Chapter 3: Chemical Kinetics

Qualitative effect. When the temperature is raised, a larger fraction of the molecules acquires kinetic energy equal to or greater than the activation energy $$E_a$$. The frequency of effective collisions therefore increases and the rate constant $$k$$ rises. As a rule of thumb, for most chemical reactions $$k$$ becomes nearly two to three times its original value for every $$10\,\mathrm{K}$$ rise in temperature.

Quantitative representation – the Arrhenius equation. The temperature dependence of $$k$$ was expressed empirically by Svante Arrhenius as

$$k = A\,\exp\!\left(-\dfrac{E_a}{RT}\right)$$

where

• $$k$$ = rate constant of the reaction,
• $$A$$ = frequency (or pre-exponential) factor, characteristic of the reaction,
• $$E_a$$ = activation energy,
• $$R = 8.314\;\mathrm{J\,mol^{-1}\,K^{-1}}$$ (universal gas constant),
• $$T$$ = absolute temperature in kelvin.

Taking natural logarithms gives the linear form

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

or, using common logarithms,

$$\log k = \log A - \dfrac{E_a}{2.303\,R}\cdot\dfrac{1}{T}$$

A plot of $$\ln k$$ (or $$\log k$$) versus $$1/T$$ is therefore a straight line with slope $$-E_a/R$$ (or $$-E_a/(2.303\,R)$$) and intercept $$\ln A$$ (or $$\log A$$). From the slope the activation energy can be evaluated experimentally.

Alternatively, if the rate constants $$k_1$$ and $$k_2$$ are known at two different temperatures $$T_1$$ and $$T_2$$, $$E_a$$ can be obtained from

$$\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303\,R}\,\dfrac{T_2-T_1}{T_1\,T_2}.$$

The rate of a reaction is defined as

$$\text{rate}= -\frac{\Delta [\mathrm{A}]}{\Delta t}$$

where the negative sign compensates for the decrease in concentration of the reactant.

For the given time interval 30 s to 60 s:

• Initial concentration (at 30 s): $$[\mathrm{A}]_1 = 0.31\,\mathrm{mol\,L^{-1}}$$
• Final concentration (at 60 s): $$[\mathrm{A}]_2 = 0.17\,\mathrm{mol\,L^{-1}}$$
• Time interval: $$\Delta t = 60\,\mathrm{s} - 30\,\mathrm{s} = 30\,\mathrm{s}$$

Change in concentration:

$$\Delta [\mathrm{A}] = [\mathrm{A}]_2 - [\mathrm{A}]_1 = 0.17 - 0.31 = -0.14\,\mathrm{mol\,L^{-1}}$$

Substituting in the rate expression:

$$\text{rate}= -\frac{-0.14\,\mathrm{mol\,L^{-1}}}{30\,\mathrm{s}} = \frac{0.14}{30}\,\mathrm{mol\,L^{-1}\,s^{-1}}$$

$$\text{rate}= 0.00467\,\mathrm{mol\,L^{-1}\,s^{-1}} \;\approx\; 4.7\times10^{-3}\,\mathrm{mol\,L^{-1}\,s^{-1}}$$

Let the rate law be $$r_0 = k[\mathrm{A}]^{m}[\mathrm{B}]^{n}$$, where $$m$$ and $$n$$ are the orders with respect to $$\mathrm{A}$$ and $$\mathrm{B}$$, and $$k$$ is the rate constant.

The experimental data are:

1. Order with respect to $$\mathrm{B}$$ (compare Exps. 1 and 2, $$[\mathrm{A}]$$ constant):

$$\frac{r_1}{r_2}=\left(\frac{[\mathrm{B}]_1}{[\mathrm{B}]_2}\right)^{n}$$

$$\frac{5.07\times10^{-5}}{5.07\times10^{-5}}=\left(\frac{0.30}{0.10}\right)^{n}\;\Rightarrow\;1=3^{n}\;\Rightarrow\;n=0$$

Hence the reaction is zero order in $$\mathrm{B}$$.

2. Order with respect to $$\mathrm{A}$$ (compare Exps. 2 and 3, using $$n=0$$):

With $$n=0$$ the rate law simplifies to $$r_0=k[\mathrm{A}]^{m}$$, so

$$\frac{r_3}{r_2}=\left(\frac{[\mathrm{A}]_3}{[\mathrm{A}]_2}\right)^{m}$$

$$\frac{1.43\times10^{-4}}{5.07\times10^{-5}}=\left(\frac{0.40}{0.20}\right)^{m}$$

$$2.82=2^{m}\;\Rightarrow\;m=\frac{\log(2.82)}{\log 2}=1.5$$

Thus the reaction is $$\tfrac{3}{2}$$ order in $$\mathrm{A}$$.

3. Overall order

Overall order $$=m+n=1.5+0=1.5$$.

