If `Delta_(f)G^(@)` for `NH_(3)(g)` is `-16.4 kJ "mol"^(-1)`, then`DeltaG^(@)` for the reaction: `N_(2)(g) + 3H_(2)(g) to 2NH_(3)(g)` is

Options

A. `32.8 kJ "mol^(1)`
B. `16.4 kJ "mol"^(-1)`
C. `-16.4 kJ "mol"^(-1)`
D. `-32.8 kJ "mol"^(-1)`

Answer

To find the value of \(\Delta G^\circ\) for the reaction:

\[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \]

we start with the given standard Gibbs free energy of formation for ammonia (\(NH_3(g)\)), which is:

\[ \Delta_f G^\circ (NH_3) = -16.4 \, \text{kJ/mol} \]

This value represents the change in Gibbs free energy when one mole of \(NH_3\) is formed from its elements in their standard states.

### Step 1: Understand the Reaction Stoichiometry
The reaction we are considering produces 2 moles of \(NH_3\). Since the given \(\Delta_f G^\circ\) is for the formation of 1 mole of \(NH_3\), we need to adjust for the formation of 2 moles.

### Step 2: Multiply the Gibbs Free Energy of Formation
To find the \(\Delta G^\circ\) for the formation of 2 moles of \(NH_3\), we multiply the Gibbs free energy of formation for 1 mole by 2:

\[ \Delta G^\circ = 2 \times \Delta_f G^\circ (NH_3) \]

### Step 3: Calculate the Result
Substituting the value we have:

\[ \Delta G^\circ = 2 \times (-16.4 \, \text{kJ/mol}) = -32.8 \, \text{kJ} \]

### Final Answer
Thus, the \(\Delta G^\circ\) for the reaction \(N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)\) is:

\[ \Delta G^\circ = -32.8 \, \text{kJ} \]

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