The enthalpy of formation of ammonia is `-46.0 KJ mol^(-1)` . The enthalpy change for the reaction `2NH_(3)(g)rarr N_(2)(g)+3H_(2)(g)` is :

Options

A. `46.0 KJ mol^(-1)`
B. `92.0 KJ mol^(-1)`
C. `-23.0 KJ mol^(-1)`
D. `-92.0 KJ mol^(-1)`

Answer

To find the enthalpy change for the reaction \( 2NH_3(g) \rightarrow N_2(g) + 3H_2(g) \), we can use the enthalpy of formation of ammonia provided in the question.

### Step-by-step Solution:

1. **Understand the Enthalpy of Formation**: The enthalpy of formation of ammonia (\( NH_3 \)) is given as \(-46.0 \, \text{kJ/mol}\). This value indicates the energy change when one mole of ammonia is formed from its elements in their standard states.

2. **Write the Formation Reaction**: The formation reaction for ammonia can be represented as:
\[ \frac{1}{2} N_2(g) + \frac{3}{2} H_2(g) \rightarrow NH_3(g) \]
The enthalpy change for this reaction is \(-46.0 \, \text{kJ}\) for the formation of 1 mole of \( NH_3 \).

3. **Determine the Reaction for 2 Moles of Ammonia**: The reaction we are interested in is:
\[ 2NH_3(g) \rightarrow N_2(g) + 3H_2(g) \]
This reaction is the reverse of the formation reaction, so we need to consider the enthalpy change for the reverse process.

4. **Calculate the Enthalpy Change for the Reverse Reaction**: Since the reaction is reversed, the sign of the enthalpy change will change:
\[ \text{Enthalpy change for } 2NH_3 = 2 \times (-(-46.0 \, \text{kJ})) = 2 \times 46.0 \, \text{kJ} = 92.0 \, \text{kJ} \]

5. **Final Answer**: The enthalpy change for the reaction \( 2NH_3(g) \rightarrow N_2(g) + 3H_2(g) \) is:
\[ +92.0 \, \text{kJ} \]

### Summary:
The enthalpy change for the reaction \( 2NH_3(g) \rightarrow N_2(g) + 3H_2(g) \) is \( +92.0 \, \text{kJ} \).