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The Chemical Industry

The ln k against 1/T plot and finding the activation enthalpy

  • 1 thing to know
  • Components 01, 02 and 03

Overview

Recall questions on which quantities are plotted, what the gradient and intercept give, and calculating Ea and A from them.

What you need to know

Answer each one before you reveal it.

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Numbers

What is A when the intercept ln A is 24.0, what temperature does 1/T = 3.20 × 10⁻³ K⁻¹ correspond to, and how does a ln k against 1/T plot show k rising with temperature?

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Which two quantities are plotted to obtain a straight line from the Arrhenius equation?

    Model answer

    ln k is plotted on the vertical axis against 1/T on the horizontal axis.

  2. 2

    What does the gradient of a plot of ln k against 1/T equal?

    Model answer

    The gradient equals −Ea/R.

  3. 3

    What does the intercept on the ln k axis equal?

    Model answer

    The intercept equals ln A.

  4. 4

    Why must the intercept be read where 1/T is zero rather than at the edge of the plotted data?

    Model answer

    The intercept is defined at 1/T = 0, so the line must be extrapolated back to that axis, or ln A calculated from a point on the line.

  5. 5

    Rearrange the gradient relationship to give the activation enthalpy.

    Model answer

    Ea = −gradient × R.

  6. 6

    An Arrhenius plot has a gradient of −6.50 × 10³ K. Calculate the activation enthalpy in kJ mol⁻¹.

    Model answer

    Ea = 6.50 × 10³ × 8.314 = 5.40 × 10⁴ J mol⁻¹, which is 54.0 kJ mol⁻¹.

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