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Question1. For the reaction R → P, the concentration of a reactant changes from 0.03 M to 0.02 M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.
Solution :
Average rate of reaction
= 6.67 × 10^{−6 M s−1}
Question2. In a reaction, 2A → Products, the concentration of A decreases from 0.5 mol L^{−1} to 0.4 mol L^{−1} in 10 minutes. Calculate the rate during this interval?
Solution :
= 0.005 mol L^{−1 min−1}
= 5 × 10^{−3 }M min^{−1}
Question3. For a reaction, A + B → Product; the rate law is given by, r = k [A]^{½ }[B]^{2}. What is the order of the reaction?
Solution :
The order of the reaction r = k [A]^{½ }[B]^{2}
= 1 / 2 + 2
= 2.5
Question4. The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y?
Solution :
The reaction X → Y follows second order kinetics.
Therefore, the rate equation for this reaction will be:
Rate = k[X]^{2 (1)}
Let [X] = a mol L^{−1}, then equation (1) can be written as:
Rate_{1} = k .(a)^{2}
= ka^{2}
If the concentration of X is increased to three times, then [X] = 3a mol L^{−1}
Now, the rate equation will be:
Rate = k (3a)^{2}
= 9(ka^{2)}
Hence, the rate of formation will increase by 9 times.
Question5. A first order reaction has a rate constant 1.15 10^{−3 }s^{−1}. How long will 5 g of this reactant take to reduce to 3 g?
Solution :
From the question, we can write down the following information:
Initial amount = 5 g
Final concentration = 3 g
Rate constant = 1.15 10^{−3} s^{−1}
We know that for a 1^{st }order reaction,
= 444.38 s
= 444 s (approx)
Question6. Time required to decompose SO_{2}Cl_{2} to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.
Solution :
We know that for a 1^{st} order reaction,
t_{½ }= 0.693 / k
It is given that t_{1/2} = 60 min
k = 0.693 / t½
= 0.693 / 60
= 0.01155 min^{-1}
= 1.155 min^{-1}
Or
k = 1.925 x 10-2 s^{-1}
Question7. What will be the effect of temperature on rate constant?
Solution :
The rate constant of a reaction is nearly doubled with a 10° rise in temperature. However, the exact dependence of the rate of a chemical reaction on temperature is given by Arrhenius equation,
k = Ae - E_{a} / RT
Where,
A is the Arrhenius factor or the frequency factor
T is the temperature
R is the gas constant
E_{a} is the activation energy
Question8. The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate E_{a}.
Solution :
It is given that T_{1} = 298 K
∴T_{2} = (298 + 10) K
= 308 K
We also know that the rate of the reaction doubles when temperature is increased by 10°.
Therefore, let us take the value of k_{1} = k and that of k_{2} = 2k
Also, R = 8.314 J K^{−1 mol−1}
Now, substituting these values in the equation:
= 52897.78 J mol^{−1}
= 52.9 kJ mol^{−1}
Question9. The activation energy for the reaction 2HI_{(}_{g}_{)} → H_{2} + I_{2(}_{g}_{)} is 209.5 kJ mol^{−1} at 581K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?
Solution :
In the given case:
E_{a} = 209.5 kJ mol^{−1} = 209500 J mo^{l−1}
T = 581 K
R = 8.314 JK^{−1} mol^{−1}
Now, the fraction of molecules of reactants having energy equal to or greater than activation energy is given as:
x = e^{−Ea / RT}
⇒Inx= −Ea / RT
⇒logx=−Ea / 2.303RT
⇒logx= −209500Jmol^{−1} / 2.303 × 8.314JK^{−1}mol^{−1}×581
=−18.8323
Now,x= Antilog (−18.8323)
=1.471×10^{−19}
Question10. From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.
(i) 3 NO(g) → N_{2}O_{ }(g) Rate = k[NO]^{2}
(ii) H_{2}O_{2 }(aq) + 3 I^{−}_{ (aq) }+ 2 H^{+} → 2 H_{2}O (l) + I_{3}^{-} Rate = k[H_{2}O_{2}][I^{−}]
(iii) CH_{3}CHO(g) → CH_{4}(g) + CO(g) Rate = k [CH_{3}CHO]^{3/2}
(iv) C_{2}H_{5}Cl(g) → C_{2}H_{4}(g) + HCl(g) Rate = k [C_{2}H_{5}Cl]
Solution :
(i) Given rate = k [NO]^{2}
Therefore, order of the reaction = 2
Dimension of k = Rate / [NO]^{2}
= mol L^{-1} s^{-1 }/ (mol L-1)^{2}
= mol L^{-1} s^{-1} / mol^{2} L^{-2}
= L mol^{-1}s^{-1}
(ii) Given rate = k [H_{2}O_{2}] [I^{−}]
Therefore, order of the reaction = 2
Dimension of
k = Rate / [H_{2}O_{2}][I ^{-} ]
= mol L^{-1} s^{-1} / (mol L^{-1}) (mol L^{-1})
= L mol^{-1} s^{-1}
(iii) Given rate = k [CH_{3}CHO]^{3/2}
Therefore, order of reaction = 3 / 2
Dimension of k = Rate / [CH3CHO]^{3/2}
= mol L^{-1} s^{-1} / (mol L^{-1})^{3/2}
= mol L^{-1} s^{-1} / mol^{3/2} L^{-3/2}
= L½ mol-½ s-1
(iv) Given rate = k [C_{2}H_{5}Cl]
Therefore, order of the reaction = 1
Dimension of k = Rate / [C_{2}H_{5}Cl]
= mol L^{-1} s^{-1 } / mol L^{-1}
= s^{-1}
Question11. For the reaction:
2A + B → A_{2}B
the rate = k[A][B]^{2} with k = 2.0 × 10^{−6} mol^{−2} L_{2} 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}.
