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 find \(\Delta G\) for the reaction \(A(l) \rightarrow 2B(g)\), we can use the following steps:

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

### Step 2: Calculate \(\Delta N_g\)
\(\Delta N_g\) is the change in the number of moles of gas during the reaction. Here, we have:
- Products: 2 moles of \(B(g)\)
- Reactants: 0 moles of \(A(l)\)

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

### Step 3: Convert \(\Delta S\) to kcal
Since \(\Delta S\) is given in calories, we need to convert it to kilocalories:
\[ \Delta S = 20 \, \text{Cal K}^{-1} = 20 \times 10^{-3} \, \text{kcal K}^{-1} = 0.02 \, \text{kcal K}^{-1} \]

### Step 4: Calculate \(\Delta H\) using the relation
Using the relation \(\Delta H = \Delta U + \Delta N_g \cdot R \cdot T\):
- \(R = 2 \times 10^{-3} \, \text{kcal K}^{-1}\) (as we are using kcal)
- Now, substituting the values:
\[ \Delta H = \Delta U + \Delta N_g \cdot R \cdot T \]
\[ \Delta H = 2.1 \, \text{kcal} + (2) \cdot (2 \times 10^{-3} \, \text{kcal K}^{-1}) \cdot (300 \, \text{K}) \]
\[ \Delta H = 2.1 \, \text{kcal} + (2 \cdot 0.002 \cdot 300) \]
\[ \Delta H = 2.1 \, \text{kcal} + 1.2 \, \text{kcal} = 3.3 \, \text{kcal} \]

### Step 5: Calculate \(\Delta G\) using the Gibbs Helmholtz equation
Using the equation:
\[ \Delta G = \Delta H - T \Delta S \]
Substituting the values:
\[ \Delta G = 3.3 \, \text{kcal} - (300 \, \text{K} \cdot 0.02 \, \text{kcal K}^{-1}) \]
\[ \Delta G = 3.3 \, \text{kcal} - 6 \, \text{kcal} \]
\[ \Delta G = -2.7 \, \text{kcal} \]

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

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