Result: Order w.r.t. $$\mathrm{A}$$ is $$\tfrac{3}{2}$$; order w.r.t. $$\mathrm{B}$$ is 0.

Let the rate law be written in the general form

$$r = k [\mathrm{A}]^{m}[\mathrm{B}]^{n}$$

where $$m$$ and $$n$$ are the orders with respect to $$\mathrm{A}$$ and $$\mathrm{B}$$, respectively, and $$k$$ is the rate constant.

Step 1 – Determine the order with respect to $$\mathrm{B}$$

Choose two experiments in which $$[\mathrm{A}]$$ is the same (II and III):

Form the ratio

$$\frac{r_{\mathrm{III}}}{r_{\mathrm{II}}}=\frac{k(0.3)^{m}(0.4)^{n}}{k(0.3)^{m}(0.2)^{n}}=\left(\frac{0.4}{0.2}\right)^{n}=2^{n}$$

Numerically,

$$\frac{2.88\times10^{-1}}{7.2\times10^{-2}}=4.0=2^{n}\;\Rightarrow\;n=2$$

Step 2 – Determine the order with respect to $$\mathrm{A}$$

Choose two experiments in which $$[\mathrm{B}]$$ is the same (I and IV):

Form the ratio

$$\frac{r_{\mathrm{IV}}}{r_{\mathrm{I}}}=\frac{k(0.4)^{m}(0.1)^{2}}{k(0.1)^{m}(0.1)^{2}}=\left(\frac{0.4}{0.1}\right)^{m}=4^{m}$$

Numerically,

$$\frac{2.40\times10^{-2}}{6.0\times10^{-3}}=4.0=4^{m}\;\Rightarrow\;m=1$$

Step 3 – Write the rate law

Therefore,

$$r = k[\mathrm{A}]^{1}[\mathrm{B}]^{2}$$

Step 4 – Evaluate the rate constant $$k$$

Use experiment I (any set gives the same result):

$$6.0\times10^{-3}=k(0.1)(0.1)^{2}=k(0.1)(0.01)=k(1.0\times10^{-3})$$

$$k=\frac{6.0\times10^{-3}}{1.0\times10^{-3}}=6.0$$

Verify with the other three experiments:

• Experiment II: $$k=\frac{7.2\times10^{-2}}{0.3\times0.2^{2}}=\frac{7.2\times10^{-2}}{0.012}=6.0$$
• Experiment III: $$k=\frac{2.88\times10^{-1}}{0.3\times0.4^{2}}=\frac{2.88\times10^{-1}}{0.048}=6.0$$
• Experiment IV: $$k=\frac{2.40\times10^{-2}}{0.4\times0.1^{2}}=\frac{2.40\times10^{-2}}{0.004}=6.0$$

All give the same value, confirming consistency.

Step 5 – Units of $$k$$

The overall order is $$m+n=1+2=3$$. Hence,

$$[k]=\frac{\mathrm{mol\,L^{-1}\,min^{-1}}}{(\mathrm{mol\,L^{-1}})^{3}}=\mathrm{L^{2}\,mol^{-2}\,min^{-1}}$$

Final result

Rate law: $$r = 6.0\,\mathrm{L^{2}\,mol^{-2}\,min^{-1}}\,[\mathrm{A}][\mathrm{B}]^{2}$$

Rate law for the reaction
Because the reaction is first order with respect to $$\mathrm{A}$$ and zero order with respect to $$\mathrm{B}$$, the differential rate law is

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

Thus the initial rate depends only on $$[\mathrm{A}]$$; the concentration of $$\mathrm{B}$$ is irrelevant.

Determination of the rate constant k

Using Experiment I, where the concentrations and rate are completely known:

$$k=\dfrac{\text{Rate}}{[\mathrm{A}]}=\dfrac{2.0\times10^{-2}\;\mathrm{mol\,L^{-1}\,min^{-1}}}{0.1\;\mathrm{mol\,L^{-1}}}=0.20\;\mathrm{min^{-1}}.$$

Filling the blanks

1. Experiment II — missing [A]
     $$[\mathrm{A}]_{\text{II}}=\dfrac{\text{Rate}_{\text{II}}}{k}=\dfrac{4.0\times10^{-2}}{0.20}=0.20\;\mathrm{mol\,L^{-1}}.$$
2. Experiment III — missing rate
     $$\text{Rate}_{\text{III}}=k\,[\mathrm{A}]_{\text{III}}=0.20\times0.40=0.080\;\mathrm{mol\,L^{-1}\,min^{-1}}=8.0\times10^{-2}\;\mathrm{mol\,L^{-1}\,min^{-1}}.$$
3. Experiment IV — missing [A]
     $$[\mathrm{A}]_{\text{IV}}=\dfrac{\text{Rate}_{\text{IV}}}{k}=\dfrac{2.0\times10^{-2}}{0.20}=0.10\;\mathrm{mol\,L^{-1}}.$$

Completed table

The blanks have been filled consistently with a first-order dependence on $$\mathrm{A}$$ and zero-order dependence on $$\mathrm{B}$$.

