Edexcel International A Level (IAL) Chemistry (YCH11) - Unit 4 - 12.8 ΔS_surroundings = −ΔH/T-Study Notes - New Syllabus
Edexcel International A Level (IAL) Chemistry (YCH11) -Unit 4 – 12.8 ΔS_surroundings = −ΔH/T- Study Notes- New syllabus
Edexcel International A Level (IAL) Chemistry (YCH11) -Unit 4 – 12.8 ΔS_surroundings = −ΔH/T- Study Notes -International A Level (IAL) Chemistry (YCH11) – per latest Syllabus.
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Edexcel International A Level (IAL) Chemistry (YCH11) -Concise Summary Notes- All Topics
12.8 Calculation of Entropy Change of the Surroundings and Total Entropy
The entropy change of the surroundings depends on the heat energy exchanged during a reaction. By combining this with the entropy change of the system, the total entropy change can be determined, which indicates whether a reaction is spontaneous.
Entropy Change of the Surroundings
\( \mathrm{\Delta S_{surroundings} = -\frac{\Delta H}{T}} \)
Where:
- \( \mathrm{\Delta H} \) = enthalpy change (in \( \mathrm{J\ mol^{-1}} \))
- \( \mathrm{T} \) = temperature in Kelvin
Key Points
- Temperature must always be in Kelvin.
- \( \mathrm{\Delta H} \) must be converted to \( \mathrm{J\ mol^{-1}} \) if given in \( \mathrm{kJ\ mol^{-1}} \).
- The negative sign is essential as it reflects heat transfer between system and surroundings.
Interpretation
- Exothermic reaction (\( \mathrm{\Delta H < 0} \)) → \( \mathrm{\Delta S_{surroundings} > 0} \)
- Endothermic reaction (\( \mathrm{\Delta H > 0} \)) → \( \mathrm{\Delta S_{surroundings} < 0} \)
Total Entropy Change
\( \mathrm{\Delta S_{total} = \Delta S_{system} + \Delta S_{surroundings}} \)
- \( \mathrm{\Delta S_{total} > 0} \): reaction is spontaneous
- \( \mathrm{\Delta S_{total} < 0} \): reaction is not spontaneous
- \( \mathrm{\Delta S_{total} = 0} \): system at equilibrium
Method
- Calculate \( \mathrm{\Delta S_{system}} \) using entropy values.
- Convert \( \mathrm{\Delta H} \) to \( \mathrm{J\ mol^{-1}} \).
- Calculate \( \mathrm{\Delta S_{surroundings}} \) using \( \mathrm{-\frac{\Delta H}{T}} \).
- Add the two values to find \( \mathrm{\Delta S_{total}} \).
Example 1:
A reaction has \( \mathrm{\Delta H = -50\ kJ\ mol^{-1}} \) at \( \mathrm{298\ K} \). Calculate \( \mathrm{\Delta S_{surroundings}} \).
▶️ Answer/Explanation
Convert \( \mathrm{\Delta H} \): \( \mathrm{-50\ kJ = -50000\ J} \)
\( \mathrm{\Delta S_{surroundings} = -\frac{-50000}{298}} \)
\( \mathrm{\Delta S_{surroundings} = \frac{50000}{298} \approx 168\ J\ mol^{-1}K^{-1}} \)
The positive value reflects increased disorder in the surroundings due to heat release.
Example 2:
A reaction has \( \mathrm{\Delta S_{system} = +120\ J\ mol^{-1}K^{-1}} \) and \( \mathrm{\Delta H = +30\ kJ\ mol^{-1}} \) at \( \mathrm{298\ K} \). Determine whether the reaction is spontaneous.
▶️ Answer/Explanation
Convert \( \mathrm{\Delta H} \): \( \mathrm{30\ kJ = 30000\ J} \)
\( \mathrm{\Delta S_{surroundings} = -\frac{30000}{298} \approx -101\ J\ mol^{-1}K^{-1}} \)
\( \mathrm{\Delta S_{total} = 120 + (-101) = 19\ J\ mol^{-1}K^{-1}} \)
Since \( \mathrm{\Delta S_{total} > 0} \), the reaction is spontaneous.