For the reaction at `25^(@)C, C_(2)O_(4)(l) rarr 2XO_(2)(g)`
`Delta H =2.1 kcal` and `DeltaS=20 cal K^(-1)` . The reaction would be `:`

Options

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

Explanation

To determine whether the reaction \( C_2O_4(l) \rightarrow 2CO_2(g) \) is spontaneous at \( 25^\circ C \), we will calculate the Gibbs free energy change (\( \Delta G \)) using the formula:

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

### Step 1: Convert the given values
1. **Convert \( \Delta H \) from kilocalories to calories**:
\[ \Delta H = 2.1 \, \text{kcal} = 2.1 \times 1000 \, \text{cal} = 2100 \, \text{cal} \]

2. **Convert the temperature from Celsius to Kelvin**:
\[ T = 25^\circ C + 273 = 298 \, \text{K} \]

3. **Given \( \Delta S \)**:
\[ \Delta S = 20 \, \text{cal K}^{-1} \]

### Step 2: Substitute the values into the Gibbs free energy equation
Now, substitute the values into the Gibbs free energy equation:
\[ \Delta G = \Delta H - T \Delta S \]
\[ \Delta G = 2100 \, \text{cal} - (298 \, \text{K} \times 20 \, \text{cal K}^{-1}) \]

### Step 3: Calculate \( T \Delta S \)
Calculate \( T \Delta S \):
\[ T \Delta S = 298 \times 20 = 5960 \, \text{cal} \]

### Step 4: Calculate \( \Delta G \)
Now substitute back to find \( \Delta G \):
\[ \Delta G = 2100 \, \text{cal} - 5960 \, \text{cal} \]
\[ \Delta G = 2100 - 5960 = -3860 \, \text{cal} \]

### Step 5: Analyze the result
Since \( \Delta G \) is negative (\( -3860 \, \text{cal} \)), the reaction is spontaneous.

### Conclusion
The reaction \( C_2O_4(l) \rightarrow 2CO_2(g) \) is spontaneous at \( 25^\circ C \).

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