For the reaction :-
`N_(2)O_(4(g))hArr2NO_(2(g))`
`DeltaU=2.0Kcal,DeltaS=50CalK^(-1)at300K` calculate `DeltaG`?

Options

A. `+12.4K Cal`
B. `-12.4K Cal`
C. `-6.4K Cal`
D. `+6.4K Cal`

Explanation

To calculate the Gibbs free energy change (ΔG) for the reaction \( N_2O_4(g) \rightleftharpoons 2NO_2(g) \), we will follow these steps:

### Step 1: Understand the given values
We have:
- \( \Delta U = 2.0 \, \text{kcal} \)
- \( \Delta S = 50 \, \text{cal K}^{-1} \)
- Temperature \( T = 300 \, \text{K} \)

### Step 2: Convert units where necessary
Since \( \Delta S \) is given in calories, we need to convert it to kilocalories for consistency with \( \Delta U \):
\[ \Delta S = 50 \, \text{cal K}^{-1} = \frac{50}{1000} \, \text{kcal K}^{-1} = 0.05 \, \text{kcal K}^{-1} \]

### Step 3: Calculate \( \Delta H \)
We can use the relation between \( \Delta H \) and \( \Delta U \):
\[ \Delta H = \Delta U + \Delta N_g \cdot R \cdot T \]
Where:
- \( \Delta N_g = N_p - N_r \) (moles of products - moles of reactants)
- For the reaction, \( N_p = 2 \) (from \( 2NO_2 \)) and \( N_r = 1 \) (from \( N_2O_4 \)), thus:
\[ \Delta N_g = 2 - 1 = 1 \]

The gas constant \( R \) in calories is \( 2 \, \text{cal K}^{-1} \, \text{mol}^{-1} \). Therefore:
\[ \Delta H = 2.0 \, \text{kcal} + (1 \cdot 2 \, \text{cal K}^{-1} \cdot 300 \, \text{K}) \cdot \frac{1 \, \text{kcal}}{1000 \, \text{cal}} \]
Calculating the second term:
\[ 1 \cdot 2 \cdot 300 = 600 \, \text{cal} = 0.6 \, \text{kcal} \]
Thus:
\[ \Delta H = 2.0 \, \text{kcal} + 0.6 \, \text{kcal} = 2.6 \, \text{kcal} \]

### Step 4: Calculate \( \Delta G \)
Now we can calculate \( \Delta G \) using the formula:
\[ \Delta G = \Delta H - T \Delta S \]
Substituting the values:
\[ \Delta G = 2.6 \, \text{kcal} - (300 \, \text{K} \cdot 0.05 \, \text{kcal K}^{-1}) \]
Calculating \( T \Delta S \):
\[ 300 \cdot 0.05 = 15 \, \text{kcal} \]
Thus:
\[ \Delta G = 2.6 \, \text{kcal} - 15 \, \text{kcal} = -12.4 \, \text{kcal} \]

### Final Answer
The value of \( \Delta G \) is:
\[ \Delta G = -12.4 \, \text{kcal} \]

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