Bond energy of `N-H,H-H` and `N-=N` are `a,b,c` respectively. The `DeltaH` for the reaction,
`2NH_(3)[g] rarr N_(2)[g]+3H_(2)[g]` is `:`

Options

A. `6a - 3b -c`
B. `6a + 3b + c`
C. `a + 6b -c`
D. `6a + b -3c`

Answer

To calculate the change in enthalpy (ΔH) for the reaction:

\[ 2 \text{NH}_3(g) \rightarrow \text{N}_2(g) + 3 \text{H}_2(g) \]

we will use the bond energies provided for the bonds involved in the reaction. The bond energies are given as follows:
- Bond energy of N-H = a
- Bond energy of H-H = b
- Bond energy of N≡N = c

### Step-by-Step Solution:

1. **Identify the Bonds in Reactants and Products:**
- In the reactants (2 NH₃), each NH₃ molecule has 3 N-H bonds. Thus, for 2 NH₃, there are a total of:
\[ \text{Total N-H bonds} = 2 \times 3 = 6 \text{ N-H bonds} \]
- In the products, there is 1 N₂ molecule which has 1 N≡N bond, and 3 H₂ molecules, each having 1 H-H bond. Thus, for 3 H₂, there are:
\[ \text{Total H-H bonds} = 3 \times 1 = 3 \text{ H-H bonds} \]

2. **Calculate the Total Bond Energy for Reactants:**
- The total bond energy for the reactants (2 NH₃) is:
\[ \text{Bond Energy of Reactants} = 6 \times a \]

3. **Calculate the Total Bond Energy for Products:**
- The total bond energy for the products (N₂ and 3 H₂) is:
\[ \text{Bond Energy of Products} = c + 3 \times b \]

4. **Apply the Formula for ΔH:**
- The change in enthalpy (ΔH) for the reaction can be calculated using the formula:
\[ \Delta H = \text{Bond Energy of Reactants} - \text{Bond Energy of Products} \]
- Substituting the values we calculated:
\[ \Delta H = (6a) - (c + 3b) \]
- Simplifying this gives:
\[ \Delta H = 6a - 3b - c \]

### Final Answer:
Thus, the change in enthalpy (ΔH) for the reaction is:
\[ \Delta H = 6a - 3b - c \]