Calculating Enthalpy Change using Enthalpy of Formation
HSC Chemistry Syllabus
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Explain the enthalpy changes in a reaction in terms of breaking and reforming bonds, and relate this to:
– The Law of Conservation of Energy
Calculating Enthalpy Change using Enthalpy of Formation
This video will explain how to calculate enthalpy change of reaction using enthalpy of formation values.
Enthalpy Change vs Enthalpy of Formation
Enthalpy is a measurement of energy in a chemical system that changes as chemical reactions proceed. Enthalpy change (`\Delta H`) is the change of this energy throughout a chemical reaction, most of which is attributed to the change in chemical potential energy (bond energy).
Enthalpy change of formation (`\Delta H_f`) is the enthalpy change when a compound is formed from its constituent elements in standard states. For example, the following equation demonstrates the formation of methane from carbon and hydrogen in their standard states. The enthalpy change of formation of methane is –75 kJ per mole of methane formed.
$$C(s) + 2H_2(g) \rightarrow CH_4(g) \hspace{2cm} \Delta H_f^{\circ} = -75 \text{ kJ/mol}$$
The specific reaction is exothermic, meaning energy is released.
Relationship between Enthalpy Change and Enthalpy of Formation
The following formula can be used to calculate enthalpy change of a reaction using enthalpy of formation values of its reactants and products.
$$\Delta H_{\text{reaction}} = \Sigma \Delta H_f^{\circ} \text{(products)} - \Sigma \Delta H_f^{\circ} \text{(reactants)}$$
The enthalpy change of a reaction equals the difference between the sum of enthalpy of formation values of products and that of reactants. In a chemical reaction, the number of atoms of each element must be equal before and after the reaction (law of conservation of mass). Before atoms can be assembled to form products, they must be first disassembled from reactants.
energy is released when bonds are formed in the products. This amount of energy is the enthalpy of formation
Consider the formation of carbon dioxide from carbon monoxide and oxygen gas.
$$CO(g) + \frac{1}{2} O_2(g) \rightarrow CO_2(g)$$
Before carbon dioxide molecules can be formed, carbon and oxygen atoms must be produced by breaking bonds in carbon monoxide and oxygen molecules. The enthalpy change of the former reaction is the negative enthalpy change of formation of carbon monoxide.
$$CO(g) \rightarrow C(s) + \frac{1}{2} O_2(g) \hspace{2 cm} \Delta H = -\Delta H_f (CO)$$
Enthalpy change of formation of oxygen gas is zero because it is already in its standard state.
This is then followed by the reaction between solid carbon and oxygen gas to form carbon dioxide. The enthalpy change of this reaction is simply the enthalpy of formation of carbon dioxide.
$$C(s) + O_2(g) \rightarrow CO_2(g) \hspace{2cm} \Delta H = \Delta H_f (CO_2)$$
Therefore, the enthalpy change of the reaction between carbon monoxide and oxygen gas to form carbon dioxide is given by the sum of enthalpy changes of these two reactions.
$$\Delta H_{\text{reaction}} = -\Delta H_f(CO) + \Delta H_f(CO_2)$$
$$\Delta H_{\text{reaction}} = \Delta H_f \text{(product)} - \Delta H_f \text{(reactants)}$$
The enthalpy of formation values are given:
- `\Delta H_f (CO) = -110 \text{ kJ/mol}`
- `\Delta H_f (CO_2) = -395 \text{ kJ/mol}`
The enthalpy change of reaction can therefore be calculated (keeping in mind enthalpy of formation of oxygen gas is zero). The enthalpy change of formation should also be multiplied by the respective stoichiometric ratio.
$$\Delta H_{\text{reaction}} = -395 \times 1 - (-110 \times 1) = -285 \text{ kJ/mol}$$
The enthalpy changes are represented by the following diagram.
The above diagram shows that the enthalpy change of a reaction (formation of carbon dioxide) remains unchanged regardless of the pathway taken. The first pathway is a direct one (blue), carbon dioxide is formed directly from carbon and oxygen in their standard states. The second pathway is a stepwise one, carbon monoxide is formed first from carbon and oxygen in their standard states, followed by reaction with oxygen to form carbon dioxide. The net enthalpy change is the same. This phenomenon is known as Hess's Law.
Example 1
Calculate the enthalpy change for the oxidation of sulfur dioxide to form sulfur trioxide given the enthalpy of formation values for sulfur dioxide and sulfur trioxide.
$$S(s) + O_2(g) \rightarrow SO_2(g) \hspace{2cm} \Delta H_f = –297 \text{ kJ/mol}$$
$$S(s) + \frac{3}{2} O_2(g) \rightarrow SO_3(g) \hspace{2cm} \Delta H_f = –396 \text{ kJ/mol}$$
The oxidation of sulfur dioxide to form sulfur trioxide is represented by the following equation.
$$SO_2(g) + \frac{1}{2}O_2(g) \rightarrow SO_3(g)$$
The enthalpy change of this reaction is the difference in sum of enthalpy of formation values of reactants and that of products.
$$\Delta H = \Delta H_f (SO_3) - [\Delta H_f (SO2) + \Delta H_f (O_2)]$$
$$\Delta H = -396 \times 1 - (-297 \times 1 + 0) = -99 \text{ kJ/mol of } SO_3$$
Example 2
Calculate the standard enthalpy change of formation for nitrogen dioxide given the enthalpy of formation of nitrogen monoxide and the enthalpy change for the combustion of nitrogen monoxide.
$$2NO(g) + O_2(g) \rightarrow 2NO_2(g) \hspace{2cm} \Delta H = -114.1 \text{ kJ}$$
Enthalpy of formation of oxygen is 0 since it is in elemental form:
$$–114.1 = 2 \times (\Delta H_f^{\circ}(NO_2)) - 2 \times 90.3$$
$$–114.1 = 2 \times (\Delta H_f^{\circ}(NO_2)) -180.6$$
$$2 \times \Delta H_f^{\circ} (NO_2) = 66.5$$
$$\Delta H_f^{\circ}(NO_2) = 33.25 \text{ kJ/mol}$$
Calculate the enthalpy change for the combustion of methane using the following enthalpy of formation values.
Answer: (detailed explanation in video)
Combustion of methane equation:
$$CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$$
$$\Sigma \Delta H_f^{\circ} \text{(products)} = -395 + 2 \times -286 = -967 \text{ kJ}$$
$$\Sigma \Delta H_f^{\circ} \text{(reactants)} = -75 + 2 \times 0 = -75 \text{ kJ}$$
$$\Delta H_{\text{reaction}} = -967 - (-75) = -892 \text{ kJ}$$
Since this is calculated using 1 mole of methane gas:
$$\Delta H_{\text{reaction}} = -892 \text{ kJ/mol}$$
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