Dinitrogen monoxide, N₂O, can decompose to nitrogen and oxygen gas:
2 N2O(g) → 2 N2(g) + O2(g)
Use bond dissociation enthalpies (see Table on last page) to estimate the enthalpy change for this reaction. (Hint: As in problem #6, first draw the Lewis structures of the reactants and products that obey the octet rule so you knov the types of bonds involved to determine their bond dissociation energies. N₂O has the following stable Lewis
structure: :N=N-0:
The Correct Answer and Explanation is :
To estimate the enthalpy change for the decomposition of dinitrogen monoxide (N₂O) into nitrogen (N₂) and oxygen (O₂) using bond dissociation enthalpies, follow these steps:
1. Write the balanced reaction:
The decomposition of dinitrogen monoxide can be written as:
[
2 \text{N}_2\text{O}(g) \rightarrow 2 \text{N}_2(g) + \text{O}_2(g)
]
2. Draw the Lewis Structures:
We will use the following Lewis structures for the molecules:
- N₂O (dinitrogen monoxide): The stable structure of N₂O is :N=N-O:, where:
- N-N bond: single bond (N-N)
- N-O bond: single bond (N-O)
- N₂ (nitrogen): Each nitrogen molecule has a triple bond between two nitrogen atoms:
- N≡N bond: triple bond (N≡N)
- O₂ (oxygen): Oxygen gas has a double bond between the two oxygen atoms:
- O=O bond: double bond (O=O)
3. Bond Dissociation Enthalpies:
To estimate the enthalpy change for the reaction, use bond dissociation enthalpies for each bond involved.
- N₂O has:
- 1 N-N bond: approximately 160 kJ/mol
- 1 N-O bond: approximately 200 kJ/mol
- N₂ has:
- 1 N≡N bond: approximately 945 kJ/mol
- O₂ has:
- 1 O=O bond: approximately 498 kJ/mol
4. Calculate the Enthalpy Change:
To find the total enthalpy change, we need to consider both the bonds broken (reactants) and the bonds formed (products).
- Bonds broken:
- 2 N-N bonds (in two N₂O molecules): 2 × 160 kJ/mol = 320 kJ/mol
- 2 N-O bonds (in two N₂O molecules): 2 × 200 kJ/mol = 400 kJ/mol
- Total energy to break bonds: 320 kJ/mol + 400 kJ/mol = 720 kJ/mol
- Bonds formed:
- 2 N≡N bonds (in two N₂ molecules): 2 × 945 kJ/mol = 1890 kJ/mol
- 1 O=O bond (in one O₂ molecule): 1 × 498 kJ/mol = 498 kJ/mol
- Total energy to form bonds: 1890 kJ/mol + 498 kJ/mol = 2388 kJ/mol
5. Enthalpy Change for the Reaction:
The enthalpy change (ΔH) can be calculated by subtracting the total energy required to break bonds from the energy released when new bonds are formed:
[
\Delta H = \text{Bonds formed} – \text{Bonds broken}
]
[
\Delta H = 2388 \, \text{kJ/mol} – 720 \, \text{kJ/mol} = 1668 \, \text{kJ/mol}
]
Thus, the enthalpy change for the decomposition of 2 moles of N₂O into nitrogen and oxygen is approximately +1668 kJ/mol. Since the reaction is endothermic (energy is absorbed), the positive value indicates that the decomposition requires energy input to proceed.
Conclusion:
The enthalpy change for the reaction is positive, meaning the decomposition of N₂O absorbs energy, making it an endothermic process.