6.1.3Unit 6. Practical Skills in Chemistry II
Activation Energy — IAL Edexcel Linear (Chemistry) Revision Notes & Topical Questions
5 min read 6 parts Lesson 7 of 14 Unit quiz
Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
Section 1 of 6Arrhenius Equation
Rate constants increase with temperature in a predictable exponential way.
Form
- Form k = Ae^(–Ea/RT), where A is a constant, Ea the activation energy, R the gas constant and T the temperature in kelvin.
Logarithmic form
- Logarithmic form ln k = –Ea/RT + ln A, which has the form y = mx + c.
Graph
- Graph Plotting ln k against 1/T gives a straight line with gradient –Ea/R and intercept ln A.
Units
- Units R = 8.31 J mol⁻¹ K⁻¹, so Ea is found in J mol⁻¹ and then divided by 1000 for kJ mol⁻¹.
Section 2 of 6Practical Method
1
Choose a clock reaction, such as the iodine clock, that gives a clear colour change at a convenient time.
2
Place the reactants in a water bath and record the temperature of the mixture just before and after mixing.
3
Mix, start the stopwatch, and record the time for the end point, then repeat at about five temperatures from 10 °C to 50 °C.
4
Convert each temperature to kelvin by adding 273 and calculate 1/T.
5
Use 1/time as a measure of rate, calculate ln(1/time), and plot against 1/T.
Section 3 of 6Interpreting the Graph
Interpreting the Graph
Gradient
- Gradient A steeper negative gradient means a larger activation energy.
Calculation
- Calculation Ea = –gradient × R, so a gradient of –5200 K gives Ea = 5200 × 8.31 ≈ 43 200 J mol⁻¹.
Intercept
- Intercept The intercept on the y-axis gives ln A, related to collision frequency and orientation.
Best-fit line
- Best-fit line A line of best fit reduces the influence of random errors in individual results.
Section 4 of 6Accuracy Points
Temperature control
- Temperature control Small errors in T cause large errors in 1/T, so keep the mixture at constant temperature.
Units
- Units Use kelvin in all Arrhenius calculations.
Range
- Range A wider temperature range makes the gradient more accurate.
Approximation
- Approximation Using 1/time as the rate is valid only if the extent of reaction at the end point is the same each time.
Section 5 of 6Worked Example
Question
- A graph of ln(rate) against 1/T has gradient –5200 K.
- Calculate Ea in kJ mol⁻¹ (R = 8.31 J mol⁻¹ K⁻¹).
- Gradient = –Ea ÷ R, so Ea = 5200 × 8.31.
- 5200 × 8.31 = 43 212 J mol⁻¹.
- Divide by 1000 to give kJ mol⁻¹.
Answer Ea = 43.2 kJ mol⁻¹
Section 6 of 6Exam Checklist & Tips
6.1.3 – Key Points for the Exam
- Activation energy The minimum energy that colliding particles need for a reaction to occur.
- Arrhenius equation The equation relating the rate constant to temperature and activation energy.
- Gradient The slope of a line, found by dividing the change in y by the change in x.
- Rate constant The proportionality constant k in the rate equation.
Common Mistakes
- Using temperatures in °C for 1/T, which gives a wrong gradient.
- Forgetting the minus sign in gradient = –Ea/R.
- Quoting Ea in J mol⁻¹ with units of kJ mol⁻¹ without dividing by 1000.
- Plotting ln k against T rather than 1/T.
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