The enthalpy of formaion of ammonia gas at 298 K is -46.11 kJ `mol^(-1)`. The equaiton to which this value realtes is

Options

A. `1/2 N_2(g) + 3/2 H_2(g) rarr NH_3(g)`
B. `N_2(g)+3H(g) rarrr NH_3(g)`
C. `N_2(g) + 3H_2(g)rarr 2NH_3(g)`
D. `1/2 N_2(g) +3/2 H_2(g) rarr 2 NH_3(l)`

Explanation

To determine the correct equation that relates to the enthalpy of formation of ammonia gas at 298 K, we need to understand the definition of standard enthalpy of formation.

### Step-by-Step Solution:

1. **Understand the Definition**: The standard enthalpy of formation (ΔH°_f) is defined as the enthalpy change when one mole of a compound is formed from its elements in their most stable states (reference states) at standard conditions (298 K and 1 atm).

2. **Identify the Elements of Ammonia**: Ammonia (NH₃) is composed of nitrogen (N) and hydrogen (H). Therefore, we need to consider the most stable forms of these elements:
- The most stable form of nitrogen is N₂ (g).
- The most stable form of hydrogen is H₂ (g).

3. **Write the Formation Reaction**: Based on the definition, the formation reaction for one mole of ammonia from its elements in their reference states is:
\[ \frac{1}{2} N_2 (g) + \frac{3}{2} H_2 (g) \rightarrow NH_3 (g) \]

4. **Check the Enthalpy Change**: The enthalpy change associated with this reaction is given as -46.11 kJ/mol, which indicates that the formation of ammonia from its elements is an exothermic process.

5. **Conclusion**: The correct equation that relates to the enthalpy of formation of ammonia gas at 298 K is:
\[ \frac{1}{2} N_2 (g) + \frac{3}{2} H_2 (g) \rightarrow NH_3 (g) \quad \Delta H° = -46.11 \, \text{kJ/mol} \]