For radioactive decay the rate law is first order.

Let

• $$N_0$$ = number of $$\mathrm{^{14}C}$$ nuclei present in the living wood,
• $$N$$ = number remaining in the artefact.

Given $$\dfrac{N}{N_0}=0.80$$.

The first-order integrated rate law is

$$kt = \ln\!\left(\dfrac{N_0}{N}\right).$$

Step 1 — rate constant. Half-life $$t_{1/2}=5730\,\text{years}$$, and for a first-order process

$$k = \dfrac{0.693}{t_{1/2}} = \dfrac{0.693}{5730\,\text{year}} = 1.21\times10^{-4}\,\text{year}^{-1}.$$

Step 2 — substitute in the rate equation.

$$t = \dfrac{1}{k}\,\ln\!\left(\dfrac{N_0}{N}\right) = \dfrac{1}{1.21\times10^{-4}\,\text{year}^{-1}}\,\ln\!\left(\dfrac{1}{0.80}\right).$$

Evaluate the logarithm:

$$\ln\!\left(\dfrac{1}{0.80}\right)=\ln(1.25)=0.2231.$$

Therefore

$$t = \dfrac{0.2231}{1.21\times10^{-4}}\,\text{year} = 1.845\times10^{3}\,\text{year} \approx 1.85\times10^{3}\,\text{year}.$$

Step 3 — result. The wood in the archaeological sample is approximately $$1.85\times10^{3}$$ years old, i.e. about 1850 years.

The integrated rate law for a first-order reaction is

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

Given data:

• Rate constant, $$k = 60\,\mathrm{s^{-1}}$$
• Final concentration, $$[R]=\dfrac{[R]_0}{16}$$

Hence the concentration ratio is

$$\frac{[R]_0}{[R]} = \frac{[R]_0}{[R]_0/16} = 16$$

Substituting these values,

$$t = \frac{2.303}{k}\,\log 16$$

Since $$\log 16 = 1.2041$$,

$$t = \frac{2.303}{60}\,(1.2041)$$

$$t = 0.0463\,\mathrm{s}$$

Therefore, the time required for the reactant concentration to fall to one-sixteenth of its original value is approximately $$4.6\times10^{-2}\,\mathrm{s}$$.

Given data

• Half-life of $$\mathrm{^{90}Sr}$$ : $$t_{1/2}=28.1\,\text{years}$$
• Initial mass incorporated in bones : $$m_0 = 1\,\mu\text{g}$$

For a first-order nuclear decay the amount left after time $$t$$ is

$$m = m_0\,e^{-\lambda t}$$   with   $$\lambda = \dfrac{0.693}{t_{1/2}}$$.

1. Decay constant

$$\lambda = \frac{0.693}{28.1\,\text{y}} = 2.47\times10^{-2}\,\text{y}^{-1}$$ (to three significant figures)

2. Mass left after 10 years

Time $$t = 10\,\text{y}$$

Exponent: $$\lambda t = (2.47\times10^{-2})(10) = 0.247$$

$$m_{10} = 1\,\mu\text{g}\times e^{-0.247} = 1\,\mu\text{g}\times 0.781 = 0.781\,\mu\text{g}$$

3. Mass left after 60 years

Time $$t = 60\,\text{y}$$

Exponent: $$\lambda t = (2.47\times10^{-2})(60) = 1.48$$

$$m_{60} = 1\,\mu\text{g}\times e^{-1.48} = 1\,\mu\text{g}\times 0.227 = 0.227\,\mu\text{g}$$

Result

Assuming no metabolic loss:

• After 10 years: $$\boxed{\;0.78\,\mu\text{g}\;}$$ of $$\mathrm{^{90}Sr}$$ will remain.
• After 60 years: $$\boxed{\;0.23\,\mu\text{g}\;}$$ of $$\mathrm{^{90}Sr}$$ will remain.

