Corrected exercises: chemical kinetics (202-SN2-RE)

$v_{N_{2}O_{5}}=-\frac{\Delta[N_{2}O_{5}]}{\Delta t}=-\frac{0.088-0.100}{100}=1.2\times 10^{-4}$ mol/(L·s).

The ratios follow the coefficients: for every 2 $N_{2}O_{5}$ consumed, 4 $NO_{2}$ and 1 $O_{2}$ form. $v_{NO_{2}}=\frac{4}{2}\times 1.2\times 10^{-4}=2.4\times 10^{-4}$ mol/(L·s) and $v_{O_{2}}=\frac{1}{2}\times 1.2\times 10^{-4}=6.0\times 10^{-5}$ mol/(L·s). $NO_{2}$ appears twice as fast as $N_{2}O_{5}$ disappears: speaking of “THE rate” without naming the species is ambiguous.

$v=\frac{1}{2}v_{N_{2}O_{5}}=\frac{1}{4}v_{NO_{2}}=\frac{1}{1}v_{O_{2}}=6.0\times 10^{-5}$ mol/(L·s): the three quotients give the same number. Dividing by the coefficient cancels the stoichiometry and leaves a single rate, a property of the REACTION and not of one species.

Smaller. The rate depends on the frequency of effective collisions, hence on the concentration of the reactants: as $N_{2}O_{5}$ is consumed, its molecules become scarcer, collisions become less frequent and the reaction slows down. This is why concentration-time curves flatten out: the initial rate is the greatest of the whole history of the reaction.

From experiment 1 to experiment 2: $[A]$ doubles with $[B]$ constant, and $v$ is multiplied by $\frac{8.0}{2.0}=4=2^{2}$: order 2 in $A$. From experiment 1 to experiment 3: $[B]$ doubles with $[A]$ constant, and $v$ is multiplied by 2: order 1 in $B$. Always change only ONE concentration at a time: that is what isolates each exponent.

$v=k[A]^{2}[B]$, overall order $2+1=3$.

Using experiment 1: $k=\frac{v}{[A]^{2}[B]}=\frac{2.0\times 10^{-3}}{(0.10)^{2}(0.10)}=2.0$ L²/(mol²·s). Units: $v$ is in mol/(L·s) and $[A]^{2}[B]$ in mol³/L³; the quotient forces L²·mol⁻²·s⁻¹. The units of $k$ encode the overall order: getting them right proves you understood the rate law.

$v=2.0\times(0.30)^{2}(0.05)=9.0\times 10^{-3}$ mol/(L·s). The balanced equation says NOTHING about the exponents: it summarizes the bookkeeping of atoms, not the molecular story of the reaction. Here the orders (2 and 1) do not even match the coefficients (1 and 1), which is experimental proof that the reaction does not proceed through a single $A$-$B$ collision.

$\ln[H_{2}O_{2}]$ as a function of time: the integrated rate law $\ln[A]_{t}=\ln[A]_{0}-kt$ is the equation of a straight line with slope $-k$. It is also the TEST for first order: if the points line up in $\ln[A]$, the reaction is first order.

$t_{1/2}=\frac{\ln 2}{k}=\frac{0.693}{0.0231}=30.0$ min.

90 minutes is exactly 3 half-lives: $0.500\rightarrow 0.250\rightarrow 0.125\rightarrow 0.0625$ mol/L. Check: $[A]=0.500\,e^{-(0.0231)(90)}=0.500\,e^{-2.079}=0.0625$ mol/L. The two routes agree, as they must.

$t=\frac{1}{k}\ln\frac{[A]_{0}}{[A]}=\frac{1}{0.0231}\ln\frac{0.500}{0.100}=\frac{\ln 5}{0.0231}\approx 69.7$ min. Consistent: 0.100 mol/L lies between 2 half-lives (0.125) and 3 half-lives (0.0625), hence between 60 and 90 min.

It is CONSTANT: going from 0.500 to 0.250 takes 30 min, but going from 0.0625 to 0.03125 ALSO takes 30 min: the time needed to halve the concentration does not depend on the starting point. This is the exclusive property of first order, where the rate is proportional to what remains: exactly the law of radioactive decay, with the same equations: $k$ plays the role of the decay constant, and $t_{1/2}=\frac{\ln 2}{k}$ in both worlds.

$E_{a}$ is the energy barrier that molecules must cross during a collision in order to react: only the small fraction of collisions at least that energetic counts. This fraction, $e^{-E_{a}/RT}$, is EXPONENTIAL in $T$: heating shifts the tail of the distribution of molecular speeds, and a few degrees multiply the number of effective collisions far more than the mere increase in agitation would suggest.

