Formation of ammonia is shown by the reaction,
`N_(2(g))+3H_(2(g)) rarr 2NH_(3(g)), Delta_(f)H^(@)=-"91.8 kJ mol"^(-1)`
What will be the enthalpy of reaction for the decomposition of `NH_(3)` according to the reaction ?
`2NH_(3(g)) rarr N_(2(g))+3H_(2(g)), Delta_(r)H^(@)=`

Options

A. `-"91.8 kJ mol"^(-1)`
B. `+"91.8 kJ mol"^(-1)`
C. `-"45.9 kJ mol"^(-1)`
D. `+"45.9 kJ mol"^(-1)`

Explanation

To find the enthalpy of reaction for the decomposition of ammonia, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Reaction**:
The formation of ammonia is given by the reaction:
\[ N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)} \]
with an enthalpy change (\(\Delta_f H^\circ\)) of \(-91.8 \, \text{kJ/mol}\).

2. **Understand the Meaning of the Enthalpy Change**:
The negative sign indicates that the formation of ammonia is an exothermic process, meaning heat is released when ammonia is formed from nitrogen and hydrogen.

3. **Write the Decomposition Reaction**:
The decomposition of ammonia can be represented as:
\[ 2NH_{3(g)} \rightarrow N_{2(g)} + 3H_{2(g)} \]

4. **Apply the Concept of Reversing the Reaction**:
When a reaction is reversed, the sign of the enthalpy change also reverses. Therefore, if the formation of ammonia has an enthalpy change of \(-91.8 \, \text{kJ/mol}\), the decomposition of ammonia will have the opposite sign.

5. **Calculate the Enthalpy of Reaction for Decomposition**:
Thus, the enthalpy change for the decomposition reaction will be:
\[ \Delta_r H^\circ = +91.8 \, \text{kJ/mol} \]

### Final Answer:
The enthalpy of reaction for the decomposition of ammonia is:
\[ \Delta_r H^\circ = +91.8 \, \text{kJ/mol} \]