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Enthalpy, Entropy, and Free Energy

Calculating [tex]$\Delta G_{rxn}$[/tex]

Calculate [tex]$\Delta G_{rxn}$[/tex] for this equation, rounding your answer to the nearest whole number.

[tex]\[
\begin{array}{l}
4 \text{NH}_3(g) + 5 \text{O}_2(g) \rightarrow 4 \text{NO}(g) + 6 \text{H}_2\text{O}(g) \\
\Delta G_{f}^{\circ} (\text{NH}_3) = -16.66 \, \text{kJ/mol} \\
\Delta G_{f}^{\circ} (\text{O}_2) = 0 \, \text{kJ/mol} \\
\Delta G_{f}^{\circ} (\text{NO}) = 86.71 \, \text{kJ/mol} \\
\Delta G_{f}^{\circ} (\text{H}_2\text{O}) = -228.57 \, \text{kJ/mol} \\
\end{array}
\][/tex]

[tex]\Delta G_{rxn} = \boxed{\,\,\,\,\,\,\,\,\,\,\,\,} \text{kJ}[/tex]



Answer :

GinnyAnswer
To calculate the standard Gibbs free energy change ([tex]\(\Delta G_{rxn}\)[/tex]) for the given chemical reaction, we will follow these steps:

### Step 1: Write down the balanced chemical equation
[tex]\[ 4 \text{NH}_3(g) + 5 \text{O}_2(g) \rightarrow 4 \text{NO}(g) + 6 \text{H}_2\text{O}(g) \][/tex]

### Step 2: Write down the Gibbs free energy of formation ([tex]\(\Delta G_f^\circ\)[/tex]) for each compound involved in the reaction:
- [tex]\(\Delta G_f^\circ (\text{NH}_3) = -16.66 \text{ kJ/mol}\)[/tex]
- [tex]\(\Delta G_f^\circ (\text{O}_2) = 0 \text{ kJ/mol}\)[/tex] (Since it's a diatomic molecule in its standard state)
- [tex]\(\Delta G_f^\circ (\text{NO}) = 86.71 \text{ kJ/mol}\)[/tex]
- [tex]\(\Delta G_f^\circ (\text{H}_2\text{O}) = -228.57 \text{ kJ/mol}\)[/tex]

### Step 3: Apply the formula for the standard Gibbs free energy change of the reaction
[tex]\[ \Delta G_{rxn}^\circ = \sum \nu \Delta G_f^\circ (\text{products}) - \sum \nu \Delta G_f^\circ (\text{reactants}) \][/tex]

### Step 4: Calculate the sum of the Gibbs free energy of formation for the reactants and products
#### Reactants:
[tex]\[ \Delta G_{\text{reactants}} = (4 \cdot \Delta G_f^\circ (\text{NH}_3)) + (5 \cdot \Delta G_f^\circ (\text{O}_2)) \][/tex]
[tex]\[ = (4 \cdot -16.66 \text{ kJ/mol}) + (5 \cdot 0 \text{ kJ/mol}) \][/tex]
[tex]\[ = -66.64 \text{ kJ} \][/tex]

#### Products:
[tex]\[ \Delta G_{\text{products}} = (4 \cdot \Delta G_f^\circ (\text{NO})) + (6 \cdot \Delta G_f^\circ (\text{H}_2\text{O})) \][/tex]
[tex]\[ = (4 \cdot 86.71 \text{ kJ/mol}) + (6 \cdot -228.57 \text{ kJ/mol}) \][/tex]
[tex]\[ = 346.84 \text{ kJ} + (-1371.42 \text{ kJ}) \][/tex]
[tex]\[ = -1024.580 \text{ kJ} \][/tex]

### Step 5: Calculate [tex]\(\Delta G_{rxn}^\circ\)[/tex]
[tex]\[ \Delta G_{rxn}^\circ = \Delta G_{\text{products}} - \Delta G_{\text{reactants}} \][/tex]
[tex]\[ = -1024.580 \text{ kJ} - (-66.64 \text{ kJ}) \][/tex]
[tex]\[ = -1024.580 \text{ kJ} + 66.64 \text{ kJ} \][/tex]
[tex]\[ = -957.94 \text{ kJ} \][/tex]

### Step 6: Round the answer to the nearest whole number
[tex]\[ \Delta G_{rxn}^\circ \approx -958 \text{ kJ} \][/tex]

Therefore, the standard Gibbs free energy change for the reaction, rounded to the nearest whole number, is [tex]\(\boxed{-958 \text{ kJ}}\)[/tex].