2.8.6 · D5Chemical Kinetics
Question bank — Methods to determine order — initial rates, integrated method, half-life method
True or false — justify
For a reaction , the order in must be
False. Order is experimental; the coefficient only equals the exponent if the step is elementary, which you may not assume for an overall reaction.
The three methods (initial rates, integrated, half-life) can give different orders for the same reaction
False. They are three fingerprints of one underlying rate law, so they must all agree; a disagreement means a measurement or fitting error.
A straight line in the vs plot proves the reaction is first order
True — only makes fall linearly with time, so that particular plot being straight is the unique signature of first order.
If vs is straight, then vs must be curved
True. A given data set can be linear on only one of the three plots; if is the straight one, the raw plot must bend (it decays exponentially, not linearly).
Every reaction has a constant half-life
False. Only first order has independent of ; zero-order half-life shrinks and second-order half-life grows as the reaction proceeds.
The rate constant has the same units for all orders
False. Since , matching units of rate () forces to carry units , which change with .
In the integrated method, a negative slope always means a positive
True for and , where the linear slope is ; the concentration term falls, so slope is negative and . For the plotted quantity rises, giving slope .
Spot the error
"I measured the rate at s in both experiments and used the initial-rate formula."
The formula needs initial rates. At s products have built up, is uncertain, and reverse/side reactions distort the rate — the isolation logic collapses.
"Doubling doubled the rate, so ."
The formula is . Doubling rate for double concentration is first order, not second; the student invented an extra factor of 2.
", so I'll use this to find for my zero-order reaction."
is the first-order half-life only. For zero order , so using the wrong formula gives a wrong and hides the concentration dependence entirely.
"To find the order in I doubled both and and saw the rate go ×8."
Changing two concentrations at once mixes both exponents (). The isolation method demands you hold every other reactant fixed so only 's exponent shows.
"The slope of my vs line is , so ."
Rate constants are never negative. The slope equals , so ; the minus sign just records that concentration is falling.
" vs was straight with slope ."
For second order the plot rises over time, so its slope is , not . A negative slope there would be physically impossible.
"My data fit all three plots equally well, so the reaction is zero, first, and second order."
Only one plot can be truly straight; "fits all three" means your concentration range or time window is too narrow to tell them apart. Extend the run — the curves diverge more as drops further.
Why questions
Why do we take the logarithm in the initial-rate formula?
The order sits in the exponent of . Taking pulls down to a multiplier, turning an exponent-solving problem into a simple division.
Why must all other reactant concentrations be held constant when isolating the order in one reactant?
So the only thing changing the rate is the reactant under study; any change in the others would add unknown factors and make the measured ratio ambiguous.
Why does the integrated method plot three different quantities (, , ) rather than just ?
Because each order predicts a different function of that is linear in time. Testing which transformed quantity straightens the data is exactly what reveals the order.
Why is a first-order half-life independent of starting concentration?
Because contains no — the fractional decay per unit time is constant, so halving always takes the same clock time no matter how much you started with.
Why can't you determine order just by inspecting the balanced equation?
The overall equation reflects mass balance, not mechanism. The actual rate depends on the slow (rate-determining) step, whose molecularity need not match the overall coefficients.
Why is the general rule useless exactly at ?
At the exponent , so — the dependence vanishes. That is why first order has a constant half-life, but it also means you can't extract from the exponent there and must recognise the flat behaviour directly.
Why do initial rates avoid the problem of reverse reactions?
At virtually no product exists, so there is nothing to run the reaction backwards; the measured rate is purely the forward process you want to characterise.
Edge cases
What does the half-life method report if doubling doubles ?
Then , which is the zero-order fingerprint (), so the reaction is zero order.
For a zero-order reaction, what happens to the rate as approaches zero?
The rate stays (constant) right up until runs out, then abruptly stops — zero order means rate does not depend on while any remains.
Can the order be a fraction or negative, and would these methods still work?
Yes. The initial-rate log formula returns any real (e.g. or ); fractional/negative orders signal complex mechanisms, but the isolation and log methods handle them without modification.
If a plot of vs is a straight line all the way to , what is the order and what is the slope?
Zero order, and the slope is ; the concentration drops at a constant rate independent of how much remains.
Two experiments give the same initial rate despite different — what order is that?
Zero order in that reactant: since rate does not depend on , changing leaves the rate unchanged, giving .
Why does the second-order half-life grow as the reaction proceeds?
Because uses the current concentration as the new starting point; each successive half-life starts from a smaller , so is larger and the halving takes longer.
For a first-order reaction, what is the relationship between the third half-life and the first?
They are equal. First-order half-life is constant, so the 1st, 2nd, and 3rd all take the same time — this is why radioactive decay (Radioactive Decay) has a fixed half-life.
If measured "order" from initial rates differs from the order that straightens the integrated plot, what should you suspect?
A hidden complication — accumulating product, a temperature drift (Arrhenius Equation), or a mechanism change over the run. Genuine kinetics forces the two methods to agree.
Recall One-line self-test before you close this page
Cover the answers and re-derive: (1) which plot is straight for each order, (2) how depends on for each order, (3) why "initial" and "isolate one reactant" are non-negotiable. If any of these feels shaky, revisit Integrated Rate Equations and Half-Life of Reactions.