Exercises — Chemical Kinetics

The order is the sum of the exponents in the rate law, and the units follow because $k$ is whatever is left when the rate is divided by the concentration terms, so rearranging gives

$k=\frac{\text{Rate}}{(\text{concentration}{)}^n}=\frac{{\text{mol L}}^{-1}{\text{s}}^{-1}}{({\text{mol L}}^{-1}{)}^n}=({\text{mol L}}^{-1}{)}^{1-n}{\text{s}}^{-1}$

(i) Order $=2$. Units: $({\text{mol L}}^{-1}{)}^{-1}{\text{s}}^{-1}={\text{mol}}^{-1}{\text{L s}}^{-1}$.

(ii) Order $=1+1=2$. Units: ${\text{mol}}^{-1}{\text{L s}}^{-1}$.

(iii) Order $=3/2=1.5$. Units: $({\text{mol L}}^{-1}{)}^{-1/2}{\text{s}}^{-1}={\text{mol}}^{-1/2}{\text{L}}^{1/2}{\text{s}}^{-1}$.

(iv) Order $=1$. Units: $({\text{mol L}}^{-1}{)}^0{\text{s}}^{-1}={\text{s}}^{-1}$.

Notice how far the orders are from the stoichiometric coefficients. In (i) three molecules of NO appear in the equation but the order is 2; in (ii) the coefficients are 1, 3 and 2 while the H${}^{+}$ term does not appear in the rate law at all. This is the standing warning of section 3.2.2 that a rate law is an experimental result and cannot be predicted from a balanced equation.

✦ (i) second order, mol${}^{-1}$ L s${}^{-1}$; (ii) second order, mol${}^{-1}$ L s${}^{-1}$; (iii) order 1.5, mol${}^{-1/2}$ L${}^{1/2}$ s${}^{-1}$; (iv) first order, s${}^{-1}$

Initial rate. Substituting directly:

$\text{Rate}=k[\text{A}][\text{B}{]}^2=2.0\times 10^{-6}\times 0.1\times (0.2{)}^2$

$\text{Rate}=2.0\times 10^{-6}\times 0.1\times 0.04=8.0\times 10^{-9}{\text{mol L}}^{-1}{\text{s}}^{-1}$

After A falls to 0.06 M. The concentration of A has dropped by $0.1-0.06=0.04$ mol L${}^{-1}$. The equation is $2\text{A}+\text{B}\to {\text{A}}_2\text{B}$, so B is consumed at half the rate of A:

$\Delta [\text{B}]=\frac{0.04}{2}=0.02{\text{mol L}}^{-1}$

$[\text{B}]=0.2-0.02=0.18{\text{mol L}}^{-1}$

Substituting the new concentrations:

$\text{Rate}=2.0\times 10^{-6}\times 0.06\times (0.18{)}^2=2.0\times 10^{-6}\times 0.06\times 0.0324$

$\text{Rate}=3.89\times 10^{-9}{\text{mol L}}^{-1}{\text{s}}^{-1}$

The whole difficulty of the second part is remembering to update B as well as A. Leaving B at 0.2 M gives $4.8\times 10^{-9}$, which is wrong by about 25%.

✦ Initial rate $=8.0\times 10^{-9}$ mol L${}^{-1}$ s${}^{-1}$; after A falls to 0.06 M the rate is $3.89\times 10^{-9}$ mol L${}^{-1}$ s${}^{-1}$

The balanced equation is

$2{\text{NH}}_3\to {\text{N}}_2+3{\text{H}}_2$

and for a zero order reaction the rate is independent of concentration:

$\text{Rate}=k[{\text{NH}}_3{]}^0=k=2.5\times 10^{-4}{\text{mol L}}^{-1}{\text{s}}^{-1}$

Relating the rate of reaction to each species through the coefficients:

$\text{Rate}=-\frac{1}{2}\frac{d[{\text{NH}}_3]}{dt}=\frac{d[{\text{N}}_2]}{dt}=\frac{1}{3}\frac{d[{\text{H}}_2]}{dt}$

So:

$\frac{d[{\text{N}}_2]}{dt}=2.5\times 10^{-4}{\text{mol L}}^{-1}{\text{s}}^{-1}$

$\frac{d[{\text{H}}_2]}{dt}=3\times 2.5\times 10^{-4}=7.5\times 10^{-4}{\text{mol L}}^{-1}{\text{s}}^{-1}$

Hydrogen appears three times as fast as nitrogen, exactly as the coefficients require.

A printed unit error to note. The question gives $k$ as mol${}^{-1}$ L s${}^{-1}$, which are the units of a second order rate constant. Table 3.3 of the same chapter states that a zero order rate constant has units mol L${}^{-1}$ s${}^{-1}$. Since the question itself declares the reaction zero order, the intended units are mol L${}^{-1}$ s${}^{-1}$ and the numerical answers are unaffected.

