Considering the reaction,
`C(s) + O_(2)(g) to CO_(2)(g) + 393.5 kJ`
the signs of `DeltaH, DeltaS` and `DeltaG` respectively are:

Options

A. `+,-,-`
B. `-,+,-`
C. `-,-,-`
D. `-,+,1`

Explanation

To determine the signs of ΔH, ΔS, and ΔG for the reaction:

\[ C(s) + O_2(g) \rightarrow CO_2(g) + 393.5 \, \text{kJ} \]

we will analyze each thermodynamic quantity step by step.

### Step 1: Determine ΔH (Change in Enthalpy)

1. **Identify the type of reaction**: The given reaction is a combustion reaction, where carbon (solid) reacts with oxygen (gas) to form carbon dioxide (gas) and releases energy (393.5 kJ).

2. **Sign of ΔH**: In combustion reactions, energy is released, indicating that the reaction is exothermic. Therefore, the change in enthalpy (ΔH) is negative.

\[ \Delta H < 0 \]

### Step 2: Determine ΔS (Change in Entropy)

1. **Analyze the states of matter**: In the reaction, we start with one solid (C) and one gas (O₂), and we end up with one gas (CO₂).

2. **Change in disorder**: The transition from solid and gas to only gas generally leads to an increase in disorder (entropy). However, since we are losing one mole of gas (O₂) while producing one mole of gas (CO₂), we need to consider the overall change.

3. **Sign of ΔS**: In this case, the reaction results in a decrease in the number of gas moles, which typically leads to a decrease in entropy. Therefore, ΔS is negative.

\[ \Delta S < 0 \]

### Step 3: Determine ΔG (Change in Gibbs Free Energy)

1. **Use the Gibbs Free Energy equation**: The relationship between ΔG, ΔH, and ΔS is given by the equation:

\[ \Delta G = \Delta H - T\Delta S \]

2. **Evaluate ΔG**: Since ΔH is negative and ΔS is also negative, the term \( -T\Delta S \) will be positive (as T is always positive). However, because ΔH is significantly negative due to the exothermic nature of the reaction, it will dominate the equation.

3. **Sign of ΔG**: Therefore, ΔG will also be negative, indicating that the reaction is spontaneous.

\[ \Delta G < 0 \]

### Conclusion

Based on the analysis:

- ΔH is negative,
- ΔS is negative,
- ΔG is negative.

Thus, the signs of ΔH, ΔS, and ΔG respectively are:

\[ \text{Negative, Negative, Negative} \]