Integrated rate law for a first order reaction

The integrated form (using common logarithm) is

$$k t = 2.303\,\log \frac{a}{a-x}$$

where

• $$a$$ = initial concentration (or amount)
• $$x$$ = amount reacted after time $$t$$
• $$a-x$$ = concentration left

1. Time for 90 % completion  ($$t_{90\%}$$)

90 % of the reactant has reacted, so 10 % remains:

$$a-x = 0.10\,a$$

Substitute in the rate law:

$$k\,t_{90\%} = 2.303\,\log\frac{a}{0.10\,a}$$

$$= 2.303\,\log 10$$

But $$\log 10 = 1$$, therefore

$$t_{90\%} = \frac{2.303}{k}$$

2. Time for 99 % completion  ($$t_{99\%}$$)

99 % of the reactant has reacted, so 1 % remains:

$$a-x = 0.01\,a$$

Substitute in the rate law:

$$k\,t_{99\%} = 2.303\,\log\frac{a}{0.01\,a}$$

$$= 2.303\,\log 100$$

Since $$\log 100 = 2$$,

$$t_{99\%} = \frac{2\,(2.303)}{k} = \frac{4.606}{k}$$

3. Ratio of the two times

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

Hence

$$t_{99\%} = 2\,t_{90\%}$$

Therefore, for a first-order reaction, the time required for 99 % completion is exactly twice the time required for 90 % completion.

For a first-order reaction the rate constant $$k$$ is obtained from

$$k = \frac{2.303}{t}\,\log\left(\frac{[R]_0}{[R]}\right)$$

Let the initial concentration be $$[R]_0 = a$$. After 40 min, 30 % has decomposed, so 70 % remains:

$$[R] = 0.70a$$

Substituting $$t = 40\,\text{min}$$,

$$k = \frac{2.303}{40}\,\log\left(\frac{a}{0.70a}\right) = \frac{2.303}{40}\,\log\left(\frac{1}{0.70}\right)$$

Numerically, $$\log(1/0.70) = 0.1549$$, hence

$$k = \frac{2.303\,(0.1549)}{40} = 8.92 \times 10^{-3}\;\text{min}^{-1}$$

The half-life of a first-order reaction is

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

Therefore,

$$t_{1/2} = \frac{0.693}{8.92 \times 10^{-3}} = 7.77 \times 10^{1}\;\text{min} \approx 78\;\text{min}$$

Thus, the half-life of the reaction is about 78 minutes.

Step 1 – Write the reaction
$$\mathrm{(CH_3)_2CH\!{-}N\!=\!N\!{-}CH(CH_3)_2}\;\longrightarrow\;\mathrm{C_6H_{14}}+\mathrm{N_2}$$
One mole of reactant gives two moles of products, hence the total number of moles (and therefore the pressure) increases by one mole for each mole decomposed.

Step 2 – Relate the observed pressure to the extent of decomposition
Let the initial pressure of azo-isopropane be $$P_0$$.
At time $$t$$, let $$x$$ mm Hg of the reactant have decomposed.
Remaining reactant: $$(P_0-x)$$
Products formed: $$2x$$
Total pressure at time $$t$$:
$$P_t=(P_0-x)+2x=P_0+x$$
Therefore $$x=P_t-P_0$$ and the partial pressure of the undecomposed reactant is
$$P_A=P_0-x=2P_0-P_t$$.

Step 3 – First-order rate law in terms of pressure
For a first-order reaction
$$k=\frac{2.303}{t}\log\frac{[A]_0}{[A]}=\frac{2.303}{t}\log\frac{P_0}{P_A}=\frac{2.303}{t}\log\frac{P_0}{2P_0-P_t}$$

Step 4 – Calculate $$k$$ from each set of data

Step 5 – Final rate constant
The two values are very close; their average is
$$k\approx2.2\times10^{-3}\;\mathrm{s^{-1}}$$ at 543 K.

Step 1 : Express the change in pressure in terms of extent of decomposition
At t = 0 the vessel contains only $$\mathrm{SO_2Cl_2}$$ having pressure $$P_0 = 0.5\,\text{atm}$$.
After time t, let the pressure corresponding to the amount of $$\mathrm{SO_2Cl_2}$$ that has decomposed be x (in atm).

Step 2 : Evaluate the first-order rate constant k
For a first-order reaction
$$k = \frac{2.303}{t}\,\log\!\left(\frac{[A]_0}{[A]_t}\right)$$
Replacing concentrations by the corresponding partial pressures,
$$k = \frac{2.303}{t}\,\log\!\left(\frac{P_0}{P_{\mathrm{SO_2Cl_2}}}\right)$$.

From the experimental entry (t = 100 s, $$P_t = 0.6\,\text{atm}$$):

• $$x = 0.6 - 0.5 = 0.10\,\text{atm}$$
• $$P_{\mathrm{SO_2Cl_2}} = 0.5 - 0.10 = 0.40\,\text{atm}$$

Step 3 : Find the rate when the total pressure is 0.65 atm
For $$P_t = 0.65\,\text{atm}$$:

• $$x = 0.65 - 0.50 = 0.15\,\text{atm}$$
• $$P_{\mathrm{SO_2Cl_2}} = 0.50 - 0.15 = 0.35\,\text{atm}$$

Thus the rate of decomposition when the total pressure is 0.65 atm is $$7.8\times10^{-4}\;\text{atm s}^{-1}$$.