✦ $d[{\text{N}}_2]/dt=2.5\times 10^{-4}$ mol L${}^{-1}$ s${}^{-1}$ and $d[{\text{H}}_2]/dt=7.5\times 10^{-4}$ mol L${}^{-1}$ s${}^{-1}$

When the reaction is followed by pressure, the pressure of the ether stands in for its concentration, so bar replaces mol L${}^{-1}$ throughout.

Units of rate. The rate is a change of pressure per unit time:

$\text{Rate}=\frac{\text{bar}}{\text{min}}={\text{bar min}}^{-1}$

Units of the rate constant. Rearranging the rate law with $n=3/2$:

$k=\frac{\text{Rate}}{p^{3/2}}=\frac{{\text{bar min}}^{-1}}{{\text{bar}}^{3/2}}={\text{bar}}^{-1/2}{\text{min}}^{-1}$

So the rate is measured in bar min${}^{-1}$ and the rate constant in bar${}^{-1/2}$ min${}^{-1}$.

The general pattern is the same as with concentrations: a rate constant of order $n$ carries units of $(\text{unit of the measured quantity}{)}^{1-n}$ per unit time. Here $1-\frac{3}{2}=-\frac{1}{2}$, hence bar${}^{-1/2}$.

✦ Rate in bar min${}^{-1}$ and rate constant in bar${}^{-1/2}$ min${}^{-1}$

Section 3.2 opens by naming the three factors the chapter goes on to treat quantitatively:

Concentration of the reactants. The rate generally rises as concentrations rise, and the exact dependence is given by the experimentally determined rate law. For gases the corresponding variable is partial pressure.

Temperature. The rate constant increases sharply with temperature, roughly doubling for every 10 K near room temperature, because a larger fraction of molecules carries energy above $E_a$. This is described by the Arrhenius equation.

Catalyst. A catalyst provides an alternative path of lower activation energy and so increases the rate without being consumed. It changes neither $\Delta G$ nor the equilibrium constant, only the speed at which equilibrium is reached.

Beyond these three, the following also matter in practice:

Nature of the reactants. Ionic reactions in solution are almost instantaneous, while reactions requiring covalent bonds to be broken are far slower.

Surface area. For heterogeneous reactions, powdering a solid or increasing the area of a catalyst raises the rate, because reaction occurs only at the interface.

Radiation. Some reactions, such as the chlorination of methane, are initiated by light.

✦ Concentration of reactants, temperature and catalyst are the three the chapter treats; the nature of the reactants, surface area in heterogeneous systems and exposure to radiation also affect the rate

Second order with respect to a reactant means $\text{Rate}=k[\text{A}{]}^2$ with the other concentrations held fixed.

(i) Concentration doubled.

${\text{Rate}}^{\prime }=k(2[\text{A}]{)}^2=4k[\text{A}{]}^2=4\times \text{Rate}$

The rate becomes four times as large.

(ii) Concentration halved.

${\text{Rate}}^{\prime }=k{\left(\frac{[\text{A}]}{2}\right)}^2=\frac{1}{4}k[\text{A}{]}^2=\frac{\text{Rate}}{4}$

The rate falls to one quarter of its original value.

Both follow the general rule that multiplying a concentration by $f$ multiplies the rate by $f^n$, with $n=2$ here. The effect is amplified because the concentration enters squared, which is why halving costs a factor of four rather than two.

✦ (i) The rate becomes four times as large; (ii) the rate falls to one quarter

The effect. The rate constant increases as the temperature rises. Experimentally it is found that a rise of about 10 K nearly doubles the rate constant for many reactions near room temperature. The chapter illustrates this with the decomposition of N${}_2$O${}_5$, whose half-life is 10 days at 0 °C, 5 hours at 25 °C and only 12 minutes at 50 °C.

The quantitative representation. The dependence is given by the Arrhenius equation:

$k=A{e}^{-E_a/RT}$

where $A$ is the Arrhenius or pre-exponential factor, related to collision frequency, and $E_a$ is the activation energy.

Taking natural logarithms gives a straight-line form:

$\ln k=\ln A-\frac{E_a}{RT}$

A plot of $\ln k$ against $1/T$ is therefore linear, with slope $-E_a/R$ and intercept $\ln A$, which is how both quantities are obtained experimentally.

For two temperatures the constant $A$ cancels, leaving the form used in most problems:

$\log \frac{k_2}{k_1}=\frac{E_a}{2.303R}\left[\frac{T_2-T_1}{T_1{T}_2}\right]$

The physical reason is that $e^{-E_a/RT}$ is the fraction of molecules with energy at least $E_a$. Raising the temperature broadens the Maxwell-Boltzmann distribution and shifts it to higher energies, so that fraction grows steeply.

