Q – 4.1. From the rate expressions for the following reactions, determine their order of reaction and the dimensions of the rate constants:
3 NO (g) ⟶ N2O (g) + NO2 (g) ; Rate = k [NO]2
H2O2 (aq) + 3 I- (aq) + H+ ⟶ 2 H2O (I) + I-3 ;
Rate = k [H2O2] [I-]
CH3CHO (g) ⟶ CH4 (g) + CO (g)
Rate = k [CH3CHO]3/2
C2H5Cl (g) ⟶ C2H4 (g) + HCI (g) ;
Rate = k [C2H5Cl].
3 NO (g) ⟶ N2O (g) + NO2 (g) ; Rate = k [NO]2
H2O2 (aq) + 3 I- (aq) + H+ ⟶ 2 H2O (I) + I-3 ;
Rate = k [H2O2] [I-]
CH3CHO (g) ⟶ CH4 (g) + CO (g)
Rate = k [CH3CHO]3/2
C2H5Cl (g) ⟶ C2H4 (g) + HCI (g) ;
Rate = k [C2H5Cl].
Q – 4.2. For the reaction 2 A + B ⟶ A2B, 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 and [B] = 0.2 mol L-1. Calculate the rate of reaction after [A] is reduced to 0.06 mol L-1.
Ans. Initial rate = k [A] [B]2 = (2.0 × 10-6 mol-2 L2s-1)
(0.1 mol L-1) (0.2 mol L-1)2 = 8 × 10-9 mol L-1 s-1
When [A] is reduced from 0.10 mol L -1 to 0.06 mol L -1,
i.e., 0.04 mol L-1 of A has reacted,
B reacted = 1/2 × 0.04 mol L-1= 0.02 mol L-1.
Hence, new [B] = 0.2 – 0.02 = 0.18 mol L -1.
Now, rate = (2.0 × 10-6mol -2 L 2 s-1)(0.06 mol L -1) (0.18 mol L-1)2
= 3.89 × 10-9 mol L-1 s-1.
(0.1 mol L-1) (0.2 mol L-1)2 = 8 × 10-9 mol L-1 s-1
When [A] is reduced from 0.10 mol L -1 to 0.06 mol L -1,
i.e., 0.04 mol L-1 of A has reacted,
B reacted = 1/2 × 0.04 mol L-1= 0.02 mol L-1.
Hence, new [B] = 0.2 – 0.02 = 0.18 mol L -1.
Now, rate = (2.0 × 10-6mol -2 L 2 s-1)(0.06 mol L -1) (0.18 mol L-1)2
= 3.89 × 10-9 mol L-1 s-1.
Q – 4.3. The decomposition of NH3 on platinum surface,
2 NH3 (g) □(→┴Pt ) N2 (g) + 3 H3 (g), is zero order reaction with k = 2.5 × 10-4 mol-1 L s-1 . What are the rates of production of N2 and H2?
2 NH3 (g) □(→┴Pt ) N2 (g) + 3 H3 (g), is zero order reaction with k = 2.5 × 10-4 mol-1 L s-1 . What are the rates of production of N2 and H2?
Q – 4.4. The decomposition of dimethyl ether leads to the formation of CH4, H2 and Co and reaction rate is given rate = k [P_(CH_2 OCH_3 )]3/2.
If the pressure is measured in bar time in minutes, then what are the units of the rate and rate constants?
If the pressure is measured in bar time in minutes, then what are the units of the rate and rate constants?
Ans. In terms of pressures, units of rate = bar -1/2 min-1
Q – 4.5. Mention the factors that effect the rate of a chemical reaction.
Ans. In the example discussed earlier,
i.e., PCl5 ⟶ PCl3 + Cl2, 1 mole of PCI5 dissociates to from 1 mole of PCl3 and 1 mole of Cl2, i.e., the stoichiometric coefficient of each reactant and product is the same. For such a case, the rate of reaction will be same whether we express it in terms of decrease in the concentration of PCI5 or increase in the concentration of PCI3 and Cl2. However, a difficulty arises, if the stoichiometric coefficients of a reaction, i.e., coefficients of reactants and products in the balanced equation are not same. For example, consider the reaction,
2N2O5 ⟶ 4NO2 + O2.
For this reaction, rate of decomposition (disappearance) of
i.e., PCl5 ⟶ PCl3 + Cl2, 1 mole of PCI5 dissociates to from 1 mole of PCl3 and 1 mole of Cl2, i.e., the stoichiometric coefficient of each reactant and product is the same. For such a case, the rate of reaction will be same whether we express it in terms of decrease in the concentration of PCI5 or increase in the concentration of PCI3 and Cl2. However, a difficulty arises, if the stoichiometric coefficients of a reaction, i.e., coefficients of reactants and products in the balanced equation are not same. For example, consider the reaction,
2N2O5 ⟶ 4NO2 + O2.
