water
- The density of water is 1 g/cm3
- Mass of 100 cm3 of water = 100g
Step 2: C = specific heat capacity of water (c)
- That means that it takes 4.18 joules to increase the temperature of 1 gram of water by 1 K (or 1°C)
- This value can be memorized or should be given to you in the test or exam
Step 3: Δt = change in temperature
- Δt = 40°C – 20°C = 20°C
Step 4: Apply calculation
- Equation q = m • c • Δt
- q = 100 g x 4.18 j/g°c x 20°c = 8,360 j or 8.40 k
Step 5: Divide by the number of moles to get molar enthalpy change of the propane gas reaction
- Moles of propane = Mass ÷ Molar Mass
- 0.5 ÷ 44 = 0.1135 moles
- ∆H / mol = 8.4 ÷ 0.01136 = 739 kJ/mol
Step 6: ΔH is defined as the heat flow at constant pressure. The heat absorbed or released at constant temperature is equal to ΔH, the relationship between heat (q) and ΔHrxn is:
- ΔHrxn = − m • c • Δt= -739 kJ/mol(rxn)
- rxn means reaction
- The heat is given off by the propane and absorbed by the water
- The reaction for propane is negative — exothermic — because heat is given off