✦ The rate constant rises with temperature, roughly doubling per 10 K, and is described by $k=A{e}^{-E_a/RT}$, whose logarithmic form $\ln k=\ln A-E_a/RT$ gives a straight line of slope $-E_a/R$ against $1/T$

The average rate over an interval is the change in concentration divided by the length of the interval, with a minus sign for a reactant:

$r_{\text{av}}=-\frac{\Delta [\text{A}]}{\Delta t}=-\frac{[\text{A}{]}_{60}-[\text{A}{]}_{30}}{60-30}$

$r_{\text{av}}=\frac{0.31-0.17}{30}=\frac{0.14}{30}$

$r_{\text{av}}=4.67\times 10^{-3}{\text{mol L}}^{-1}{\text{s}}^{-1}$

Only the two entries at 30 s and 60 s are used; the values at 0 s and 90 s are there for the other parts of a fuller problem.

The word "pseudo first order" tells us that water, though a reactant, is present in such large excess that its concentration is effectively constant, so the reaction behaves as first order in A alone. That fact is not needed for an average rate, but it is what makes the table decay so cleanly, with roughly equal fractional drops in equal times.

✦ $r_{\text{av}}=4.67\times 10^{-3}$ mol L${}^{-1}$ s${}^{-1}$

(i) The differential rate equation. First order in A and second order in B means the exponents are 1 and 2:

$-\frac{d[\text{A}]}{dt}=k[\text{A}][\text{B}{]}^2$

The overall order is $1+2=3$, since order is defined as the sum of the exponents.

(ii) Concentration of B tripled. Only B changes, and it enters squared:

${\text{Rate}}^{\prime }=k[\text{A}](3[\text{B}]{)}^2=9k[\text{A}][\text{B}{]}^2=9\times \text{Rate}$

The rate becomes nine times as large.

(iii) Both concentrations doubled. Each factor contributes according to its own order:

${\text{Rate}}^{\prime }=k(2[\text{A}])(2[\text{B}]{)}^2=2\times 4\times k[\text{A}][\text{B}{]}^2=8\times \text{Rate}$

The rate becomes eight times as large, which is $2^3$, the factor raised to the overall order. That shortcut only works when every concentration is multiplied by the same factor, as it is here.

✦ (i) $-d[\text{A}]/dt=k[\text{A}][\text{B}{]}^2$; (ii) the rate becomes nine times; (iii) the rate becomes eight times

Write the rate law as $r_0=k[\text{A}{]}^x[\text{B}{]}^y$ and compare experiments in pairs.

Order with respect to B. In experiments 1 and 2, $[\text{A}]$ is the same at 0.20 M while $[\text{B}]$ falls from 0.30 to 0.10 M, a threefold change. The rate is unchanged at $5.07\times 10^{-5}$. A concentration that can be tripled without affecting the rate does not appear in the rate law, so

$y=0$

Order with respect to A. Since $y=0$, the B column can now be ignored entirely. Comparing experiments 2 and 3, $[\text{A}]$ doubles from 0.20 to 0.40 M:

$\frac{1.43\times 10^{-4}}{5.07\times 10^{-5}}=2.82=2^x$

$x=\frac{\log 2.82}{\log 2}=\frac{0.4502}{0.3010}=1.50$

So the reaction is of order $1.5$ in A and zero in B, giving an overall order of $1.5$, and the rate law is

$r_0=k[\text{A}{]}^{3/2}$

The fractional order is the point of the question. It is a reminder that order is an experimental quantity, need not be a whole number, and cannot be inferred from any equation.

✦ Order $1.5$ with respect to A and zero with respect to B, so the rate law is $r_0=k[\text{A}{]}^{3/2}$ and the overall order is 1.5

Let the rate law be $\text{Rate}=k[\text{A}{]}^x[\text{B}{]}^y$.

Order in B, from experiments II and III. Here $[\text{A}]$ is fixed at 0.3 M while $[\text{B}]$ doubles from 0.2 to 0.4 M:

$\frac{2.88\times 10^{-1}}{7.2\times 10^{-2}}=4=2^y\Rightarrow y=2$

Order in A, from experiments I and IV. Here $[\text{B}]$ is fixed at 0.1 M while $[\text{A}]$ goes from 0.1 to 0.4 M, a fourfold rise:

$\frac{2.40\times 10^{-2}}{6.0\times 10^{-3}}=4=4^x\Rightarrow x=1$

The rate law is therefore

$\text{Rate}=k[\text{A}][\text{B}{]}^2$

with an overall order of 3.

The rate constant, from experiment I:

$k=\frac{6.0\times 10^{-3}}{(0.1)(0.1{)}^2}=\frac{6.0\times 10^{-3}}{1.0\times 10^{-3}}=6.0{\text{mol}}^{-2}{\text{L}}^2{\text{min}}^{-1}$

Checking against the other three experiments confirms the value, since a rate constant that varies between experiments would mean the rate law is wrong:

All four agree. Note that the tabulated quantity is the rate of formation of D, and since D has a coefficient of 1 this is also the rate of the reaction, so no factor of $\frac{1}{2}$ enters despite the 2 in front of A.