For this reaction, rate of decomposition (disappearance) of
Q – 4.6. 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 1/2?
Q – 4.7. What is the effect of temperature on the rate constant of a reaction? How can this temperature effect on the rate constant be represented quantitatively?
Ans. the rate constant of a reaction increases with increase of temperature and becomes nearly double for every 100 rise of temperature. The effect can be represented quantitatively by Arrhenius equation,
where Ea represents the activation energy of the reaction and A represents the frequency factor.
where Ea represents the activation energy of the reaction and A represents the frequency factor.
Q – 4.8. In a pseudo first order hydrolysis of an ester in water, the following results were obtained:
t/s 0 30 60 90
[Ester]/mol l-1 0.55 0.31 0.17 0.085
Calculate the average rate of reaction between the time interval 30 to 60 seconds.
Calculate the pseudo first order rate constant for the hydrolysis if ester.
t/s 0 30 60 90
[Ester]/mol l-1 0.55 0.31 0.17 0.085
Calculate the average rate of reaction between the time interval 30 to 60 seconds.
Calculate the pseudo first order rate constant for the hydrolysis if ester.
Ans. (i) Average rate during the interval 30-60 sec.
Q – 4.9. A reaction is first order in A and second order in B.
Write differential rate equation.
How is the rate affected on increasing the concentration of B three times?
How is the rate affected when concentration of both A and B is doubled?
Write differential rate equation.
How is the rate affected on increasing the concentration of B three times?
How is the rate affected when concentration of both A and B is doubled?
Q – 4.10. In reaction between A and B, the initial rate of reaction was measured for different initial concentrations of A and B as given below:
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 7.16 × 10-5
what is the order of reaction with respect to A and B ?
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 7.16 × 10-5
what is the order of reaction with respect to A and B ?
Ans. r0 = [A]α [B]β
(r0)1 = 5.07 × 10-5 = (0.20)α (0.30)β (i)
(r0)2 = 5.07 × 10-5 = (0.20)α (0.10)β (ii)
(r0)3 = 7.16 × 10-5 = (0.40)α (0.05)β (iii)
Log 1.412 = α log 2 or α = 0.1523/0.3010 = 0.5.
( or directly 1.412 = √2 = 21/2 ؞ 2α = 21/2 or α = 1/2)
Thus, order w.r.t. A = 0.5, order w.r.t. B = 0
(r0)1 = 5.07 × 10-5 = (0.20)α (0.30)β (i)
(r0)2 = 5.07 × 10-5 = (0.20)α (0.10)β (ii)
(r0)3 = 7.16 × 10-5 = (0.40)α (0.05)β (iii)
Log 1.412 = α log 2 or α = 0.1523/0.3010 = 0.5.
( or directly 1.412 = √2 = 21/2 ؞ 2α = 21/2 or α = 1/2)
Thus, order w.r.t. A = 0.5, order w.r.t. B = 0
Q – 4.11. The following rate data were obtained at 303 K for the following reaction : 2A + B ⟶ 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 2.4 × 10-2
What is the rate law? What is the order with respect to each reactant and the overall order? also calculate the rate constant and write its units.
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 2.4 × 10-2
What is the rate law? What is the order with respect to each reactant and the overall order? also calculate the rate constant and write its units.
Ans. From experiments I and IV, it may be noted that [B] is same but [A] has been made four times, the rate of reaction has also become four times. This means that w.r.t. A,
Rate ∝ [A]
From experiments II and III, it may be noted that [A] is kept same and [B] has been doubled, the rate of reaction has become four times. This means that w.r.t. B,
Rate ∝ [B]
Combining (i) and (ii), we get the rate law for the given reaction as:
Rate = k [A] [B]2
Thus, order w.r.t. A = 1, order w/r/t/ B = 2 and overall order of the reaction = 1 + 2 = 3
The rate constant and its units can be calculated from the data of each experiment using the expression.
؞ rate constant, k = 6.0 mol-2 L2 min-1
Alternatively: suppose order w.r.t. A is α and w.r.t B is β. Then the rate law will be Rate = k [A]α [B]β Our aim is to find α, β and k.
Substituting the values of expts I to IV, we get
(Rate)expt 1 = 6.0 × 10-3 = k (0.1)α (0.1)β (i)
(Rate)expt 2 = 7.2 × 10-2 = k (0.3)α (0.2)β (ii)
(Rate)expt 3 = 2.88 × 10-1 = k (0.3)α (0.4)β (iii)
(Rate)expt 4 = 2.4 × 10-2 = k (0.4)α (0.1)β (iv)
Rate ∝ [A]
From experiments II and III, it may be noted that [A] is kept same and [B] has been doubled, the rate of reaction has become four times. This means that w.r.t. B,
Rate ∝ [B]
Combining (i) and (ii), we get the rate law for the given reaction as:
Rate = k [A] [B]2
Thus, order w.r.t. A = 1, order w/r/t/ B = 2 and overall order of the reaction = 1 + 2 = 3
The rate constant and its units can be calculated from the data of each experiment using the expression.
