The enthalpy of formation of ammonia is `-46.0 kJ mol^(-1)`. The enthalpy for the reaction `2N_(2)(g)+6H_(2)(g)to4NH_(3)(g)`

Options

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

Explanation

To find the enthalpy change for the reaction:

\[ 2N_2(g) + 6H_2(g) \rightarrow 4NH_3(g) \]

we can use the enthalpy of formation of ammonia, which is given as \(-46.0 \, \text{kJ/mol}\). The enthalpy of formation refers to the energy change when one mole of a compound is formed from its elements in their standard states.

### Step-by-Step Solution:

1. **Identify the enthalpy of formation for ammonia**:
The enthalpy of formation (\( \Delta H_f \)) of ammonia (\( NH_3 \)) is given as \(-46.0 \, \text{kJ/mol}\). This means that when 1 mole of \( NH_3 \) is formed from its elements (nitrogen and hydrogen), \( 46.0 \, \text{kJ} \) of energy is released.

2. **Calculate the total enthalpy change for the formation of 4 moles of ammonia**:
Since the reaction produces 4 moles of \( NH_3 \), we need to multiply the enthalpy of formation of one mole by 4:
\[ \Delta H = 4 \times (-46.0 \, \text{kJ/mol}) = -184.0 \, \text{kJ} \]

3. **Interpret the result**:
The negative sign indicates that the reaction is exothermic, meaning it releases energy. Therefore, the enthalpy change for the reaction \( 2N_2(g) + 6H_2(g) \rightarrow 4NH_3(g) \) is \(-184.0 \, \text{kJ}\).

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

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