In which of the following reactions, standard reaction entropy changes (`DeltaS`) is positive and standard Gibbs energy change ('DeltaG`) decreases sharply with increasing temperature?

Options

A. `C ("graphite") + 1/2 O_2(g) to CO(g) `
B. `CO(g) + 1/2O_2(g) to CO_2(g)`
C. `Mg (s) + 1/2O_2(g) to MgO (g) `
D. `1/2 C ("graphite") + 1/2 O_2(g) to 1/2 CO_2 (g) `

Explanation

To determine in which of the given reactions the standard reaction entropy change (ΔS) is positive and the standard Gibbs energy change (ΔG) decreases sharply with increasing temperature, we can follow these steps:

### Step 1: Understand the relationship between ΔS and ΔG
The Gibbs free energy change (ΔG) is given by the equation:
\[ \Delta G = \Delta H - T \Delta S \]
Where:
- ΔG = Gibbs free energy change
- ΔH = Enthalpy change
- T = Temperature (in Kelvin)
- ΔS = Entropy change

For ΔG to decrease sharply with increasing temperature, ΔS must be positive. This means that as the temperature increases, the term \( -T\Delta S \) becomes more negative, leading to a decrease in ΔG.

### Step 2: Analyze the reactions for ΔS
To determine whether ΔS is positive, we need to calculate the change in the number of gaseous moles (ΔNg) for each reaction. The formula is:
\[ \Delta N_g = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} \]

### Step 3: Calculate ΔNg for each reaction
1. **Reaction 1:** \( C_{(s)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{(g)} \)
- Products: 1 mole of \( CO \) (gaseous)
- Reactants: 0 moles of gas (1/2 mole of \( O_2 \) is gaseous)
- ΔNg = 1 - 0 = 1 (positive)

2. **Reaction 2:** \( \frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)} \rightarrow NH_{3(g)} \)
- Products: 1 mole of \( NH_3 \) (gaseous)
- Reactants: 2 moles of gas (1/2 mole of \( N_2 \) + 3/2 moles of \( H_2 \))
- ΔNg = 1 - 2 = -1 (negative)

3. **Reaction 3:** \( MgO_{(s)} + \frac{1}{2} O_{2(g)} \rightarrow Mg_{(s)} + O_{2(g)} \)
- Products: 0 moles of gas
- Reactants: 1/2 mole of \( O_2 \) (gaseous)
- ΔNg = 0 - 1/2 = -1/2 (negative)

4. **Reaction 4:** \( H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_2O_{(l)} \)
- Products: 0 moles of gas (water is liquid)
- Reactants: 1 mole of \( H_2 \) + 1/2 mole of \( O_2 \) (gaseous)
- ΔNg = 0 - 1.5 = -1.5 (negative)

### Step 4: Identify the reaction with positive ΔS
From the calculations, only **Reaction 1** has a positive ΔNg, which indicates that ΔS is also positive.

### Step 5: Confirm that ΔG decreases sharply with increasing temperature
Since ΔS is positive for Reaction 1, as temperature (T) increases, the term \( -T\Delta S \) becomes more negative, which means that ΔG will decrease sharply.

### Conclusion
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: \( C_{(s)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{(g)} \)**