✦ Rate $=k[\text{A}][\text{B}{]}^2$, overall order 3, with $k=6.0$ mol${}^{-2}$ L${}^2$ min${}^{-1}$

Since the reaction is first order in A and zero order in B, the rate law is

$\text{Rate}=k[\text{A}{]}^1[\text{B}{]}^0=k[\text{A}]$

The entire B column is therefore irrelevant to the arithmetic, which is the point of the question.

Find $k$ from experiment I, the only complete row:

$k=\frac{\text{Rate}}{[\text{A}]}=\frac{2.0\times 10^{-2}}{0.1}=0.2{\text{min}}^{-1}$

Experiment II: given the rate, find $[\text{A}]$.

$[\text{A}]=\frac{4.0\times 10^{-2}}{0.2}=0.2{\text{mol L}}^{-1}$

Experiment III: given $[\text{A}]=0.4$, find the rate.

$\text{Rate}=0.2\times 0.4=8.0\times 10^{-2}{\text{mol L}}^{-1}{\text{min}}^{-1}$

Experiment IV: given the rate, find $[\text{A}]$.

$[\text{A}]=\frac{2.0\times 10^{-2}}{0.2}=0.1{\text{mol L}}^{-1}$

The completed table:

Experiments I and IV make the zero order in B visible: the same rate with $[\text{B}]$ doubled.

✦ $k=0.2$ min${}^{-1}$; II has $[\text{A}]=0.2$ mol L${}^{-1}$, III has rate $8.0\times 10^{-2}$ mol L${}^{-1}$ min${}^{-1}$, and IV has $[\text{A}]=0.1$ mol L${}^{-1}$

For a first order reaction the half-life is independent of the initial concentration:

$t_{1/2}=\frac{0.693}{k}$

The unit of $t_{1/2}$ is simply the reciprocal of the unit of $k$, so no conversion is needed in any part.

(i) $t_{1/2}=\frac{0.693}{200}=3.47\times 10^{-3}$ s

(ii) $t_{1/2}=\frac{0.693}{2}=0.347$ min, which is about 20.8 s

(iii) $t_{1/2}=\frac{0.693}{4}=0.173$ years, which is about 63 days

The three cases span more than eight orders of magnitude in time, which is a fair reflection of how wide a range of rates chemical kinetics has to describe.

✦ (i) $3.47\times 10^{-3}$ s, (ii) 0.347 min, (iii) 0.173 years

Radioactive decay follows first order kinetics, so

$k=\frac{0.693}{t_{1/2}}=\frac{0.693}{5730}=1.209\times 10^{-4}{\text{year}}^{-1}$

The integrated first order rate law, using percentages in place of concentrations since only their ratio matters:

$t=\frac{2.303}{k}\log \frac{[\text{R}{]}_0}{[\text{R}]}=\frac{2.303}{1.209\times 10^{-4}}\log \frac{100}{80}$

$t=19042\times \log (1.25)=19042\times 0.09691$

$t=1845\text{years}$

The artifact is roughly 1845 years old.

As a sanity check, only 20% of the carbon-14 has decayed, which is much less than half, so the age must be well under one half-life of 5730 years. It is, at about a third of one.

This is radiocarbon dating in its simplest form. It works because living things exchange carbon with the atmosphere and so hold the atmospheric ${}^{14}$C level, and stop doing so at death, after which the clock runs.

✦ About 1845 years old

(i) The plot of $[{\text{N}}_2{\text{O}}_5]$ against $t$ is a smooth curve falling from $1.63\times 10^{-2}$ mol L${}^{-1}$ and flattening as it goes, never reaching zero. It is not a straight line, which already rules out zero order.

(ii) Half-life from the graph. Half of the initial value is $0.815\times 10^{-2}$ mol L${}^{-1}$. Reading across the table, this lies between $t=1200$ s (0.93) and $t=1600$ s (0.78). Interpolating:

$t_{1/2}\approx 1200+\frac{0.93-0.815}{0.93-0.78}\times 400=1200+307=1507\text{s}$

about 1500 seconds, or 25 minutes.

(iii) The plot of $\log [{\text{N}}_2{\text{O}}_5]$ against $t$:

These points lie on a straight line of negative slope.

(iv) The rate law. A straight line for $\log [\text{R}]$ against $t$ is the signature of first order kinetics, since

$\log [\text{R}]=\log [\text{R}{]}_0-\frac{k}{2.303}t$

So the rate law is

$\text{Rate}=k[{\text{N}}_2{\text{O}}_5]$

(v) The rate constant. Applying $k=\frac{2.303}{t}\log \frac{[\text{R}{]}_0}{[\text{R}]}$ to each row:

The values are constant within experimental scatter, which confirms first order. Their mean is

$k=4.65\times 10^{-4}{\text{s}}^{-1}$

A least-squares line through the $\log [{\text{N}}_2{\text{O}}_5]$ points has slope $-2.083\times 10^{-4}$, giving $k=2.303\times 2.083\times 10^{-4}=4.80\times 10^{-4}$ s${}^{-1}$, so the two routes agree to within about 3%.

