4.2.4Unit 4. Rates, Equilibria and Further Organic Chemistry
Reaction Feasibility — IAL Edexcel Linear (Chemistry) Revision Notes & Topical Questions
5 min read 6 parts Lesson 22 of 56 Unit quiz
Curriculum Equation Key معادلة تفاعل تكوين الإستر (الأسترة) - Esterification Reaction EquationFast Exam Solution
R-COOH + R2-OH ⇌ (conc. H₂SO₄) R-COO-R2 + H₂O
Word Equation & Conditions: حمض كربوكسيلي + كحول (في وجود حمض الكبريتيك المركز كعامل حفاز ونازع للماء) ⇌ إستر + ماء
Location & Specifications: الكيمياء العضوية للثانوية العامة والتحصيلي و Edexcel Chemistry.
Section 1 of 6The Feasibility Criterion
Thermodynamics tells us whether a reaction can happen, but not how quickly.
Condition
- Condition A reaction is feasible when ΔS°total is greater than zero at the temperature concerned.
Two contributions
- Two contributions The system term depends on the change in disorder of the substances, and the surroundings term depends on ΔH and T.
Temperature dependence
- Temperature dependence Only ΔS°surroundings changes noticeably with T, because ΔS°system and ΔH are nearly constant.
Limit
- Limit When ΔS°total equals zero the reaction is at the boundary, which is the equilibrium position.
Section 2 of 6Temperature Effects
Temperature Effects
Exothermic and ΔS positive
- Exothermic and ΔS positive Feasible at every temperature, since both terms are positive.
Endothermic and ΔS positive
- Endothermic and ΔS positive Feasible only above a certain temperature, when the system term outweighs the surroundings term.
Exothermic and ΔS negative
- Exothermic and ΔS negative Feasible only below a certain temperature, where the surroundings term still outweighs the system term.
Endothermic and ΔS negative
- Endothermic and ΔS negative Not feasible at any temperature, because both terms are negative.
| ΔH | ΔS°system | Feasible |
|---|---|---|
| Negative | Positive | Always |
| Positive | Positive | High T |
| Negative | Negative | Low T |
| Positive | Negative | Never |
Section 3 of 6Finding the Changeover Temperature
1
Set ΔS°total = 0, so ΔS°system − ΔH ÷ T = 0.
2
Rearrange to T = ΔH ÷ ΔS°system, with ΔH in J mol⁻¹.
3
Substitute the values and calculate T in kelvin.
4
Subtract 273 to give the temperature in °C if needed.
5
Above this temperature (for positive ΔH and ΔS) the reaction is feasible.
Section 4 of 6Feasible but Slow
Kinetic barrier
- Kinetic barrier A reaction with a large activation energy may be feasible but so slow that it effectively does not occur.
Example
- Example Diamond converting to graphite and the reaction of hydrogen with oxygen at room temperature are both slow without a trigger.
Catalyst
- Catalyst A catalyst can make a feasible reaction go faster, but cannot make an infeasible reaction happen.
Conditions
- Conditions Real conditions may differ from standard conditions, so predictions are only a guide.
Section 5 of 6Worked Example
Question
- For CaCO₃(s) → CaO(s) + CO₂(g), ΔH = +178 kJ mol⁻¹ and ΔS°system = +160 J K⁻¹ mol⁻¹.
- At what temperature does the decomposition become feasible?
- Set ΔS°total = 0: 160 − 178 000 ÷ T = 0.
- T = 178 000 ÷ 160 = 1112.5 K.
- Above about 1.11 × 10³ K (839 °C) ΔS°total is positive and the decomposition is feasible.
Answer T ≈ 1.11 × 10³ K (about 839 °C)
Section 6 of 6Exam Checklist & Tips
4.2.4 – Key Points for the Exam
- Feasible A reaction for which the total entropy change is positive.
- Equilibrium A state in which total entropy change is zero and no further net change occurs.
- Kinetic stability Stability arising because a high activation energy makes a reaction very slow.
- Thermodynamic stability Stability arising because the total entropy change is not positive.
- Changeover temperature The temperature at which ΔS°total changes sign.
Common Mistakes
- Forgetting to convert ΔH to J before dividing by ΔS in J K⁻¹ mol⁻¹.
- Assuming that a feasible reaction must be fast.
- Using the temperature in °C in T = ΔH ÷ ΔS.
- Saying that a catalyst can make a non-feasible reaction occur.
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