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Mathematical Skills in Chemistry

Tangents and rates of change

  • 3 things to know
  • Component 3

Overview

Recall questions on constructing a tangent to a curve and taking its gradient, obtaining an instantaneous rate from a concentration–time graph, why the initial rate is measured at time zero, comparing gradients at different times, and what a levelling curve indicates.

What you need to know

Answer each one before you reveal it.

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Key term

Why is a rate obtained from a tangent less precise than one obtained from a straight-line graph?

Key term

Why is the gradient of a tangent negative for a reactant and positive for a product?

Numbers

A tangent at the origin of a concentration against time curve falls from 0.600 mol dm⁻³ to zero at 250 s. Calculate the initial rate.

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Why is a tangent needed to find the rate from a curved concentration against time graph?

    Model answer

    The gradient of a curve is different at every point, so a single straight line drawn to touch the curve at one point is the only way to express the rate of change at that particular moment.

  2. 2

    What does the gradient of the tangent to a concentration against time curve represent, and what are its units?

    Model answer

    It gives the rate of the reaction at that instant, in mol dm⁻³ s⁻¹, since the change in concentration is divided by the change in time.

  3. 3

    Where must a tangent touch the curve if the initial rate is wanted?

    Model answer

    It must touch the curve at the point where time is zero, so the gradient corresponds to the moment of mixing when the concentrations are still those that were measured out.

  4. 4

    A tangent at the origin of a concentration against time curve falls from 0.600 mol dm⁻³ to zero at 250 s. Calculate the initial rate.

    Model answer

    The gradient is 0.600 ÷ 250 = 2.40 × 10⁻³, so the initial rate is 2.40 × 10⁻³ mol dm⁻³ s⁻¹.

  5. 5

    The tangent to a gas volume against time curve has a gradient of 12.0 cm³ s⁻¹. Convert this to a rate in mol s⁻¹ using a molar gas volume of 24.0 dm³ mol⁻¹.

    Model answer

    One mole occupies 24 500 cm³ at room temperature and pressure, so the rate is 12.0 ÷ 24 500 = 4.90 × 10⁻⁴ mol s⁻¹.

  6. 6

    Why is the initial rate the most reliable rate to take from a curve?

    Model answer

    At the moment of mixing the concentrations of every reactant are known exactly from the volumes and concentrations measured out, and no product has yet accumulated to interfere or to drive a back reaction.

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