(vi) Half-life from $k$:

$t_{1/2}=\frac{0.693}{4.65\times 10^{-4}}=1490\text{s}$

This agrees closely with the 1507 s read off the graph in part (ii), the small difference being what interpolation between two tabulated points can be expected to cost.

✦ $t_{1/2}\approx 1500$ s from the graph; the rate law is Rate $=k[{\text{N}}_2{\text{O}}_5]$ with $k=4.65\times 10^{-4}$ s${}^{-1}$, giving $t_{1/2}=1490$ s, in agreement

By the integrated rate law:

$t=\frac{2.303}{k}\log \frac{[\text{R}{]}_0}{[\text{R}]}=\frac{2.303}{60}\log 16$

$t=0.03838\times 1.2041=4.62\times 10^{-2}\text{s}$

By half-lives, which is faster here. One sixteenth is ${\left(\frac{1}{2}\right)}^4$, so four half-lives have elapsed:

$t_{1/2}=\frac{0.693}{60}=1.155\times 10^{-2}\text{s}$

$t=4\times 1.155\times 10^{-2}=4.62\times 10^{-2}\text{s}$

Both give the same answer, as they must. The half-life route works whenever the fraction remaining is a power of one half, which is why $1/4$, $1/8$, $1/16$ and $1/32$ appear so often in kinetics questions.

✦ $t=4.62\times 10^{-2}$ s, which is four half-lives

Radioactive decay is first order, so

$k=\frac{0.693}{28.1}=2.466\times 10^{-2}{\text{year}}^{-1}$

The integrated rate law rearranged for the amount remaining:

$\log \frac{[\text{R}{]}_0}{[\text{R}]}=\frac{kt}{2.303}$

After 10 years:

$\log \frac{1}{[\text{R}]}=\frac{2.466\times 10^{-2}\times 10}{2.303}=0.1071$

$[\text{R}]=10^{-0.1071}=0.781\text{mg}$

After 60 years:

$\log \frac{1}{[\text{R}]}=\frac{2.466\times 10^{-2}\times 60}{2.303}=0.6426$

$[\text{R}]=10^{-0.6426}=0.228\text{mg}$

A check on the second figure: 60 years is a little over two half-lives, and two half-lives would leave 0.25 mg, so 0.228 mg is right.

The chemistry behind the question is that strontium sits directly below calcium in group 2, so the body handles Sr${}^{2+}$ much as it handles Ca${}^{2+}$ and deposits it in bone, where it irradiates the marrow for decades. That is why ${}^{90}$Sr from fallout is treated as one of the most dangerous fission products.

✦ 0.781 mg remains after 10 years and 0.228 mg after 60 years

For a first order reaction,

$t=\frac{2.303}{k}\log \frac{[\text{R}{]}_0}{[\text{R}]}$

For 99% completion, 1% of the reactant remains, so $[\text{R}]=0.01[\text{R}{]}_0$:

$t_{99}=\frac{2.303}{k}\log \frac{[\text{R}{]}_0}{0.01[\text{R}{]}_0}=\frac{2.303}{k}\log 100=\frac{2.303}{k}\times 2=\frac{4.606}{k}$

For 90% completion, 10% remains, so $[\text{R}]=0.1[\text{R}{]}_0$:

$t_{90}=\frac{2.303}{k}\log \frac{[\text{R}{]}_0}{0.1[\text{R}{]}_0}=\frac{2.303}{k}\log 10=\frac{2.303}{k}\times 1=\frac{2.303}{k}$

Taking the ratio, in which both $k$ and 2.303 cancel:

$\frac{t_{99}}{t_{90}}=\frac{4.606/k}{2.303/k}=2$

Hence $t_{99}=2t_{90}$, as required.

The result holds for every first order reaction whatever its rate constant, because $k$ cancels. The reason is that each successive factor of ten in the fraction remaining costs the same amount of time, so going from 10% left to 1% left takes exactly as long as going from 100% to 10%.

✦ $t_{99}=4.606/k$ and $t_{90}=2.303/k$, so $t_{99}=2t_{90}$ for any first order reaction

If 30% has decomposed then 70% remains, so $[\text{R}]=0.70[\text{R}{]}_0$.

$k=\frac{2.303}{t}\log \frac{[\text{R}{]}_0}{[\text{R}]}=\frac{2.303}{40}\log \frac{100}{70}$

$k=0.05758\times \log (1.4286)=0.05758\times 0.1549$

$k=8.92\times 10^{-3}{\text{min}}^{-1}$

Then

$t_{1/2}=\frac{0.693}{k}=\frac{0.693}{8.92\times 10^{-3}}=77.7\text{min}$

A quick check: 30% decomposition is less than half, so it must take less than one half-life, and 40 min is indeed less than 77.7 min.

