The enthalpy of formation of ammonia is -46.0 kJ `mol^(-1)` .The enthalpy for reaction `2N_2(g)+6H_2(g) rarr 4 NH_3(g)` is equal to

Options

A. `-46.0 kJ`
B. 46.0 KJ
C. `184.0 kJ`
D. `-184.0 kJ`

Explanation

To find the enthalpy change for the reaction \(2N_2(g) + 6H_2(g) \rightarrow 4NH_3(g)\), we can use the given enthalpy of formation of ammonia, which is \(-46.0 \, \text{kJ mol}^{-1}\).

### Step-by-Step Solution:

1. **Understand the Enthalpy of Formation**: The enthalpy of formation (\( \Delta H_f \)) of a compound is defined as the heat change that results when one mole of the compound is formed from its elements in their standard states. For ammonia (\(NH_3\)), the enthalpy of formation is given as \(-46.0 \, \text{kJ/mol}\).

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

3. **Scale the Reaction for 4 Moles of Ammonia**: To find the enthalpy change for the formation of 4 moles of ammonia, we need to multiply the enthalpy change of the formation reaction by 4:
\[ 4 \times \left( \frac{1}{2} N_2(g) + 3H_2(g) \rightarrow NH_3(g) \right) \Rightarrow 2N_2(g) + 6H_2(g) \rightarrow 4NH_3(g) \]
The enthalpy change for this reaction will be:
\[ \Delta H = 4 \times (-46.0 \, \text{kJ}) = -184.0 \, \text{kJ} \]

4. **Final Result**: Therefore, the enthalpy change for the reaction \(2N_2(g) + 6H_2(g) \rightarrow 4NH_3(g)\) is:
\[ \Delta H = -184.0 \, \text{kJ} \]

### Conclusion:
The enthalpy for the reaction \(2N_2(g) + 6H_2(g) \rightarrow 4NH_3(g)\) is \(-184.0 \, \text{kJ}\).

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