Enthalpy of Formation and Bond Energy
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
Enthalpy of Formation and Bond Energy Explained
This video will explore the concept of enthalpy of formation, explaining its role in understanding the energy changes involved in when forming compounds from their elemental states. It'll also explore bond energy, and demonstrate how bond energy values can be utilised to calculate estimated enthalpy of formation values.
What is Bond Energy? (Recap)
Bond energy is the average energy required to break a specific chemical bond. Consider methane (`CH_4`), which consists of four C-H bonds.
Figure 1. tetrahedral structure of methane (`CH_4`)
The bone energy of the C–H bond is 413 kJ/mol which means one mole of C–H bond requires 413 kJ to break. This means that to break all C–H bonds of one mole of `CH_4`, a total of 1652 kJ/mol of energy is needed. This information is usually presented in tables showing bond types and their respective energies.
Table 1. Bond energies of a range of different chemical bond types.
|
Bond type |
Bond energy (kJ/mol) |
|
C=O |
746 |
|
O–H |
464 |
|
C–H |
413 |
|
O=O |
495 |
What is Enthalpy Change of Formation?
The enthalpy change of formation (`\Delta H_f`) is the enthalpy change involved in the formation of one mole of a compound from its constituent elements in their standard states.
The standard enthalpy change of formation (`\Delta H°_f`) is the enthalpy change of formation under standard conditions (298.15 K, 1 atm or approximately 100 kPa). This means it is measured when both reactants (i.e. constituent elements) and product are in their standard states.
Example: Formation of Methane
A thermochemical equation representing the formation of methane (`CH_4`) is:
$$C(s) + 2H_2(g) \rightarrow CH_4(g) \hspace{2cm} \Delta H_f = \; –75 \text{ kJ/mol}$$
Here, methane is formed from its elements: carbon (solid) and hydrogen (diatomic gas), both in their standard elemental forms. A total of 75 kJ of energy is released during this exothermic process, indicated by the negative enthalpy change.
Enthalpy of formation applies to all compounds:
Formation of ionic compounds such as sodium chloride:
$$2Na(s) + Cl_2(g) \rightarrow 2NaCl(s) \; \Delta H_f = -411 \text{ kJ/mol}$$
Formation of more complex covalent compounds:
$$4C(s) + O_2(g) +6H_2(g) \rightarrow 2C_2H_6O(l) \; \Delta H_f = -270 \text{ kJ/mol}$$
The enthalpy change of formation for pure elements is zero. This is because it takes no energy to form an element from itself in the standard state. For example, the enthalpy change of formation of oxygen gas (element in standard state) is zero.
The enthalpy change of the following thermochemical equation is not an example of enthalpy of formation. The value of - 500 kJ is the bond energy of O=O bond.
$$O(g) + O(g) \rightarrow O_2(g) \hspace{2cm} \Delta H = \text{ } –500 \text{ } kJ$$
Processes in Enthalpy of Formation
The enthalpy of formation is just another form of enthalpy change and thus also involves multiple processes.
Consider the formation of methane from carbon and hydrogen gas:
$$C(s) + 2H_2(g) \rightarrow CH_4(g)$$
The following steps are involved:
Step 1: Sublimation of carbon (endothermic), turning solid carbon atoms into gaseous state
$$C(s) \rightarrow C(g) \hspace{2cm} \Delta H = +711 \; \text{kJ}$$
Step 2: Dissociation of hydrogen molecules (endothermic) to form hydrogen atoms. This requires breaking covalent bonds between hydrogen atoms.
$$2H_2(g) \rightarrow 4H(g) \hspace{2cm} \Delta H = 438 \times 2 = +866 \; \text{kJ}$$
Step 3: Formation of methane molecules (exothermic). This requires formation of covalent bonds between already vapourised carbon and hydrogen atoms.
$$C(g) + 4H(g) \rightarrow CH_4(g) \hspace{2cm} \Delta H = -413 \times 4 = -1652 \; \text{kJ}$$
Enthalpy of formation of methane is given by:
$$\Delta H_f = 711 + 866 - 1652 = 75 \; kJ \; mol^{–1}$$
Worked Examples
Example 1
Write the enthalpy of formation equation for ammonia.
$$N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$$
Example 2
(a) Write the enthalpy of formation equation for carbon monoxide.
Answer:
$$2C(s) + O_2(g) \rightarrow 2CO$$
(b) Use the bond energies provided in the table and the enthalpy of sublimation of carbon to calculate the enthalpy change for the formation of carbon monoxide.
|
Bond type |
Bond Energy (kJ/mol) |
|
O=O |
495 |
|
C≡O |
1072 |
$$2C(s) + O_2(g) \rightarrow 2CO(g)$$
$$\text{2 moles of carbon sublimated} = 714 \text{ kJ} \times 2 = 1428 \text{ kJ}$$
$$\text{Bonds broken} = 1 \times \text{O=O bonds}$$
$$\text{Energy absorbed} = 495 \text{ kJ}$$
$$\text{Bonds formed} = 2 \times \text{C}\equiv \text{O} \text{ bonds}$$
$$\text{Energy released} = 1072 \times 2= -2144 \text{ kJ}$$
$$\Delta H_f = 495 - 2144 + 1428 = -221 \text{ kJ for 2 mol CO formed}$$
$$\text{Molar } \Delta H_f = \frac{-221}{2} = -110.5 \text{ kJ/mol}$$
Use the bond energies provided in the table and the enthalpy change of formation of water to calculate the enthalpy of vaporisation of water.
$$2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \; \; \Delta H_f^{\circ} = -292 \text{ kJ/mol}$$
|
Bond type |
Bond Energy (kJ/mol) |
|
H–H |
436 |
|
O=O |
495 |
|
O–H |
464 |
Answer:
Consider the energy absorbed during bond breaking in reactants:
$$\text{Bonds broken} = 2 \times \text{H–H bonds & 1} \times \text{O=O bond}$$
$$\text{Energy absorbed} = 436 \times 2 + 495 = 1367 \text{ kJ}$$
Consider the energy released during bond formation in products:
$$\text{Bonds formed} = -464 \times 4 = -1856 \text{ kJ}$$
$$\text{Total energy} = 1367 – 1856 = \; -489 \text{ kJ for 2 moles of water in gaseous state}$$
Since the enthalpy of formation of water requires the water in liquid state, we need to also consider the enthalpy change of condensation of water:
$$H_2O(g) \rightarrow H_2O(l)$$
$$\Delta H_f^{\circ} = -292 \times 2 = -489 + \Delta H_{\text{condensation}}$$
$$\Delta H_{\text{condensation}} = -95 \text{ kJ (for 2 moles of water)}$$
$$\text{Molar } \Delta H_{\text{condensation}} = \frac{-95}{2} = -47.5 \text{ kJ/mol}$$
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