Calculate the magnitude of standard entropy change for reaction `X hArr Y` if `DeltaH^(@) = 25 KJ` and `K_("eq") "is" 10^(-7)` at `300` K.

Explanation

To calculate the magnitude of the standard entropy change (ΔS°) for the reaction \( X \rightleftharpoons Y \), we will follow these steps:

### Step 1: Write down the relevant equations
We will use the following equations:
1. \( \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \)
2. \( \Delta G^\circ = -2.303 R T \log K_{eq} \)

Where:
- \( \Delta G^\circ \) = standard Gibbs free energy change
- \( \Delta H^\circ \) = standard enthalpy change
- \( T \) = temperature in Kelvin
- \( \Delta S^\circ \) = standard entropy change
- \( R \) = universal gas constant (8.314 J/mol·K)
- \( K_{eq} \) = equilibrium constant

### Step 2: Substitute known values into the equations
Given:
- \( \Delta H^\circ = 25 \, \text{kJ} = 25000 \, \text{J} \) (since we need to convert kJ to J)
- \( K_{eq} = 10^{-7} \)
- \( T = 300 \, \text{K} \)

From the second equation:
\[ \Delta G^\circ = -2.303 \cdot R \cdot T \cdot \log K_{eq} \]
Substituting the values:
\[ \Delta G^\circ = -2.303 \cdot 8.314 \cdot 300 \cdot \log(10^{-7}) \]

### Step 3: Calculate \( \log(10^{-7}) \)
\[ \log(10^{-7}) = -7 \]
Now substitute this back into the equation:
\[ \Delta G^\circ = -2.303 \cdot 8.314 \cdot 300 \cdot (-7) \]

### Step 4: Calculate \( \Delta G^\circ \)
Calculating:
\[ \Delta G^\circ = 2.303 \cdot 8.314 \cdot 300 \cdot 7 \]
Calculating the constants:
\[ = 2.303 \cdot 8.314 \cdot 300 \cdot 7 \approx 2.303 \cdot 8.314 \cdot 2100 \approx 2.303 \cdot 17460.6 \approx 40132.5 \, \text{J} \approx 40.13 \, \text{kJ} \]

### Step 5: Substitute \( \Delta G^\circ \) back into the first equation
Now we have:
\[ \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \]
Substituting known values:
\[ 40.13 \, \text{kJ} = 25 \, \text{kJ} - 300 \Delta S^\circ \]
Convert \( 40.13 \, \text{kJ} \) to J:
\[ 40130 \, \text{J} = 25000 \, \text{J} - 300 \Delta S^\circ \]

### Step 6: Solve for \( \Delta S^\circ \)
Rearranging gives:
\[ 300 \Delta S^\circ = 25000 - 40130 \]
\[ 300 \Delta S^\circ = -15130 \]
\[ \Delta S^\circ = \frac{-15130}{300} \approx -50.43 \, \text{J/K} \]

### Step 7: Calculate the magnitude of \( \Delta S^\circ \)
The magnitude of the standard entropy change is:
\[ |\Delta S^\circ| = 50.43 \, \text{J/K} \]

### Final Answer
The magnitude of the standard entropy change for the reaction \( X \rightleftharpoons Y \) is approximately **50.43 J/K**.