`DeltaG°` for the following reaction:
`I_2(s) + H_2S(g) rarr 2HI(g) + S(s)` at 298 K is,
Given that `Delta_fG°HI(g) = 1.8 kJ mol^-1`. `Delta_fG°H_2S(g) = 33.8 kJ` `mol^-1`.

Options

A. 30200 kJ
B. -30.2 kJ
C. -30200J
D. -302 J

Answer

To calculate the standard Gibbs free energy change (ΔG°) for the reaction:

\[ \text{I}_2(s) + \text{H}_2\text{S}(g) \rightarrow 2\text{HI}(g) + \text{S}(s) \]

at 298 K, we can use the following formula:

\[ \Delta G° = \sum \Delta_f G° \text{(products)} - \sum \Delta_f G° \text{(reactants)} \]

### Step 1: Identify the components of the reaction
- **Products**: 2 moles of HI(g) and 1 mole of S(s)
- **Reactants**: 1 mole of I2(s) and 1 mole of H2S(g)

### Step 2: Write down the standard Gibbs free energy of formation values
- Given:
- \(\Delta_f G° \text{(HI)} = 1.8 \, \text{kJ/mol}\)
- \(\Delta_f G° \text{(H2S)} = 33.8 \, \text{kJ/mol}\)
- \(\Delta_f G° \text{(S)} = 0 \, \text{kJ/mol}\) (since sulfur is in its standard state)
- \(\Delta_f G° \text{(I2)} = 0 \, \text{kJ/mol}\) (since iodine is in its standard state)

### Step 3: Calculate the Gibbs free energy for products
\[ \text{For products: } \Delta_f G° \text{(products)} = 2 \times \Delta_f G° \text{(HI)} + \Delta_f G° \text{(S)} \]
\[ = 2 \times 1.8 \, \text{kJ/mol} + 0 \, \text{kJ/mol} = 3.6 \, \text{kJ/mol} \]

### Step 4: Calculate the Gibbs free energy for reactants
\[ \text{For reactants: } \Delta_f G° \text{(reactants)} = \Delta_f G° \text{(I2)} + \Delta_f G° \text{(H2S)} \]
\[ = 0 \, \text{kJ/mol} + 33.8 \, \text{kJ/mol} = 33.8 \, \text{kJ/mol} \]

### Step 5: Substitute the values into the Gibbs free energy formula
\[ \Delta G° = \Delta_f G° \text{(products)} - \Delta_f G° \text{(reactants)} \]
\[ = 3.6 \, \text{kJ/mol} - 33.8 \, \text{kJ/mol} \]
\[ = -30.2 \, \text{kJ/mol} \]

### Final Answer
Thus, the standard Gibbs free energy change (ΔG°) for the reaction is:
\[ \Delta G° = -30.2 \, \text{kJ/mol} \]