Predict the sign of `DeltaS^(@)` for the following reaction `:`
`2H_(2)S (g) + 3O_(2)(g) rarr 2H_(2)O(g) + 2SO_(2)(g)`

Options

A. `+ve`
B. `-ve`
C. zero
D. cannot be predicted

Explanation

To determine the sign of \( \Delta S^\circ \) (the change in standard entropy) for the reaction:

\[ 2H_{2}S (g) + 3O_{2}(g) \rightarrow 2H_{2}O(g) + 2SO_{2}(g) \]

we can follow these steps:

### Step 1: Count the number of moles of gaseous reactants and products.
- **Reactants**:
- \( 2H_{2}S \) contributes 2 moles.
- \( 3O_{2} \) contributes 3 moles.
- Total moles of reactants = \( 2 + 3 = 5 \) moles.

- **Products**:
- \( 2H_{2}O \) contributes 2 moles.
- \( 2SO_{2} \) contributes 2 moles.
- Total moles of products = \( 2 + 2 = 4 \) moles.

### Step 2: Compare the total number of moles of reactants and products.
- Total moles of reactants = 5
- Total moles of products = 4

### Step 3: Determine the change in entropy (\( \Delta S^\circ \)).
- Since the number of gaseous products (4 moles) is less than the number of gaseous reactants (5 moles), the randomness or disorder of the system decreases.

### Step 4: Predict the sign of \( \Delta S^\circ \).
- A decrease in the number of gas moles indicates a decrease in randomness, which means that the entropy change (\( \Delta S^\circ \)) will be negative.

Thus, we conclude that:
\[ \Delta S^\circ < 0 \]

### Final Answer:
The sign of \( \Delta S^\circ \) for the reaction is negative.

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