Using the listed [`DeltaG_(f)^(@)` values] calculate `DeltaG^(@)` for the reaction :
`3H_(2)S(g)[-33.6] +2HNO_(3)(l)[-80.6]rarr2NO(g)[+86.6]+4H_(2)O(l)[-237.1]+3S(s)[0.0]`

Options

A. `-513.2`
B. `-1037.0`
C. `+433.4`
D. `+225.0`

Explanation

To calculate the standard Gibbs free energy change (ΔG°) for the given reaction, we will use the formula:

\[ \Delta G^\circ = \sum (\Delta G_f^\circ \text{ of products}) - \sum (\Delta G_f^\circ \text{ of reactants}) \]

### Step 1: Identify the products and reactants
The reaction is:
\[ 3H_{2}S(g) + 2HNO_{3}(l) \rightarrow 2NO(g) + 4H_{2}O(l) + 3S(s) \]

**Products:**
- 2NO(g)
- 4H2O(l)
- 3S(s)

**Reactants:**
- 3H2S(g)
- 2HNO3(l)

### Step 2: Write down the ΔG_f° values
From the problem statement:
- ΔG_f° for H2S(g) = -33.6 kJ/mol
- ΔG_f° for HNO3(l) = -80.6 kJ/mol
- ΔG_f° for NO(g) = +86.6 kJ/mol
- ΔG_f° for H2O(l) = -237.1 kJ/mol
- ΔG_f° for S(s) = 0 kJ/mol (as it is in its standard state)

### Step 3: Calculate ΔG_f° for products
\[ \Delta G_f^\circ \text{ (products)} = [2 \times \Delta G_f^\circ \text{ (NO)}] + [4 \times \Delta G_f^\circ \text{ (H2O)}] + [3 \times \Delta G_f^\circ \text{ (S)}] \]

Substituting the values:
\[ = [2 \times 86.6] + [4 \times (-237.1)] + [3 \times 0] \]
\[ = 173.2 - 948.4 + 0 \]
\[ = -775.2 \text{ kJ} \]

### Step 4: Calculate ΔG_f° for reactants
\[ \Delta G_f^\circ \text{ (reactants)} = [3 \times \Delta G_f^\circ \text{ (H2S)}] + [2 \times \Delta G_f^\circ \text{ (HNO3)}] \]

Substituting the values:
\[ = [3 \times (-33.6)] + [2 \times (-80.6)] \]
\[ = -100.8 - 161.2 \]
\[ = -262.0 \text{ kJ} \]

### Step 5: Calculate ΔG°
Now, substitute the values into the ΔG° equation:
\[ \Delta G^\circ = \Delta G_f^\circ \text{ (products)} - \Delta G_f^\circ \text{ (reactants)} \]
\[ = -775.2 - (-262.0) \]
\[ = -775.2 + 262.0 \]
\[ = -513.2 \text{ kJ} \]

### Final Answer:
\[ \Delta G^\circ = -513.2 \text{ kJ} \]

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