For the reaction at `25^(@)C,X_(2)O_(4(l))to2XO_(2(g)),DeltaH=2.1kcal and DeltaS=20" cal "K^(-1)`. The reaction would be

Options

A. spontaneous
B. non-spontaneous
C. at equilibrium
D. unpredictable

Answer

To determine whether the reaction \( X_2O_4(l) \to 2XO_2(g) \) is spontaneous, we need to calculate the Gibbs free energy change (\( \Delta G \)) using the following formula:

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

### Step 1: Convert Given Values
- **Given:**
- \( \Delta H = 2.1 \, \text{kcal} \)
- \( \Delta S = 20 \, \text{cal K}^{-1} \)
- Temperature \( T = 25^\circ C = 25 + 273 = 298 \, \text{K} \)

**Conversion of \( \Delta H \) to calories:**
\[ \Delta H = 2.1 \, \text{kcal} = 2.1 \times 1000 \, \text{cal} = 2100 \, \text{cal} \]

### Step 2: Calculate \( T \Delta S \)
Now, we calculate \( T \Delta S \):
\[ T \Delta S = 298 \, \text{K} \times 20 \, \text{cal K}^{-1} = 5960 \, \text{cal} \]

### Step 3: Calculate \( \Delta G \)
Now we can substitute the values into the Gibbs free energy equation:
\[ \Delta G = \Delta H - T \Delta S \]
\[ \Delta G = 2100 \, \text{cal} - 5960 \, \text{cal} = -3860 \, \text{cal} \]

### Step 4: Analyze the Result
Since \( \Delta G < 0 \), the reaction is spontaneous.

### Final Answer
The reaction \( X_2O_4(l) \to 2XO_2(g) \) is spontaneous.

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