The commonest slip here is putting 30 into the logarithm instead of 70. The integrated rate law always works with the amount remaining, never the amount consumed.

✦ $k=8.92\times 10^{-3}$ min${}^{-1}$ and $t_{1/2}=77.7$ min

The decomposition is

$({\text{CH}}_3{)}_2\text{CHN}=\text{NCH}({\text{CH}}_3{)}_2(\text{g})\to {\text{N}}_2(\text{g})+{\text{C}}_6{\text{H}}_{14}(\text{g})$

One mole of gas becomes two, so the total pressure rises as the reaction proceeds. The pressure quoted at $t=0$ is that of the pure reactant, $p_0=35.0$ mm Hg.

Let $x$ be the pressure of azoisopropane that has decomposed at time $t$. Then the partial pressures are $(p_0-x)$, $x$ and $x$, so

$P_{\text{total}}=(p_0-x)+x+x=p_0+x$

$x=P_{\text{total}}-p_0$

and the pressure of reactant remaining is

$p_{\text{A}}=p_0-x=2p_0-P_{\text{total}}=70.0-P_{\text{total}}$

Building the table:

Now apply the first order integrated rate law, using pressures in place of concentrations since they are proportional at constant volume and temperature:

$k=\frac{2.303}{t}\log \frac{p_0}{p_{\text{A}}}$

At $t=360$ s:

$k=\frac{2.303}{360}\log \frac{35.0}{16.0}=6.397\times 10^{-3}\times 0.3401=2.18\times 10^{-3}{\text{s}}^{-1}$

At $t=720$ s:

$k=\frac{2.303}{720}\log \frac{35.0}{7.0}=3.199\times 10^{-3}\times 0.6990=2.24\times 10^{-3}{\text{s}}^{-1}$

The two values agree closely, which confirms that the reaction is first order.

$k_{\text{av}}=2.21\times 10^{-3}{\text{s}}^{-1}$

The whole difficulty is the relation $p_{\text{A}}=2p_0-P_{\text{total}}$. Using the total pressure directly in the logarithm gives an answer that is not even the right order of magnitude.

✦ $k\approx 2.2\times 10^{-3}$ s${}^{-1}$, from $2.18\times 10^{-3}$ at 360 s and $2.24\times 10^{-3}$ at 720 s

One molecule of gas becomes two, so the total pressure rises. With $p_0=0.5$ atm and $x$ the pressure decomposed,

$P_{\text{total}}=(p_0-x)+x+x=p_0+x\Rightarrow x=P_{\text{total}}-p_0$

$p_{{\text{SO}}_2{\text{Cl}}_2}=p_0-x=2p_0-P_{\text{total}}=1.0-P_{\text{total}}$

Step 1: find $k$. At $t=100$ s the total pressure is 0.6 atm, so

$p_{{\text{SO}}_2{\text{Cl}}_2}=1.0-0.6=0.4\text{atm}$

$k=\frac{2.303}{100}\log \frac{0.5}{0.4}=0.02303\times 0.09691=2.23\times 10^{-3}{\text{s}}^{-1}$

Step 2: find the reactant pressure at $P_{\text{total}}=0.65$ atm.

$p_{{\text{SO}}_2{\text{Cl}}_2}=1.0-0.65=0.35\text{atm}$

Step 3: the rate. For a first order reaction the rate is the rate constant times the reactant pressure:

$\text{Rate}=k{p}_{{\text{SO}}_2{\text{Cl}}_2}=2.23\times 10^{-3}\times 0.35$

$\text{Rate}=7.8\times 10^{-4}{\text{atm s}}^{-1}$

The question is deliberately built in two halves: the data fix $k$, and only then can the rate at a third, different pressure be found. A rate and a rate constant are not the same thing, and the difference is exactly the factor $p$.

A printed slip to note: the table header reads "Time/s${}^{-1}$", but time is measured in seconds, so it should read "Time/s".

✦ $k=2.23\times 10^{-3}$ s${}^{-1}$ and the rate at 0.65 atm total pressure is $7.8\times 10^{-4}$ atm s${}^{-1}$

Preparing the data. Convert temperatures to kelvin and strip the factor of $10^5$ from $k$:

The graph. Plotting $\ln k$ against $1/T$ gives a good straight line of negative slope, as the Arrhenius equation requires:

$\ln k=\ln A-\frac{E_a}{R}\cdot \frac{1}{T}$

Activation energy from the slope. A least-squares line through all five points has slope $-1.210\times 10^4$ K, so

$E_a=-\text{slope}\times R=1.210\times 10^4\times 8.314=1.006\times 10^5{\text{J mol}}^{-1}$

$E_a\approx 100.6{\text{kJ mol}}^{-1}$

Taking only the two end points instead gives $E_a=102.3$ kJ mol${}^{-1}$, so a hand-drawn graph should be expected to land somewhere near 100 to 102 kJ mol${}^{-1}$.