Solution :
The initial rate of the reaction is
Rate = k [A][B]^{2}
= (2.0 × 10^{−6 mol−2} L^{2} s^{−1}) (0.1 mol L^{−1}) (0.2 mol L^{−1})^{2}
= 8.0 × 10^{−9 }mol^{−2} L^{2} s^{−1}
When [A] is reduced from 0.1 mol L^{−1 }to 0.06 mol^{−1}, the concentration of A reacted = (0.1 − 0.06) mol L^{−1 }= 0.04 mol L^{−1}
Therefore, concentration of B reacted 1/2 x 0.04 mol L^{-1} = 0.02 mol L^{−1}
Then, concentration of B available, [B] = (0.2 − 0.02) mol L^{−1}
= 0.18 mol L^{−1}
After [A] is reduced to 0.06 mol L^{−1}, the rate of the reaction is given by,
Rate = k [A][B]^{2}
= (2.0 × 10^{−6 }mol^{−2} L^{2} s^{−1}) (0.06 mol L^{−1}) (0.18 mol L^{−1})^{2}
= 3.89 mol L^{−1} s^{−1}
Question12. The decomposition of NH_{3} on platinum surface is zero order reaction. What are the rates of production of N_{2} and H_{2} if k = 2.5 × 10^{−4} mol^{−1} L s^{−1}?
Solution :
The decomposition of NH_{3} on platinum surface is represented by the following equation.
= 7.5 × 10^{−4 }mol L^{−1 }s^{−1}
Question13. The decomposition of dimethyl ether leads to the formation of CH_{4}, H_{2} and CO and the reaction rate is given by
Rate = k [CH_{3}OCH_{3}]^{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 (P_{CH3OCH3})^{3/2}
If the pressure is measured in bar andtime in minutes, then what are the units of rate and rate constants?
Solution :
If pressure is measured in bar and time in minutes, then
Unit of rate = bar min^{−1}
Rate = k [CH_{3}OCH_{3}]^{3/2}
⇒ k =Rate / [CH_{3}OCH_{3}]^{3/2}
Therefore, unit of rate constants(k) = bar min^{−1 /} bar^{3/2}
= bar^{-½} min ^{-1}
Question14. Mention the factors that affect the rate of a chemical reaction.
Solution :
The factors that affect the rate of a reaction are as follows.
(i) Concentration of reactants (pressure in case of gases)
(ii) Temperature
(iii) Presence of a catalyst
15. 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?
Solution :
Letthe concentration of the reactant be [A] = a
Rate of reaction, R = k [A]^{2}
= ka^{2}
(i)If the concentration of the reactant is doubled, i.e. [A] = 2a, then the rate of the reaction would be
R = k(2a)^{2}
= 4ka^{2}
= 4 R
Therefore, the rate of the reaction would increase by 4 times.
(ii) If the concentration of the reactant is reduced to half, i.e. [A] = 1/2 a , then the rate of the reaction would be
R = k(1/2a)^{2}
= 1/4 Ka^{2}
= 1/4 R
Therefore, the rate of the reaction would be reduced to
16. What change would happen in the rate constant of a reaction when there is a change in its temperature? How can this temperature effect on rate constant be represented quantitatively?
Solution :
When a temperature of 10∘ rises for a chemical reaction then the rate constant increases and becomes near to double of its original value.
The temperature effect on the rate constant can be represented quantitatively by Arrhenius equation,
k=Ae^{−Ea/RT}
Where,
k = rate constant,
A = Frequency factor / Arrhenius factor,
R = gas constant
T = temperature
Ea = activation energy for the reaction.
Question17. In a pseudo first order hydrolysis of 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 |
(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.
Solution :
(i) Average rate of reaction between the time interval, 30 to 60 seconds, d[ester] / dt
= (0.31-0.17) / (60-30)
= 0.14 / 30
= 4.67 × 10^{−3} mol L^{−1} s^{−1}
(ii) For a pseudo first order reaction,
k = 2.303/ t log [R]º / [R]
For t = 30 s, k_{1}
= 1.911 × 10^{−2 }s^{−1}
For t = 60 s, k1 = 2.303/ 30 log 0.55 / 0.31
= 1.957 × 10^{−2 }s^{−1}
For t = 90 s, k3 = 2.303/ 90 log 0.55 / 0.085
= 2.075 × 10 - 2s - 1
= 2.075 × 10^{−2 }s^{−1}
Then, average rate constant, k = k_{1} + k_{2}+ k_{3} / 3
= 1.911 × 10 ^{- 2 } + 1.957 × 10 ^{- 2} + 2.075 × 10^{ - 2 / 3}
= 1.981 x 10^{-2} s - 1
Question18. 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 are doubled?
Solution :
(i) The differential rate equation will be
Question19. In a reaction between A and B, the initial rate of reaction (r_{0}) 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 |
r_{0}/ 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?