Step 1 — Convert the data into convenient form. Convert temperatures to kelvin and divide the tabulated values of $$10^{5}k$$ by $$10^{5}$$ to obtain $$k$$ in $$\mathrm{s^{-1}}$$, and then evaluate $$\ln k$$ and $$1/T$$:

Step 2 — Arrhenius plot. Plotting $$\ln k$$ (on the $$y$$-axis) against $$1/T$$ (on the $$x$$-axis) gives an essentially straight line that slopes downwards.

Step 3 — Slope and activation energy. Taking the two extreme points,

$$\text{slope} = \dfrac{\ln k_2-\ln k_1}{(1/T_2)-(1/T_1)} = \dfrac{-3.843-(-14.05)}{(2.833-3.663)\times10^{-3}} = \dfrac{10.21}{-0.830\times10^{-3}} = -1.23\times10^{4}\,\mathrm{K}.$$

The Arrhenius equation in linear form is $$\ln k = \ln A - \dfrac{E_a}{R}\,\dfrac{1}{T}$$, so $$\text{slope}=-E_a/R$$ and

$$E_a = -\text{slope}\times R = (1.23\times10^{4}\,\mathrm{K})(8.314\,\mathrm{J\,K^{-1}\,mol^{-1}}) \approx 1.02\times10^{5}\,\mathrm{J\,mol^{-1}} = 1.0\times10^{2}\,\mathrm{kJ\,mol^{-1}}.$$

Step 4 — Arrhenius factor $$A$$. Using the data at $$T = 313\,\mathrm{K}$$ (any point will do):

$$\ln A = \ln k + \dfrac{E_a}{RT} = -8.266 + \dfrac{1.02\times10^{5}}{8.314\times313} = -8.266 + 39.33 = 31.07.$$

Therefore $$A = e^{31.07} \approx 3.3\times10^{13}\,\mathrm{s^{-1}}.$$

Step 5 — Predicted rate constants. Substituting into $$\ln k = 31.07 - \dfrac{1.02\times10^{5}}{8.314\,T}$$:

• At $$30^{\circ}\mathrm{C}$$ ($$T = 303\,\mathrm{K}$$): $$\ln k = 31.07 - 40.62 = -9.55$$, so $$k \approx 7.1\times10^{-5}\,\mathrm{s^{-1}}.$$
• At $$50^{\circ}\mathrm{C}$$ ($$T = 323\,\mathrm{K}$$): $$\ln k = 31.07 - 38.10 = -7.03$$, so $$k \approx 8.9\times10^{-4}\,\mathrm{s^{-1}}.$$

Final results:

• Activation energy $$E_a \approx 1.0\times10^{2}\,\mathrm{kJ\,mol^{-1}}$$
• Arrhenius factor $$A \approx 3.3\times10^{13}\,\mathrm{s^{-1}}$$
• $$k_{30^{\circ}\mathrm{C}} \approx 7.1\times10^{-5}\,\mathrm{s^{-1}}$$ and $$k_{50^{\circ}\mathrm{C}} \approx 8.9\times10^{-4}\,\mathrm{s^{-1}}$$

Step 1 : Write the Arrhenius equation
$$k = A \, e^{-E_a/(RT)}$$

Step 2 : Isolate the pre-exponential factor
$$A = k\,e^{E_a/(RT)}$$

Step 3 : Insert the given data

• Rate constant : $$k = 2.418 \times 10^{-5}\,\mathrm{s^{-1}}$$
• Temperature : $$T = 546\,\mathrm{K}$$
• Activation energy : $$E_a = 179.9\,\mathrm{kJ\,mol^{-1}} = 179.9 \times 10^{3}\,\mathrm{J\,mol^{-1}}$$
• Gas constant : $$R = 8.314\,\mathrm{J\,mol^{-1}\,K^{-1}}$$

Step 4 : Evaluate the exponent
$$RT = 8.314\,\mathrm{J\,mol^{-1}\,K^{-1}} \times 546\,\mathrm{K} = 4539.4\,\mathrm{J\,mol^{-1}}$$
$$\frac{E_a}{RT} = \frac{179.9 \times 10^{3}}{4539.4} = 39.64$$

Step 5 : Calculate the exponential term
$$e^{39.64} = 10^{\,\frac{39.64}{2.302585}} = 10^{17.22} \approx 1.66 \times 10^{17}$$

Step 6 : Determine the pre-exponential factor
$$A = 2.418 \times 10^{-5}\,\mathrm{s^{-1}} \times 1.66 \times 10^{17}$$
$$A \approx 4.0 \times 10^{12}\,\mathrm{s^{-1}}$$

The required pre-exponential (frequency) factor is therefore about $$4.0 \times 10^{12}\,\mathrm{s^{-1}}$$.