؞ rate constant, k = 6.0 mol-2 L2 min-1
Alternatively: suppose order w.r.t. A is α and w.r.t B is β. Then the rate law will be Rate = k [A]α [B]β Our aim is to find α, β and k.
Substituting the values of expts I to IV, we get
(Rate)expt 1 = 6.0 × 10-3 = k (0.1)α (0.1)β (i)
(Rate)expt 2 = 7.2 × 10-2 = k (0.3)α (0.2)β (ii)
(Rate)expt 3 = 2.88 × 10-1 = k (0.3)α (0.4)β (iii)
(Rate)expt 4 = 2.4 × 10-2 = k (0.4)α (0.1)β (iv)
Q – 4.12. The reaction between A and B is first order w.r.t. A and zero order w.r.t. 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
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. The rate expression will be : Rate : k [A]1 [B]0 = k [A]
For expt. I, 2.0 × 10-2 = mol L-1 min-1 = k (0.1 M) or k = 0.2 min-1
For expt. II , 4.0 × 10-2 mol L-1 = 0.2 min-1 [A] or [A] = 0.2 mol L-1
For expt. III, rate = (0.2 min-) (0.4 mol L-1) = 0.08 mol L-1 min-1
For expt. IV, 2.0 × 10-2 mol L-1 min-1 = 0.2 min-1 [A] or [A] = 0.1 mol L-1
For expt. I, 2.0 × 10-2 = mol L-1 min-1 = k (0.1 M) or k = 0.2 min-1
For expt. II , 4.0 × 10-2 mol L-1 = 0.2 min-1 [A] or [A] = 0.2 mol L-1
For expt. III, rate = (0.2 min-) (0.4 mol L-1) = 0.08 mol L-1 min-1
For expt. IV, 2.0 × 10-2 mol L-1 min-1 = 0.2 min-1 [A] or [A] = 0.1 mol L-1
Q – 4.13. Calculate the half- life of a first order reaction from their rate constants given below :
200 s-1 (b) 2 min -1 (c) 4 year-1
200 s-1 (b) 2 min -1 (c) 4 year-1
Q – 4.14. The half – life for radioactive decay of 14 C is 5730 y. an archaeological artifact contained wood that had only 80 % of the 14 C found in living tree. Estimate the age of the sample.
Ans. Radioactive decay follows first order kinetics.
Q – 4.15. The rate of formation of a dimer in a second order dimerization reaction is 9.5 × 10-5 mol L-1 s-1 at 0.01 mol L-1
Monomer concentration. Calculate the rate constant.
Monomer concentration. Calculate the rate constant.
Ans. If the monomer is represented by A, then the reaction is :
2A ⟶ (A)2
As the reaction is of second order, the rate of reaction will be given by : Rate = k [A]2
2A ⟶ (A)2
As the reaction is of second order, the rate of reaction will be given by : Rate = k [A]2
Q – 4.16. the rate constant of 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?
Q – 4.17. During nuclear explosion, one of the products is90 Sr with half- life of 28.1 years. If 1 µg of 90Sr was absorbed in 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. As radioactive disintegrations follow first order kinetics,
Q – 4.18. For a first order reaction, that the time required for 99 % completion of a first order reaction is twice the required for the completion of 90%.
Q – 4.19. A first order reaction takes 40 min for 30% decomposition. Calculate t1/2.
Ans. 30% decomposition means that x = 30% of a = 0.30 a
As reaction is of 1st order,
Ans. 30% decomposition means that x = 30% of a = 0.30 a
As reaction is of 1st order,
Q – 4.20. For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained:
T(sec) (mm of Hg)
0 35.0
360 54.0
720 63.0
Calculate the rate constant.
T(sec) (mm of Hg)
0 35.0
360 54.0
720 63.0
Calculate the rate constant.
Ans. (CH3)2CH N = N CH(CH3)2 (g) ⟶ N2 (g) + C6H14 (g)
Initial pressure P0 0 0
After time t P0 – P p p
Total pressure after time t (Pt) = (P0 – p) + p + p = P0 + p or p = Pt – Po
a ∝ P0 and (a – x) ∝ P0 – p or substituting the value of p,
a – x ∝ P0 – (Pt – P0), i.e., (a – x) ∝ P0 – Pt
As decomposition of azoisoproapne is a first order reaction,
Initial pressure P0 0 0
After time t P0 – P p p
Total pressure after time t (Pt) = (P0 – p) + p + p = P0 + p or p = Pt – Po
a ∝ P0 and (a – x) ∝ P0 – p or substituting the value of p,
a – x ∝ P0 – (Pt – P0), i.e., (a – x) ∝ P0 – Pt
As decomposition of azoisoproapne is a first order reaction,
Q – 4.21. The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume:
SO2Cl2 (g) ⟶ SO2 (g) + Cl2 (g)
Experiment Time/s Total pressure/atm
1 0 0.5
2 100 0.6
Calculate the rate of reaction when total pressure is 0.65 atm.
