For the synthesis of ammonia at 300 K :
`N_2(g)+3H_2(g)to2NH_3(g)`
Calculate the value of `DeltaG^@` in Kcal and give your answer in magnitude by using the following data :
`{:(( ),N_2,H_2,NH_3),(DeltaH_f(Kcal//"mole"),0,0,-10),(S^@(Cal//"K-mole"),40,30,45):}`

Explanation

To calculate the value of ΔG° for the synthesis of ammonia at 300 K, we will follow these steps:

### Step 1: Write the Reaction
The balanced chemical equation for the synthesis of ammonia is:
\[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \]

### Step 2: Calculate ΔH°
Using the given standard enthalpy of formation (ΔH_f) values:
- ΔH_f(N2) = 0 kcal/mol
- ΔH_f(H2) = 0 kcal/mol
- ΔH_f(NH3) = -10 kcal/mol

The expression for ΔH° for the reaction is:
\[ \Delta H° = \left(2 \times \Delta H_f(NH3)\right) - \left(3 \times \Delta H_f(H2) + \Delta H_f(N2)\right) \]
Substituting the values:
\[ \Delta H° = \left(2 \times -10\right) - \left(3 \times 0 + 0\right) = -20 \text{ kcal} \]

### Step 3: Calculate ΔS°
Using the given standard entropy (S°) values:
- S°(N2) = 40 cal/(K·mol)
- S°(H2) = 30 cal/(K·mol)
- S°(NH3) = 45 cal/(K·mol)

The expression for ΔS° for the reaction is:
\[ \Delta S° = \left(2 \times S°(NH3)\right) - \left(3 \times S°(H2) + S°(N2)\right) \]
Substituting the values:
\[ \Delta S° = \left(2 \times 45\right) - \left(3 \times 30 + 40\right) = 90 - (90 + 40) = 90 - 130 = -40 \text{ cal/K} \]

### Step 4: Convert ΔS° to kcal
Since ΔS° is given in cal/K, we need to convert it to kcal:
\[ \Delta S° = -40 \text{ cal/K} = -0.04 \text{ kcal/K} \]

### Step 5: Calculate ΔG°
Using the Gibbs free energy equation:
\[ \Delta G° = \Delta H° - T \Delta S° \]
Where T = 300 K. Substituting the values:
\[ \Delta G° = -20 \text{ kcal} - (300 \text{ K} \times -0.04 \text{ kcal/K}) \]
\[ \Delta G° = -20 \text{ kcal} + 12 \text{ kcal} = -8 \text{ kcal} \]

### Final Answer
The value of ΔG° is:
\[ \Delta G° = -8 \text{ kcal} \]
In magnitude, the answer is \(8 \text{ kcal}\).