The rate constant shows units $$\mathrm{s^{-1}}$$, so the reaction is first order in $$\mathrm{A}$$.

For a first-order reaction, the integrated rate law is

$$\ln \frac{[\mathrm{A}]_0}{[\mathrm{A}]_t}=kt$$

where

• $$[\mathrm{A}]_0=1.0\,\mathrm{mol\,L^{-1}}$$   (initial concentration)
• $$k=2.0\times10^{-2}\,\mathrm{s^{-1}}$$   (rate constant)
• $$t=100\,\mathrm{s}$$   (time elapsed)
• $$[\mathrm{A}]_t$$   (concentration after time $$t$$)

Substitute the data:

$$\ln \frac{1.0}{[\mathrm{A}]_t}=(2.0\times10^{-2}\,\mathrm{s^{-1}})(100\,\mathrm{s})$$

$$\ln \frac{1.0}{[\mathrm{A}]_t}=2.0$$

Remove the natural logarithm by taking antilog (exponential) on both sides:

$$\frac{1.0}{[\mathrm{A}]_t}=e^{2.0}$$

$$[\mathrm{A}]_t=\frac{1.0}{e^{2.0}}$$

Numerical value:

$$e^{2.0}\approx7.39$$

$$[\mathrm{A}]_t=\frac{1.0}{7.39}\;\mathrm{mol\,L^{-1}}\approx0.135\,\mathrm{mol\,L^{-1}}$$

The concentration of $$\mathrm{A}$$ remaining after 100 s is therefore about $$0.14\,\mathrm{mol\,L^{-1}}$$ (to two significant figures).

The decomposition of sucrose is first order, so the integrated rate law (in the base-10 form used in NCERT) is

$$k = \dfrac{2.303}{t}\log\dfrac{[\mathrm{R}]_0}{[\mathrm{R}]}$$

Step 1: Rate constant from the half-life. For a first order reaction the half-life is independent of the initial concentration:

$$k = \dfrac{0.693}{t_{1/2}} = \dfrac{0.693}{3.00\,\mathrm{h}} = 0.231\,\mathrm{h^{-1}}$$

Step 2: Apply the integrated rate equation at $$t = 8\,\mathrm{h}$$.

$$0.231\,\mathrm{h^{-1}} = \dfrac{2.303}{8\,\mathrm{h}}\log\dfrac{[\mathrm{R}]_0}{[\mathrm{R}]}$$

$$\log\dfrac{[\mathrm{R}]_0}{[\mathrm{R}]} = \dfrac{0.231\times 8}{2.303} = 0.8025$$

$$\dfrac{[\mathrm{R}]_0}{[\mathrm{R}]} = \text{antilog}(0.8025) = 6.346$$

Step 3: Fraction of sucrose remaining.

$$\dfrac{[\mathrm{R}]}{[\mathrm{R}]_0} = \dfrac{1}{6.346} = 0.158$$

Hence about $$0.158$$ (i.e. $$15.8\%$$) of the original sucrose remains after 8 hours.

Step 1 : Start with the Arrhenius equation

The temperature–rate constant relation for any reaction is given by

$$k = A\,e^{-E_a/(R T)}$$

where

• $$A$$ is the frequency (pre-exponential) factor,
• $$E_a$$ is the activation energy,
• $$R = 8.314\,\mathrm{J\,mol^{-1}\,K^{-1}}$$ is the gas constant,
• $$T$$ is the absolute temperature.

Step 2 : Identify the exponential term in the given rate expression

The rate constant for the hydrocarbon decomposition is expressed as

$$k = (4.5 \times 10^{11}\,\mathrm{s^{-1}})\,e^{-28000\,\mathrm{K}/T}$$

Comparing the exponent with $$e^{-E_a/(R T)}$$ gives

$$\frac{-E_a}{R}\;=\;-28000\,\mathrm{K}$$

Step 3 : Solve for $$E_a$$

$$E_a = (28000\,\mathrm{K})(R)$$

Substitute $$R = 8.314\,\mathrm{J\,mol^{-1}\,K^{-1}}$$:

$$E_a = 28000\,\mathrm{K}\;\times\;8.314\,\mathrm{J\,mol^{-1}\,K^{-1}}$$

$$E_a = 232\,792\,\mathrm{J\,mol^{-1}}$$

Step 4 : Convert joules to kilojoules

$$E_a = \frac{232\,792\,\mathrm{J\,mol^{-1}}}{1000} = 232.792\,\mathrm{kJ\,mol^{-1}}$$

Rounded to three significant figures:

Activation energy

$$E_a \approx 2.33 \times 10^{2}\,\mathrm{kJ\,mol^{-1}}$$

Step 1 : Identify the Arrhenius form
For a first-order reaction the Arrhenius equation in base 10 is
$$\log k = \log A - \dfrac{E_a}{2.303\,R}\,\dfrac1T$$ The given relation is
$$\log k = 14.34 - \dfrac{1.25\times10^{4}}{T}$$ Hence the slope equals $$-\dfrac{E_a}{2.303 R}= -1.25\times10^{4}.$$

