Consider the following spontaneous reaction `3X_(2)(g)rarr2X_(3)(g)`. What are the sign of `DeltaH`, `DeltaS` and `DeltaG` for the reaction ?

Options

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

Explanation

To determine the signs of ΔH (enthalpy change), ΔS (entropy change), and ΔG (Gibbs free energy change) for the spontaneous reaction \(3X_2(g) \rightarrow 2X_3(g)\), we can follow these steps:

### Step 1: Analyze the Reaction
The reaction shows that 3 moles of \(X_2\) gas are converting into 2 moles of \(X_3\) gas.

### Step 2: Determine ΔG
For a spontaneous reaction, the change in Gibbs free energy (ΔG) must be negative:
\[ \Delta G = \Delta H - T \Delta S < 0 \]
This implies that either ΔH must be negative, ΔS must be positive, or both must be true to ensure ΔG is negative.

### Step 3: Determine ΔS
Entropy (ΔS) is associated with the disorder of a system. In this reaction:
- We start with 3 moles of gas and end with 2 moles of gas.
- A decrease in the number of gas moles typically indicates a decrease in entropy.

Thus, we conclude:
\[ \Delta S < 0 \quad (\text{negative}) \]

### Step 4: Determine ΔH
Since we have established that ΔS is negative, we need to determine the sign of ΔH to ensure that ΔG remains negative. The equation for ΔG can be rearranged:
\[ \Delta G = \Delta H - T \Delta S \]
Substituting ΔS:
\[ \Delta G = \Delta H - T(\text{negative value}) \]
This means that \( -T \Delta S \) is positive. Therefore, for ΔG to remain negative, ΔH must also be negative:
\[ \Delta H < 0 \quad (\text{negative}) \]

### Step 5: Summarize the Signs
From the analysis:
- ΔG is negative (since the reaction is spontaneous).
- ΔS is negative (as the number of gas moles decreases).
- ΔH is negative (to keep ΔG negative given that ΔS is negative).

### Final Answer
- ΔH: Negative
- ΔS: Negative
- ΔG: Negative