The Arrhenius Equation and Finding Activation Energy · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

Activation energy from a straight line

Science · Chemistry · ages 17-18
Name ______________________   Date ____________

  1. What do you plot against what to find the activation energy?

    • k against T in Celsius
    • ln k against 1/T with T in kelvin
    • Ea against the rate constant

  2. Why does a small temperature rise change the rate so much?

    • Every particle gains exactly the same energy
    • The share of particles past the barrier grows fast
    • The activation energy itself rises with temperature

  3. T in the Arrhenius plot must be in kelvin, since Celsius would shift 1/T.

    Circle one:   True   False

  4. What does adding a catalyst do to the Arrhenius line?

    • It gets less steep, and the intercept stays the same
    • It gets steeper, and the intercept stays the same
    • It keeps its slope, and the intercept moves up

  5. An Arrhenius plot has gradient -5000 kelvin. With R = 8.31, work out Ea in kJ per mole.

    Answer: ______________

  6. A student plots ln k against T instead of 1/T and gets a curve. What went wrong?

    • The data must be wrong, since the plot is always straight
    • The catalyst must have run out during the runs
    • The wrong pair was plotted, since the straight form needs 1/T

  7. A catalysed run and the original run give lines that meet at the same intercept. A student says the catalyst left everything unchanged. What is the error?

    • The shared intercept is normal, since ln A stays put while the slope flattens
    • The shared intercept proves the catalyst did nothing at all
    • The shared intercept means the barrier must have grown

  8. Two reactions have Arrhenius slopes of -4000 K and -9000 K. Which has the larger Ea, and why?

    • The -9000 K one, since a steeper fall means a bigger barrier
    • The -4000 K one, since a flatter line means a bigger barrier
    • They are equal, since the slope only sets the intercept
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Answer key

For grown-ups. Fold this page away before handing over the rest.

Activation energy from a straight line W1-mt_QBQDSNwbXK-s1

  1. ln k against 1/T with T in kelvin · The straight-line form needs ln k against 1/T, with T in kelvin.
  2. The share of particles past the barrier grows fast · Warming fattens the fast tail, so many more particles clear the barrier.
  3. True · Celsius shifts every 1/T value and wrecks the gradient. The sentence is true.
  4. It gets less steep, and the intercept stays the same · Lower Ea flattens the slope, while ln A and its intercept do not move.
  5. 41.55 · Ea = 5000 times 8.31 = 41550 J per mole, which is 41.55 kJ per mole.
  6. The wrong pair was plotted, since the straight form needs 1/T · Only ln k against 1/T straightens the equation. Plotting against T bends it.
  7. The shared intercept is normal, since ln A stays put while the slope flattens · The catalyst lowers Ea and flattens the slope but leaves ln A alone, so one shared intercept is expected.
  8. The -9000 K one, since a steeper fall means a bigger barrier · Slope is -Ea/R, so the steeper -9000 K slope hides the larger Ea.
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