NCERT Solutions for Class 12 Chemistry Chapter 4 - Chemical Kinetics
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1. From the rate expression for the following reactions determine their order of reaction and the dimensions of the rate constants:
$$\textbf{(i) 3NO(g)}\xrightarrow{}\textbf{N}_2\textbf{O(g)}+\textbf{NO}_2\textbf{(g)}\\\textbf{Rate = k[NO]}^{2}\\\textbf{(ii) H}_2\textbf{O}_2(aq)+3\textbf{I}^{\normalsize-}(aq)+2\textbf{H}^{\normalsize+}\xrightarrow{}2\textbf{H}_2\textbf{O(l)}+\textbf{I}_{3}^{\normalsize-}\\\textbf{Rate = k[H}_2\textbf{O}_2][I^{\normalsize–}]\\\textbf{(iii) CH}_3\textbf{CHO(g)}\xrightarrow{}\textbf{CH}_4\textbf{(g)}+\textbf{CO(g)}\\\textbf{Rate = k}[\textbf{CH}_3\textbf{CHO}]^{3/2}\\\textbf{(iv) C}_2\textbf{H}_5\textbf{Cl(g)}\xrightarrow{}\textbf{C}_2\textbf{H}_4\textbf{(g)}+\textbf{HCl(g)}$$
Ans. (i) Order of reaction = 2
Dimension of rate constant
$$k=\frac{\text{Rate}}{[\text{NO}]^{2}}=\frac{\text{mol L}^{\normalsize-1}s^{\normalsize-1}}{(\text{mol L}^{\normalsize-1})^{2}}=\text{mol L}^{\normalsize-1}s^{\normalsize-1}$$
(ii) Order or reaction = 2
Dimension of rate constant
$$k=\frac{\text{Rate}}{[\text{H}_2\text{O}_2][\text{I}^{\normalsize-}]}=\frac{\text{mol L}^{\normalsize-1}s^{\normalsize-1}}{(\text{mol}\space L^{\normalsize-1})(\text{mol L}^{\normalsize-1})}=\text{mol L}^{\normalsize-1}s^{\normalsize-1}$$
(iii) Order of reaction = 3/2
$$k=\frac{\text{Rate}}{[\text{CH}_3\text{CHO}]^{3/2}}=\frac{\text{mol}^{\normalsize-1}s^{\normalsize-1}}{(\text{mol L}^{\normalsize-1})(\text{mol L}^{\normalsize-1})}=\text{mol}^{\normalsize-1/2 }\text{L}^{1/2}s^{\normalsize-1}\\\text{(iv) Order of reaction = 1}\\\text{Dimension of rate constant}\\k=\frac{\text{Rate}}{\text{C}_2\text{H}_5 \text{Cl}}=\frac{\text{mol L}^{\normalsize-1}s^{\normalsize-1}}{\text{mol L}^{\normalsize-1}}=s^{\normalsize-1}$$
2. For the reaction; $$2\textbf{A + B}\xrightarrow{}\textbf{A}_2\textbf{B},$$the rate = k [A][B]2 with k = 2·0 × 10–6 mol–2 L2 s–1. Calculate the initial rate of the reaction when [A] = 0·1 mol L–1; [B] = 0·2 mol L–1. Calculate the rate of reaction after [A] is reduced to 0·06 mol L–1.
Ans. Given : k = 2 × 10–6 mol–2 L2 s–1
To find : Rate when [A] = 0.1 mol L–1, [B] = 0.2 mol L–1
Rate when [A] = 0.1 mol L–1, [B] = 0.06 mol L–1
Formula : Rate = k[A][B]2
First case :
Rate = k[A][B]2
= (2.0 × 10–6 mol–2 L2 s–1) × (0.1 mol L–1) × (0.2 mol L–1)2
= 8 × 10–9 mol L–1 s–1 = 8 × 10–9 Ms–1.
Second case :
The concentration of A after taking part in the reaction = 0.06 mol L–1.
Amount of A reacted = (0.1 – 0.06) = 0.04 mol L–1
$$\text{Amount of B reacted =}\frac{1}{2}×0.04\space\text{mol L}^{\normalsize-1}=0.02\space\text{mol L}^{\normalsize-1}.$$
The concentration of B after taking part in the reaction = (0.2 – 0.02) = 0.18 mol L–1
Rate = k[A][B]2
= (2.0 × 10–6 mol–2 L2 s–1) × (0.06 mol L–1) × (0.18 mol L–1)2.
= 3.89 × 10–9 mol L–1 s–1 = 3.90 × 10–9 Ms–1.
Hence, the rate of reaction is 3.90 × 10–9 Ms–1.
4. The decomposition of dimethyl ether leads to the formation of CH4, H2 and CO and the reaction, rate is given by Rate = k [CH3OCH3]3/2. The rate of reaction is followed by increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether, i.e., Rate = k (PCH3OCH3)3/2
If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants?