Solution :
Let the order of the reaction with respect to A be x and with respect to B be y.
Therefore,
Dividing equation (iii) by (ii), we obtain
= 1.496
= 1.5 (approximately)
Hence, the order of the reaction with respect to A is 1.5 and with respect to B is zero.
Question20. The following results have been obtained during the kinetic studies of the 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 |
2.40 × 10^{−2} |
Determine the rate law and the rate constant for the reaction.
Solution :
Let the order of the reaction with respect to A be x and with respect to B be y.
Therefore, rate of the reaction is given by,
Dividing equation (iv) by (i), we obtain
Dividing equation (iii) by (ii), we obtain
Question21. 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} |
Solution :
The given reaction is of the first order with respect to A and of zero order with respect to B.
Therefore, the rate of the reaction is given by,
Rate = k [A]^{1 }[B]^{0}
⇒ Rate = k [A]
From experiment I, we obtain
2.0 × 10^{−2 }mol^{ }L^{−1} min^{−1} = k (0.1 mol L^{−1})
⇒ k = 0.2 min^{−1}
From experiment II, we obtain
4.0 × 10^{−2 mol }L^{−1} min^{−1} = 0.2 min^{−1} [A]
⇒ [A] = 0.2 mol L^{−1}
From experiment III, we obtain
Rate = 0.2 min^{−1} × 0.4 mol L^{−1}
= 0.08 mol L^{−1} min^{−1}
From experiment IV, we obtain
2.0 × 10^{−2 }mol^{ }L^{−1} min^{−1} = 0.2 min^{−1} [A]
⇒ [A] = 0.1 mol L^{−1}
Question22. Calculate the half-life of a first order reaction from their rate constants given below:
(i) 200 s^{−1} (ii) 2 min^{−1} (iii) 4 years^{−1}
Solution :
(i) Half life, t _{1/2} = 0.693 / k
= 0.693 / 200 s^{-1}
= 3.47×10 ^{-3 }s (approximately)
(ii) Half life, t _{1/2} = 0.693 / k
= 0.693 / 2 min-1
= 0.35 min (approximately)
(iii) Half life, t _{1/2} = 0.693 / k
= 0.693 / 4 years^{-1}
= 0.173 years (approximately)
Question23. The half-life for radioactive decay of 14C 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.
Solution :
Here, k = 0.693 / t_{1/2}
= 0.693 / 5730 years^{-1}
It is known that,
= 1845 years (approximately)
Hence, the age of the sample is 1845 years.
Question24. The experimental data for decomposition of N_{2}O_{5}
[2N2O5 → 4NO2 + O2] in gas phase at 318K are given below:
t(s) |
0 |
400 |
800 |
1200 |
1600 |
2000 |
2400 |
2800 |
3200 |
10^{2} × [N_{2}O_{5}] mol L^{-1} |
1.63 |
1.36 |
1.14 |
0.93 |
0.78 |
0.64 |
0.53 |
0.43 |
0.35 |
(i) Plot [N_{2}O_{5}] against t.
(ii) Find the half-life period for the reaction.
(iii) Draw a graph between log [N_{2}O_{5}] 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).
Solution :
(ii) Time corresponding to the concentration, 1630x10^{2} / 2 mol L^{-1} = 81.5 mol L^{-1} is the half life. From the graph, the half life is obtained as 1450 s.
(iii)
t(s) |
10^{2} × [N_{2}O_{5}] mol L^{-1} |
Log[N_{2}O_{5}] |
0 |
1.63 |
− 1.79 |
400 |
1.36 |
− 1.87 |
800 |
1.14 |
− 1.94 |
1200 |
0.93 |
− 2.03 |
1600 |
0.78 |
− 2.11 |
2000 |
0.64 |
− 2.19 |
2400 |
0.53 |
− 2.28 |
2800 |
0.43 |
− 2.37 |
3200 |
0.35 |
− 2.46 |
(iv) The given reaction is of the first order as the plot, Log[N_{2}O_{5}] v/s t, is a straight line. Therefore, the rate law of the reaction is
Rate = k [N_{2}O_{5}]
(v) From the plot, Log[N_{2}O_{5}] v/s t, we obtain
- k /2.303
Again, slope of the line of the plot Log[N_{2}O_{5}] v/s t is given by
- k / 2.303. = -0.67 / 3200
Therefore, we obtain,
- k / 2.303 = - 0.67 / 3200
⇒ k = 4.82 x 10-4 s-1
(vi) Half-life is given by,
t_{½} = 0.693 / k
= 0.639 / 4.82x10^{-4} s
=1.438 x 103
This value, 1438 s, is very close to the value that was obtained from the graph.
25. 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 1/16^{th} value?
Solution :
It is known that,
Question26. During nuclear explosion, one of the products is 90Sr with half-life of 28.1 years. If 1μg 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.
Solution :
,
Therefore, 0.7814 μg of 90^{Sr} will remain after 10 years.
Again,
Therefore, 0.2278 μg of 90^{Sr} will remain after 60 years.
Question27. For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
Solution :
For a first order reaction, the time required for 99% completion is
t_{1} = 2.303/k Log 100/100-99
= 2.303/k Log 100
= 2x 2.303/k
For a first order reaction, the time required for 90% completion is
t_{2} = 2.303/k Log 100 / 100-90
= 2.303/k Log 10
= 2.303/k
Therefore, t_{1} = 2t_{2}
Hence, the time required for 99% completion of a first order reaction is twice the time required for the completion of 90% of the reaction.
