2.8.8 · D5Chemical Kinetics
Question bank — Activation energy from Arrhenius plot; effect of catalyst
This is the "gotcha" deck for the parent topic. Every item below targets a way students misread the Arrhenius equation or the energy diagram of a catalyst. Read the prompt, answer out loud with a reason, then reveal.
Now use the picture below as your reference for every "slope" and "intercept" item that follows.
Recall The three reflexes that defuse 90% of the traps
- is on the x-axis, not . Move right → grows → temperature falls.
- The slope is — it is negative, and steeper down means bigger .
- A catalyst changes (kinetics), never or (thermodynamics).
True or false — justify
Every prompt is a statement. Answer "true" or "false" and give the one-sentence reason.
A higher activation energy makes the Arrhenius line slope upward more steeply.
False — the slope is , so a larger makes the line drop more steeply as grows; the line goes down-right, never up.
On an Arrhenius plot, moving to the right means the reaction is getting hotter.
False — the x-axis is , so moving right means is increasing, i.e. temperature is decreasing and the reaction is getting colder.
A catalyst lowers the activation energy of the forward reaction only.
False — the catalyst offers a new pathway with a lower peak that both directions must cross, so both forward and reverse drop, which is exactly why is unchanged.
Two reactions with the same must have the same rate constant .
False — also depends on the pre-exponential factor ; equal slopes on the plot can sit at different heights (different intercepts ).
At infinite temperature, the rate constant approaches the pre-exponential factor .
True — as , so and ; on the plot this is , the y-intercept .
Adding a catalyst shifts the equilibrium toward more product.
False — a catalyst speeds forward and reverse equally, so equilibrium is reached faster but sits in the same place; depends only on , which the catalyst never touches.
The units of are always .
False — carries the same units as , which depend on reaction order; only for a first-order reaction is it .
If a reaction has , its rate constant does not depend on temperature.
True — with the exponential is for all , so is flat; the Arrhenius line becomes horizontal (zero slope).
Doubling the temperature always doubles the rate constant.
False — depends on through , which is exponential and non-linear, so doubling can multiply by far more (or, near saturation, far less) than 2.
A catalyzed reaction has a smaller y-intercept on the Arrhenius plot.
False (usually) — the y-intercept is ; assuming the intercept is roughly unchanged, but at every real temperature the catalyzed line sits higher because its slope is less steep.
Spot the error
Each line contains a flawed statement. Reveal the correction.
" where is the slope."
The slope is negative (), so ; forgetting the minus sign gives a negative, physically impossible activation energy.
"Since , plotting against gives a straight line."
The linear term is , which is linear in , not in ; you must plot against to get a straight line.
"A catalyst makes an endothermic reaction exothermic by lowering the energy barrier."
Lowering the barrier changes only the peak height (); the reactant-to-product energy gap is fixed, so an endothermic reaction stays endothermic.
"The catalyzed and uncatalyzed lines on an Arrhenius plot are parallel."
They are parallel only if the two slopes are equal, i.e. — but a catalyst lowers , so the catalyzed slope is less steep and the lines are not parallel; they cross or diverge.
"Because the catalyst is consumed to lower the barrier, you must keep adding more."
A catalyst is regenerated each cycle and not consumed overall, so a tiny amount can turn over vast quantities of reactant.
"A larger means the reaction is more temperature-sensitive."
Temperature sensitivity is set by (it lives in the exponent); only scales up or down uniformly and shifts the intercept, not the slope.
"We take of the Arrhenius equation because logs make the numbers smaller."
We take to turn the exponential into a linear relationship () so the slope can be read off a straight line — the point is the shape, not the size.
Why questions
Give the reasoning, not just a label.
Why does the fraction of successful collisions rise exponentially with temperature rather than linearly?
Because the Maxwell–Boltzmann distribution puts the fraction of molecules above into an exponential tail , so small shifts in move a disproportionately large chunk of molecules past the barrier.
Why does a modest 25 kJ/mol drop in produce a rate boost of tens of thousands?
The rate ratio is ; at 300 K, kJ/mol, so is about ten "" units and — the exponential amplifies small energy changes enormously.
Why must the slope of an Arrhenius plot be negative for any real reaction?
Real activation energies are positive (), and the slope equals ; a positive over a positive forces the slope negative, meaning falls as rises (as falls).
Why can a catalyst speed up a reaction without changing ?
is the height difference between reactants and products; the catalyst only lowers the transition-state peak between them, which is a kinetics quantity, leaving the two endpoint energies (and hence , ) untouched.
Why do we plot two points at the extremes of the data to estimate the slope rather than two adjacent points?
The extreme points span the largest range in , so the same measurement scatter causes proportionally less error in the "rise over run" slope (though a best-fit line over all points is better still).
Why does raising temperature increase collision frequency only slightly but the rate a lot?
Collision frequency scales with molecular speed, which goes as (a weak dependence), whereas the successful-collision fraction is exponential in and dominates the rate change.
Edge cases
Boundary and degenerate scenarios the topic quietly assumes away.
What does the Arrhenius plot look like if a reaction proceeds through two mechanisms with different over a temperature range?
The plot curves (bends) instead of being straight, because the lower- path dominates at low and the higher- path takes over at high , each contributing its own slope.
If temperature K, what happens to ?
, so and — the reaction effectively stops because no molecules have enough energy; on the plot sends .
Can a catalyst make a thermodynamically unfavourable () reaction go to completion?
No — the catalyst only accelerates the approach to equilibrium; if the equilibrium still favours reactants, so you get the same small product amount, just faster.
What if the measured (y-intercept) comes out negative — is that an error?
Not necessarily — simply means in its units, which is fine; it says nothing is broken, only that the pre-exponential factor is a small number.
For an enzyme near substrate saturation, does the Arrhenius plot still stay straight?
Often not — at saturation the observed rate is limited by turnover steps (and enzymes denature at high ), so the plot can flatten or even bend downward, violating the simple single-barrier assumption.
If the pre-exponential factors differ () between catalyzed and uncatalyzed paths, can the catalyzed reaction ever be slower?
In principle yes at some temperatures — if is much smaller than , the intercept penalty could outweigh the lower- advantage in a narrow window, though for genuine catalysts the barrier drop normally wins.