Ans. As the concentration in the rate law equation is given in terms of pressure,
∴ Unit of rate = bar min–1
$$\therefore\space\text{Unit of k =}\frac{\text{Rate}}{[\text{P}_{\text{CH}_3\text{OCH}_3}]^{3/2}}=\frac{\text{bar min}^{\normalsize-1}}{\text{ bar}^{3/2}}\\\text{= bar}^{-1/2}\text{min}^{\normalsize-1}.$$
5. Mention the factors that affect the rate of a chemical reaction.
Ans. The factors that affects the rate of chemical reaction are as follows :
(i) Nature of reactants: Nature or reactivity of reactants affects the rate of chemical reaction.
E.g. Al is more reactive than Zn. Therefore, the rate of reaction of Al with HCl is higher than that of Zn.
(ii) Size of the particles of the reactants: Smaller the size of reactant particles, higher the rate of reaction.
E.g. When HCl is added to pieces of Sahabad tiles, the CO2 effervescence is formed slowly. However, when HCl is added to sahabad powder, the CO2 effervescence is formed at a faster rate.
(iii) Concentration of reactants: The rate of chemical reaction increases with increase in the concentration of reactants.
E.g. Dilute HCl reacts slowly with CaCO3 whereas concentrated HCl reacts rapidly with CaCO3.
(iv) Temperature of the reaction: Higher the temperature, faster the rate of chemical reaction.
E.g. The temperature in the refrigerator is low. So, the rate of decomposition of perishable foodstuffs is low and it remains fresh for a longer time.
(v) Catalyst: Presence of catalyst increase the rate of a chemical reaction.
E.g. The decomposition of H2O2 into H2O and O2 takes place slowly at room temperature. However, the same reaction occurs at a faster rate on adding MnO2 powder in it.
6. A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is:
(i) Doubled
(ii) Reduced to half ?
Ans. Rate = k[A]2 = ka2
(i) If [A] = 2a, rate = k(2a)2 = 4 ka2 = 4ka2 = 4 times.
Rate of reaction becomes 4 times.
$$\text{(ii) If [A] = a/2, rate = k}\bigg(\frac{a}{2}\bigg)=\frac{1}{4}\text{ka}^{2}=\frac{1}{4}\text{th.}$$
Rate of reaction reduced to one fourth.
7. What is the effect of temperature on the rate constant of reaction? How can this effect of on rate constant be represented quantitatively?
Ans. The rate constant (k) of a reaction depends on temperature and it increases with rise in temperature and becomes nearly double with about every 10° rise in temperature. The effect of temperature on the rate constant is expressed quantitatively by Arrhenius equation.
k = Ae−Ea/Rt
$$\text{log}\frac{k_2}{k_1}=\frac{\text{E}_a}{2.303 \text{R}}\bigg[\frac{1}{\text{T}_1}-\frac{1}{\text{T}_2}\bigg]=\frac{0}{2.303\space \text{R}}\bigg[\frac{\text{T}_2-\text{T}_1}{\text{T}_1\text{T}_2}\bigg]=0\\\frac{k_2}{k_1}=\text{Antilog 0 = 1 or}\space k_2=k_1=1.6×10^{6}s^{\normalsize-1}$$
8. In a pseudo first order reaction in water, the following results were obtained :
| t/s | 0 | 30 | 60 | 90 |
| [A]/mol L–1 | 0.55 | 0.31 | 0.17 | 0.085 |
(i) Calculate the average rate of reaction between the time interval 30 to 60 seconds.
(ii) Calculate the pseudo first order rate constant for the hydrolysis of ester.
$$\textbf{Ans.}\space\text{(i) Average rate of reaction between the time interval 30 – 60 sec =}\space\frac{\text{C}_2-\text{C}_1}{t_2-t_1}\\=\frac{0.31-0.17}{60-30}=\frac{0.14}{30}=4.6×10^{\normalsize-3}\text{mol L}^{\normalsize-1}s^{\normalsize-1}\\\text{(ii) For a pseudo first order reaction, k’ =}\frac{2.303}{t}\text{log}\frac{[\text{A}_0]}{\text{A}}\\\text{when initial concentration of A, [A}_0] = 0.55\text{M}\\\text{when t = 30 sec k}_1′ =\frac{2.303}{30}\text{log}\frac{0.55}{0.31}=1.92×10^{\normalsize-2}s^{\normalsize-1}\\\text{when t = 60 sec k}_2=\frac{2.303}{60}\text{log}\frac{0.55}{0.17}=1.92×10^{\normalsize-2}s^{\normalsize-1}\\\therefore\text{Average rate constant,}\space k'=\frac{k'_1+k'_2+k'_3}{3}=\frac{(1.91+1.96+2.07)×10^{\normalsize-2}}{3}$$
= 1.98 × 10–2 s–1.
9. A reaction is first order in A and second order in B.
(i) Write the differential rate equation.
(ii) How is the rate affected on increasing the concentration of B three times?
(iii) How is the rate affected when the concentrations of both A and B is doubled?
$$\textbf{Ans.}(i)\space\text{Rate of reaction =}\frac{-d[\text{A}]}{dt}=\frac{1}{2}=\frac{-d[B]}{dt}$$
= k[A][B]2
(ii) Rate = k[A][B]2
When [B] becomes three times, rate
= k[A][3B]2 = 9k[A][B]2
= 9 times
Hence, when concentration of B is tripled, rate of reaction increased by 9 times.
