Consider the reacton
`C(s) + (1)/(2) O_(2)(g) rarr CO(g) + 200 kJ`
The signs of `Delta S, Delta H and Delta G` respectively are

Options

A. `+, -, -`
B. `-, +, +`
C. `-, -, -`
D. `-, +, -`

Answer

To determine the signs of ΔS (change in entropy), ΔH (change in enthalpy), and ΔG (change in Gibbs free energy) for the reaction:

\[ C(s) + \frac{1}{2} O_2(g) \rightarrow CO(g) + 200 \text{ kJ} \]

we will analyze each component step by step.

### Step 1: Analyze ΔS (Change in Entropy)
1. **Identify the states of the reactants and products**:
- Reactants: 1 solid (C) and 0.5 moles of gas (O₂).
- Products: 1 mole of gas (CO).

2. **Determine the change in the number of gas moles**:
- Reactants have 0.5 moles of gas.
- Products have 1 mole of gas.
- The number of gaseous moles increases from 0.5 to 1.

3. **Conclusion about ΔS**:
- Since the number of gas moles increases, the disorder (entropy) of the system increases.
- Therefore, ΔS is positive.

### Step 2: Analyze ΔH (Change in Enthalpy)
1. **Identify the nature of the reaction**:
- The reaction releases 200 kJ of energy, indicating it is exothermic.

2. **Conclusion about ΔH**:
- For exothermic reactions, ΔH is negative.
- Therefore, ΔH is negative.

### Step 3: Analyze ΔG (Change in Gibbs Free Energy)
1. **Use the Gibbs free energy equation**:
\[ \Delta G = \Delta H - T \Delta S \]

2. **Substitute the signs of ΔH and ΔS**:
- We have ΔH < 0 (negative) and ΔS > 0 (positive).

3. **Analyze the equation**:
- Since ΔH is negative and ΔS is positive, the term \( -T \Delta S \) will also be negative (as T is always positive).
- Therefore, the overall expression for ΔG will be negative.

### Conclusion
- The signs of ΔS, ΔH, and ΔG are:
- ΔS: Positive (+)
- ΔH: Negative (−)
- ΔG: Negative (−)

Thus, the answer is: **ΔS: +, ΔH: −, ΔG: −**.