Step 1: Understanding the Concept:
According to the second law of thermodynamics, the entropy change of the surroundings (\( \Delta S_{\text{surr}} \)) depends on the heat exchanged with the system.
If the system undergoes an exothermic reaction, it releases heat into the surroundings, which increases the thermal disorder and thus increases the entropy of the surroundings.
Step 2: Key Formula or Approach:
The entropy change of the surroundings at constant temperature and pressure is given by:
\[ \Delta S_{\text{surr}} = \frac{q_{\text{surr}}}{T} \]
Because energy is conserved, the heat absorbed by the surroundings (\( q_{\text{surr}} \)) is equal in magnitude but opposite in sign to the heat released by the system (\( q_{\text{sys}} \)).
At constant pressure, \( q_{\text{sys}} = \Delta H_{\text{sys}} \).
Therefore:
\[ \Delta S_{\text{surr}} = -\frac{\Delta H_{\text{sys}}}{T} \]
Step 3: Detailed Explanation:
The problem states that the reaction releases \( 525 \text{ kJ} \) of heat to the surroundings.
This means the reaction is exothermic, and the enthalpy change of the system is negative:
\( \Delta H_{\text{sys}} = -525 \text{ kJ} \)
Therefore, the heat gained by the surroundings is positive:
\( q_{\text{surr}} = +525 \text{ kJ} \)
Entropy is standardly expressed in Joules per Kelvin (\( \text{J K}^{-1} \)), so we must convert kilojoules to joules:
\( q_{\text{surr}} = 525 \text{ kJ} \times 1000 \text{ J/kJ} = 525,000 \text{ J} \)
The constant temperature \( T \) is \( 300 \text{ K} \).
Substitute these values into the entropy equation:
\[ \Delta S_{\text{surr}} = \frac{525,000 \text{ J}}{300 \text{ K}} \]
\[ \Delta S_{\text{surr}} = \frac{5250}{3} \text{ J K}^{-1} \]
\[ \Delta S_{\text{surr}} = 1750 \text{ J K}^{-1} \]
The value is positive, which makes sense since the surroundings absorbed heat.
Step 4: Final Answer:
The entropy change of the surrounding is \( 1750 \text{ J K}^{-1} \).