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: Identify the spontaneity of the reaction
Since the reaction is stated to be spontaneous, we know that:
- **ΔG < 0** (Gibbs free energy change is negative)

### Step 2: Calculate ΔN (change in moles of gas)
To find ΔN, we need to calculate the difference between the moles of products and the moles of reactants:
- Moles of products = 2 (from \(2X_{3}\))
- Moles of reactants = 3 (from \(3X_{2}\))

Now, we can calculate ΔN:
\[ \Delta N = \text{Moles of products} - \text{Moles of reactants} = 2 - 3 = -1 \]

### Step 3: Determine the sign of ΔS
The change in entropy (ΔS) is related to ΔN. If ΔN is negative, then ΔS will also be negative:
- Since ΔN = -1, we have:
\[ \Delta S < 0 \]

### Step 4: Relate ΔG, ΔH, and ΔS
Using the Gibbs free energy equation:
\[ \Delta G = \Delta H - T\Delta S \]
Since we already established that ΔG < 0 and ΔS < 0, we can analyze the sign of ΔH.

### Step 5: Determine the sign of ΔH
Given that ΔG is negative and ΔS is negative, we can rearrange the equation to find ΔH:
\[ \Delta H = \Delta G + T\Delta S \]
- Since ΔG is negative and TΔS (where T is a positive temperature) is also negative, the sum of two negative values (ΔG and TΔS) will also be negative:
\[ \Delta H < 0 \]

### Conclusion
Based on the analysis:
- **ΔH < 0** (Enthalpy change is negative)
- **ΔS < 0** (Entropy change is negative)
- **ΔG < 0** (Gibbs free energy change is negative)

### Final Answer
The signs for the reaction \(3X_{2}(g) \rightarrow 2X_{3}(g)\) are:
- ΔH: Negative
- ΔS: Negative
- ΔG: Negative