The enthalpy of formation of ammonia when calculated from the following bond energy data is (B.E. of `N-H, H-H, N-=N` is `"389 kJ mol"^(-1), "435 kJ mol"^(-1),"945.36 kJ mol"^(-1)` respectively)

Options

A. `-"41.82 kJ mol"^(-1)`
B. `+"83.64 kJ mol"^(-1)`
C. `-"945.36 kJ mol"^(-1)`
D. `-"833 kJ mol"^(-1)`

Explanation

To calculate the enthalpy of formation of ammonia (NH₃) using bond energy data, we will follow these steps:

### Step 1: Write the balanced equation for the formation of ammonia.
The formation of ammonia from nitrogen and hydrogen can be represented as:
\[ \text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g) \]

### Step 2: Identify the bonds broken and formed.
In this reaction:
- Bonds broken:
- 1 N≡N bond in N₂
- 3 H–H bonds in 3 H₂ molecules
- Bonds formed:
- 6 N–H bonds in 2 NH₃ molecules (since each NH₃ has 3 N–H bonds)

### Step 3: Write down the bond energies.
From the data provided:
- Bond energy of N≡N = 945.36 kJ/mol
- Bond energy of H–H = 435 kJ/mol
- Bond energy of N–H = 389 kJ/mol

### Step 4: Calculate the total bond energies for bonds broken and formed.
- Total bond energy of reactants (bonds broken):
\[ \text{Bond energy of N}_2 + 3 \times \text{Bond energy of H}_2 = 945.36 \, \text{kJ/mol} + 3 \times 435 \, \text{kJ/mol} \]
\[ = 945.36 + 1305 = 2250.36 \, \text{kJ/mol} \]

- Total bond energy of products (bonds formed):
\[ 6 \times \text{Bond energy of N–H} = 6 \times 389 \, \text{kJ/mol} = 2334 \, \text{kJ/mol} \]

### Step 5: Apply Hess's law to find the enthalpy change (ΔH).
Using the formula:
\[ \Delta H = \text{Total bond energy of reactants} - \text{Total bond energy of products} \]
Substituting the values:
\[ \Delta H = 2250.36 \, \text{kJ/mol} - 2334 \, \text{kJ/mol} = -83.64 \, \text{kJ/mol} \]

### Step 6: Adjust for the formation of 1 mole of NH₃.
Since the reaction produces 2 moles of NH₃, we need to divide the enthalpy change by 2 to find the enthalpy of formation for 1 mole of NH₃:
\[ \Delta H_f = \frac{-83.64}{2} = -41.82 \, \text{kJ/mol} \]

### Final Answer:
The enthalpy of formation of ammonia (NH₃) is:
\[ \Delta H_f = -41.82 \, \text{kJ/mol} \]

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