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

Answer

To find the change in Gibbs free energy (ΔG) for the reaction given, we can follow these steps:

### Step 1: Write down the Gibbs free energy equation
The Gibbs free energy (ΔG) is related to the change in enthalpy (ΔH) and the change in entropy (ΔS) by the equation:
\[ \Delta G = \Delta H - T \Delta S \]

### Step 2: Relate ΔH to ΔU
Since we are given ΔU (change in internal energy) instead of ΔH, we can use the relationship:
\[ \Delta H = \Delta U + P \Delta V \]
For gases, we can express \(P \Delta V\) in terms of the change in the number of moles of gas:
\[ \Delta H = \Delta U + \Delta n RT \]
where \(\Delta n\) is the change in the number of moles of gas, \(R\) is the universal gas constant, and \(T\) is the temperature in Kelvin.

### Step 3: Calculate Δn
From the reaction \(A (l) \rightarrow 2 B (g)\), we see that:
- The number of moles of gaseous products (B) = 2
- The number of moles of gaseous reactants (A) = 0 (since A is a liquid)

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

### Step 4: Substitute ΔH into the Gibbs equation
Now we can substitute ΔH into the Gibbs free energy equation:
\[ \Delta G = \Delta U + \Delta n RT - T \Delta S \]

### Step 5: Substitute the known values
We have:
- ΔU = 2.1 kcal
- Δn = 2
- T = 300 K
- ΔS = 20 Cal/K = 0.02 kcal/K (since 1 kcal = 1000 Cal)

Now, we need to use the value of R in kcal:
\[ R = 2 \text{ Cal/(mol K)} = 0.002 \text{ kcal/(mol K)} \]

### Step 6: Calculate the terms
Now we can calculate \( \Delta n RT \):
\[ \Delta n RT = 2 \times 0.002 \text{ kcal/(mol K)} \times 300 \text{ K} = 1.2 \text{ kcal} \]

Now calculate \(T \Delta S\):
\[ T \Delta S = 300 \text{ K} \times 20 \text{ Cal/K} = 6000 \text{ Cal} = 6 \text{ kcal} \]

### Step 7: Substitute everything back into the Gibbs equation
Now substituting everything back into the Gibbs equation:
\[ \Delta G = 2.1 \text{ kcal} + 1.2 \text{ kcal} - 6 \text{ kcal} \]
\[ \Delta G = 3.3 \text{ kcal} - 6 \text{ kcal} = -2.7 \text{ kcal} \]

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

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