​Standard enthalpy changes

​​In a nutshell

There are four key standard enthalpy changes to learn. Using this knowledge and the understanding of the specific heat formula, enthalpy changes can be calculated from experimental data. Data can be recorded using calorimetry, a method to measure changes in heat. 

Equations

Constants

​

Variable units

Types of standard enthalpy changes

Examples

Standard enthalpy change of reaction, $$\varDelta_r H^\theta$$:

$$2 H_2 + O_2 \rightarrow 2 H_2O$$

Formation of water

Standard enthalpy change of formation,$$\varDelta_f H^\theta$$:

$$C + 2H_2 \rightarrow CH_4$$

Formation of methane

​

Standard enthalpy change of combustion, $$\varDelta_c H^\theta$$:

$$C_3H_8 + 5 O_2 \rightarrow 3 CO_2 +4 H_2O$$

Combustion of propane

Standard enthalpy change of neutralisation, $$\varDelta_{neut} H^\theta$$:

$$CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$$

Reaction between ethanoic acid and sodium hydroxide

Measuring enthalpy changes

Calorimetry can be used to measure enthalpy changes during a reaction.

When the reactants are solids and liquids a calorimetry experiment can be carried out by adding the reaction mixture to a polystyrene cup. A thermometer should be in the mixture with a lid fastened to the top of the cup. Once the reaction has started, the thermometer is read at time intervals to measure changes in temperature.

When one of the reactants is a gas a calorimetry experiment can be carried out using a calorimeter. This will measure enthalpy changes for reactions where gases are produced e.g. combustion. The fuel will be burnt and there is a surrounding chamber of water with a stirrer and a thermometer. The temperature changes of the water are measured to determine the changes in heat for a reaction.

​

Specific heat formula

DEFINITION

The specific heat capacity of water is the heat energy, $$q$$ required to increase the temperature of $$1\ g$$​ of water by $$1\ K$$. This has a value of $$4.18\ J g^{-1} K^{-1}$$.

To calculate an enthalpy change during a reaction, you need to know the change in temperature, the mass of the solution and the specific heat capacity. This can be input into the specific heat formula. You also need to know the number of moles of reactants and products to work out the enthalpy change. Heat change is equal to enthalpy change when carried out at constant pressure so you can convert between the two.

$$q=mc\varDelta T$$

​​Note: $$\varDelta T$$ can be measured in $$K$$ or $$^o C$$ as the difference will be the same for both units of temperature.

​

Calculating standard enthalpy changes

Standard enthalpy changes can be calculated from experimental data using the specific heat formula ($$q=mc\varDelta T$$).

Note: Enthalpy changes are required in units of $$kJ\ mol^{-1}$$ so $$q$$ must be converted into $$kJ$$.​

Example:

$$50\ cm^3$$​ of $$2\ mol\ dm^{-3}$$​ $$CH_3COOH$$ was added to a beaker and neutralised using $$50\ cm^3$$ of $$2\ mol\ dm^{-3}$$ $$NaOH$$. There was a temperature increase from $$17.2\ ^oC$$ to $$28.4\ ^oC$$. Assuming that $$c=4.18\ Jg^{-1}K^{-1}$$ and $$d=1\ g\, cm^{-3}$$, calculate $$\varDelta_{neut} H^\theta$$ for the reaction in $$kJ\ mol^{-1}$$.

PROCEDURE

1. $$\varDelta_{neut} H^\theta$$ forms one mole of water so reactants and products in 1:1 ratio:

$$CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$$

​​

2. Add volumes of reactants together and work out mass:

 $$50\ cm^3+ 50\ cm^3=100\ cm^3$$​​

$$100\ cm^3\times 1\ g\ cm^{-3}=100\ g$$

3. Plug in numbers and convert from $$J$$ to $$kJ$$:

$$q=mc\varDelta T$$​​

$$q= 100\times4.18\times(28.4-17.2)$$​​

$$q= 4681.6\ J/1000$$

$$q=4.68\ kJ$$

4. Work out moles of NaOH as $$1:1$$​ ratio with water:

$$n (NaOH)= 2\times\frac{50}{1000}$$​​

$$n(NaOH)= 0.1\ moles$$

5. Work out $$\varDelta_{neut} H^\theta$$​ in units of $$kJ\ mol^{-1}$$​:

$$\varDelta_{neut} H^\theta=\frac{q}{n}$$​​

$$\varDelta_{neut} H^\theta=\frac{4.68}{0.1}$$

$$\varDelta_{neut} H^\theta=46.8\ kJ\ mol^{-1}$$

​​

6. Temperature increased so this is an exothermic reaction and enthalpy change is negative:

Therefore the answer is $$\varDelta_{neut} H^\theta=\underline{-46.8\ kJ\ mol^{-1}}$$

Note: An endothermic reaction would have a positive $$\varDelta H^\theta$$ instead.​​