Question28. A first order reaction takes 40 min for 30% decomposition. Calculate t_{1/2}.
Solution :
For a first order reaction,
t = 2.303/k Log [R] º / [R]
k = 2.303/40min Log 100 / 100-30
= 2.303/40min Log 10 / 7
= 8.918 x 10-3 min^{-1}
Therefore, t_{1/2} of the decomposition reaction is
t1/2 = 0.693/k
= 0.693 / 8.918 x 10-3 min
= 77.7 min (approximately)
= 77.7 min (approximately)
Question29. 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.
Solution :
The decomposition of azoisopropane to hexane and nitrogen at 543 K is represented by the following equation.
After time, t, total pressure, Pt = (P_{º} - p) + p + p
⇒ Pt = (P_{º }+ p)
⇒ p = P_{t} - P_{0}
therefore, P_{º} - p = P_{0} - Pt - P_{0}
= 2P_{0} − P_{t}
For a first order reaction,
k = 2.303/t Log P_{0} /P_{0} - p
= 2.303/t Log P_{0} / 2 P_{0} - P_{t}
When t = 360 s, k = 2.303 / 360s log 35.0 / 2x35.0 - 54.0
= 2.175 × 10^{−3 }s^{−1}
When t = 720 s, k = 2.303 / 720s log 35.0 / 2x35.0 - 63.0
= 2.235 × 10^{−3} s^{−1}
Hence, the average value of rate constant is
k = (2.175 × 10^{ - 3} + 2.235 × 10 ^{- 3} ) / 2 s ^{- 1}
= 2.21 × 10^{−3 }s^{−1}
Question30. The following data were obtained during the first order thermal decomposition of SO_{2}Cl_{2} at a constant volume.
SO_{2}Cl_{2}(g) → SO_{2}(g) + Cl_{2}(g)
Experiment |
Time/s^{−1} |
Total pressure/atm |
1 |
0 |
0.5 |
2 |
100 |
0.6 |
Calculate the rate of the reaction when total pressure is 0.65 atm.
Solution :
The thermal decomposition of SO_{2}Cl_{2} at a constant volume is represented by the following equation.
After time, t, total pressure, P_{t} = (P_{º} - p) + p + p
⇒ P_{t} = (P_{º} + p)
⇒ p = Pt - Pº
therefore, Pº - p = Pº - Pt - Pº
= 2 P_{º }- P_{t}
For a first order reaction,
k = 2.303/t Log P_{º }/ P_{º} - p
= 2.303/t Log P_{º} / 2 P_{º} - Pt
When t= 100 s,
k = 2.303 / 100s log 0.5 / 2x0.5 - 0.6
= 2.231 × 10 - 3s - 1
When Pt= 0.65 atm,
P0+ p= 0.65
⇒ p= 0.65 - P0
= 0.65 - 0.5
= 0.15 atm
Therefore, when the total pressure is 0.65 atm, pressure of SOCl2 is
_{PSOCL2} = P_{0} - p
= 0.5 - 0.15
= 0.35 atm
Therefore, the rate of equation, when total pressure is 0.65 atm, is given by,
Rate = k(p_{SOCL2})
= (2.23 × 10 - 3s - 1) (0.35 atm)
= 7.8 × 10^{ - 4} atm s^{ - 1}
Question31. The rate constant for the decomposition of N_{2}O_{5} at various temperatures is given below:
T/°C |
0 |
20 |
40 |
60 |
80 |
10^{5} X K /S^{-1} |
0.0787 |
1.70 |
25.7 |
178 |
2140 |
Draw a graph between ln k and 1/T and calculate the values of A and E_{a}.
Predict the rate constant at 30º and 50ºC.
Solution :
From the given data, we obtain
T/°C |
0 |
20 |
40 |
60 |
80 |
T/K |
273 |
293 |
313 |
333 |
353 |
1/T / k^{-1} |
3.66×10^{−3} |
3.41×10^{−3} |
3.19×10^{−3} |
3.0×10^{−3} |
2.83 ×10^{−3} |
10^{5} X K /S^{-1} |
0.0787 |
1.70 |
25.7 |
178 |
2140 |
ln k |
−7.147 |
− 4.075 |
−1.359 |
−0.577 |
3.063 |
Slope of the line,
In k= - 2.8
Therefore, k = 6.08x10-2s-1
Again when T = 50 + 273K = 323K,
1/T = 3.1 x 10-3 K
In k = - 0.5
Therefore, k = 0.607 s-1
Question32. The rate constant for the decomposition of hydrocarbons 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.
Solution :
k = 2.418 × 10^{−5} s^{−1}
T = 546 K
E_{a} = 179.9 kJ mol^{−1} = 179.9 × 10^{3 }J mol^{−1}
According to the Arrhenius equation,
= (0.3835 − 5) + 17.2082
= 12.5917
Therefore, A = antilog (12.5917)
= 3.9 × 10^{12} s^{−1} (approximately)
Question33. 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}.
Solution :
k = 2.0 × 10^{−2} s^{−1}
T = 100 s
[A]_{o} = 1.0 moL^{−1}
Since the unit of k is s^{−1}, the given reaction is a first order reaction.
