Bond Breaking and Forming, Bond Enthalpy, and Calculating Enthalpy Change ($ΔH$)
Bond Breaking and Bond Forming: The Energy Perspective
Breaking Bonds: An Endothermic Process
- Breaking a chemical bond requires energy.
- Why? Because bonds represent stable arrangements of atoms, and separating them disrupts this stability.
- Since energy is absorbed, bond breaking is an endothermic process, and the enthalpy change is positive ($+ΔH$).
Bond enthalpy
The energy required to break one mole of a specific bond in the gaseous state is called its bond enthalpy (or bond dissociation energy)
Breaking a hydrogen molecule (H₂) into two hydrogen atoms can be represented as:
$$\text{H}_2(g) \to 2\text{H}(g) \quad \Delta H = +436 \, \text{kJ mol}^{-1}$$
Forming Bonds: An Exothermic Process
- In contrast, forming a new chemical bond releases energy.
- This occurs because atoms achieve a more stable, lower-energy arrangement when bonded.
- As a result, bond formation is an exothermic process, and the enthalpy change is negative ($−ΔH$).
When two hydrogen atoms combine to form a hydrogen molecule:
$$2\text{H}(g) \to \text{H}_2(g) \quad \Delta H = -436 \, \text{kJ mol}^{-1}$$
The Balance of Bond Breaking and Forming
- In any chemical reaction, energy is absorbed to break bonds in the reactants and released when new bonds form in the products.
- The overall energy change of the reaction depends on the balance between these two processes:
- If more energy is released in bond formation than is absorbed in bond breaking, the reaction is exothermic ($ΔH< 0$).
- If more energy is absorbed in bond breaking than is released in bond formation, the reaction is endothermic ($ΔH >0$).
To determine whether a reaction is exothermic or endothermic, compare the total energy of the bonds broken with the total energy of the bonds formed.
Bond Enthalpy: A Measure of Bond Strength
Average Bond Enthalpy
- In reality, the energy required to break a particular bond can vary depending on the molecular environment.
- To account for these variations, bond enthalpies are often reported as average bond enthalpies.
Average bond enthalpy
Average bond enthalpy represents the average energy required to break a given bond across a range of compounds.
The bond enthalpy of a $C–H$ bond in methane (CH₄) is slightly different from that in ethane (C₂H₆).
- For methane (CH₄), breaking each successive $C–H$ bond requires different amounts of energy:
- First $C–H$ bond: $439 \text{ kJ mol}^{-1}$
- Second $C–H$ bond: $462 \text{ kJ mol}^{-1}$
- Third $C–H$ bond: $424 \text{ kJ mol}^{-1}$
- Fourth $C–H$ bond: $338 \text{ kJ mol}^{-1}$
- The average $C–H$ bond enthalpy is approximately $414 \text{ kJ mol}^{-1}$.
- Bond enthalpy values are provided in the IB Chemistry data booklet (Section 12).
- These values are crucial for calculating the enthalpy changes of reactions.
Calculating Enthalpy Change (ΔH) Using Bond Enthalpies
The enthalpy change of a reaction ($ΔH$) can be estimated using bond enthalpy data with the following formula:
$$\Delta H = \sum \text{(Bond enthalpies of bonds broken)} - \sum \text{(Bond enthalpies of bonds formed)}$$
Step-by-Step Approach
- Draw the Full Structural Formulas: Write the structural formulas of all reactants and products to identify the bonds involved.
- List Bonds to Be Broken and Formed: Count the number of each type of bond broken in the reactants and formed in the products.
- Substitute Bond Enthalpy Values: Use bond enthalpy values from the data booklet to calculate the total energy for breaking and forming bonds.
- Apply the Formula: Subtract the total energy of the bonds formed from the total energy of the bonds broken.
Ensure that the units of bond enthalpy ($\text{kJ mol}^{-1}$) are consistent throughout your calculation.
Calculating ΔH for a Reaction
Ethene reacts with hydrogen bromide to form bromoethane.
$$\text{C}_2\text{H}_4(g) + \text{HBr}(g) \to \text{C}_2\text{H}_5\text{Br}(g)$$
- Structural Formulas:
- Reactants: Ethene (C=C, 4 C–H bonds), HBr (H–Br bond)
- Products: Bromoethane (C–C, 5 C–H bonds, C–Br bond)
- Bonds Broken:
- 1 C=C bond: 614 kJ/mol
- 1 H–Br bond: 366 kJ/mol
- Total: $614 + 366 = 980 \, \text{kJ/mol}$
- Bonds Formed:
- 1 C–C bond: $346 \text{ kJ mol}^{-1}$
- 1 C–Br bond: $285 \text{ kJ mol}^{-1}$
- 5 C–H bonds: $5 \times 414 = 2070 \, \text{kJ mol}^{-1}$
- Total: $346 + 285 + 2070 = 2701 \, \text{kJ mol}^{-1}$
- Calculate ΔH:
$$
\Delta H = 980 - 2701 = -1721 \, \text{kJ mol}^{-1}
$$ - Conclusion: The reaction is exothermic, as indicated by the negative $ΔH$ value.
- Students sometimes overlook all bonds in the products, such as newly formed single bonds.
- Always double-check your bond count!
Limitations of Bond Enthalpy Calculations
While bond enthalpy calculations are useful for estimating ΔH, they have limitations:
- Average Values: Bond enthalpies are averages and may not match the exact bonds in a given molecule.
- State of Matter: Bond enthalpy values apply to gaseous molecules and do not account for intermolecular forces in liquids or solids.
- Experimental vs. Theoretical Values: Calculated $ΔH$ values may differ from experimental values due to simplifications in the bond enthalpy model.
- Methane reacts with chlorine to form chloromethane and hydrogen chloride:
$$\text{CH}_4(g) + \text{Cl}_2(g) \to \text{CH}_3\text{Cl}(g) + \text{HCl}(g)$$
Using bond enthalpy values from the data booklet, calculate the enthalpy change ($ΔH$) for this reaction. - How does the bond enthalpy model connect to the law of conservation of energy?
- How does it help us predict whether a reaction will release or absorb energy?
