For the reaction,
`X_(2)O_(4)(l)rarr2XO_(2)(g)`,`DeltaE=3.1Kcal` ,
`DeltaS=30cal//K` at 300K . Hence` DeltaG `is

Answer

To calculate ΔG for the reaction \( X_2O_4(l) \rightarrow 2XO_2(g) \), we will follow these steps:

### Step 1: Identify Given Values
- ΔE (or ΔH) = 3.1 kcal
- ΔS = 30 cal/K
- Temperature (T) = 300 K

### Step 2: Convert ΔS to kcal
Since ΔS is given in cal/K, we need to convert it to kcal/K:
\[ \Delta S = 30 \, \text{cal/K} = 30 \times 10^{-3} \, \text{kcal/K} = 0.03 \, \text{kcal/K} \]

### Step 3: Calculate ΔH
Using the relationship between ΔE and ΔH:
\[ \Delta H = \Delta E + \Delta n_g RT \]
Where:
- \( R = 2 \times 10^{-3} \, \text{kcal/K} \) (gas constant)
- \( \Delta n_g = \text{moles of products} - \text{moles of reactants} = 2 - 1 = 1 \)

Now substituting the values:
\[ \Delta H = 3.1 \, \text{kcal} + (1)(2 \times 10^{-3} \, \text{kcal/K})(300 \, \text{K}) \]
Calculating the second term:
\[ \Delta H = 3.1 \, \text{kcal} + (1)(0.6 \, \text{kcal}) = 3.1 + 0.6 = 3.7 \, \text{kcal} \]

### Step 4: Calculate ΔG
Using the Gibbs free energy equation:
\[ \Delta G = \Delta H - T \Delta S \]
Substituting the values:
\[ \Delta G = 3.7 \, \text{kcal} - (300 \, \text{K})(0.03 \, \text{kcal/K}) \]
Calculating the second term:
\[ \Delta G = 3.7 \, \text{kcal} - 9 \, \text{kcal} = 3.7 - 9 = -5.3 \, \text{kcal} \]

### Final Answer
\[ \Delta G = -5.3 \, \text{kcal} \]

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