Therefore, k = 2.303/t Log [A]º / [A]
⇒2.0 × 110-2 s-1 = 2.303/100s Log 1.0 / [A]
⇒2.0 × 110-2 s-1 = 2.303/100s ( - Log [A] )
⇒ - Log [A] = - (2.0 x 10-2 x 100) / 2.303
⇒ [A] = antilog [- (2.0 x 10-2 x 100) / 2.303]
= 0.135 mol L^{−1 (approximately)}
Hence, the remaining concentration of A is 0.135 mol L^{−1.}
Question34. Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t_{1/2 }= 3.00 hours. What fraction of sample of sucrose remains after 8 hours?
Solution :^{For a first order reaction, k = 2.303/t Log [R]º / [R] It is given that, t1/2 = 3.00 hours Therefore, k = 0.693 / t1/2 = 0.693 / 3 h-1 = 0.231 h - 1 Then, 0.231 h - 1 = 2.303 / 8h Log [R]º / [R]}
Hence, the fraction of sample of sucrose that remains after 8 hours is 0.158.
Question35. The decomposition of hydrocarbon follows the equation k = (4.5 × 10^{11} s^{−1}) e^{−28000} K/T Calculate E_{a}.
Solution :
The given equation is
k = (4.5 × 10^{11 }s^{−1}) e^{−28000} K/T (i)
Arrhenius equation is given by,
k= Ae^{ -Ea/RT} (ii)
From equation (i) and (ii), we obtain
Ea / RT = 28000K / T
⇒ Ea = R x 28000K
= 8.314 J K^{−1} mol^{−1} × 28000 K
= 232792 J mol^{−1}
= 232.792 kJ mol^{−1}
Question36. The rate constant for the first order decomposition of H_{2}O_{2} is given by the following equation:
log k = 14.34 − 1.25 × 10^{4 K/T}
Calculate E_{a} for this reaction and at what temperature will its half-period be 256 minutes?
Solution :
Arrhenius equation is given by,
k= Ae -Ea/RT
⇒In k = In A - Ea/RT
⇒In k = Log A - Ea/RT
⇒ Log k = Log A - Ea/2.303RT (i)
The given equation is
Log k = 14.34 - 1.25 104 K/T (ii)
From equation (i) and (ii), we obtain
Ea/2.303RT = 1.25 104 K/T
= 1.25 × 10^{4} K × 2.303 × 8.314 J K^{−1} mol^{−1}
= 239339.3 J mol^{−1 }(approximately)
= 239.34 kJ mol^{−1}
Also, when t_{1/2} = 256 minutes,
k = 0.693 / t^{1/2}
= 0.693 / 256
= 2.707 × 10^{ - 3 }min^{ - 1}
= 4.51 × 10 - 5s^{ - 1}
= 2.707 × 10^{−3} min^{−1}
= 4.51 × 10^{−5} s^{−1}
It is also given that, log k = 14.34 − 1.25 × 10^{4 }K/T
= 668.95 K
= 669 K (approximately)
Question37. The decomposition of A into product has value of k as 4.5 × 10^{3 }s^{−1} at 10°C and energy of activation 60 kJ mol^{−1}. At what temperature would k be 1.5 × 10^{4} s^{−1}?
Solution :
From Arrhenius equation, we obtain
log k2/k1 = Ea / 2.303 R (T_{2} - T_{1}) / T_{1}T_{2}
Also, k_{1} = 4.5 × 10^{3 s−1}
T_{1} = 273 + 10 = 283 K
k_{2} = 1.5 × 10^{4} s^{−1}
E_{a} = 60 kJ mol^{−1} = 6.0 × 10^{4} J mol^{−1}
Then,
= 297 K
= 24°C
Hence, k would be 1.5 × 10^{4 }s^{−1} at 24°C.
Question38. 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 × 10^{10 }s^{−1}. Calculate k at 318 K and E_{a}.
Solution :
For a first order reaction,
t = 2.303 / k log a / a - x
At 298 K, t = 2.303 / k log 100 / 90
= 0.1054 / k
At 308 K, t' = 2.303 / k' log 100 / 75
= 2.2877 / k'
According to the question,
t = t'
⇒ 0.1054 / k = 2.2877 / k'
⇒ k' / k = 2.7296
From Arrhenius equation, we obtain
To calculate k at 318 K,
It is given that, A = 4 x 1010 s-1, T = 318K
Again, from Arrhenius equation, we obtain
Therefore, k = Antilog (-1.9855)
= 1.034 x 10-2 s -1
Question39. The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.
Solution :
From Arrhenius equation, we obtain
Hence, the required energy of activation is 52.86 kJmol^{−1.}
Question4. The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y?
Solution :
The reaction X → Y follows second order kinetics.
Therefore, the rate equation for this reaction will be:
Rate = k[X]^{2 (1)}
Let [X] = a mol L^{−1}, then equation (1) can be written as:
Rate_{1} = k .(a)^{2}
= ka^{2}
If the concentration of X is increased to three times, then [X] = 3a mol L^{−1}
Now, the rate equation will be:
Rate = k (3a)^{2}
= 9(ka^{2)}
Hence, the rate of formation will increase by 9 times.
Question5. A first order reaction has a rate constant 1.15 10^{−3 }s^{−1}. How long will 5 g of this reactant take to reduce to 3 g?
