For the formation of `C(g) "at" 300` K.
`A(g) + 3B(g) rarr 2C(g)`
Calculate the magnitude of `DeltaG^(@)` (Kcal) if given data:
`{:(,A,B,C),(DeltaH_(f)^(@)("Kcal mol"^(-1)),0,0,-10),(DeltaS_(f)^(@) ("Cal K"^(-1) "mol"^(-1)), 40,30,45):}`

Answer

To solve the problem, we need to calculate the standard Gibbs free energy change (ΔG°) for the reaction:

\[ A(g) + 3B(g) \rightarrow 2C(g) \]

Given data:
- Enthalpy of formation (ΔH_f°) in Kcal/mol:
- A: 0
- B: 0
- C: -10
- Entropy of formation (ΔS_f°) in Cal/K·mol:
- A: 40
- B: 30
- C: 45

### Step 1: Calculate ΔH° of the reaction
The enthalpy change for the reaction (ΔH°) can be calculated using the formula:

\[ \Delta H^\circ_{\text{reaction}} = \sum \Delta H^\circ_f \text{(products)} - \sum \Delta H^\circ_f \text{(reactants)} \]

Substituting the values:

\[ \Delta H^\circ_{\text{reaction}} = [2 \times (-10)] - [1 \times 0 + 3 \times 0] \]

Calculating this gives:

\[ \Delta H^\circ_{\text{reaction}} = -20 \text{ Kcal} \]

### Step 2: Calculate ΔS° of the reaction
The entropy change for the reaction (ΔS°) can be calculated using the formula:

\[ \Delta S^\circ_{\text{reaction}} = \sum \Delta S^\circ_f \text{(products)} - \sum \Delta S^\circ_f \text{(reactants)} \]

Substituting the values:

\[ \Delta S^\circ_{\text{reaction}} = [2 \times 45] - [1 \times 40 + 3 \times 30] \]

Calculating this gives:

\[ \Delta S^\circ_{\text{reaction}} = 90 - (40 + 90) = 90 - 130 = -40 \text{ Cal/K·mol} \]

### Step 3: Convert ΔS° to Kcal
Since ΔS° is in Cal/K·mol, we need to convert it to Kcal:

\[ \Delta S^\circ_{\text{reaction}} = -40 \text{ Cal/K·mol} = -0.04 \text{ Kcal/K·mol} \]

### Step 4: Calculate ΔG° using the Gibbs free energy equation
The Gibbs free energy change (ΔG°) can be calculated using the formula:

\[ \Delta G^\circ_{\text{reaction}} = \Delta H^\circ_{\text{reaction}} - T \Delta S^\circ_{\text{reaction}} \]

Where \( T = 300 \text{ K} \).

Substituting the values we have:

\[ \Delta G^\circ_{\text{reaction}} = -20 \text{ Kcal} - (300 \text{ K}) \times (-0.04 \text{ Kcal/K·mol}) \]

Calculating this gives:

\[ \Delta G^\circ_{\text{reaction}} = -20 + 12 = -8 \text{ Kcal} \]

### Final Answer
The magnitude of ΔG° is:

\[ \Delta G^\circ = -8 \text{ Kcal} \]