Pre-exponential factor from the intercept. The least-squares intercept is $\ln A=30.29$, so

$A=e^{30.29}=1.4\times 10^{13}{\text{s}}^{-1}$

using the end points instead gives $2.9\times 10^{13}$ s${}^{-1}$. The intercept is a long extrapolation to $1/T=0$, so it is far more sensitive to how the line is drawn than the slope is, and an answer anywhere in the range $10^{13}$ to $10^{14}$ s${}^{-1}$ is reasonable.

Predicting $k$ at 30 °C and 50 °C. Substituting $T=303$ K and $T=323$ K into the fitted line:

$k(303\text{K})=6.4\times 10^{-5}{\text{s}}^{-1}$

$k(323\text{K})=7.6\times 10^{-4}{\text{s}}^{-1}$

Both fall neatly between the tabulated neighbours, which is the reassuring sign that the line has been fitted sensibly. Note how steeply the rate constant climbs: over the 80 degrees of the table it rises by a factor of about 27,000, which is what an activation energy of 100 kJ mol${}^{-1}$ does.

✦ $E_a\approx 100$ to 102 kJ mol${}^{-1}$ and $A\approx 1.4\times 10^{13}$ s${}^{-1}$; the predicted rate constants are about $6.4\times 10^{-5}$ s${}^{-1}$ at 30 °C and $7.6\times 10^{-4}$ s${}^{-1}$ at 50 °C

The Arrhenius equation is $k=A{e}^{-E_a/RT}$, so

$A=k{e}^{E_a/RT}$

Working in logarithms to base ten:

$\log A=\log k+\frac{E_a}{2.303RT}$

$\log A=\log (2.418\times 10^{-5})+\frac{179900}{2.303\times 8.314\times 546}$

$\log A=-4.6162+\frac{179900}{10454.5}=-4.6162+17.2085$

$\log A=12.592$

$A=3.9\times 10^{12}{\text{s}}^{-1}$

The units of $A$ are always the same as those of $k$, since the exponential factor is dimensionless. Here both are s${}^{-1}$, so the reaction is first order.

The size of $A$ is worth noting. It is roughly the collision frequency, so molecules are colliding some $10^{12}$ times a second, yet the rate constant is only $2.4\times 10^{-5}$ s${}^{-1}$. The whole of that gap, seventeen powers of ten, is the exponential factor $e^{-E_a/RT}$.

✦ $A=3.9\times 10^{12}$ s${}^{-1}$

The rate constant is quoted in s${}^{-1}$, and from Table 3.3 those are the units of a first order rate constant. So the integrated law to use is

$[\text{A}]=[\text{A}{]}_0{e}^{-kt}$

or equivalently

$\log \frac{[\text{A}{]}_0}{[\text{A}]}=\frac{kt}{2.303}$

Substituting:

$\log \frac{1.0}{[\text{A}]}=\frac{2.0\times 10^{-2}\times 100}{2.303}=\frac{2.0}{2.303}=0.8685$

$\frac{1.0}{[\text{A}]}=10^{0.8685}=7.39$

$[\text{A}]=0.135{\text{mol L}}^{-1}$

A check: $kt=2.0$, and $e^{-2}=0.135$, which is the same result reached directly.

Reading the order off the units is a habit worth building. Nothing in the wording of this question says "first order", and the s${}^{-1}$ is the only clue given.

✦ $[\text{A}]=0.135$ mol L${}^{-1}$

From the half-life,

$k=\frac{0.693}{t_{1/2}}=\frac{0.693}{3.00}=0.231{\text{h}}^{-1}$

The integrated first order rate law gives the fraction remaining directly:

$\log \frac{[\text{R}{]}_0}{[\text{R}]}=\frac{kt}{2.303}=\frac{0.231\times 8}{2.303}=\frac{1.848}{2.303}=0.8025$

$\frac{[\text{R}{]}_0}{[\text{R}]}=10^{0.8025}=6.35$

$\frac{[\text{R}]}{[\text{R}{]}_0}=\frac{1}{6.35}=0.158$

So about 0.158, or 15.8%, of the sucrose remains after 8 hours.

As a check, 8 hours is between two and three half-lives, which would leave between 25% and 12.5%; 15.8% sits properly between them.

The reaction is the inversion of cane sugar mentioned in section 3.4. It is really second order, since water is a reactant, but water is present in such vast excess that its concentration never changes appreciably, making the reaction pseudo first order.

✦ About 0.158, or 15.8%, of the sucrose remains

Compare the given expression with the Arrhenius equation:

$k=A{e}^{-E_a/RT}\text{against}k=(4.5\times 10^{11}{\text{s}}^{-1})e^{-28000\text{K}/T}$

Matching the exponents term by term:

$\frac{E_a}{RT}=\frac{28000\text{K}}{T}\Rightarrow \frac{E_a}{R}=28000\text{K}$

Therefore

$E_a=28000\times R=28000\times 8.314$

$E_a=232792{\text{J mol}}^{-1}=232.8{\text{kJ mol}}^{-1}$

The unit K attached to the 28000 in the printed expression is the clue that makes the comparison work: it is there precisely so that dividing by $T$ in kelvin leaves a dimensionless exponent. Reading it off also gives $A=4.5\times 10^{11}$ s${}^{-1}$ at no extra cost.

