For the reaction `A hArr B+C` at equilibrium, the concentration of A is `1xx10^(-3)MB` is 0.15 M and C is 0.05 M. The `DeltaG^(@)` for the hydrolysis of A at 300 K is `-X" kJ/mole"`. The value of X is ?
Report your answer by rounding it upto nearest integer.

Answer

To find the value of \( X \) in the given reaction \( A \rightleftharpoons B + C \) at equilibrium, we will follow these steps:

### Step 1: Write the expression for the equilibrium constant \( K_{eq} \)
The equilibrium constant \( K_{eq} \) for the reaction can be expressed as:

\[ K_{eq} = \frac{[B][C]}{[A]} \]

### Step 2: Substitute the given concentrations into the equilibrium expression
From the question, we have:
- \([A] = 1 \times 10^{-3} \, \text{M}\)
- \([B] = 0.15 \, \text{M}\)
- \([C] = 0.05 \, \text{M}\)

Substituting these values into the equilibrium expression:

\[ K_{eq} = \frac{(0.15)(0.05)}{(1 \times 10^{-3})} \]

### Step 3: Calculate \( K_{eq} \)
Calculating the numerator:

\[ 0.15 \times 0.05 = 0.0075 \]

Now substituting back into the equation:

\[ K_{eq} = \frac{0.0075}{1 \times 10^{-3}} = 7.5 \]

### Step 4: Use the relationship between \( \Delta G^\circ \) and \( K_{eq} \)
The relationship between standard Gibbs free energy change \( \Delta G^\circ \) and the equilibrium constant \( K_{eq} \) is given by:

\[ \Delta G^\circ = -RT \ln K_{eq} \]

Where:
- \( R = 8.314 \, \text{J/(mol K)} \)
- \( T = 300 \, \text{K} \)

### Step 5: Substitute the values into the equation
Substituting \( R \), \( T \), and \( K_{eq} \):

\[ \Delta G^\circ = - (8.314 \, \text{J/(mol K)}) \times (300 \, \text{K}) \times \ln(7.5) \]

### Step 6: Calculate \( \ln(7.5) \)
Using a calculator, we find:

\[ \ln(7.5) \approx 2.014 \]

### Step 7: Calculate \( \Delta G^\circ \)
Now substituting \( \ln(7.5) \) back into the equation:

\[ \Delta G^\circ = - (8.314) \times (300) \times (2.014) \]

Calculating this step by step:

1. \( 8.314 \times 300 = 2494.2 \, \text{J/mol} \)
2. \( 2494.2 \times 2.014 \approx 5025.5 \, \text{J/mol} \)

Thus:

\[ \Delta G^\circ \approx -5025.5 \, \text{J/mol} \]

### Step 8: Convert to kJ/mol
Since \( \Delta G^\circ \) is often expressed in kJ/mol:

\[ \Delta G^\circ \approx -5.0255 \, \text{kJ/mol} \]

### Step 9: Find \( X \)
Since \( \Delta G^\circ = -X \, \text{kJ/mol} \), we have:

\[ X \approx 5.0255 \]

Rounding to the nearest integer gives:

\[ X \approx 5 \]

### Final Answer:
The value of \( X \) is **5**.

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