Mind the kelvins: $T_{1}=298$ K, $T_{2}=308$ K. $\ln\frac{k_{2}}{k_{1}}=\frac{E_{a}}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right)$ gives $E_{a}=\frac{R\ln 2}{\frac{1}{298}-\frac{1}{308}}=\frac{(8.314)(0.693)}{1.09\times 10^{-4}}\approx 5.3\times 10^{4}$ J/mol, that is about 53 kJ/mol: a typical value for ordinary reactions.

$\frac{k_{cat}}{k}=e^{\Delta E_{a}/RT}=e^{20\,000/(8.314)(298)}=e^{8.07}\approx 3.2\times 10^{3}$: the reaction becomes three thousand times faster, without heating and without changing $\Delta H$ or the position of equilibrium. The whole chemical industry (and every enzyme in living things) summed up in one exponential.

Cooling plays the exponential in reverse: at 4 °C, the fraction of effective collisions for spoilage reactions drops several-fold compared with 25 °C, so food “ages” in slow motion. As for the 10 °C rule: the exact factor is $e^{\frac{E_{a}}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right)}$, which depends on $E_{a}$ AND on the temperature range; it equals 2 only for $E_{a}\approx 53$ kJ/mol around room temperature (exactly our part b)). A reaction with a large $E_{a}$ triples or quadruples; one with a small $E_{a}$ barely gains 30%: the “rule” is just our result in part b) elevated to a proverb.

Zero order: $[A]_{t}=[A]_{0}-kt$, straight line of $[A]$ versus $t$, slope $-k$. First order: $\ln[A]_{t}=\ln[A]_{0}-kt$, straight line of $\ln[A]$ versus $t$, slope $-k$. Second order: $\frac{1}{[A]_{t}}=\frac{1}{[A]_{0}}+kt$, straight line of $\frac{1}{[A]}$ versus $t$, slope $+k$ (the only positive slope of the three). The general method is therefore: plot all three, and the one that lines up gives the order.

The first half-life (from 0.100 to 0.050) lasts 100 s; the second (from 0.050 to 0.025) lasts 200 s: it DOUBLES when the starting concentration is halved. This is the signature of second order, where $t_{1/2}=\frac{1}{k[A]_{0}}$ is inversely proportional to the initial concentration. (First order: constant half-life. Zero order: $t_{1/2}=\frac{[A]_{0}}{2k}$, it would be HALVED.) Successive half-lives are enough to decide, without any regression.

$k=\frac{1}{t}\left(\frac{1}{[A]_{t}}-\frac{1}{[A]_{0}}\right)=\frac{1}{100}\left(\frac{1}{0.050}-\frac{1}{0.100}\right)=\frac{20-10}{100}=0.10$ L/(mol·s). Units: $\frac{1}{[A]}$ is in L/mol, divided by seconds, hence L·mol⁻¹·s⁻¹. Check on the second interval: $\frac{40-20}{200}=0.10$: same $k$, so second order is confirmed.

First order with $t_{1/2}=100$ s: 300 s is exactly 3 half-lives, so $0.100\rightarrow 0.050\rightarrow 0.025\rightarrow 0.0125$ mol/L. Reality gives 0.025 mol/L, which is TWICE as much. A second-order reaction runs out of steam much faster than a first-order one: its rate falls as the SQUARE of what remains, so the end of the reaction drags on interminably.

The classmate is right on that one point, and on nothing else. A straight line of $[A]$ versus $t$ characterizes zero order; a curve therefore does rule it out. But first order and second order BOTH give a curved decay that flattens out, visually very similar: you have to linearize ($\ln[A]$ or $1/[A]$) or compare half-lives to tell them apart. Drawing a conclusion from a quick look at $[A](t)$ is the classic error of the chapter.

An elementary step IS allowed to display its coefficients: the direct collision $NO_{2}+CO$ would give $v=k[NO_{2}][CO]$: first order in each. But experiment measures $v=k[NO_{2}]^{2}$: no dependence on $[CO]$ at all. Nature contradicts the one-step mechanism: the experimental rate law is the final judge of mechanisms.

Sum: $NO_{2}+NO_{2}+NO_{3}+CO\rightarrow NO_{3}+NO+NO_{2}+CO_{2}$. Cancelling $NO_{3}$ (on both sides) and one $NO_{2}$ leaves $NO_{2}+CO\rightarrow NO+CO_{2}$: the overall equation is recovered. $NO_{3}$ is the INTERMEDIATE: created in step 1, destroyed in step 2, absent from the overall equation and (almost) undetectable in the flask.