Solution :
From the question, we can write down the following information:
Initial amount = 5 g
Final concentration = 3 g
Rate constant = 1.15 10^{−3} s^{−1}
We know that for a 1^{st }order reaction,
= 444.38 s
= 444 s (approx)
Question6. Time required to decompose SO_{2}Cl_{2} to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.
Solution :
We know that for a 1^{st} order reaction,
t_{½ }= 0.693 / k
It is given that t_{1/2} = 60 min
k = 0.693 / t½
= 0.693 / 60
= 0.01155 min^{-1}
= 1.155 min^{-1}
Or
k = 1.925 x 10-2 s^{-1}
Question7. What will be the effect of temperature on rate constant?
Solution :
The rate constant of a reaction is nearly doubled with a 10° rise in temperature. However, the exact dependence of the rate of a chemical reaction on temperature is given by Arrhenius equation,
k = Ae - E_{a} / RT
Where,
A is the Arrhenius factor or the frequency factor
T is the temperature
R is the gas constant
E_{a} is the activation energy
Question8. The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate E_{a}.
Solution :
It is given that T_{1} = 298 K
∴T_{2} = (298 + 10) K
= 308 K
We also know that the rate of the reaction doubles when temperature is increased by 10°.
Therefore, let us take the value of k_{1} = k and that of k_{2} = 2k
Also, R = 8.314 J K^{−1 mol−1}
Now, substituting these values in the equation:
= 52897.78 J mol^{−1}
= 52.9 kJ mol^{−1}
Question9. The activation energy for the reaction 2HI_{(}_{g}_{)} → H_{2} + I_{2(}_{g}_{)} is 209.5 kJ mol^{−1} at 581K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?
Solution :
In the given case:
E_{a} = 209.5 kJ mol^{−1} = 209500 J mo^{l−1}
T = 581 K
R = 8.314 JK^{−1} mol^{−1}
Now, the fraction of molecules of reactants having energy equal to or greater than activation energy is given as:
x = e^{−Ea / RT}
⇒Inx= −Ea / RT
⇒logx=−Ea / 2.303RT
⇒logx= −209500Jmol^{−1} / 2.303 × 8.314JK^{−1}mol^{−1}×581
=−18.8323
Now,x= Antilog (−18.8323)
=1.471×10^{−19}
Question10. From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.
(i) 3 NO(g) → N_{2}O_{ }(g) Rate = k[NO]^{2}
(ii) H_{2}O_{2 }(aq) + 3 I^{−}_{ (aq) }+ 2 H^{+} → 2 H_{2}O (l) + I_{3}^{-} Rate = k[H_{2}O_{2}][I^{−}]
(iii) CH_{3}CHO(g) → CH_{4}(g) + CO(g) Rate = k [CH_{3}CHO]^{3/2}
(iv) C_{2}H_{5}Cl(g) → C_{2}H_{4}(g) + HCl(g) Rate = k [C_{2}H_{5}Cl]
Solution :
(i) Given rate = k [NO]^{2}
Therefore, order of the reaction = 2
Dimension of k = Rate / [NO]^{2}
= mol L^{-1} s^{-1 }/ (mol L-1)^{2}
= mol L^{-1} s^{-1} / mol^{2} L^{-2}
= L mol^{-1}s^{-1}
(ii) Given rate = k [H_{2}O_{2}] [I^{−}]
Therefore, order of the reaction = 2
Dimension of
k = Rate / [H_{2}O_{2}][I ^{-} ]
= mol L^{-1} s^{-1} / (mol L^{-1}) (mol L^{-1})
= L mol^{-1} s^{-1}
(iii) Given rate = k [CH_{3}CHO]^{3/2}
Therefore, order of reaction = 3 / 2
Dimension of k = Rate / [CH3CHO]^{3/2}
= mol L^{-1} s^{-1} / (mol L^{-1})^{3/2}
= mol L^{-1} s^{-1} / mol^{3/2} L^{-3/2}
= L½ mol-½ s-1
(iv) Given rate = k [C_{2}H_{5}Cl]
Therefore, order of the reaction = 1
Dimension of k = Rate / [C_{2}H_{5}Cl]
= mol L^{-1} s^{-1 } / mol L^{-1}
= s^{-1}
Question11. For the reaction:
2A + B → A_{2}B
the rate = k[A][B]^{2} with k = 2.0 × 10^{−6} mol^{−2} L_{2} 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}.
Solution :
The initial rate of the reaction is
Rate = k [A][B]^{2}
= (2.0 × 10^{−6 mol−2} L^{2} s^{−1}) (0.1 mol L^{−1}) (0.2 mol L^{−1})^{2}
= 8.0 × 10^{−9 }mol^{−2} L^{2} s^{−1}
When [A] is reduced from 0.1 mol L^{−1 }to 0.06 mol^{−1}, the concentration of A reacted = (0.1 − 0.06) mol L^{−1 }= 0.04 mol L^{−1}
Therefore, concentration of B reacted 1/2 x 0.04 mol L^{-1} = 0.02 mol L^{−1}
Then, concentration of B available, [B] = (0.2 − 0.02) mol L^{−1}
= 0.18 mol L^{−1}
After [A] is reduced to 0.06 mol L^{−1}, the rate of the reaction is given by,
Rate = k [A][B]^{2}
= (2.0 × 10^{−6 }mol^{−2} L^{2} s^{−1}) (0.06 mol L^{−1}) (0.18 mol L^{−1})^{2}
= 3.89 mol L^{−1} s^{−1}
Question12. The decomposition of NH_{3} on platinum surface is zero order reaction. What are the rates of production of N_{2} and H_{2} if k = 2.5 × 10^{−4} mol^{−1} L s^{−1}?
