OCR A-Level Chemistry B H433
Development of Practical Skills in Chemistry
Elements of Life
Developing Fuels
Elements From the Sea
The Ozone Story
What's in a Medicine?
The Chemical Industry
Polymers and Life
Oceans
Developing Metals
Colour by Design
Chemical Literacy
Mathematical Skills in Chemistry
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.
Questions from this subtopic
Each one with the model answer underneath.
- 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
What does the gradient of a plot of ln k against 1/T equal?
Model answer
The gradient equals −Ea/R.
- 3
What does the intercept on the ln k axis equal?
Model answer
The intercept equals ln A.
- 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
Rearrange the gradient relationship to give the activation enthalpy.
Model answer
Ea = −gradient × R.
- 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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