AQA · A-Level · Chemistry
Course version: 7405 · Version from Sept 2016
Students should be able to: calculate entropy changes from absolute…
Full specification wording
Gibbs free-energy change, ∆G, and entropy change, ∆S (A-level only): Students should be able to: calculate entropy changes from absolute entropy values use the relationship ∆G = ∆H – T∆S to determine how ∆G varies with temperature use the relationship ∆G = ∆H – T∆S to determine the temperature at which a reaction becomes feasible.
Three calculations come out of one equation.
Checked against the AQA 7405 specification (opens in a new tab) on .
- Point 6 of 6
- 4 key terms
Study notes
From the checked teaching notes for this specification point.
The idea
Three calculations come out of one equation. To find ∆S, subtract the sum of the reactant entropies from the sum of the product entropies, each multiplied by its balancing number. To find how ∆G varies with temperature, treat ∆G = ∆H − T∆S as a straight line of ∆G against T with intercept ∆H and gradient −∆S, so a positive ∆S makes ∆G fall as the temperature rises. To find the temperature at which a reaction just becomes feasible, set ∆G to zero, giving T = ∆H/∆S. For a decomposition with ∆H of +123 kJ mol⁻¹ and ∆S of +144 J K⁻¹ mol⁻¹ this is 123000/144 = 854 K. Convert the units first: that step alone carries a mark.
What earns the marks
- Converting ∆H into joules or ∆S into kilojoules so the two are compatible.
- ∆S calculated as ΣS(products) − ΣS(reactants), including the balancing numbers.
- Setting ∆G to zero and rearranging to T = ∆H/∆S.
- The temperature quoted to at least two significant figures with the unit K.
Common mistakes and how to put them right
Dividing 123 by 144 without converting, giving 0.854 K instead of 854 K.
Inverting the fraction to ∆S/∆H, producing an answer near 0.001 K.
Quoting the feasibility temperature in degrees Celsius.
Forgetting the balancing numbers when summing absolute entropies.
Reporting a temperature above which a reaction is feasible when ∆S is negative, where the inequality runs the other way.
Exam tips
- Do the unit conversion on its own line and label it; it is an independent mark and protects the rest.
- Sanity check the answer: a feasibility temperature should be a plausible number of kelvin, not a fraction of one.
- State whether the reaction becomes feasible above or below the temperature you calculate, using the sign of ∆S to decide.
Try the lesson
The start of the Gibbs free-energy change, ∆G, and entropy change, ∆S (A-level only): Students should be able to: calculate entropy changes from absolute entropy values use the relationship ∆G = ∆H – T∆S to determine how ∆G varies with temperature use the relationship ∆G = ∆H – T∆S to determine the temperature at which a reaction becomes feasible. lesson, made from checked teaching notes for this specification point.
Have a go first
Which expression is used to calculate the entropy change of a reaction from absolute entropy values?
Listen
Three calculations come out of one equation. To find ∆S, subtract the sum of the reactant entropies from the sum of the product entropies, each multiplied by its balancing number. To find how ∆G varies with temperature, treat ∆G = ∆H − T∆S as a straight line of ∆G against T with intercept ∆H and gradient −∆S, so a positive ∆S makes ∆G fall as the temperature rises. To find the temperature at which a reaction just becomes feasible, set ∆G to zero, giving T = ∆H/∆S. For a decomposition with ∆H of +123 kJ mol⁻¹ and ∆S of +144 J K⁻¹ mol⁻¹ this is 123000/144 = 854 K. Convert the units first: that step alone carries a mark. A strong answer earns marks for these points. converting ∆H into joules or ∆S into kilojoules so the two are compatible. ∆S calculated as ΣS(products) − ΣS(reactants), including the balancing numbers. setting ∆G to zero and rearranging to T = ∆H/∆S. the temperature quoted to at least two significant figures with the unit K. Exam tip: Do the unit conversion on its own line and label it; it is an independent mark and protects the rest.
Key words
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calculating ∆S from absolute entropies
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variation of ∆G with temperature
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setting ∆G to zero
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unit conversion
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See every definition
- calculating ∆S from absolute entropies
- Subtract the sum of the reactant entropies from the sum of the product entropies, each multiplied by its balancing number.
- variation of ∆G with temperature
- Treat ∆G = ∆H − T∆S as a straight line of ∆G against T with intercept ∆H and gradient −∆S, so a positive ∆S makes ∆G fall as the temperature rises.
- setting ∆G to zero
- This gives the temperature at which a reaction just becomes feasible, rearranging to T = ∆H/∆S.
- unit conversion
- Convert ∆H into joules or ∆S into kilojoules so the two are compatible, and this step alone carries a mark.
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Test your understanding of Gibbs free-energy change, ∆G, and entropy change, ∆S (A-level only): Students should be able to: calculate entropy changes from absolute entropy values use the relationship ∆G = ∆H – T∆S to determine how ∆G varies with temperature use the relationship ∆G = ∆H – T∆S to determine the temperature at which a reaction becomes feasible.
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