The heat of formation of `NH_(3)(g)` is `-46 " kJ mol"^(-1)`. The `DeltaH` (in `" kJ mol"^(-1)`) of the reaction, `2NH_(3)(g)rarrN_(2)(g)+3H_(2)(g)` is

Options

A. 46
B. `-46`
C. 92
D. `-92`

Explanation

To find the ΔH (enthalpy change) of the reaction \(2NH_3(g) \rightarrow N_2(g) + 3H_2(g)\), we can use the heat of formation of ammonia \(NH_3(g)\), which is given as \(-46 \, \text{kJ mol}^{-1}\).

### Step-by-Step Solution:

1. **Understand the Heat of Formation**: The heat of formation of \(NH_3(g)\) is \(-46 \, \text{kJ mol}^{-1}\). This means that when 1 mole of \(NH_3(g)\) is formed from its elements in their standard states (i.e., \(N_2(g)\) and \(H_2(g)\)), 46 kJ of energy is released.

\[ N_2(g) + \frac{3}{2} H_2(g) \rightarrow NH_3(g) \quad \Delta H = -46 \, \text{kJ} \]

2. **Reverse the Reaction**: To find the ΔH for the decomposition of ammonia, we need to reverse the formation reaction. When we reverse a reaction, the sign of ΔH also changes.

\[ NH_3(g) \rightarrow N_2(g) + \frac{3}{2} H_2(g) \quad \Delta H = +46 \, \text{kJ} \]

3. **Multiply the Reaction**: The given reaction involves 2 moles of \(NH_3(g)\). Therefore, we need to multiply the entire reversed reaction by 2.

\[ 2NH_3(g) \rightarrow 2N_2(g) + 3H_2(g) \quad \Delta H = 2 \times 46 \, \text{kJ} = +92 \, \text{kJ} \]

4. **Final Result**: The ΔH for the reaction \(2NH_3(g) \rightarrow N_2(g) + 3H_2(g)\) is thus:

\[ \Delta H = +92 \, \text{kJ mol}^{-1} \]

### Answer:
\[ \Delta H = +92 \, \text{kJ mol}^{-1} \]

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