Bond dissociation enthalpies of `H_(2(g)) and N_(2(g))` are `"426.0 kJ mol"^(-1)` and `"941.8 kJ mol"^(-1)`, respectively, and enthalpy of formation of `NH_(3(g))" is "-"46 kJ mol"^(-1)`. What are the enthalpy of atomisation of `NH_(3(g))` and the average bond enthalpy of `N-H` bond respectively ( in kJ `"mol"^(-1)`)?

Options

A. 1170.9, 390.3
B. 117, 300
C. 300, 200
D. 2000, 1975

Explanation

To solve the problem, we need to find the enthalpy of atomization of NH₃ and the average bond enthalpy of the N-H bond. We will use the given bond dissociation enthalpies and the enthalpy of formation of NH₃.

### Step 1: Write the reaction for the formation of NH₃
The balanced reaction for the formation of ammonia (NH₃) from nitrogen (N₂) and hydrogen (H₂) is:
\[ \text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g) \]
The enthalpy change (ΔH) for this reaction is given as:
\[ \Delta H = -46 \text{ kJ/mol} \]

### Step 2: Write the bond dissociation energies
From the problem, we know:
- Bond dissociation enthalpy of H₂: \( D(H-H) = 426.0 \text{ kJ/mol} \)
- Bond dissociation enthalpy of N₂: \( D(N≡N) = 941.8 \text{ kJ/mol} \)

### Step 3: Write the equation for the enthalpy of formation
The enthalpy of formation can be expressed in terms of bond dissociation energies:
\[ \Delta H = \text{(Bond energies of reactants)} - \text{(Bond energies of products)} \]

### Step 4: Calculate the bond energies of reactants
For the reactants:
- 1 mole of N₂ contributes \( 941.8 \text{ kJ} \)
- 3 moles of H₂ contribute \( 3 \times 426.0 \text{ kJ} = 1278.0 \text{ kJ} \)

Total bond energy of reactants:
\[ \text{Total bond energy of reactants} = 941.8 + 1278.0 = 2219.8 \text{ kJ} \]

### Step 5: Calculate the bond energies of products
For the products:
- 2 moles of NH₃ have 6 N-H bonds (3 N-H bonds per NH₃):
\[ \text{Bond energy of products} = 6 \times D(N-H) \]

### Step 6: Set up the equation
Substituting into the enthalpy formation equation:
\[ -46 = 2219.8 - 6 \times D(N-H) \]

### Step 7: Solve for D(N-H)
Rearranging the equation gives:
\[ 6 \times D(N-H) = 2219.8 + 46 \]
\[ 6 \times D(N-H) = 2265.8 \]
\[ D(N-H) = \frac{2265.8}{6} \]
\[ D(N-H) = 377.63 \text{ kJ/mol} \]

### Step 8: Calculate the enthalpy of atomization of NH₃
The enthalpy of atomization of NH₃ is the energy required to break all the bonds in 1 mole of NH₃:
\[ \text{Enthalpy of atomization of NH}_3 = 3 \times D(N-H) \]
\[ \text{Enthalpy of atomization of NH}_3 = 3 \times 377.63 \]
\[ \text{Enthalpy of atomization of NH}_3 = 1132.89 \text{ kJ/mol} \]

### Final Answers
- Enthalpy of atomization of NH₃: **1132.89 kJ/mol**
- Average bond enthalpy of N-H bond: **377.63 kJ/mol**