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?

Options

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

Explanation

To solve the problem, we need to analyze the given reactions to determine which one has a positive standard reaction entropy change (ΔS°) and also causes the standard Gibbs energy change (ΔG°) to decrease sharply with increasing temperature.

### Step-by-Step Solution:

1. **Understanding ΔS°**:
- The standard reaction entropy change (ΔS°) is positive when there is an increase in the number of gaseous moles from reactants to products. This is because an increase in gaseous moles generally indicates an increase in randomness or disorder.

2. **Analyzing the Reactions**:
- Let's denote the reactions as follows:
1. \( C_{(s)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{(g)} \)
2. \( A_{(g)} + B_{(g)} \rightarrow C_{(g)} \)
3. \( D_{(s)} + \frac{1}{2} E_{(g)} \rightarrow F_{(s)} \)
4. \( G_{(s)} + \frac{1}{2} H_{(g)} \rightarrow I_{(g)} + J_{(g)} \)

- We will calculate Δn (change in moles of gas) for each reaction:
- **Reaction 1**:
- Reactants: 0.5 moles of \( O_2 \) (g) + 1 mole of \( C \) (s) = 0.5 gaseous moles
- Products: 1 mole of \( CO \) (g) = 1 gaseous mole
- Δn = 1 - 0.5 = 0.5 (positive)

- **Reaction 2**:
- Reactants: 1 mole of \( A \) (g) + 1 mole of \( B \) (g) = 2 gaseous moles
- Products: 1 mole of \( C \) (g) = 1 gaseous mole
- Δn = 1 - 2 = -1 (negative)

- **Reaction 3**:
- Reactants: 0.5 moles of \( E \) (g) + 1 mole of \( D \) (s) = 0.5 gaseous moles
- Products: 0 moles of gas (solid \( F \)) = 0 gaseous moles
- Δn = 0 - 0.5 = -0.5 (negative)

- **Reaction 4**:
- Reactants: 0.5 moles of \( H \) (g) + 1 mole of \( G \) (s) = 0.5 gaseous moles
- Products: 2 moles of gas (1 mole of \( I \) and 1 mole of \( J \)) = 2 gaseous moles
- Δn = 2 - 0.5 = 1.5 (positive)

3. **Identifying Positive ΔS°**:
- From our analysis:
- Reaction 1 has Δn = 0.5 (positive)
- Reaction 4 has Δn = 1.5 (positive)
- Therefore, both reactions 1 and 4 have positive ΔS°.

4. **Analyzing ΔG°**:
- The Gibbs free energy change is given by the equation:
\[ ΔG° = ΔH° - TΔS° \]
- For ΔG° to decrease sharply with increasing temperature, ΔS° must be positive, and the term \( TΔS° \) must grow larger than ΔH° as temperature increases.

5. **Conclusion**:
- Reaction 1 has a positive ΔS° and will have a ΔG° that decreases with increasing temperature.
- Reaction 4 also has a positive ΔS°, but we need to ensure that the enthalpy change (ΔH°) is not too large to negate the effect of \( TΔS° \).
- Since the problem specifically asks for a reaction where ΔG° decreases sharply, we conclude that **Reaction 1** is the most likely candidate.

### Final Answer:
The reaction in which the standard reaction entropy change (ΔS°) is positive and the standard Gibbs energy change (ΔG°) decreases sharply with increasing temperature is **Reaction 1**.