The slow step is the bottleneck: the overall rate is its rate. Elementary step 1: $v=k_{1}[NO_{2}][NO_{2}]=k_{1}[NO_{2}]^{2}$: exactly the observed law. $[CO]$ does not appear because $CO$ only comes into play AFTER the bottleneck: the fast step 2 absorbs the $NO_{3}$ as soon as it forms, however abundant $CO$ is. Adding $CO$ does not speed up the reaction any more than adding crates speeds up a factory whose oven is saturated.

Step 1: two molecules collide: BIMOLECULAR. Step 2: two species as well: bimolecular. (Termolecular steps are extremely rare: three bodies at the same place at the same instant.)

No. The criterion is the order of appearance. An intermediate is PRODUCED first, consumed afterwards ($NO_{3}$: created in step 1, destroyed in step 2). A catalyst is CONSUMED first, regenerated afterwards: it enters the mechanism before coming out of it unchanged, and so is present in the reaction mixture from the start. Both are absent from the overall equation, hence the confusion; but one is a fleeting creature of the reaction, the other a tool lent to it.

$k=\frac{\ln 2}{t_{1/2}}=\frac{0.693}{6.0}\approx 0.116$ h$^{-1}$.

24 h $=$ 4 half-lives: $80\rightarrow 40\rightarrow 20\rightarrow 10\rightarrow 5.0$ mg. (Integrated rate law: $80\,e^{-(0.1155)(24)}=80\,e^{-2.77}=5.0$ mg: confirmed.)

$80\rightarrow 10$ mg is a division by 8 $=2^{3}$: exactly 3 half-lives, that is $t=18$ h. Integrated rate law: $t=\frac{1}{k}\ln\frac{80}{10}=\frac{\ln 8}{0.1155}=18.0$ h. The two methods agree: when the ratio is a power of 2, half-lives do the calculation in your head.

Exponential decay is faster (in absolute value) the larger the amount: a double dose of 160 mg loses 80 mg in the first 6 hours, whereas a single dose loses only 40. Doubling the dose adds only ONE half-life of coverage (the time to fall from 160 to 80), that is 6 h more, while doubling the peak concentration, with its side effects. Regular doses, timed to the half-life, keep the concentration inside the therapeutic window, between the effectiveness threshold and the toxicity threshold: dosing is first-order kinetics put to work.

$E_{a,\text{fwd}}=130-40=90$ kJ/mol: this is the climb from the REACTANTS to the top. $E_{a,\text{rev}}=130-10=120$ kJ/mol: the climb from the PRODUCTS to the same top. The top is shared by both directions, only the starting point changes.

$\Delta H=E_{\text{products}}-E_{\text{reactants}}=10-40=-30$ kJ/mol: negative, so the reaction is EXOTHERMIC. The same value comes out from the difference of the barriers: $\Delta H=E_{a,\text{fwd}}-E_{a,\text{rev}}=90-120=-30$ kJ/mol. This relation is general and serves as a check: if your two barriers do not give back your $\Delta H$, one reading is wrong.

Catalyzed path: $E_{a,\text{cat}}=85-40=45$ kJ/mol, that is 45 kJ/mol less. Factor: $\frac{k_{cat}}{k}=e^{\Delta E_{a}/RT}=e^{45\,000/((8.314)(298))}=e^{18.16}\approx 7.7\times 10^{7}$: the forward reaction is accelerated almost a hundred million times.

Catalyzed reverse barrier: $85-10=75$ kJ/mol, against 120 kJ/mol without a catalyst: it also drops by exactly 45 kJ/mol, since it is the SAME top that is lowered and the level of the products has not moved. The factor is therefore the same: $e^{45\,000/RT}\approx 7.7\times 10^{7}$. Both directions are accelerated in the same proportion.

The equilibrium constant is the ratio of the two rate constants, $K=\frac{k_{\text{fwd}}}{k_{\text{rev}}}$. The catalyst multiplies the numerator and the denominator by the SAME factor: the ratio is rigorously unchanged, and so are $K$ and the equilibrium composition. This is the kinetic translation of a thermodynamic fact: $K$ depends only on $\Delta G^{\circ}$, that is, on the STARTING and ENDING levels, over which the catalyst has no hold, since it changes only the path between them. And yet industry pays: an equilibrium yield is worth nothing if it takes a thousand years to reach. The catalyst does not buy a better yield, it buys TIME, and often allows working at a lower temperature, hence at a lower energy cost.

Because the order is an EXPERIMENTAL fact and molecularity is a hypothesis about the mechanism: the two coincide only for an elementary step. A TERMOLECULAR step would require three molecules to meet at the same place at the same instant with the right orientation: so improbable that such steps are almost nonexistent in the gas phase. A third-order law must therefore almost always be read as the signature of a multistep mechanism, not of a triple collision.

