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

Options

A. (a) -1.7 kcal
B. (b) 4.2 kcal
C. (c) -2.7 kcal
D. (d) 0.7 kcal

Answer

To solve the problem, we need to calculate the change in Gibbs free energy (ΔG) for the reaction given the change in internal energy (ΔU), change in entropy (ΔS), and temperature (T).

### Step-by-Step Solution:

1. **Identify the Given Values:**
- ΔU = 2.1 kcal
- ΔS = 20 Cal/K
- T = 300 K

2. **Convert ΔS from Cal to kcal:**
- Since 1 kcal = 1000 Cal,
\[ \Delta S = \frac{20 \text{ Cal/K}}{1000} = 0.02 \text{ kcal/K} \]

3. **Calculate Δn (Change in Moles of Gas):**
- The reaction shows that 1 mole of A (liquid) converts to 2 moles of B (gas).
- Therefore, the change in the number of moles (Δn) is:
\[ \Delta n = \text{moles of gaseous products} - \text{moles of gaseous reactants} = 2 - 0 = 2 \]

4. **Calculate ΔH (Change in Enthalpy):**
- The formula for ΔH is:
\[ \Delta H = \Delta U + \Delta n \cdot R \cdot T \]
- Here, R (the gas constant) in kcal is approximately 0.001987 kcal/(K·mol). However, in this case, we can use R = 2 Cal/(K·mol) = 0.002 kcal/(K·mol) for our calculations.
- Now substituting the values:
\[ \Delta H = 2.1 \text{ kcal} + (2 \cdot 0.002 \text{ kcal/(K·mol)} \cdot 300 \text{ K}) \]
\[ \Delta H = 2.1 \text{ kcal} + (2 \cdot 0.002 \cdot 300) \]
\[ \Delta H = 2.1 \text{ kcal} + 1.2 \text{ kcal} = 3.3 \text{ kcal} \]

5. **Calculate ΔG (Change in Gibbs Free Energy):**
- The formula for ΔG is:
\[ \Delta G = \Delta H - T \cdot \Delta S \]
- Now substituting the values:
\[ \Delta G = 3.3 \text{ kcal} - (300 \text{ K} \cdot 0.02 \text{ kcal/K}) \]
\[ \Delta G = 3.3 \text{ kcal} - 6 \text{ kcal} = -2.7 \text{ kcal} \]

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