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 will use the given enthalpy of formation of ammonia and apply the principles of thermodynamics.

### Step-by-Step Solution:

1. **Understand the Enthalpy of Formation**:
The enthalpy of formation (\(ΔH_f\)) of ammonia (\(NH_3\)) is given as \(-46.0 \, kJ/mol\). This means that when one mole of ammonia is formed from its elements (nitrogen and hydrogen), \(46.0 \, kJ\) of energy is released.

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

3. **Scale the Reaction**:
Since the reaction we are interested in involves 2 moles of ammonia, we need to multiply the entire formation reaction by 2:
\[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \]
The enthalpy change for this scaled reaction will be:
\[ ΔH = 2 \times (-46.0 \, kJ) = -92.0 \, kJ \]

4. **Reverse the Reaction**:
The reaction we need is the decomposition of ammonia:
\[ 2NH_3(g) \rightarrow N_2(g) + 3H_2(g) \]
When we reverse a reaction, the sign of the enthalpy change also reverses. Therefore, the enthalpy change for the decomposition reaction will be:
\[ ΔH = +92.0 \, kJ \]

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

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