Sum: $NO+NO+N_{2}O_{2}+O_{2}\rightarrow N_{2}O_{2}+2NO_{2}$. Cancelling $N_{2}O_{2}$ on both sides gives $2NO+O_{2}\rightarrow 2NO_{2}$, the overall equation. The intermediate is the dimer $N_{2}O_{2}$: produced in step 1, consumed in step 2.

The slow step, being elementary, imposes $v=k_{2}[N_{2}O_{2}][O_{2}]$. We cannot settle for this because it involves $[N_{2}O_{2}]$: an intermediate is not a quantity the experimenter controls or measures (it is neither weighed out at the start nor present at the end). A usable rate law must contain only the concentrations of the REACTANTS.

Step 1 being a fast equilibrium, it stays balanced at all times: $K_{1}=\frac{[N_{2}O_{2}]}{[NO]^{2}}$, hence $[N_{2}O_{2}]=K_{1}[NO]^{2}$. Substituting: $v=k_{2}K_{1}[NO]^{2}[O_{2}]$. This is exactly the observed form, with $k_{obs}=k_{2}K_{1}$: the measured constant is a PRODUCT, a rate constant times an equilibrium constant. The mechanism is therefore consistent with experiment.

Since $k_{obs}=k_{2}K_{1}$, the temperature dependence is that of the product: $E_{a,\text{app}}=E_{a,2}+\Delta H_{1}=20+(-40)=-20$ kJ/mol, a NEGATIVE apparent activation energy. Factor between 298 K and 348 K: $\ln\frac{k_{348}}{k_{298}}=\frac{E_{a,\text{app}}}{R}\left(\frac{1}{298}-\frac{1}{348}\right)=\frac{-20\,000}{8.314}(4.82\times 10^{-4})=-1.16$, that is $\frac{k_{348}}{k_{298}}=e^{-1.16}\approx 0.31$: heating by 50 K drops the rate to 31% of its value. The paradox is only apparent: heating does speed up step 2 (like any elementary step), but it shifts the exothermic equilibrium 1 toward DISSOCIATION of the dimer. Much less $N_{2}O_{2}$ is available, and this shortage largely outweighs the gain in rate. An apparent activation energy that is negative is never that of an elementary step: it is always the sign of an exothermic pre-equilibrium upstream of the bottleneck.

First order: converting 90% means 10% of the reactant remains, so $t=\frac{1}{k}\ln\frac{[A]_{0}}{0.10[A]_{0}}=\frac{\ln 10}{k}=\frac{2.303}{5.0\times 10^{-5}}\approx 4.6\times 10^{4}$ s, about 12.8 hours. The specification (2.0 h) is nowhere near met. Note that the time does NOT depend on the initial concentration: this is specific to first order.

We need $\frac{\ln 10}{k}\leq 7200$ s, so $k\geq \frac{2.303}{7200}=3.2\times 10^{-4}$ s$^{-1}$: about 6.4 times the current constant.

Arrhenius between 300 K and $T_{2}$: $\ln\frac{k_{2}}{k_{1}}=\frac{E_{a}}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right)$. Here $\ln(6.40)=1.856$ and $\frac{E_{a}}{R}=\frac{80\,000}{8.314}=9622$. Hence $\frac{1}{300}-\frac{1}{T_{2}}=\frac{1.856}{9622}=1.93\times 10^{-4}$, so $\frac{1}{T_{2}}=3.333\times 10^{-3}-1.93\times 10^{-4}=3.140\times 10^{-3}$ and $T_{2}\approx 318$ K, about 45 °C. Heating by 18 degrees is enough: that is the full weight of the exponential.

$\frac{k_{cat}}{k}=e^{(80\,000-60\,000)/((8.314)(300))}=e^{8.02}\approx 3.0\times 10^{3}$, so $k_{cat}\approx 0.15$ s$^{-1}$ and $t=\frac{\ln 10}{0.15}\approx 15$ s. The catalyst does not merely reach the target: it blows past it, bringing 12.8 hours down to a quarter of a minute, and without heating. The process is then no longer limited by the chemistry but by the chemical engineering: feed, heat transfer, removal of the heat released.

No, and for two different reasons. The catalyst changes absolutely nothing about the equilibrium: it lowers both barriers, forward and reverse, by the same amount and leaves $K$ intact. Heating, on the other hand, is NOT neutral: $K$ depends on temperature (van 't Hoff equation), and for an exothermic reaction, heating DECREASES the equilibrium yield. That is the fundamental difference between the two options: the catalyst buys rate for free, while heating buys rate at the cost of yield (and of energy). This is exactly the dilemma of the Haber-Bosch process, settled by a compromise: moderate temperature, an iron catalyst and high pressure.