For the reaction `A hArr B+C` at equilibrium, the concentration of A is `1xx10^(-3)M` B 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 solve the problem, we need to determine the value of \( X \) in the equation \( \Delta G^\circ = -X \) kJ/mol for the hydrolysis of \( A \) at 300 K. We will use the equilibrium concentrations given to find the equilibrium constant \( K_c \) and then relate it to \( \Delta G^\circ \).

### Step-by-Step Solution:

1. **Write the Reaction and Given Concentrations**:
The reaction is:
\[ A \rightleftharpoons B + C \]
Given concentrations at equilibrium:
- \([A] = 1 \times 10^{-3} \, \text{M}\)
- \([B] = 0.15 \, \text{M}\)
- \([C] = 0.05 \, \text{M}\)

2. **Calculate the Equilibrium Constant \( K_c \)**:
The expression for the equilibrium constant \( K_c \) is:
\[ K_c = \frac{[B][C]}{[A]} \]
Substituting the given concentrations:
\[ K_c = \frac{(0.15)(0.05)}{1 \times 10^{-3}} = \frac{0.0075}{0.001} = 7.5 \]

3. **Use the Relationship Between \( \Delta G^\circ \) and \( K_c \)**:
The relationship is given by:
\[ \Delta G^\circ = -RT \ln K_c \]
Where:
- \( R = 8.314 \, \text{J/mol·K} \)
- \( T = 300 \, \text{K} \)

4. **Calculate \( \Delta G^\circ \)**:
First, convert \( R \) to kJ:
\[ R = 0.008314 \, \text{kJ/mol·K} \]
Now, substituting the values:
\[ \Delta G^\circ = - (0.008314)(300) \ln(7.5) \]

5. **Calculate \( \ln(7.5) \)**:
Using a calculator:
\[ \ln(7.5) \approx 2.014 \]

6. **Substituting Back to Find \( \Delta G^\circ \)**:
\[ \Delta G^\circ = - (0.008314)(300)(2.014) \approx -5.02 \, \text{kJ/mol} \]

7. **Determine the Value of \( X \)**:
Since \( \Delta G^\circ = -X \):
\[ -X = -5.02 \implies X = 5.02 \]
Rounding \( X \) to the nearest integer gives:
\[ X \approx 5 \]

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