In which of the following reactions,standard reaction entropy change`(DeltaS^(@))`is positive and standard Gibb,s energy change`(DeltaG^(@))`decreases sharply with increasing temperature?
(a)`Mg(s)+(1)/(2)O_(2)(g)to MgO(s)`
(b)`(1)/(2)C " graphite " +(1)/(2)O_(2)(g)to (1)/(2)CO_(2)(g)`
(c)`C " graphite " +(1)/(2)O_(2)(g)to CO(g)`
(d)`CO(g)+(1)/(2)O_(2)(g)to CO_(2)(g)`

Options

A. `Mg(s)+(1)/(2)O_(2)(g)to MgO(s)`
B. `(1)/(2)C " graphite " +(1)/(2)O_(2)(g)to (1)/(2)CO_(2)(g)`
C. `C " graphite " +(1)/(2)O_(2)(g)to CO(g)`
D. `CO(g)+(1)/(2)O_(2)(g)to CO_(2)(g)`

Explanation

To determine which reaction has a positive standard reaction entropy change (ΔS°) and a standard Gibbs energy change (ΔG°) that decreases sharply with increasing temperature, we need to analyze each reaction step by step.

### Step 1: Analyze the reactions

1. **Reaction (a)**:
\[ \text{Mg(s)} + \frac{1}{2} \text{O}_2(g) \rightarrow \text{MgO(s)} \]
- Reactants: 1 solid (Mg) + 0.5 gas (O₂)
- Products: 1 solid (MgO)
- **Entropy Change (ΔS°)**: Decreases (since we start with gas and end with solids).

2. **Reaction (b)**:
\[ \frac{1}{2} \text{C (graphite)} + \frac{1}{2} \text{O}_2(g) \rightarrow \frac{1}{2} \text{CO}_2(g) \]
- Reactants: 0.5 solid (C) + 0.5 gas (O₂)
- Products: 0.5 gas (CO₂)
- **Entropy Change (ΔS°)**: No change (since both sides have the same amount of gas).

3. **Reaction (c)**:
\[ \text{C (graphite)} + \frac{1}{2} \text{O}_2(g) \rightarrow \text{CO(g)} \]
- Reactants: 1 solid (C) + 0.5 gas (O₂)
- Products: 1 gas (CO)
- **Entropy Change (ΔS°)**: Increases (since we go from 1 solid + 0.5 gas to 1 gas).

4. **Reaction (d)**:
\[ \text{CO(g)} + \frac{1}{2} \text{O}_2(g) \rightarrow \text{CO}_2(g) \]
- Reactants: 1 gas (CO) + 0.5 gas (O₂)
- Products: 1 gas (CO₂)
- **Entropy Change (ΔS°)**: Decreases (since we go from 1.5 gas to 1 gas).

### Step 2: Determine ΔG° behavior with temperature

- The Gibbs free energy change is given by the equation:
\[ \Delta G° = \Delta H° - T\Delta S° \]
- For ΔG° to decrease sharply with increasing temperature, ΔS° must be positive. This means that the reaction must produce more disorder (increase in entropy).

### Step 3: Identify the correct reaction

From the analysis:
- Reaction (a): ΔS° is negative.
- Reaction (b): ΔS° is zero.
- Reaction (c): ΔS° is positive (this is a candidate).
- Reaction (d): ΔS° is negative.

### Conclusion

The only reaction with a positive ΔS° is reaction (c):
\[ \text{C (graphite)} + \frac{1}{2} \text{O}_2(g) \rightarrow \text{CO(g)} \]
This reaction also leads to a decrease in ΔG° as temperature increases, making it the correct answer.

### Final Answer
**(c) C (graphite) + (1/2) O₂(g) → CO(g)**