Enthalpy of Combustion Calculations

This is part of the HSC Chemistry course under the topic Alcohols.

HSC Chemistry Syllabus

• conduct a practical investigation to measure and reliably compare the enthalpy of combustion for a range of alcohols

• write equations, state conditions and predict products to represent the reactions of alcohols, including but not limited to (ACSCH128, ACSCH136):

Calculating Enthalpy of Combustion in HSC Chemistry

This video walks through common calculation questions for enthalpy of combustion in HSC Chemistry.

 

Enthalpy of combustion is an essential concept in HSC Chemistry. Mastering the calculations will boost your exam performance. Here, we'll focus on top tips and common mistakes to avoid in enthalpy of combustion calculations.

Top Tips for HSC Chemistry Students

• Always start with a balanced chemical equation for the combustion reaction.
• Double-check the stoichiometric coefficients in the equation.

• Make sure the molecular formula of the alcohol or hydrocarbon is correct in order to calculate the correct molecular mass.

• Make sure units for energy, specific heat capacity and mass are consistent.

• Don't forget to express the result in correct units e.g. kilojoules per mole (kJ/mol).

• Consider energy losses to surrounding if applicable. 

Useful Formula for Calculations Involving Heat of Combustion

The heat absorbed by the water is given by

 

$$q = mc\Delta T$$

 

where 

• `q` = energy absorbed by water in Joules. This value should be positive as it is absorbed by water
• `m` = mass of water in kg
• `c` = specific heat capacity of water (`4.18 \times 10^3 \text{ J kg}^{-1} \text{  K}^{-1}`)
• `\Delta T` = change in temperature (in either Kelvins or degree Celsius)

 

It is important to understand that for change in temperature (`\Delta T`), the units Kelvins and degree Celsius are equivalent. For example, when temperature is increased from 25 ºC to 80 ºC (`\Delta T = 55^{\circ} \text{C}`), this is equivalent to an increase of temperature from 298 K to 353 K (`\Delta T = 55 \text{ K}`). 

 

Some amount of energy released by combustion is inevitably lost to the surrounding i.e. not absorbed by water. Therefore, energy absorbed by water (`q_{\text{absorbed}}`) is realistically never equal to energy released by combustion of the alcohol (`q_{\text{combustion}}`). The latter and former are only equal if assuming there is no energy lost to the surrounding. 

The enthalpy change or heat of combustion is calculated by dividing `q_{\text{combustion}}` by the number of moles of alcohol consumed during complete combustion.

 

$$n(\text{alcohol}) = \frac{m}{\text{molecular mass of alcohol}}$$

 

$$\Delta{H} = - \frac{q_{\text{combustion}}}{n}$$  

 

The negative sign in the formula reflects the fact that energy is released from the combustion reaction (exothermic). 

 

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