Chemical kinetics: how do I find the order of a reaction and read the zero, first and second order graphs?

4 answers & replies

• [+][–]CCCommons Chemistry DeskTeam· 17d ago

  Learn the three integrated cases as a single table, and the graph questions become lookups.

  Order	Rate law	Straight-line plot	Slope	Unit of $k$	$t_{1/2}$
  0	$r = k$	$[A]$ vs $t$	$-k$	mol L$^{-1}$ s$^{-1}$	$\dfrac{[A]_0}{2k}$
  1	$r = k[A]$	$\ln[A]$ vs $t$	$-k$	s$^{-1}$	$\dfrac{0.693}{k}$
  2	$r = k[A]^2$	$\dfrac{1}{[A]}$ vs $t$	$+k$	L mol$^{-1}$ s$^{-1}$	$\dfrac{1}{k[A]_0}$

  General unit of $k$ for order $n$: $(\text{mol L}^{-1})^{1-n}\,\text{s}^{-1}$.

  Traps worth naming:

  • Order is found by experiment, not read off the balanced equation. Molecularity applies only to elementary steps and is always a whole number.
  • Only first order has a half-life that does not depend on the starting concentration.
  • Rate and rate constant are not the same thing: rate depends on concentration, $k$ depends on temperature (and catalyst).

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• [+][–]CMCommons Mock & PYQ DeskTeam· 17d ago

  Initial-rate tables look long but take under a minute if you only compare two rows at a time.

  Say the data are:

  Exp	$[A]$	$[B]$	Rate
  1	0.1	0.1	$r$
  2	0.2	0.1	$4r$
  3	0.1	0.2	$2r$

  Experiments 1 and 2: $[B]$ is fixed and $[A]$ doubles, so $2^x = 4$ and the order in $A$ is $x = 2$. Experiments 1 and 3: $[A]$ fixed, $[B]$ doubles, so $2^y = 2$ and $y = 1$. Overall order is $3$, and the rate law is $r = k[A]^2[B]$.

  The rule: pick two rows where only one concentration changes, set up the ratio, solve for the exponent. You never need to calculate $k$ unless the question asks for it.

  In a mock, mark kinetics table questions as quick wins and do them early. If the numbers are not "nice" doublings, use logs, but check the options first because often one comparison is enough to eliminate three of them.

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• [+][–]CPCommons Physics DeskTeam· 17d ago

  Graph questions are slope-and-intercept questions in disguise. Convert the law to $y = mx + c$ and read off what the slope means.

  • First order: $\ln[A] = \ln[A]_0 - kt$. Plot $\ln[A]$ against $t$, slope $= -k$.
  • Zero order: $[A] = [A]_0 - kt$. A straight descending line that hits zero at $t = [A]_0/k$.
  • Arrhenius: $\ln k = \ln A - \dfrac{E_a}{R}\cdot\dfrac{1}{T}$. Plot $\ln k$ against $1/T$, slope $= -E_a/R$, so the line falls as $1/T$ increases.
  • Rate vs concentration: $r = k[A]^n$ gives $\log r = \log k + n\log[A]$, so a $\log r$ vs $\log[A]$ plot has slope $n$.

  Two checks that save time. If the axis is not simply $[A]$ or $t$, ask "which transformation makes this linear?" before reading anything. And if the question gives a slope, note the sign before the unit: a negative slope in a $\ln$ plot means $k$ is the positive magnitude of that slope.

  Same habit works for every graph in Physics: name the axes, write the straight-line form, interpret slope and intercept.

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• [+][–]CDCommons Dropper DeskTeam· 17d ago

  Kinetics is a good chapter to rebuild in a second year because it is small: about five relationships, plus the Arrhenius idea.

  Try this. Without opening a book, write the table from the first answer on a blank page. Compare. Whatever you got wrong goes at the top of tomorrow's page. Do this on separate days for a week or so, instead of spending a whole afternoon re-reading the chapter. Recall is slower and more uncomfortable than re-reading, and it is much better at showing you where the gaps really are.

  Then solve a few mixed questions where you have to decide the order yourself. Many repeat-year students have only practised numericals where the order is given in the problem statement, so the exam version feels unfamiliar.

  If you have done kinetics before, tell us what the chapter finally clicked on for you. A line or two is enough.

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