(iii) When both [A] and [B] are doubled, rate = k[2A][2B]2 = 8[A][B]2 = 8 times.
Hence, rate of reaction increases by 8 times.
| A/mol L–1 | 0.20 | 0.20 | 0.40 |
| B/mol L–1 | 0.30 | 0.10 | 0.05 |
| r0/mol L–1 s–1 | 5.07× 10–5 | 5.07 × 10–5 | 1.43 × 10–4 |
What is the order of the reaction with respect to A and B?
Ans. Rate = [A]α[B]β
r1 = 5.07 × 10–5 = (0.20)α(0.30)β ...(i)
r2 = 5.07 × 10–5 = (0.20)α(0.10)β ...(ii)
r3 = 1.43 × 10–4 = (0.40)α(0.05)β ...(iii)
$$\frac{r_1}{r_2}=\frac{5.07×10^{\normalsize-5}}{5.07×10^{\normalsize-5}}=\frac{(0.2)^{\alpha}}{(0.2)^{\alpha}}\frac{(0.3)^{\beta}}{(0.10)^{\beta}} =1=(3)^{\beta}\\\beta=0\\\frac{r_3}{r_2}=\frac{1.43×10^{\normalsize-4}}{5.06×10^{\normalsize-5}}=\frac{(0.4)^{\alpha}}{(0.20)^{\alpha}}\frac{(0.05)^{\beta}}{(0.10)^{\beta}}\\=2.826=2^\alpha\bigg(\frac{1}{2}\bigg)^\beta$$
2.826 = 2α (as β = 0)
Taking log on both the sides, we get
log 2.826 = α log 2
⇒ 0.4511 = α × 0.3010
$$\alpha=\frac{0.4511}{0.300}=1.498=1.5$$
∴ Hence, Order with respect to A = 1.5 and order with respect to B = 0
11. The following results have been obtained during the kinetic studies of the reaction.
$$2\text{A + B}\xrightarrow{}\text{C + D}$$
| Experiment | [A] mol L-1 | [B] mol L-1 | Initial rate of formation of D/mol L–1 min–1 |
| I | 0.1 | 0.1 | 6.0 × 10–3 |
| II | 0.3 | 0.2 | 7.2 × 10–2 |
| III | 0.3 | 0.4 | 2.88 × 10–1 |
| IV | 0.4 | 0.1 | 2.40 × 10–2 |
Determine the rate law and the rate constant for the reaction.
Ans. Rate = k[A]α[B]β
r1 = k[0.1]α[0.1]β = 6.0 × 10–3
r2 = k[0.3]α[0.2]β = 7.2 × 10–2
r3 = k[0.3]α[0.4]β = 2.88 × 10–1
r4 = k[0.4]α[0.1]β = 2.40 × 10–2
$$\frac{r_1}{r_2}=\frac{6.0×10^{\normalsize-3}}{2.40×10^{\normalsize-2}}\frac{(0.1)^{\alpha}}{(0.4)^{\alpha}}=\bigg(\frac{1}{4}\bigg)^{\alpha}\\\frac{1}{4}=\bigg(\frac{1}{4}\bigg)^{\alpha}\\\alpha=1\\\frac{r_2}{r_3}=\frac{7.2×10^{\normalsize-2}}{2.88×10^{\normalsize-1}}=\frac{k(0.3)^{\alpha}(0.2)^\beta}{k(0.3)^\alpha(0.4)^{\beta}}\\\frac{1}{4}=\bigg(\frac{1}{2}\bigg)^\beta\\\bigg(\frac{1}{2}\bigg)^{2}=\bigg(\frac{1}{2}\bigg)^\beta$$
∴ β = 2
∴ Rate law expression, is Rate = k[A][B]2
∴ Order with respect to A = 1 and
Order with respect to B = 2
Overall order of the reaction = 1 + 2 = 3
Calculation of rate constant k
Substituting values of experiment I, we get
6.0 × 10–3 = k(0.1)(0.1)2
= k × 1 × 10–3
∴ k = 6 mol–2 L2 min–1
12. The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table:
| Experiment | [A] mol L-1 | [B] mol L-1 | Initial rate mol L–1 min–1 |
| I | 0.1 | 0.1 | 2.0×10-2 |
| II | – | 0.2 | 4.0 × 10–2 |
| III | 0.4 | 0.4 | – |
| IV | – | 0.2 | 2.0 × 10–2 |
Ans. Rate law expression :
Rate = k[A]1[B]0 = k[A]
For 1st experiment R1 = 2.0 × 10–2 mol L–1 min–1
= k[0.1] mol L–1
k = 0.2 min–1
For IInd experiment R2 = 4.0 × 10–2 mol L–1 min–1
= k[0.2 min–1] [A]
[A] = 0.2 mol L–1
For IIInd experiment R3 = Rate = k[A]
= (0.2 min–1) (0.4 mol L–1)
= 0.08 mol L–1 min–1
For IVth experiment R4 = 2.0 × 10–2 mol L–1 min–1
= k[A] = 0.2 min–1[A]
∴ [A] = 0.1 mol L–1
[A] = 0.2 mol L–1
13. Calculate the half-life of a first order reaction from their rate constants given below:
(i) 200 –1 (ii) 2 min–1 (iii) 4 years–1
Ans. Given : k1 = 200 s–1, k2 = 2 min–1, k3 = 4 years–1
To find t1/2 in all three cases
$$\text{Formula : Half life for a first order reaction is, t}_{1/2} =\frac{0.693}{k}\\\text{Calculation :}\\\text{(i)}\space\text{t}_{1/2}=\frac{0.693}{200 s ^{\normalsize-1}}=0.346×10^{\normalsize-2}s=3.46×10^{\normalsize-3}s\\\text{(ii)}\space\text{t}_{1/2}=\frac{0.693}{2\space\text{min}^{\normalsize-1}}=0.346\space\text{min}=3.46×10^{\normalsize-1}\space\text{min.}\\\text{(iii)}\space t_{1/2}=\frac{0.693}{4 \text{Year}^{\normalsize-1}}=0.173\space\text{year}=1.73×10^{\normalsize-1}\text{year}$$