Solution :
The decomposition of NH_{3} on platinum surface is represented by the following equation.
= 7.5 × 10^{−4 }mol L^{−1 }s^{−1}
Question13. The decomposition of dimethyl ether leads to the formation of CH_{4}, H_{2} and CO and the reaction rate is given by
Rate = k [CH_{3}OCH_{3}]^{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 (P_{CH3OCH3})^{3/2}
If the pressure is measured in bar andtime in minutes, then what are the units of rate and rate constants?
Solution :
If pressure is measured in bar and time in minutes, then
Unit of rate = bar min^{−1}
Rate = k [CH_{3}OCH_{3}]^{3/2}
⇒ k =Rate / [CH_{3}OCH_{3}]^{3/2}
Therefore, unit of rate constants(k) = bar min^{−1 /} bar^{3/2}
= bar^{-½} min ^{-1}
Question14. Mention the factors that affect the rate of a chemical reaction.
Solution :
The factors that affect the rate of a reaction are as follows.
(i) Concentration of reactants (pressure in case of gases)
(ii) Temperature
(iii) Presence of a catalyst
15. 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?
Solution :
Letthe concentration of the reactant be [A] = a
Rate of reaction, R = k [A]^{2}
= ka^{2}
(i)If the concentration of the reactant is doubled, i.e. [A] = 2a, then the rate of the reaction would be
R = k(2a)^{2}
= 4ka^{2}
= 4 R
Therefore, the rate of the reaction would increase by 4 times.
(ii) If the concentration of the reactant is reduced to half, i.e. [A] = 1/2 a , then the rate of the reaction would be
R = k(1/2a)^{2}
= 1/4 Ka^{2}
= 1/4 R
Therefore, the rate of the reaction would be reduced to
16. What change would happen in the rate constant of a reaction when there is a change in its temperature? How can this temperature effect on rate constant be represented quantitatively?
Solution :
When a temperature of 10∘ rises for a chemical reaction then the rate constant increases and becomes near to double of its original value.
The temperature effect on the rate constant can be represented quantitatively by Arrhenius equation,
k=Ae^{−Ea/RT}
Where,
k = rate constant,
A = Frequency factor / Arrhenius factor,
R = gas constant
T = temperature
Ea = activation energy for the reaction.
Question17. In a pseudo first order hydrolysis of 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 |
(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.
Solution :
(i) Average rate of reaction between the time interval, 30 to 60 seconds, d[ester] / dt
= (0.31-0.17) / (60-30)
= 0.14 / 30
= 4.67 × 10^{−3} mol L^{−1} s^{−1}
(ii) For a pseudo first order reaction,
k = 2.303/ t log [R]º / [R]
For t = 30 s, k_{1}
= 1.911 × 10^{−2 }s^{−1}
For t = 60 s, k1 = 2.303/ 30 log 0.55 / 0.31
= 1.957 × 10^{−2 }s^{−1}
For t = 90 s, k3 = 2.303/ 90 log 0.55 / 0.085
= 2.075 × 10 - 2s - 1
= 2.075 × 10^{−2 }s^{−1}
Then, average rate constant, k = k_{1} + k_{2}+ k_{3} / 3
= 1.911 × 10 ^{- 2 } + 1.957 × 10 ^{- 2} + 2.075 × 10^{ - 2 / 3}
= 1.981 x 10^{-2} s - 1
Question18. 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 are doubled?
Solution :
(i) The differential rate equation will be
Question19. In a reaction between A and B, the initial rate of reaction (r_{0}) 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 |
r_{0}/ 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?
Solution :
Let the order of the reaction with respect to A be x and with respect to B be y.
Therefore,
Dividing equation (iii) by (ii), we obtain
= 1.496
= 1.5 (approximately)
Hence, the order of the reaction with respect to A is 1.5 and with respect to B is zero.
Question20. The following results have been obtained during the kinetic studies of the 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 |
2.40 × 10^{−2} |
Determine the rate law and the rate constant for the reaction.
Solution :
Let the order of the reaction with respect to A be x and with respect to B be y.
Therefore, rate of the reaction is given by,
Dividing equation (iv) by (i), we obtain
Dividing equation (iii) by (ii), we obtain
Question21. 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} |
Solution :
The given reaction is of the first order with respect to A and of zero order with respect to B.
