4.1.6Unit 4. Rates, Equilibria and Further Organic Chemistry
Continuous Monitoring Method — IAL Edexcel Linear (Chemistry) Revision Notes & Topical Questions
5 min read 6 parts Lesson 15 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 6Following One Experiment
The continuous method uses the shape of a single curve to identify the order.
Principle
- Principle One reaction mixture is monitored over time and a concentration–time curve is plotted.
Excess reagents
- Excess reagents All other reactants are in large excess, so their concentrations stay almost constant and the rate depends on one reactant only.
Pseudo-first order
- Pseudo-first order A reaction that is first order in the limiting reactant under these conditions behaves as a first-order reaction overall.
Advantages
- Advantages One run yields all the data needed for that reactant, and less material is used.
Section 2 of 6Analysing the Curve
Analysing the Curve
Half-lives
- Half-lives Measure successive half-lives and see if they are constant, which is the signature of first order.
Tangents
- Tangents Draw tangents at several points and find the rate at each concentration.
Rate–concentration plot
- Rate–concentration plot Plot the tangent gradients against concentration to see if the line is horizontal, straight or curved.
Rate constant
- Rate constant For first order, k = ln 2 ÷ t½ or equals the gradient of the rate–concentration line.
Section 3 of 6Procedure and Data
1
Mix the reactants and begin recording the concentration, or a property proportional to it, at regular intervals.
2
Plot the graph of concentration against time and draw a smooth curve.
3
Choose several concentrations on the curve and measure the half-lives between them.
4
Decide the order from the half-life pattern, and confirm with tangents.
5
Calculate k from the half-life or from the gradient of the rate–concentration graph.
Section 4 of 6Limitations
Tangents
- Tangents Drawing tangents by eye introduces error, particularly on steep curves.
Single reactant
- Single reactant Only one reactant's order is found in each experiment, so several runs are needed for a full rate equation.
Excess assumption
- Excess assumption The concentrations of the excess reagents are assumed to be constant, which is only approximately true.
Precision
- Precision A data logger and software fit reduce the subjective error of drawing tangents.
Section 5 of 6Worked Example
Question
- Concentration of a reactant falls from 0.600 mol dm⁻³ to 0.300 mol dm⁻³ in 120 s, and from 0.300 to 0.150 mol dm⁻³ in a further 120 s.
- Find the order and k.
- The first half-life is 120 s and the second is also 120 s.
- A constant half-life shows first order.
- k = ln 2 ÷ 120 = 0.693 ÷ 120 = 5.78 × 10⁻³ s⁻¹.
Answer First order; k = 5.78 × 10⁻³ s⁻¹
Section 6 of 6Exam Checklist & Tips
4.1.6 – Key Points for the Exam
- Pseudo-first order Behaviour of a reaction that appears first order because other reactants are in large excess.
- Half-life The time for a reactant concentration to fall to half its value.
- Excess reagent A reactant present in much larger amount so that its concentration hardly changes.
- Tangent A line touching a curve at a point, giving the gradient there.
- Continuous method Following the course of one reaction mixture over time.
Common Mistakes
- Using a mixture where no reagent is in excess and then assuming the rate depends on one reactant only.
- Reading half-lives from only one part of the curve.
- Drawing a tangent that crosses the curve instead of just touching it.
- Reporting k without its unit of s⁻¹ for a first-order reaction.
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