14. The half-life for radioactive decay of HC is 5730 years. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample.
Ans. Radioactive decay follows first order kinetics.
Given: [A0] = 100, [A] = 80, t1/2 = 5730 year
To Find: Age of sample?
$$\text{Formula}:\space k=\frac{0.639}{t_{1/2}}\\t=\frac{2.303}{k}\text{log}\frac{[\text{A}_0]}{[\text{A}]}\\\text{Calculation: Radioactive decay follows first order kinetics. Therefore,}\\\text{Decay constant (k)}\space t=\frac{0.693}{t_{1/2}}=\frac{0.693}{5730}\text{Year}^{\normalsize-1}\\\text{t}=\frac{2.303}{k}\text{log}\frac{[\text{A}_0]}{[\text{A}]}\\=\frac{2.303}{(0.693/5730\space \text{years}^{\normalsize-1})}\text{log}\frac{100}{8}\\=\frac{2.303×5730}{0.693}×0.0969=1845\space\text{years}$$
Hence, the age of wood will be 1845 years.
15. The experimental data for decomposition of N2O5:
$$[\text{2N}_2\text{O}_5\xrightarrow{}4\text{NO}_2+\text{O}_2]$$
in gas phase at 318 K are given below :
| t/s | 0 | 400 | 800 | 1200 | 1600 | 2000 | 2400 | 2800 | 3200 |
| 102 × [N2O5]/mol L–1 | 1.63 | 1.36 | 1.14 | 0.93 | 0.78 | 0.64 | 0.53 | 0.43 | 0.35 |
(i) Plot [N2O5] against t.
(ii) Find the half-life period for the reaction.
(iii) Draw a graph between log [N2O5] and t.
(iv) What is the rate law?
(v) Calculate the rate constant.
(vi) Calculate the half-life period from k and compare it with (ii).
Ans. (i) Plot of [N2O5] Vs time
(ii) Initial concentration of N2O5 = 1.63 × 102 M
Time corresponding to this concentration
$$\text{N}_2\text{O}_5=\frac{1.63×10^{2}}{2}=81.5\space\text{molL}^{\normalsize-1}\text{is half life.}$$
From the plot, half-life corresponding to this concentration is, t2 = 1450 S.
(iii) Plot of log [N2O5] Vs time.
| t(s) | 0 | 400 | 800 | 1200 | 1600 | 2000 | 2400 | 2800 | 3200 |
| 102 × [N2O5]/mol L–1 | 1.63 | 1.36 | 1.14 | 0.93 | 0.78 | 0.64 | 0.53 | 0.43 | 0.35 |
| log[N2O5] | –1.79 | –1.87 | –1.94 | –2.03 | –2.11 | –2.19 | –2.28 | –2.37 | –2.46 |
(iv) As the plot obtained for log [N2O5] Vs time is a straight line, hence it is a first order reaction. Therefore, rate law for the reaction is:
∴ Rate = k[N2O5]
$$\text{(v)}\space\text{Slope of the line =}\frac{k}{2.303}\\\text{Slope}=\frac{-2.46-(-1.79)}{3200-0}=-\frac{0.67}{3200}\\k=\frac{0.67×2.303}{3200}=4.82×10^{\normalsize-4}\text{mol L}^{\normalsize-1}\text{s}^{\normalsize-1}\\\text{(vi)}\space\text{Half life period}\space t_{1/2}=\frac{0.693}{k}=\frac{0.693}{4.82×10^{\normalsize-4}}=1438\space\text{sec.}$$
Half life period (t1/2) is calculated from the formula and slope are approximately the same.
16. The rate constant for a first order reaction is 60 s–1. How much time will it take to reduce the initial concentration of the reactant to its 1/16th value?
$$\textbf{Ans.}\space \text{Given: k = 60 s}^{\normalsize-1},\text{R}=\frac{\text{R}_0}{16}\\\text{To find: time t}\\\text{For 1}^{\text{st}} \text{order reaction}\\\text{Formula:}\space t=\frac{2.303}{k}\text{log}\frac{\text{R}_0}{\text{R}}\\\text{Calculation:}\space t=\frac{2.303}{60}\text{log}\frac{\text{R}_0}{\text{R}_{0/16}}=\frac{2.303}{60}\text{log}\space16=4.62×10^{\normalsize-2}s$$
17. During nuclear explosion, one of the products is 90Sr with half-life of 28.1 years. If 1 μg of 90Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically?
