For the reaction.
`A(l)rarr 2B(g)`
`DeltaU="2.1 k Cal, "DeltaS="20 Cal K"^(-1)" at 300K`
Hence `DeltaG" in kcal "` is ?

Options

A. 2.7
B. `-2.7`
C. 5.4
D. 1.35

Answer

To solve the given problem, we need to calculate the change in Gibbs free energy (ΔG) for the reaction \( A(l) \rightarrow 2B(g) \) using the provided values for ΔU and ΔS.

### Step 1: Identify the given values
- ΔU = 2.1 kcal (change in internal energy)
- ΔS = 20 cal/K (change in entropy)
- 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:
\[ \Delta S = \frac{20 \text{ cal}}{1000} = 0.02 \text{ kcal/K} \]

### Step 3: Calculate ΔN
ΔN is the change in the number of moles of gas:
- Products: 2 moles of B (g)
- Reactants: 0 moles of A (l)

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

### Step 4: Calculate ΔH using the formula
The relationship between ΔH, ΔU, and ΔN is given by:
\[ \Delta H = \Delta U + \Delta N \cdot R \cdot T \]
Where R (universal gas constant) in kcal is approximately 0.001987 kcal/K·mol. For our calculation, we can use R = 2 cal/K·mol = 0.002 kcal/K·mol.

Now substituting the values:
\[ \Delta H = 2.1 \text{ kcal} + (2) \cdot (0.002 \text{ kcal/K·mol}) \cdot (300 \text{ K}) \]
Calculating the second term:
\[ = 2.1 \text{ kcal} + (2 \cdot 0.002 \cdot 300) \text{ kcal} = 2.1 \text{ kcal} + 1.2 \text{ kcal} = 3.3 \text{ kcal} \]

### Step 5: Calculate ΔG using the Gibbs free energy equation
The Gibbs free energy change is calculated using the formula:
\[ \Delta G = \Delta H - T \cdot \Delta S \]
Substituting the values:
\[ \Delta G = 3.3 \text{ kcal} - (300 \text{ K} \cdot 0.02 \text{ kcal/K}) \]
Calculating the second term:
\[ = 3.3 \text{ kcal} - 6 \text{ kcal} = -2.7 \text{ kcal} \]

### Final Answer
Thus, the change in Gibbs free energy (ΔG) is:
\[ \Delta G = -2.7 \text{ kcal} \]