OCR A-Level Chemistry B H433
Development of Practical Skills in Chemistry
Elements of Life
Developing Fuels
Elements From the Sea
The Ozone Story
What's in a Medicine?
The Chemical Industry
Polymers and Life
Oceans
Developing Metals
Colour by Design
Chemical Literacy
Mathematical Skills in Chemistry
Oceans
Total entropy change and the condition for feasibility
- 2 things to know
- Components 01, 02 and 03
Overview
Recall questions on the entropy change of the surroundings, adding the two contributions, and the sign that makes a reaction feasible.
What you need to know
Answer each one before you reveal it.
How is the total entropy change worked out, and why must ΔH be in J mol⁻¹ and the temperature in kelvin?
Questions from this subtopic
Each one with the model answer underneath.
- 1
How is the total entropy change of a reaction worked out?
Model answer
ΔtotS = ΔsysS + ΔsurrS.
- 2
How is the entropy change of the surroundings related to the enthalpy change?
Model answer
ΔsurrS = −ΔH/T, where ΔH is the enthalpy change and T is the temperature in kelvin.
- 3
What condition on ΔtotS makes a reaction feasible?
Model answer
ΔtotS must be positive.
- 4
What does a total entropy change of zero indicate?
Model answer
The system is at the point of balance, which corresponds to equilibrium and to the temperature at which feasibility changes over.
- 5
Why does an exothermic reaction increase the entropy of the surroundings?
Model answer
Energy is released into the surroundings, giving their particles more energy quanta to distribute and therefore more possible arrangements.
- 6
Why must an enthalpy change in kJ mol⁻¹ be converted to J mol⁻¹ before it is used in ΔsurrS = −ΔH/T?
Model answer
Entropy values are quoted in J K⁻¹ mol⁻¹, so the enthalpy must be in joules for the two terms to be added.
Other ways to work this
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