Ans. Given : t1/2 = 28.1 years
[R0] = 1 mg
To find: (i) Amount left after 10 years
Amount left after 60 years
$$\text{Formula}:\space\lambda=\frac{0.693}{t_{1/2}}\\\lambda=\frac{2.303}{t}\text{log}\frac{[\text{R}_0]}{[\text{R}]}\\\text{Calculation: As radioactive disintegration follows first order kinetics. Hence}\\\text{Decay constant of}\space^{90}\text{Sr},(\lambda)=\frac{0.693}{t_{1/2}}=\frac{0.693}{28.1}$$
= 2.466 × 10–2 yr–1
(i) To calculate the amount left after 10 year
$$\text{Using formula,}\space\lambda=\frac{2.303}{t}\text{log}\frac{[\text{R}_0]}{[\text{R}]}$$
$$\text{Or}\qquad 2.466×10^{\normalsize-2}=\frac{2.303}{10}\text{log}\frac{1}{[\text{R}]}\\\frac{2.466×10^{\normalsize-2}×10}{2.303}=-\text{log[R]}$$
Or, log [R] = – 0.1071
Or, [R] = Antilog (– 0.1071) = 0.7814 µg
Amount left after 10 years = 0.7814 µg
(ii) To calculate the amount left after 60 years,
$$\text{Or},\space 2.466×10^{\normalsize-2}=\frac{2.303}{60}\text{log}\frac{1}{[\text{R}]}\\\text{Or,}\qquad\frac{2.466×10^{\normalsize-2}×60}{2.303}=-\text{log}[\text{R}]$$
Or, log [R] = – 0.6425
Or, [R] = Antilog (–0.6425) = 0.2278 µg
Amount left after 60 years = 0.2278 µg
18. For a first order reaction show that time required for 99% completion is twice the time required for the complete 90% of the reaction.
$$\textbf{Ans.}\space\text{For a first order reaction;}\space t=\frac{2.303}{k}\text{log}\frac{[\text{R}_0]}{[\text{R}]}\\\textbf{I}^{\textbf{st}} \textbf{case:}\space \text{R}_0 = 100\%, \text{R} = 100 – 99 = 1\%\\t_{99\%}=\frac{2.303}{k}\text{log}\frac{100}{1}=\frac{2.303}{k}\text{log}10^{2}\\=\frac{2.303×2}{k}=\frac{4.606}{k}\\\textbf{II}^{\textbf{nd}}\textbf{case:}\space \text{R}_0 = 100\%; \text{R} = 100 – 90 = 10\%\\t_{90\%}=\frac{2.303}{k}\text{log}\frac{100}{10}=\frac{2.303}{k}\text{log \space 10}=\frac{2.303}{k}\\\text{Dividing eq. (ii) by eqn. (i),}\\\frac{t_{(99\%)}}{t_{(90\%)}}=\frac{4.606}{k}×\frac{k}{2.303}=2$$
It means that time required for 99% completion of reaction is twice the time required to complete 90% of the reaction.
19. A first order reaction takes 40 min for 30% decomposition. Calculate t1/2.
Ans. Given: 30% decomposition means that x = 30% of [R0] or, [R] = [R0] –3[R0] = 0.7[R0]
To Find: t1/2
$$\text{Formula: k}=\frac{2.303}{t}\text{log}\frac{[\text{R}_0]}{[\text{R}]}\\\text{Calculation: For reaction of 1}^{st} \text{order,}\\k=\frac{2.303}{t}\text{log}\frac{[\text{R}_0]}{[\text{R}]}=\frac{2.303}{40}\text{log}\frac{[\text{R}_0]}{0.70[\text{R}_0]}=\frac{2.303}{40}\text{log}\frac{10}{7}\text{min}^{\normalsize-1}\\=\frac{2.303}{40}×0.1549\space\text{min}^{\normalsize-1}=8.918×10^{\normalsize-3}\text{min}^{\normalsize-1}\\\text{For a 1}^\text{st} \text{order reaction half life is,}\\t_{1/2}=\frac{0.693}{k}=\frac{0.693}{8.918×10^{\normalsize-3}\text{min}^{\normalsize-1}}=77.7\space\text{min}$$
Hence, half-life is 77.7 min.
20. For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained:
| t(sec) | P(mm of Hg) |
| 0 | 35.0 |
| 360 | 54.0 |
| 720 | 63.0 |
Calculate the rate constant.
