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

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

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

What does adding a catalyst do to the slope of the Arrhenius line?

• It gets less steep
• It gets steeper
• Slope is unrelated to the catalyst

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

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

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

• The shared intercept is normal here: same A for these runs while the slope flattens
• The shared intercept proves the catalyst did nothing at all
• The shared intercept means the barrier must have grown

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