✦ $E_a=232.8$ kJ mol${}^{-1}$

Part 1: the activation energy. Taking logarithms to base ten of the Arrhenius equation:

$\log k=\log A-\frac{E_a}{2.303RT}$

Comparing with the given $\log k=14.34-\frac{1.25\times 10^4\text{K}}{T}$:

$\frac{E_a}{2.303R}=1.25\times 10^4\text{K}$

$E_a=1.25\times 10^4\times 2.303\times 8.314=239339{\text{J mol}}^{-1}$

$E_a=239.3{\text{kJ mol}}^{-1}$

The same comparison gives $\log A=14.34$, so $A=2.2\times 10^{14}$ s${}^{-1}$.

Part 2: the temperature at which $t_{1/2}=256$ min. First convert the half-life to the rate constant:

$t_{1/2}=256\times 60=15360\text{s}$

$k=\frac{0.693}{15360}=4.512\times 10^{-5}{\text{s}}^{-1}$

$\log k=\log (4.512\times 10^{-5})=-4.346$

Now substitute into the given equation and solve for $T$:

$-4.346=14.34-\frac{1.25\times 10^4}{T}$

$\frac{1.25\times 10^4}{T}=14.34+4.346=18.686$

$T=\frac{1.25\times 10^4}{18.686}=669\text{K}$

The question is really two questions bolted together, and the link between them is the pair of conversions $t_{1/2}\to k\to \log k$. Attempting to put the half-life into the equation directly is the usual way to go wrong.

✦ $E_a=239.3$ kJ mol${}^{-1}$, and the half-life is 256 minutes at $T=669$ K

Convert the temperature: $T_1=10+273=283$ K, with $k_1=4.5\times 10^3$ s${}^{-1}$ and $k_2=1.5\times 10^4$ s${}^{-1}$.

The two-temperature form of the Arrhenius equation:

$\log \frac{k_2}{k_1}=\frac{E_a}{2.303R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right]$

$\log \frac{1.5\times 10^4}{4.5\times 10^3}=\log (3.333)=0.5229$

$\frac{E_a}{2.303R}=\frac{60000}{2.303\times 8.314}=3133.6\text{K}$

Substituting:

$0.5229=3133.6\left[\frac{1}{283}-\frac{1}{T_2}\right]$

$\frac{1}{283}-\frac{1}{T_2}=1.669\times 10^{-4}$

$\frac{1}{T_2}=3.5336\times 10^{-3}-1.669\times 10^{-4}=3.3667\times 10^{-3}$

$T_2=297\text{K}=24{}^{\circ }\text{C}$

A rise of only 14 K triples the rate constant, which is consistent with the rule of thumb that 10 K roughly doubles it for an activation energy of this size.

The arithmetic trap is the final step: having found $1/T_2$, the answer still needs one reciprocal. Stopping at $3.3667\times 10^{-3}$ is easier to do than it sounds.

✦ $T_2=297$ K, that is 24 °C

Step 1: find the ratio of rate constants. For a first order reaction $kt=2.303\log \frac{[\text{R}{]}_0}{[\text{R}]}$.

At 298 K, 10% is complete so 90% remains:

$k_{1}t=2.303\log \frac{100}{90}=2.303\times 0.04576=0.1054$

At 308 K, 25% is complete so 75% remains:

$k_{2}t=2.303\log \frac{100}{75}=2.303\times 0.12494=0.2877$

The times are stated to be equal, so dividing eliminates $t$ entirely:

$\frac{k_2}{k_1}=\frac{0.2877}{0.1054}=2.73$

Step 2: find $E_a$.

$\log (2.73)=\frac{E_a}{2.303\times 8.314}\left[\frac{308-298}{298\times 308}\right]$

$0.4363=\frac{E_a}{19.147}\times \frac{10}{91784}$

$E_a=\frac{0.4363\times 19.147\times 91784}{10}=76664{\text{J mol}}^{-1}=76.7{\text{kJ mol}}^{-1}$

Step 3: find $k$ at 318 K, using the Arrhenius equation with the given $A$:

$\log k=\log A-\frac{E_a}{2.303RT}=\log (4\times 10^{10})-\frac{76664}{2.303\times 8.314\times 318}$

$\log k=10.602-\frac{76664}{6088.7}=10.602-12.591=-1.989$

$k=1.02\times 10^{-2}{\text{s}}^{-1}$

The elegant part of this problem is step 1: the time never has to be known, because it is the same on both sides and cancels when the ratio is taken. Trying to find $t$ first leads nowhere, since no rate constant is given to start from.