The standard enthalpy of formation of `NH_(3)` is `-46.0KJ mol^(-1)` . If the enthalpy of formation of `H_(2)` from its atoms is `-436KJ mol^(-1)` and that of `N_(2)` is `-712KJ mol^(-1)` , the average bond enthalpy of `N-H` bond in `NH_(3)` is

Options

A. `-964KJmol^(-1)`
B. `+352KJmol^(-1)`
C. `+105KJmol^(-1)`
D. `-1102KJmol^(-1)`

Explanation

To find the average bond enthalpy of the N-H bond in NH₃, we can use the given standard enthalpy of formation values and apply Hess's law. Here’s the step-by-step solution:

### Step 1: Write the reaction for the formation of ammonia
The formation of ammonia (NH₃) from its elements can be represented as:
\[ \frac{1}{2} N_2(g) + \frac{3}{2} H_2(g) \rightarrow NH_3(g) \]
The standard enthalpy change (ΔH) for this reaction is given as:
\[ \Delta H_f = -46.0 \, \text{kJ/mol} \]

### Step 2: Write the enthalpy of formation for the reactants
The enthalpy of formation for the reactants can be expressed in terms of bond enthalpies:
- Breaking 1 mole of N₂ into 2 moles of N atoms requires an energy of:
\[ \Delta H(N_2) = +712 \, \text{kJ/mol} \] (since it is the enthalpy of formation, we take it as positive when breaking bonds)
- Breaking 1 mole of H₂ into 2 moles of H atoms requires an energy of:
\[ \Delta H(H_2) = +436 \, \text{kJ/mol} \]

### Step 3: Set up the equation using bond enthalpies
The total enthalpy change for breaking the bonds in the reactants and forming the bonds in the products can be expressed as:
\[ \Delta H = \text{Bond enthalpy of N-H bonds} - \text{Energy to break N₂ and H₂} \]

Since we are forming 1 mole of NH₃, we have:
- 3 N-H bonds formed in NH₃.

Let \( BE(N-H) \) be the bond enthalpy of the N-H bond. The equation can be set up as:
\[ -46.0 = 3 \times BE(N-H) - \left( \frac{1}{2} \times 712 + \frac{3}{2} \times 436 \right) \]

### Step 4: Calculate the energy required to break N₂ and H₂
Calculating the energy to break N₂ and H₂:
\[ \text{Energy to break N₂} = \frac{1}{2} \times 712 = 356 \, \text{kJ/mol} \]
\[ \text{Energy to break H₂} = \frac{3}{2} \times 436 = 654 \, \text{kJ/mol} \]
So, the total energy required to break the bonds is:
\[ 356 + 654 = 1010 \, \text{kJ/mol} \]

### Step 5: Substitute back into the equation
Now substituting back into our equation:
\[ -46.0 = 3 \times BE(N-H) - 1010 \]

### Step 6: Solve for BE(N-H)
Rearranging the equation gives:
\[ 3 \times BE(N-H) = -46.0 + 1010 \]
\[ 3 \times BE(N-H) = 964 \]
\[ BE(N-H) = \frac{964}{3} \]
\[ BE(N-H) = 321.33 \, \text{kJ/mol} \]

### Final Answer
The average bond enthalpy of the N-H bond in NH₃ is approximately:
\[ \boxed{321.33 \, \text{kJ/mol}} \]