` A _ ("(l)") to 2 B _ ( "(g)") `
` Delta U = 2. 1 ` kcal ,` Delta S = 20 ` Cal/k, T = 300 K.
Find ` Delta G ` ( in kcal )

Explanation

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

### Step 1: Identify the given values
We have:
- \( \Delta U = 2.1 \) kcal
- \( \Delta S = 20 \) Cal/K (which is equal to \( 0.020 \) kcal/K when converted)
- \( T = 300 \) K

### Step 2: Calculate \( \Delta H \)
To calculate \( \Delta H \), we use the equation:
\[ \Delta H = \Delta U + \Delta n_g \cdot R \cdot T \]
where \( \Delta n_g \) is the change in the number of moles of gas, \( R \) is the gas constant (in kcal), and \( T \) is the temperature in Kelvin.

### Step 3: Determine \( \Delta n_g \)
In the reaction, we have:
- Products: 2 moles of \( B(g) \)
- Reactants: 1 mole of \( A(l) \) (which does not contribute to gas moles)

Thus, \( \Delta n_g = 2 - 0 = 2 \).

### Step 4: Substitute the values into the equation for \( \Delta H \)
Using \( R = 0.001987 \) kcal/K·mol (approximately \( 2 \times 10^{-3} \) kcal/K for calculations):
\[ \Delta H = 2.1 + (2 \cdot 0.001987 \cdot 300) \]
Calculating the second term:
\[ 2 \cdot 0.001987 \cdot 300 = 1.1922 \text{ kcal} \]
Now substituting back:
\[ \Delta H = 2.1 + 1.1922 = 3.2922 \text{ kcal} \approx 3.3 \text{ kcal} \]

### Step 5: Calculate \( \Delta G \)
Now we can find \( \Delta G \) using the Gibbs free energy equation:
\[ \Delta G = \Delta H - T \Delta S \]
Substituting the values:
\[ \Delta G = 3.3 - (300 \cdot 0.020) \]
Calculating \( T \Delta S \):
\[ 300 \cdot 0.020 = 6 \text{ kcal} \]
Now substituting back:
\[ \Delta G = 3.3 - 6 = -2.7 \text{ kcal} \]

### Final Answer
\[ \Delta G = -2.7 \text{ kcal} \]