For the reaction `X_(2)Y_(4)(l) to 2XY_(2)(g)` at 300 K the values of `DeltaU` and `DeltaS` are 2.2 kcal and 20 cal `K^(-1)` respectively.The value of `DeltaG` for the reaction is

Options

A. `-3400` cal
B. 3400 cal
C. `-2800` cal
D. 2000 cal

Explanation

To find the value of ΔG for the reaction \( X_{2}Y_{4}(l) \to 2XY_{2}(g) \) at 300 K, we will use the Gibbs free energy equation:

\[ \Delta G = \Delta H - T \Delta S \]

### Step 1: Calculate ΔN
First, we need to calculate ΔN, which is the change in the number of moles of gas:

\[ \Delta N = \text{moles of gaseous products} - \text{moles of gaseous reactants} \]

In this reaction, we have:
- Products: 2 moles of \( XY_{2}(g) \)
- Reactants: 0 moles of gaseous reactants (since \( X_{2}Y_{4} \) is a liquid)

Thus,

\[ \Delta N = 2 - 0 = 2 \]

### Step 2: Convert ΔU to Calories
Given that \( \Delta U = 2.2 \) kcal, we convert this to calories:

\[ \Delta U = 2.2 \, \text{kcal} \times 1000 \, \text{cal/kcal} = 2200 \, \text{cal} \]

### Step 3: Calculate ΔH
Using the relation:

\[ \Delta H = \Delta U + \Delta N \cdot R \cdot T \]

Where:
- \( R = 2 \, \text{cal K}^{-1} \) (the gas constant in appropriate units)
- \( T = 300 \, \text{K} \)

Now substituting the values:

\[ \Delta H = 2200 \, \text{cal} + (2 \cdot 2 \, \text{cal K}^{-1} \cdot 300 \, \text{K}) \]

Calculating the second term:

\[ 2 \cdot 2 \cdot 300 = 1200 \, \text{cal} \]

Now, substituting back:

\[ \Delta H = 2200 + 1200 = 3400 \, \text{cal} \]

### Step 4: Calculate ΔG
Now we can calculate ΔG using the Gibbs free energy equation:

\[ \Delta G = \Delta H - T \Delta S \]

Where \( \Delta S = 20 \, \text{cal K}^{-1} \). Thus:

\[ \Delta G = 3400 \, \text{cal} - (300 \, \text{K} \cdot 20 \, \text{cal K}^{-1}) \]

Calculating the second term:

\[ 300 \cdot 20 = 6000 \, \text{cal} \]

Now substituting back:

\[ \Delta G = 3400 - 6000 = -2800 \, \text{cal} \]

### Final Answer
The value of \( \Delta G \) for the reaction is:

\[ \Delta G = -2800 \, \text{cal} \]

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