Step 2 : Calculate the activation energy
$$E_a = 1.25\times10^{4}\times 2.303\times R$$ Using $$R = 8.314\;\mathrm{J\,mol^{-1}\,K^{-1}}$$,
$$E_a = 1.25\times10^{4}\times 2.303 \times 8.314\;\mathrm{J\,mol^{-1}}$$ $$E_a = 2.39\times10^{5}\;\mathrm{J\,mol^{-1}} \;\approx\; 2.39\times10^{2}\;\mathrm{kJ\,mol^{-1}}$$

Step 3 : Determine the rate constant that gives a half-life of 256 min
For a first-order reaction
$$t_{1/2}=\dfrac{0.693}{k}$$ Given $$t_{1/2}=256\;\text{min}=256\times60=1.536\times10^{4}\;\text{s}$$,
$$k = \dfrac{0.693}{1.536\times10^{4}}=4.51\times10^{-5}\;\text{s}^{-1}$$

Step 4 : Insert this k in the empirical equation to find T
$$\log k = \log(4.51\times10^{-5}) = \log4.51 - 5 = 0.654 - 5 = -4.346$$ Set equal to the given expression:
$$-4.346 = 14.34 - \dfrac{1.25\times10^{4}}{T}$$ $$\dfrac{1.25\times10^{4}}{T} = 14.34 + 4.346 = 18.686$$ $$T = \dfrac{1.25\times10^{4}}{18.686} = 6.69\times10^{2}\;\text{K} \approx 6.7\times10^{2}\;\text{K}$$

Result
$$E_a \approx 2.39\times10^{2}\;\mathrm{kJ\,mol^{-1}}$$
Half-life of 256 min is obtained at
$$T \approx 6.7\times10^{2}\;\text{K} \;\;(\approx 669\;\text{K}).$$

Given data

• Rate constant at the first temperature: $$k_1 = 4.5 \times 10^{3}\,\mathrm{s^{-1}}$$
• First temperature: $$T_1 = 10^{\circ}\mathrm{C} = 10 + 273 = 283\,\mathrm{K}$$
• Rate constant at the second temperature: $$k_2 = 1.5 \times 10^{4}\,\mathrm{s^{-1}}$$
• Activation energy: $$E_a = 60\,\mathrm{kJ\,mol^{-1}} = 60\,000\,\mathrm{J\,mol^{-1}}$$
• Gas constant: $$R = 8.314\,\mathrm{J\,mol^{-1}K^{-1}}$$

Step 1 : Arrhenius relation for two temperatures

$$\ln\!\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$

Step 2 : Insert numerical values

First evaluate the ratio of rate constants:

$$\frac{k_2}{k_1} = \frac{1.5 \times 10^{4}}{4.5 \times 10^{3}} = 3.3333$$

Hence

$$\ln(3.3333) = 1.20397$$

Substituting into the Arrhenius equation:

$$1.20397 = \frac{60\,000}{8.314}\left(\frac{1}{283} - \frac{1}{T_2}\right)$$

Step 3 : Solve for $$T_2$$

First compute the factor $$\dfrac{R}{E_a}$$ because we need its reciprocal:

$$\frac{R}{E_a} = \frac{8.314}{60\,000} = 1.3857 \times 10^{-4}\,\mathrm{K^{-1}}$$

Hence

$$\frac{1}{T_2} = \frac{1}{T_1} - \frac{R}{E_a}\,\ln\!\left(\frac{k_2}{k_1}\right)$$

$$\frac{1}{T_2} = \frac{1}{283} - (1.3857 \times 10^{-4})(1.20397)$$

$$\frac{1}{T_2} = 0.00353357 - 0.00016684 = 0.00336673\,\mathrm{K^{-1}}$$

Taking the reciprocal:

$$T_2 = \frac{1}{0.00336673} \approx 297\,\mathrm{K}$$

Step 4 : Convert to degree Celsius

$$T_2 (^{\circ}\mathrm{C}) = 297 - 273 = 24^{\circ}\mathrm{C}$$

Therefore, the rate constant $$k = 1.5 \times 10^{4}\,\mathrm{s^{-1}}$$ will be reached at about $$24^{\circ}\mathrm{C}$$ (\(\approx 297\,\mathrm{K}\)).

Step 1 : Integrated rate law for a first-order reaction
For a first-order reaction the relation between concentration and time is

$$t = \frac{2.303}{k}\;\log\!\frac{[A]_0}{[A]}$$

where $$k$$ is the rate constant at the temperature concerned.