$$\textbf{Ans.}\space\underset{\text{Azoisopropane}}{(\text{CH}_3)_2\text{CHN = NCH(CH}_3)_2(\text{g})}\xrightarrow{}\underset{\text{Hexane}}{\text{N}_2(g)+\text{C}_6\text{H}_{14}\text{(g)}}$$
| Initial pressure P0 | 0 | 0 |
| Pressure P0 – P | p | p |
After time t
Total pressure after time t(Pt) = (P0–p) + p + p = P0 + p or p = Pt – P0
[R]0 ∝ P0 and [R] ∝ P0 – p
On substituting the value of p,
[R] ∝ P0 – (Pt – P0), i.e. [R] ∝ 2P0 – Pt
As decomposition of azoisopropane is a first order reaction
$$\therefore\space k=\frac{2.303}{t}\text{log}\frac{[\text{R}_0]}{[\text{R}]}=\frac{2.303}{t}\text{log}\frac{\text{P}_0}{2\text{P}_0-\text{P}_t}\\\text{When t = 360 sec,}\\\text{Rate constant k}=\frac{2.303}{360}\text{log}\frac{35.0}{2×35.0-54.0}=\frac{2.303}{360}\text{log}\frac{35.0}{16}=2.175×10^{\normalsize-3}s^{\normalsize-1}\\\text{When t = 720 sec,}\\\text{Rate constant k =}\frac{2.303}{720}\text{log}\frac{35.0}{2×35.0-63}=\frac{2.303}{720}\text{log}5=2.235×10^{\normalsize-3}s^{\normalsize-1}\\\therefore\space\text{Average value of k =}\frac{2.175×2.235}{2}×10^{\normalsize-3}s^{\normalsize-1}=2.20×10^{\normalsize-3}s^{\normalsize-1}$$
21. The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume.
$$\text{SO}_2\text{Cl}_2\text{(g)}\xrightarrow{}\text{SO}_{2(g)}+\text{Cl}_{2(g)}$$
Calculate the rate of the reaction when total pressure is 0.65 atm.
$$\textbf{Ans.} \space\text{SO}_2\text{Cl}_2\text{(g)}\xrightarrow{}\text{SO}_{2(g)}+\text{Cl}_{2(g)}$$
| Experiment | Time/s–1 | Total pressure/atm |
| 1 | 0 | 0.5 |
| 2 | 100 | 0.6 |
| Let initial pressure | P0 | 0 | 0 |
| Pressure at time t | P0 – p | p | p |
Let initial pressure P0 ∝ R0
Pressure at time t, Pt = P0 – p + p + p = P0 + p
∴ Pressure of reactant at time Pt = P0 –p
= 2P0 – Pt ∝ R
$$\text{Using formula:}\qquad k=\frac{2.303}{t}\text{log}\frac{\text{P}_0}{2\text{P}_0-\text{P}_t}\\\text{When t = 100 s,}\space k=\frac{2.303}{100}\text{log}\frac{0.5}{2×0.5-0.6}=\frac{2.303}{100}\text{log}(1.25)\\=\frac{2.303}{100}(0.0969)=2.2316×10^{\normalsize-3}s^{\normalsize-1}$$
When Pt = 0.65 atm,
∴ Pressure of SO2Cl2 at time t (PSO2Cl2),
R = 2P0 – pt = 2 × 0.50 – 0.65 atm = 0.35 atm
Rate at that time = k × PSO2Cl2
= (2.2316 × 10–3) × (0.35)
= 7.8 × 10–4 atm s–1
Hence, the rate of the reaction when total pressure is 0.65 atm is 7.8 × 10–4 atm s–1.
| T/°C | 105 × k/s–1 |
| 0 | 0.0787 |
| 20 | 1.70 |
| 40 | 25.7 |
| 60 | 178 |
| 80 | 2140 |
Draw a graph between ln k and 1T and calculate the value of A and Ea. Predict the rate constant at 30°C and 50°C.
Ans. For the given data, we obtain,
| T/°C | 0 | 20 | 40 | 60 | 80 |
| T/(K) | 273 | 293 | 313 | 333 | 353 |
| $$\frac{1}{\text{T}}/\text{K}^{\normalsize-1}$$ | 3.66 × 10–3 | 3.41 × 10–3 | 3.19 × 10–3 | 3.0 × 10–3 | 2.83 × 10–3 |
| 105× k/s–1 | 0.0787 | 1.70 | 25.7 | 178 | 2140 |
| ln k | –7.147 | – 4.075 | – 1.359 | – 0.577 | 3.063 |
$$\text{Slope of the line,}\\\frac{y_2-y_1}{x_2-x_1}=-12.301\space\text{K}\\\text{According to Arrhenius equation,}\\\text{Slope}=\frac{\text{E}_a}{\text{R}}$$
⇒ Ea = –Slope × R
= –(–12.301 K) × (8.314 J K–1 mol–1)
= 102.27 kJ mol–1
Again,
$$\text{In k = In A =}\frac{\text{E}_a}{\text{RT}}\\\text{In A = In k =}\frac{\text{E}_a}{\text{RT}}\\\text{When}\space\text{ T = 273 K.}\\\text{In k = –7.147}\\\text{Then, In}\space \text{A}=-7.147+\frac{102.27×10^{3}}{8.314×273}=37.911\\\text{Therefore, A = 2.91 × 10}^{6}\\\text{When T = 30 + 273 K = 303 K,}\\\frac{1}{\text{T}}=0.0033\space\text{K}=3.3×10^{\normalsize-3}\text{K}\\\text{Then,}\space\text{at}\space \frac{1}{\text{T}}=3.1×10^{\normalsize-3}\text{K}$$
In k = –0.5
Therefore, k = 0.607 s–1
23. The rate constant for the decomposition of a hydrocarbon is 2·418 × 10–5 s–1 at 546 K. If the energy of activation is 179·9 kJ mol–1, what will be the value of pre-exponential factor?
