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

Options

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

Answer

To find the value of ΔG for the given reaction, we can follow these steps:

### Step 1: Understand the relationship between ΔG, ΔH, and ΔS
The Gibbs free energy change (ΔG) can be calculated using the equation:
\[ \Delta G = \Delta H - T \Delta S \]
where:
- ΔG = change in Gibbs free energy
- ΔH = change in enthalpy
- T = temperature in Kelvin
- ΔS = change in entropy

### Step 2: Calculate ΔH using the relationship with ΔU
We know that:
\[ \Delta H = \Delta U + \Delta n_g RT \]
where:
- ΔU = change in internal energy
- Δn_g = change in the number of moles of gas (products - reactants)
- R = universal gas constant (in appropriate units)
- T = temperature in Kelvin

### Step 3: Determine Δn_g
From the reaction:
\[ X_2O_4(l) \rightarrow 2XO_2(g) \]
- The number of moles of gaseous products (2XO2) = 2
- The number of moles of gaseous reactants = 0 (since X2O4 is a liquid)

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

### Step 4: Substitute the values into the ΔH equation
Given:
- ΔU = 2.1 kcal (convert to cal: 2.1 kcal = 2100 cal)
- R = 1.987 cal/(K·mol) (use this value for calculations)
- T = 300 K

Now we can calculate ΔH:
\[ \Delta H = \Delta U + \Delta n_g RT \]
Substituting the values:
\[ \Delta H = 2100 \text{ cal} + (2)(1.987 \text{ cal/(K·mol)})(300 \text{ K}) \]
Calculating the second term:
\[ \Delta H = 2100 \text{ cal} + (2)(1.987)(300) \]
\[ = 2100 \text{ cal} + 11922 \text{ cal} \]
\[ = 14022 \text{ cal} \]

### Step 5: Calculate ΔG
Now we can substitute ΔH and ΔS into the ΔG equation:
Given ΔS = 20 cal/K, we can now calculate ΔG:
\[ \Delta G = \Delta H - T \Delta S \]
Substituting the values:
\[ \Delta G = 14022 \text{ cal} - (300 \text{ K})(20 \text{ cal/K}) \]
Calculating:
\[ \Delta G = 14022 \text{ cal} - 6000 \text{ cal} \]
\[ = 8022 \text{ cal} \]

### Step 6: Convert ΔG to kilocalories
To convert cal to kcal:
\[ \Delta G = \frac{8022 \text{ cal}}{1000} = 8.022 \text{ kcal} \]

### Final Answer
Thus, the value of ΔG is approximately:
\[ \Delta G \approx 8.022 \text{ kcal} \]