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II-04: Graph slopes and linearization

IntoChemistry | General Chemistry II Math Readiness

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Module II-04 · Recommended before: Before kinetic plots and Arrhenius graphs

Graph slopes and linearization

A nonlinear law can be rearranged into y = mx+b so measured data become a straight line. Interpret slopes with units.

Learn the method

y = mx + b; first-order slope = −k

  1. Identify y and x in the transformed equation.
  2. Match equation slope to graph slope.
  3. Use two points or regression.
  4. Calculate the physical parameter from the slope.
  5. Note the units on both axes.

Common mistake: Do not report a negative rate constant from a negative first-order slope.

Worked example

Problem: A plot of ln[A] versus time has slope −0.20 min⁻¹. Determine k for first-order kinetics.

Step-by-step solution: Because ln[A] = −kt + ln[A]₀, slope = −k and k = +0.20 min⁻¹.

Check the units and whether the answer is reasonable before continuing.

More applications of this mathematics

Try each extension before expanding its solution.

Arrhenius plot

For slope −5000 K on a graph of ln k vs 1/T, Ea = −slope × R = 5000×8.314=41.6 kJ/mol.

Interpreting R²

R² describes the fraction of variation captured by a chosen fitted model; a high R² alone does not prove causation or validate all model assumptions.

Independent practice — five questions

Suggested mastery threshold: four out of five. Incorrect answers display a worked explanation after submission.

Choose one response for each question, then select “Check my answers.” No timer. Hints are optional. You can retry freely. Questions are formative, not proctored.

Question 1. A graph of ln[A] versus time has slope -0.4 s⁻¹. What is first-order k?
Hint for question 1

The slope is negative but k is positive.

Question 2. A graph of ln[A] versus time has slope -0.25 s⁻¹. What is first-order k?
Hint for question 2

The slope is negative but k is positive.

Question 3. A graph of ln[A] versus time has slope -0.5 s⁻¹. What is first-order k?
Hint for question 3

The slope is negative but k is positive.

Question 4. A graph of ln[A] versus time has slope -0.6 s⁻¹. What is first-order k?
Hint for question 4

The slope is negative but k is positive.

Question 5. A graph of ln[A] versus time has slope -0.15 s⁻¹. What is first-order k?
Hint for question 5

The slope is negative but k is positive.

Supporting reference

OpenStax Chemistry 2e / NIST — reference for this subject. This lesson’s wording, examples and questions are original.

W3C WCAG 2.2 accessibility guidance.

Applied practice: Finding chemical constants from line slopes

These ungraded questions extend the scored five-question check. Work each problem before opening its solution; progress records are not changed.

  1. Challenge 1. A plot of ln k versus 1/T has slope −6000 K. Determine activation energy (R = 8.314 J/(mol·K)).

    Show worked solution for challenge 1

    The Arrhenius linear form has slope = −Eₐ/R, so Eₐ = −(−6000 K)(8.314 J/(mol·K)) = 49,884 J/mol ≈ 49.9 kJ/mol.

  2. Challenge 2. A graph of ln[A] versus time has slope −0.045 min⁻¹. Identify the kinetic order and rate constant.

    Show worked solution for challenge 2

    A straight ln[A]-versus-time line indicates first-order kinetics. Its slope is −k, so k = 0.045 min⁻¹.

IntoChemistry original instructional content, 2026. Formative feedback is not an official course grade. Sources are linked for independent reading. Accessibility testing with your institution and assistive technology is still recommended.