Ans. Given: k = 2.418 × 10–5 s–1; Ea = 179900 mol–1; R = 8.314 JK–1 mol–1; T = 546 K
To find: Pre-exponential Factor (A)
Formula: k = AeEa/RT
$$\text{log k = log A}-\frac{\text{E}_a}{2.303\space\text{RT}}\\\text{log A}=\text{log k}+\frac{\text{E}_a}{2.303\text{RT}}\\=\text{log}(2.418×10^{5}s^{\normalsize-1})+\frac{(179900 \text{J mol}^{\normalsize-1})}{2.303×(8.314\space\text{J K}^{\normalsize-1} \text{mol}^{\normalsize-1})×546\text{K}}$$
= – 4.6184 + 17.21 = 12.5916
A = Antilog 12.5916 = 3.9 × 1012 s–1.
Hence, the value of pre-exponential factor is 3.9 × 1012 s–1.
24. Consider a certain reaction A → Products with k = 2.0 × 10–2 s–1. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L–1.
Ans. Given: k = 2.10–2 s–1
A0 = 1.0 mol–1
t = 100 sec
To find: A = ?
$$\text{Formula: k =}\frac{2.303}{t}\text{log}\frac{[\text{A}_0]}{[\text{A}]}\\\text{Calculation : The unit of k show that the reaction is of first order.}\\\text{Hence, k}=\frac{2.303}{t}\text{log}\frac{[\text{A}_0]}{[\text{A}]}\\\text{Or,}\qquad 2.0×10^{\normalsize-2}=\frac{2.303}{100}\text{log}\frac{1.0}{[A]}$$
Or, log [A] = –0.8684
∴ [A] = Antilog (–0.8684) = 0.1354 mol L–1
Hence, the concentration of A remaining after 100 s is 0.1354 mol L–1.
25. Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t1/2 = 3.00 hours. What fraction of sample of sucrose remains after 8 hours?
Ans. Given: t1/2 = 3 hours, t = 8 hours
$$\text{To find:}\qquad\frac{A}{A}=?\\\text{Formula:}\qquad k=\frac{0.693}{t_{1/2}}\\k=\frac{2.303}{t}\text{log}\frac{\text{[A}_0]}{[\text{A}]}\\\text{Calculation : Sucrose decomposes according to first order rate law , hence}\\k=\frac{0.693}{t_{1/2}}=\frac{0.693}{3}=0.231\space \text{hr}^{\normalsize-1}\\\text{Hence, 0.231}=\frac{2.303}{8}\text{log}\frac{[\text{A}_0]}{\text{A}}\\\text{Or,}\qquad\text{log}\frac{[\text{A}_0]}{[\text{A}]}=0.8024\\\text{Or,}\qquad\frac{[\text{A}_0]}{\text{[A]}}=\text{Antilog (0.8024)}=6.345\\\text{Or,}\qquad\frac{[\text{A}]}{[\text{A}_0]}=\frac{1}{6.345}=0.158$$
Hence, the fraction of sample of sucrose remains after 8 hours is 0.158.
26. The decomposition of a hydrocarbon follows the equation
k = (4.5 × 1011 s–1)e–28000K/T. Calculate Ea.
Ans. Given: k0 (4.5 × 1011 s–1)e–28000K/T
To Find: Ea
Formula: k = A e–Ea/RT
Given equation is k = (4.5 × 1011 s–1)e–28000K/T
Arrhenius equation, k = Ae–Ea/RT
Comparing both the equations, we get
$$-\frac{\text{E}_a}{\text{RT}}=\frac{28000 \text{K}}{\text{T}}$$
Or, Ea = 28000 K × R = 28000 × 8.314
= 232.79 kJ mol–1
27. The rate constant for the first order decomposition of H2O2 is given by the following equation:
log k = 14.34 – 1.25 × 104 K/T
Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes?
Ans. Given: log k = 14.34 – 1.25 × 104 K/T
To find: Ea
Formula: k = Ae–Ea/RT
Calculation: According to Arrhenius equation, k = Ae–Ea/RT
$$\text{Or,}\qquad\text{ln k = In A –}\frac{\text{E}_a}{\text{RT}}\\\text{Or,}\qquad\text{log K}=\text{log A}-\frac{\text{E}_a}{2.303\text{RT}}\qquad...\text(i)\\\text{Given equation is}\\\text{log k = 14.34 – 1.25 × 10}^{4} \text{K/T}\space...\text{(ii)}\\\text{Comparing (i) with (ii),}\space\frac{\text{E}_a}{2.303\text{RT}}=\frac{1.25×10^{4}\text{K}}{\text{T}}$$
or , Ea = 2.303 R × 1.25 × 104 K
= 2.303 × (8.314) × 1.25 × 104
= 239.34 kJ mol–1
$$\text{When}\qquad\text{t}_{1/2}=256\space\text{min},k=\frac{0.693}{256×60}=4.51×10^{\normalsize-5}s^{\normalsize-1}\\\text{Substituting this value in the given equation,}\\\text{log (4.51 × 10}^{\normalsize–5}) = 14.34-\frac{1.25×10^{4}\text{K}}{\text{T}}\\\text{i.e}\qquad(-5+0.6542)=14.34-\frac{1.25×10^{4}}{\text{T}}\\\text{Or}\qquad\frac{1.25×10^{4}\text{K}}{\text{T}}=18.6858$$
Or , T = 669 K
28. The decomposition of A into product has value of k as 4.5 × 103 s–1 at 10°C and energy of activation 60 kJ mol–1. At what temperature would k be 1.5 × 104 s–1 ?
