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Mathematical Skills in Chemistry

Graphs, gradients and intercepts

  • 3 things to know
  • Component 3

Overview

Recall questions on choosing which variable goes on each axis, selecting a scale and when a false origin is allowed, drawing a best-fit line, calculating a gradient with its units, what an intercept represents, and reading a chemical quantity off a calibration line.

What you need to know

Answer each one before you reveal it.

Your recallNot started
Key term

Which variable is placed on each axis of an experimental graph, and why?

Key term

How should the scales of a graph be chosen?

Key term

Why is a smooth best-fit line drawn through experimental points rather than joining them point to point?

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Which variable is placed on each axis of an experimental graph, and why?

    Model answer

    The independent variable, the one deliberately altered by the experimenter, goes on the horizontal axis and the dependent variable that responded to it goes on the vertical axis, so the gradient reads as the change in the response per unit change in the cause.

  2. 2

    How should the scales of a graph be chosen?

    Model answer

    They should be chosen so that the plotted points occupy more than half the length of each axis, using divisions that are easy to subdivide, and the origin is included only where it is a genuine data point or where the intercept is needed.

  3. 3

    Why is a smooth best-fit line drawn through experimental points rather than joining them point to point?

    Model answer

    Every point carries measurement uncertainty, so the individual values scatter about the true relationship. A best-fit line averages that scatter, while joining the points reproduces the errors as if they were real features of the chemistry.

  4. 4

    Describe the shape of a concentration against time graph for a zero-order reactant and state what the gradient gives.

    Model answer

    It is a straight line of constant negative gradient until the reactant is exhausted, and the magnitude of that gradient is the rate constant in mol dm⁻³ s⁻¹.

  5. 5

    On a straight-line concentration against time graph the concentration falls from 0.500 to 0.200 mol dm⁻³ over 150 s. Calculate the rate constant.

    Model answer

    The gradient is (0.200 − 0.500) ÷ 150 = −2.00 × 10⁻³ mol dm⁻³ s⁻¹, so the rate constant for this zero-order reaction is 2.00 × 10⁻³ mol dm⁻³ s⁻¹.

  6. 6

    What must be plotted against time to give a straight line for a first-order reaction, and what do the gradient and intercept represent?

    Model answer

    The natural logarithm of the concentration is plotted against time. The gradient is the rate constant with a negative sign and the intercept on the vertical axis is the natural logarithm of the starting concentration.

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