Graphs: plotting, gradients, intercepts and tangents — Arrhenius plots, zero-order graphs and tangents

Overview

Finding activation energy and the pre-exponential factor from Arrhenius plots, zero-order concentration-time graphs, gas-volume graphs, and how to draw an accurate tangent to a curve.

What you need to know

Answer each one before you reveal it.

Questions from this subtopic

Each one with the model answer underneath.

1. 1

   An Arrhenius plot passes through (3.00 × 10⁻³ K⁻¹, ln k = −9.00) and (3.40 × 10⁻³ K⁻¹, ln k = −12.40). Calculate the gradient and the activation energy. (R = 8.31 J K⁻¹ mol⁻¹)

   Model answer

   Gradient = (−12.40 − (−9.00)) ÷ (3.40 × 10⁻³ − 3.00 × 10⁻³) = −3.40 ÷ (4.00 × 10⁻⁴) = −8.50 × 10³ K. Since the gradient equals −Ea/R, Ea = 8.50 × 10³ × 8.31 = 7.06 × 10⁴ J mol⁻¹, that is 70.6 kJ mol⁻¹.

2. 2

   For an Arrhenius plot with gradient −8.50 × 10³ K passing through (3.00 × 10⁻³ K⁻¹, ln k = −9.00), determine the intercept and hence the pre-exponential factor.

   Model answer

   Using ln k = m(1/T) + ln A, −9.00 = (−8.50 × 10³ × 3.00 × 10⁻³) + ln A, so −9.00 = −25.5 + ln A and ln A = 16.5. Then A = e¹⁶·⁵ = 1.46 × 10⁷, in the units of k, here s⁻¹.

3. 3

   An Arrhenius plot has an intercept on the ln k axis of 24.0. Calculate the pre-exponential factor for this first-order reaction.

   Model answer

   The intercept equals ln A, so A = e²⁴·⁰ = 2.65 × 10¹⁰ s⁻¹. The intercept is the value of ln k extrapolated to 1/T = 0, which is not a physically attainable temperature, so it must be obtained from the equation of the line rather than read directly from the plotted region.

4. 4

   Why is the gradient of an Arrhenius plot negative, and what does a steeper line indicate?

   Model answer

   The gradient is −Ea/R and activation energies are positive, so the gradient is negative: as 1/T increases the temperature falls and the rate constant decreases. A steeper line corresponds to a larger activation energy, meaning the rate constant is more sensitive to a change in temperature.

5. 5

   On a plot of ln k against 1/T, a student reads the intercept directly off the plotted region, which spans 1/T = 3.0 × 10⁻³ to 3.5 × 10⁻³ K⁻¹. Explain the error.

   Model answer

   The intercept is defined at 1/T = 0, which lies far outside the plotted range. Reading a value where the line meets the edge of the drawn axes gives ln k at the smallest plotted 1/T, not ln A. The intercept must be obtained by substituting a point and the gradient into the equation of the line.

6. 6

   On a graph of concentration against time for a zero-order reaction, the concentration falls linearly from 0.500 to 0.100 mol dm⁻³ over 200 s. Calculate the rate constant, with units.

   Model answer

   Gradient = (0.100 − 0.500) ÷ 200 = −0.400 ÷ 200 = −2.00 × 10⁻³ mol dm⁻³ s⁻¹. For a zero-order reaction, rate = k, so k = 2.00 × 10⁻³ mol dm⁻³ s⁻¹, the magnitude of the gradient.

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