Ans. Given: k1 = 4.5 × 103 s–1,
T1 = 10 + 273 K = 283 K; k2 = 1.5 × 104 s–1
Ea = 60 kJ mol–1
To find : T2 = ?
Formula : k = Ae–Ea/RT
Calculation:
Applying Arrhenius equation,
$$\text{log}\frac{k_2}{k_1}=\frac{\text{E}_a}{2.303\text{R}}\bigg(\frac{\text{T}_2-\text{T}_1}{\text{T}_1\text{T}_2}\bigg)\\\text{log}\frac{1.5×10^{4}}{4.5×10^{3}}=\frac{60000}{2.303×8.314}\bigg(\frac{\text{T}_2-283}{283\text{T}_2}\bigg)\\\text{Or,}\qquad\text{log}3.333=3133.63\bigg(\frac{\text{T}_2-283}{283\text{T}_2}\bigg)\\\text{Or,}\qquad\frac{0.5228}{3133.63}=\bigg(\frac{\text{T}_2-283}{283\text{T}_2}\bigg)\\\text{Or, 0.0472 T}_2 = \text{T}_2 – 283\\\text{Or}\qquad 0.9528 \text{T}_2 = 283\\\text{Or,}\qquad\text{T}_2=\frac{283}{0.9528}$$
= 297 – 273 = 24°C
Hence, the temperature would be 24°C.
29. The time required for 10% completion of the first order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the value of A is 4 × 1010s–1, calculated at 318 K and Ea.
Ans. Given: T1 = 298 K T2 = 308 K
A = 4 × 1010 s–1
To Find: Ea and k at 318 K
$$\text{Formula:}\qquad t=\frac{2.303}{k}\text{log}\frac{[\text{A}]_0}{\text{[A]}}\\k=\text{Ae}^{-\text{E}_a/\text{RT}}\\\text{(a) Calculation of activation energy (E}_a)\\\text{For I}^{\text{st}} \text{order reaction:}\space k=\frac{2.303}{t}\text{log}\frac{[\text{A}]_0}{[\text{A}]}\\\text{At 298 K;}\space k_1=\frac{2.303}{t}\text{log}\frac{100}{90}\qquad...\text{(i)}\\\text{At 308 K;}\space k_2=\frac{2.303}{t}\text{log}\frac{100}{75}\qquad...(\text{ii})\\\text{Dividing equation (ii) by (i)}\\\frac{k_2}{k_1}=\frac{\text{log}\frac{100}{75}}{\text{log}\frac{100}{90}}=\frac{0.1249}{0.0458}=2.73$$
According to Arrhenius theory;
$$\text{log}\frac{k_2}{k_1}=\frac{\text{E}_a}{2.303\text{R}}×\frac{\text{T}_2-\text{T}_1}{\text{T}_1\text{T}_2}\\\text{log}\space2.73=\frac{\text{E}_a}{2.303\text{R}}\bigg[\frac{308-298}{298×308}\bigg]\\\text{E}_a=\frac{0.4361×2.303×(8.314\text{J mol}^{\normalsize-1})×298×308}{10}\\\text{E}_a=76640\space\text{J mol}^{\normalsize-1}=76.640\space\text{kJ mol}^{\normalsize-1}\\\text{(b)Calculation of rate constant (k)}\\\text{According to Arrhenius equation}\\\text{logk}=\text{log A}\frac{\text{E}_a}{2.303\text{RT}}\\\text{log k}=\text{log(4×10}^{10})-\frac{76640\text{J mol}^{\normalsize-1}}{2.303×(8.314\text{J mol}^{\normalsize-1}\text{K}^{\normalsize-1})×(318\text{K})}$$
log k = 10.6021 – 12.5870 = –1.9849
k = Antilog (–1.9849)
Ea = 76.640 kJ mol–1
k = 1.035 × 10–2 s–1.
30. The rate of a reaction quadruples when the temperature changes from 293K to 313K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.
Ans. Given: r2 = 4r1, T1 = 293 K, T2 = 313 K,
To find: Ea
Formula: k = Ae–Ea/RT
Using Arrhenius equation,
$$\text{log}\frac{k_2}{k_1}=\frac{\text{E}_a}{2.303\text{R}}\bigg(\frac{1}{\text{T}_1}-\frac{1}{\text{T}_2}\bigg)\\\text{log 4}=\frac{\text{E}_a}{2.303×8.314}\bigg(\frac{1}{293}-\frac{1}{313}\bigg)\\\text{Or,}\qquad\text{E}_a=\text{log 4×}\frac{2.303×8.314×293×313}{20}$$
Or , Ea = 52.864 kJ
Hence, activation energy is 52.864 kJ.
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