Edexcel International A Level (IAL) Chemistry (YCH11) - Unit 4 - 13.8 Entropy relationship (ΔS_total = R ln K)-Study Notes - New Syllabus
Edexcel International A Level (IAL) Chemistry (YCH11) -Unit 4 – 13.8 Entropy relationship (ΔS_total = R ln K)- Study Notes- New syllabus
Edexcel International A Level (IAL) Chemistry (YCH11) -Unit 4 – 13.8 Entropy relationship (ΔS_total = R ln K)- 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
13.8 Effect of Temperature on \( \mathrm{\Delta S_{total}} \) and Equilibrium Constant
The feasibility of a reaction and the magnitude of the equilibrium constant are directly linked to entropy. Temperature plays a crucial role because it affects both the entropy of the surroundings and the value of the equilibrium constant.
(i) Effect of Temperature on \( \mathrm{\Delta S_{total}} \)
The total entropy change is given by:
\( \mathrm{\Delta S_{total} = \Delta S_{system} + \Delta S_{surroundings}} \)
and:
\( \mathrm{\Delta S_{surroundings} = -\frac{\Delta H}{T}} \)
Key Insight
- \( \mathrm{\Delta S_{system}} \) is independent of temperature (approx.).
- \( \mathrm{\Delta S_{surroundings}} \) depends on temperature through \( \mathrm{\frac{1}{T}} \).
Effect for Exothermic Reactions (\( \mathrm{\Delta H < 0} \))
- \( \mathrm{\Delta S_{surroundings}} \) is positive.
- Increasing temperature decreases \( \mathrm{\frac{1}{T}} \), so:
- \( \mathrm{\Delta S_{surroundings}} \) becomes less positive.
- Therefore, \( \mathrm{\Delta S_{total}} \) decreases.
Effect for Endothermic Reactions (\( \mathrm{\Delta H > 0} \))
- \( \mathrm{\Delta S_{surroundings}} \) is negative.
- Increasing temperature reduces magnitude of \( \mathrm{\frac{1}{T}} \):
- \( \mathrm{\Delta S_{surroundings}} \) becomes less negative.
- Therefore, \( \mathrm{\Delta S_{total}} \) increases.
Summary
- Exothermic: ↑T → \( \mathrm{\Delta S_{total}} \) decreases.
- Endothermic: ↑T → \( \mathrm{\Delta S_{total}} \) increases.
(ii) Effect of Temperature on Equilibrium Constant
The relationship between entropy and equilibrium constant is:
\( \mathrm{\Delta S_{total} = R \ln K} \)
Key Insight
- If \( \mathrm{\Delta S_{total}} \) increases → \( \mathrm{K} \) increases.
- If \( \mathrm{\Delta S_{total}} \) decreases → \( \mathrm{K} \) decreases.
Effect for Exothermic Reactions
- Increasing temperature decreases \( \mathrm{\Delta S_{total}} \).
- Therefore, \( \mathrm{K} \) decreases.
Effect for Endothermic Reactions
- Increasing temperature increases \( \mathrm{\Delta S_{total}} \).
- Therefore, \( \mathrm{K} \) increases.
Magnitude of \( \mathrm{K} \)
- Large \( \mathrm{K} \) → equilibrium lies towards products.
- Small \( \mathrm{K} \) → equilibrium lies towards reactants.
- Temperature shifts this balance by changing \( \mathrm{\Delta S_{total}} \).
Key Features
- Temperature affects \( \mathrm{\Delta S_{surroundings}} \), not \( \mathrm{\Delta S_{system}} \).
- This changes \( \mathrm{\Delta S_{total}} \).
- Since \( \mathrm{\Delta S_{total} = R \ln K} \), \( \mathrm{K} \) changes accordingly.
- Exothermic: ↑T → ↓\( \mathrm{K} \); Endothermic: ↑T → ↑\( \mathrm{K} \).
Example 1:
Explain why increasing temperature decreases \( \mathrm{K} \) for an exothermic reaction using entropy.
▶️ Answer/Explanation
For an exothermic reaction, \( \mathrm{\Delta H < 0} \), so \( \mathrm{\Delta S_{surroundings}} \) is positive.
Increasing temperature reduces the value of \( \mathrm{\frac{1}{T}} \), making \( \mathrm{\Delta S_{surroundings}} \) less positive.
Therefore, \( \mathrm{\Delta S_{total}} \) decreases.
Since \( \mathrm{\Delta S_{total} = R \ln K} \), a decrease in \( \mathrm{\Delta S_{total}} \) leads to a decrease in \( \mathrm{K} \).
Example 2:
A reaction becomes more feasible at higher temperature. Explain this using \( \mathrm{\Delta S_{total}} \) and \( \mathrm{K} \).
▶️ Answer/Explanation
Increasing temperature increases \( \mathrm{\Delta S_{total}} \), making it more positive.
This indicates the reaction becomes more thermodynamically favourable.
Since \( \mathrm{\Delta S_{total} = R \ln K} \), an increase in \( \mathrm{\Delta S_{total}} \) increases \( \mathrm{K} \).
A larger \( \mathrm{K} \) means equilibrium lies further towards products.
