For the reaction:
`X_(2)O_(4)(l)rarr2XO_(2)(g)`
`DeltaU=2.1kcal , DeltaS =20 "cal" K^(-1) "at" 300 K`
Hence `DeltaG` is

Options

A. 2.7 kcal
B. `-2.7` kcal
C. 9.3 kcal
D. `-9.3` kcal

Explanation

To find the value of \(\Delta G\) for the reaction \(X_{2}O_{4}(l) \rightarrow 2XO_{2}(g)\), we will follow these steps:

### Step 1: Identify the given data
- \(\Delta U = 2.1 \, \text{kcal}\)
- \(\Delta S = 20 \, \text{cal K}^{-1}\)
- Temperature \(T = 300 \, \text{K}\)

### Step 2: Convert \(\Delta S\) to kilocalories
Since \(\Delta S\) is given in calories, we need to convert it to kilocalories:
\[ \Delta S = 20 \, \text{cal K}^{-1} = \frac{20}{1000} \, \text{kcal K}^{-1} = 0.02 \, \text{kcal K}^{-1} \]

### Step 3: Calculate \(\Delta n\)
For the reaction, the change in the number of moles of gas (\(\Delta n\)) is calculated as follows:
- Moles of gaseous products = 2 (from \(2XO_{2}\))
- Moles of gaseous reactants = 0 (from \(X_{2}O_{4}\))

Thus,
\[ \Delta n = 2 - 0 = 2 \]

### Step 4: Calculate \(\Delta H\) using the formula
We will use the formula:
\[ \Delta H = \Delta U + \Delta n \cdot R \cdot T \]
Where \(R\) in kilocalories is \(2 \times 10^{-3} \, \text{kcal K}^{-1} \text{mol}^{-1}\).

Substituting the values:
\[ \Delta H = 2.1 \, \text{kcal} + (2) \cdot (2 \times 10^{-3} \, \text{kcal K}^{-1}) \cdot (300 \, \text{K}) \]
Calculating the second term:
\[ \Delta H = 2.1 \, \text{kcal} + (2 \cdot 2 \cdot 300 \cdot 10^{-3}) \, \text{kcal} \]
\[ = 2.1 \, \text{kcal} + (1200 \cdot 10^{-3}) \, \text{kcal} \]
\[ = 2.1 \, \text{kcal} + 1.2 \, \text{kcal} = 3.3 \, \text{kcal} \]

### Step 5: Calculate \(\Delta G\) using the formula
Now we will use the formula:
\[ \Delta G = \Delta H - T \cdot \Delta S \]
Substituting the values:
\[ \Delta G = 3.3 \, \text{kcal} - (300 \, \text{K}) \cdot (0.02 \, \text{kcal K}^{-1}) \]
Calculating the second term:
\[ \Delta G = 3.3 \, \text{kcal} - (300 \cdot 0.02) \, \text{kcal} \]
\[ = 3.3 \, \text{kcal} - 6 \, \text{kcal} \]
\[ = -2.7 \, \text{kcal} \]

### Final Answer
Thus, \(\Delta G = -2.7 \, \text{kcal}\).

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