For the reaction given below the value of standard Gibbs free energy of formation at 298 K are given. What is the nature of the reaction?
`I_(2)+H_(2)Srarr2HI+S`
`DeltaG_(f)^(@)(HI)="1.8 kJ mol"^(-1), DeltaG_(f)^(@)(H_(2)S)="33.8 kJ mol"^(-1)`

Options

A. Non-spontaneous in forward direction.
B. Spontaneous in forward direction.
C. Spontaneous in backward direction.
D. Non-spontaneous in both forward and backward directions.

Explanation

To determine the nature of the reaction given by:

\[ I_2 + H_2S \rightarrow 2 HI + S \]

we need to calculate the standard Gibbs free energy change (\( \Delta G^\circ \)) for the reaction using the provided Gibbs free energy of formation values at 298 K.

### Step-by-Step Solution:

1. **Identify the Gibbs Free Energy of Formation Values**:
- For \( HI \): \( \Delta G_f^\circ (HI) = 1.8 \, \text{kJ/mol} \)
- For \( H_2S \): \( \Delta G_f^\circ (H_2S) = 33.8 \, \text{kJ/mol} \)
- For \( I_2 \) and \( S \) (as they are in their elemental forms): \( \Delta G_f^\circ (I_2) = 0 \, \text{kJ/mol} \) and \( \Delta G_f^\circ (S) = 0 \, \text{kJ/mol} \)

2. **Write the Equation for \( \Delta G^\circ \) of the Reaction**:
The standard Gibbs free energy change for the reaction can be calculated using the formula:
\[ \Delta G^\circ_{\text{reaction}} = \sum \Delta G_f^\circ \text{(products)} - \sum \Delta G_f^\circ \text{(reactants)} \]

3. **Calculate the Gibbs Free Energy for Products**:
- For the products \( 2 HI + S \):
\[ \Delta G_f^\circ \text{(products)} = 2 \times \Delta G_f^\circ (HI) + \Delta G_f^\circ (S) = 2 \times 1.8 \, \text{kJ/mol} + 0 = 3.6 \, \text{kJ/mol} \]

4. **Calculate the Gibbs Free Energy for Reactants**:
- For the reactants \( I_2 + H_2S \):
\[ \Delta G_f^\circ \text{(reactants)} = \Delta G_f^\circ (I_2) + \Delta G_f^\circ (H_2S) = 0 + 33.8 \, \text{kJ/mol} = 33.8 \, \text{kJ/mol} \]

5. **Substitute into the Gibbs Free Energy Change Equation**:
\[ \Delta G^\circ_{\text{reaction}} = 3.6 \, \text{kJ/mol} - 33.8 \, \text{kJ/mol} = -30.2 \, \text{kJ/mol} \]

6. **Interpret the Result**:
Since \( \Delta G^\circ_{\text{reaction}} = -30.2 \, \text{kJ/mol} \) is negative, this indicates that the reaction is spontaneous under standard conditions.

### Conclusion:
The nature of the reaction is spontaneous.