SO2Cl2 (g) ⟶ SO2 (g) + Cl2 (g)
Experiment Time/s Total pressure/atm
1 0 0.5
2 100 0.6
Calculate the rate of reaction when total pressure is 0.65 atm.
Ans. SO2Cl2 (g) ⟶ So2 (g) + Cl2 (g)
Proceeding exactly similar to Q.20,
When Pt = 0.65 atm, i.e., P0 + p = 0.65 atm
؞ p = 0.65 – P0 = 0.65 – 0.50 = 0.15 atm
؞ Pressure of SO2Cl2 at time t (Pso2cl2)
= P0 – p = 0.50 – 0.15 atm = 0.35 atm
Rate at that time = k × Pso2cl2 = (2.2316 × 10-3 s-1) (0.35 atm) = 7.8 × 10-5 atm s-1.
Proceeding exactly similar to Q.20,
When Pt = 0.65 atm, i.e., P0 + p = 0.65 atm
؞ p = 0.65 – P0 = 0.65 – 0.50 = 0.15 atm
؞ Pressure of SO2Cl2 at time t (Pso2cl2)
= P0 – p = 0.50 – 0.15 atm = 0.35 atm
Rate at that time = k × Pso2cl2 = (2.2316 × 10-3 s-1) (0.35 atm) = 7.8 × 10-5 atm s-1.
Q – 4.22. The values of rate constant for the decomposition of HI into H2 and I2 at different temperatures are given below :
T/K 633 667 710 738
104 k/M-1s-1 0.19 1.00 8.31 25.1
Draw a graph between in k and 1/T and calculate the values of Arrhenius parameters.
T/K 633 667 710 738
104 k/M-1s-1 0.19 1.00 8.31 25.1
Draw a graph between in k and 1/T and calculate the values of Arrhenius parameters.
Ans. From the given data, we have
T(K) 633 667 710 738
1/T (K-1) 1.58 × 10-3 1.50 × 10-3 1.41 × 10-3 1.36 × 10-3
k(M-1s-1) 0.19 × 10-4 = 1.9 × 10-5 1.00 × 10-4 8.31 × 10-4 25.1 × 10-4 = 2.51 × 10-3
In k -10.87 -9.21 -7.09 -5.99
(=2.303 log k)
Graph of in k vs 1/T. the plot obtained is as shown in the Fig.
T(K) 633 667 710 738
1/T (K-1) 1.58 × 10-3 1.50 × 10-3 1.41 × 10-3 1.36 × 10-3
k(M-1s-1) 0.19 × 10-4 = 1.9 × 10-5 1.00 × 10-4 8.31 × 10-4 25.1 × 10-4 = 2.51 × 10-3
In k -10.87 -9.21 -7.09 -5.99
(=2.303 log k)
Graph of in k vs 1/T. the plot obtained is as shown in the Fig.
Q – 4.23. 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, what will be the value of pre- exponential factor?
Ans. here, k = 2.418 × 10-5 s-1, Ea = 179.9 kJ mol-1, T = 546 K.
According to Arrhenius equation,
According to Arrhenius equation,
Q – 4.24. 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.
Q – 4.25. 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 the sample of sucrose remains after 8 hours?
Ans. As sucrose decomposes according to first order rate law,
Q – 4.26. The decomposition of a hydrocarbon follows the equation k = (4.5 × 1011 s-1) e-28000 K/T calculate Ea.
Or E a = 28000 K × R = 28000 K × 8.314 JK-1 mol-1 = 232.79 kJ mol-1.
Q – 4.27. The rate constant for the first order decomposition of H2O2 is given by the following equation:
Calculate Ea for this reaction and at what temperature will its half- period be 256 minutes?
Calculate Ea for this reaction and at what temperature will its half- period be 256 minutes?
Q – 4.28. The decomposition of A into products has value of k as 4.5 × 103 s-1 at 100C and energy of activation 60 kJ MOL-1. At what temperature would k be 1.5 × 104 s-1?
Ans. k1 = 4.5 × 103 s-1, T1 = 10 + 273 K = 283 K ;
k2 = 1.5 × 104 s-1 , T2 = ?, Ea = 60 kJ mol-1
applying Arrhenius equation,
k2 = 1.5 × 104 s-1 , T2 = ?, Ea = 60 kJ mol-1
applying Arrhenius equation,
Q – 4.29. the time required for 10% completion of a 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 × 1010 s-1, Calculate k at 318 K and Ea.