Step 2 : Write the two given times

• 10 % decomposition at $$298\,\mathrm K$$ : after the time required, $$[A] = 0.90[A]_0$$.

$$t_{10\%} = \frac{2.303}{k_{298}}\;\log \frac{[A]_0}{0.90[A]_0} = \frac{2.303}{k_{298}}\;\log 1.1111$$

• 25 % decomposition at $$308\,\mathrm K$$ : after the time required, $$[A] = 0.75[A]_0$$.

$$t_{25\%} = \frac{2.303}{k_{308}}\;\log \frac{[A]_0}{0.75[A]_0} = \frac{2.303}{k_{308}}\;\log 1.3333$$

Step 3 : Equate the two times (given to be equal)

$$\frac{2.303}{k_{298}}\,\log 1.1111 = \frac{2.303}{k_{308}}\,\log 1.3333$$
$$\Rightarrow \;\frac{k_{308}}{k_{298}} = \frac{\log 1.3333}{\log 1.1111}$$

Numerical values (common logarithms):
$$\log 1.3333 = 0.124939,\quad \log 1.1111 = 0.045757$$

$$\therefore \;\frac{k_{308}}{k_{298}} = \frac{0.124939}{0.045757} = 2.730$$

Step 4 : Determine the activation energy $$E_a$$

The Arrhenius equation in ratio form is
$$\ln \frac{k_{308}}{k_{298}} = -\frac{E_a}{R}\left(\frac{1}{308} - \frac{1}{298}\right)$$

Because $$k_{308} > k_{298}$$, the right side must be positive; hence we may write
$$\ln 2.730 = \frac{E_a}{R}\left(\frac{1}{298} - \frac{1}{308}\right)$$

Compute the two factors:

$$\ln 2.730 = 1.004$$
$$\frac{1}{298} - \frac{1}{308} = \frac{308-298}{298\times308} = \frac{10}{91784}=1.089\times10^{-4}\,\mathrm{K^{-1}}$$

Insert the gas constant $$R = 8.314\,\mathrm{J\,mol^{-1}\,K^{-1}}$$:

$$E_a = \frac{1.004\times8.314}{1.089\times10^{-4}}\;\mathrm{J\,mol^{-1}} = 7.66\times10^{4}\,\mathrm J\,\mathrm{mol^{-1}}$$

$$E_a \approx 7.7\times10^{4}\,\mathrm{J\,mol^{-1}} = 76.6\,\mathrm{kJ\,mol^{-1}}$$

Step 5 : Calculate $$k$$ at $$318\,\mathrm K$$

Using the Arrhenius form with the pre-exponential factor $$A = 4.0\times10^{10}\,\mathrm{s^{-1}}$$:

$$k_{318}=A\,e^{-E_a/(R T)}$$
$$E_a/(RT)=\frac{7.66\times10^{4}}{8.314\times318}=28.96$$

$$e^{-28.96}=2.69\times10^{-13}$$

$$k_{318}=4.0\times10^{10}\times2.69\times10^{-13}=1.08\times10^{-2}\,\mathrm{s^{-1}}$$

Result

$$E_a\approx76.6\,\mathrm{kJ\,mol^{-1}},\qquad k_{318}\approx1.1\times10^{-2}\,\mathrm{s^{-1}}$$

Given $$T_1 = 293\,\mathrm{K}$$, $$T_2 = 313\,\mathrm{K}$$, and the rate (hence the rate constant) becomes four times its original value, so $$\dfrac{k_2}{k_1} = 4$$.

The temperature dependence of the rate constant is given by the Arrhenius equation in two-temperature logarithmic form:

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

Step 1: Substitute the data.

$$\log 4 = \dfrac{E_a}{2.303\times 8.314}\left[\dfrac{313-293}{293\times 313}\right]$$

Step 2: Evaluate the numerical factors.

• $$\log 4 = 0.6021$$
• $$2.303 \times 8.314 = 19.147\,\mathrm{J\,K^{-1}\,mol^{-1}}$$
• $$T_1 T_2 = 293 \times 313 = 91\,709\,\mathrm{K^2}$$
• $$\dfrac{T_2-T_1}{T_1 T_2} = \dfrac{20}{91\,709} = 2.181\times10^{-4}\,\mathrm{K^{-1}}$$

Step 3: Solve for $$E_a$$.

$$0.6021 = \dfrac{E_a}{19.147}\times 2.181\times10^{-4}$$

$$E_a = \dfrac{0.6021 \times 19.147}{2.181\times10^{-4}}\,\mathrm{J\,mol^{-1}}$$

$$E_a = \dfrac{11.530}{2.181\times10^{-4}} = 5.286\times10^{4}\,\mathrm{J\,mol^{-1}}$$

Step 4: Convert to kilojoules.

$$E_a \approx 52.86\,\mathrm{kJ\,mol^{-1}}$$