For the reaction `2A (g) + B(g) to C (g)`
`Delta U^(@) = 20` kcal/mole, `Delta S^(@) = 100` cal/kmole at 300 K. Hence `Delta G^(@)` in kcal is __________
`[R = 2 cal mol^(-1) K^(-1)]`

Explanation

To find the value of \(\Delta G^\circ\) for the reaction \(2A(g) + B(g) \rightarrow C(g)\), we will follow these steps:

### Step 1: Identify the given values
- \(\Delta U^\circ = 20\) kcal/mole
- \(\Delta S^\circ = 100\) cal/kmole
- Temperature \(T = 300\) K
- Gas constant \(R = 2\) cal/mol·K

### Step 2: Calculate \(\Delta N_g\)
\[ \Delta N_g = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} \]
- Moles of gaseous products = 1 (from C)
- Moles of gaseous reactants = 2 (from 2A) + 1 (from B) = 3
\[ \Delta N_g = 1 - 3 = -2 \]

### Step 3: Calculate \(\Delta H^\circ\)
Using the formula:
\[ \Delta H^\circ = \Delta U^\circ + \Delta N_g \cdot R \cdot T \]
First, convert \(R\) to kcal:
\[ R = 2 \text{ cal/mol·K} = 2 \times 10^{-3} \text{ kcal/mol·K} \]
Now substitute the values:
\[ \Delta H^\circ = 20 \text{ kcal} + (-2) \cdot (2 \times 10^{-3} \text{ kcal/mol·K}) \cdot (300 \text{ K}) \]
Calculating the second term:
\[ = 20 + (-2) \cdot (0.002) \cdot (300) = 20 - 1.2 = 18.8 \text{ kcal} \]

### Step 4: Calculate \(\Delta G^\circ\)
Using the formula:
\[ \Delta G^\circ = \Delta H^\circ - T \cdot \Delta S^\circ \]
Convert \(\Delta S^\circ\) to kcal:
\[ \Delta S^\circ = 100 \text{ cal/kmole} = 0.1 \text{ kcal/kmole} \]
Now substitute the values:
\[ \Delta G^\circ = 18.8 \text{ kcal} - (300 \text{ K}) \cdot (0.1 \text{ kcal/kmole}) \]
Calculating:
\[ = 18.8 - 30 = -11.2 \text{ kcal} \]

### Final Answer
\(\Delta G^\circ = -11.2\) kcal

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