Therefore, the rate of the reaction is given by,
Rate = k [A]^{1 }[B]^{0}
⇒ Rate = k [A]
From experiment I, we obtain
2.0 × 10^{−2 }mol^{ }L^{−1} min^{−1} = k (0.1 mol L^{−1})
⇒ k = 0.2 min^{−1}
From experiment II, we obtain
4.0 × 10^{−2 mol }L^{−1} min^{−1} = 0.2 min^{−1} [A]
⇒ [A] = 0.2 mol L^{−1}
From experiment III, we obtain
Rate = 0.2 min^{−1} × 0.4 mol L^{−1}
= 0.08 mol L^{−1} min^{−1}
From experiment IV, we obtain
2.0 × 10^{−2 }mol^{ }L^{−1} min^{−1} = 0.2 min^{−1} [A]
⇒ [A] = 0.1 mol L^{−1}
Question22. Calculate the half-life of a first order reaction from their rate constants given below:
(i) 200 s^{−1} (ii) 2 min^{−1} (iii) 4 years^{−1}
Solution :
(i) Half life, t _{1/2} = 0.693 / k
= 0.693 / 200 s^{-1}
= 3.47×10 ^{-3 }s (approximately)
(ii) Half life, t _{1/2} = 0.693 / k
= 0.693 / 2 min-1
= 0.35 min (approximately)
(iii) Half life, t _{1/2} = 0.693 / k
= 0.693 / 4 years^{-1}
= 0.173 years (approximately)
Question23. The half-life for radioactive decay of 14C 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.
Solution :
Here, k = 0.693 / t_{1/2}
= 0.693 / 5730 years^{-1}
It is known that,
= 1845 years (approximately)
Hence, the age of the sample is 1845 years.
Question24. The experimental data for decomposition of N_{2}O_{5}
[2N2O5 → 4NO2 + O2] in gas phase at 318K are given below:
t(s) |
0 |
400 |
800 |
1200 |
1600 |
2000 |
2400 |
2800 |
3200 |
10^{2} × [N_{2}O_{5}] mol L^{-1} |
1.63 |
1.36 |
1.14 |
0.93 |
0.78 |
0.64 |
0.53 |
0.43 |
0.35 |
(i) Plot [N_{2}O_{5}] against t.
(ii) Find the half-life period for the reaction.
(iii) Draw a graph between log [N_{2}O_{5}] 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).
Solution :
(ii) Time corresponding to the concentration, 1630x10^{2} / 2 mol L^{-1} = 81.5 mol L^{-1} is the half life. From the graph, the half life is obtained as 1450 s.
(iii)
t(s) |
10^{2} × [N_{2}O_{5}] mol L^{-1} |
Log[N_{2}O_{5}] |
0 |
1.63 |
− 1.79 |
400 |
1.36 |
− 1.87 |
800 |
1.14 |
− 1.94 |
1200 |
0.93 |
− 2.03 |
1600 |
0.78 |
− 2.11 |
2000 |
0.64 |
− 2.19 |
2400 |
0.53 |
− 2.28 |
2800 |
0.43 |
− 2.37 |
3200 |
0.35 |
− 2.46 |
(iv) The given reaction is of the first order as the plot, Log[N_{2}O_{5}] v/s t, is a straight line. Therefore, the rate law of the reaction is
Rate = k [N_{2}O_{5}]
(v) From the plot, Log[N_{2}O_{5}] v/s t, we obtain
- k /2.303
Again, slope of the line of the plot Log[N_{2}O_{5}] v/s t is given by
- k / 2.303. = -0.67 / 3200
Therefore, we obtain,
- k / 2.303 = - 0.67 / 3200
⇒ k = 4.82 x 10-4 s-1
(vi) Half-life is given by,
t_{½} = 0.693 / k
= 0.639 / 4.82x10^{-4} s
=1.438 x 103
This value, 1438 s, is very close to the value that was obtained from the graph.
25. 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 1/16^{th} value?
Solution :
It is known that,
Question26. During nuclear explosion, one of the products is 90Sr with half-life of 28.1 years. If 1μg 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.
Solution :
,
Therefore, 0.7814 μg of 90^{Sr} will remain after 10 years.
Again,
Therefore, 0.2278 μg of 90^{Sr} will remain after 60 years.
Question27. For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
Solution :
For a first order reaction, the time required for 99% completion is
t_{1} = 2.303/k Log 100/100-99
= 2.303/k Log 100
= 2x 2.303/k
For a first order reaction, the time required for 90% completion is
t_{2} = 2.303/k Log 100 / 100-90
= 2.303/k Log 10
= 2.303/k
Therefore, t_{1} = 2t_{2}
Hence, the time required for 99% completion of a first order reaction is twice the time required for the completion of 90% of the reaction.
Question28. A first order reaction takes 40 min for 30% decomposition. Calculate t_{1/2}.
Solution :
For a first order reaction,
t = 2.303/k Log [R] º / [R]
k = 2.303/40min Log 100 / 100-30
= 2.303/40min Log 10 / 7
= 8.918 x 10-3 min^{-1}
Therefore, t_{1/2} of the decomposition reaction is
t1/2 = 0.693/k
= 0.693 / 8.918 x 10-3 min
= 77.7 min (approximately)
= 77.7 min (approximately)
Question29. 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.
Solution :
The decomposition of azoisopropane to hexane and nitrogen at 543 K is represented by the following equation.
After time, t, total pressure, Pt = (P_{º} - p) + p + p
⇒ Pt = (P_{º }+ p)
⇒ p = P_{t} - P_{0}
therefore, P_{º} - p = P_{0} - Pt - P_{0}
= 2P_{0} − P_{t}
For a first order reaction,
k = 2.303/t Log P_{0} /P_{0} - p
= 2.303/t Log P_{0} / 2 P_{0} - P_{t}
When t = 360 s, k = 2.303 / 360s log 35.0 / 2x35.0 - 54.0
= 2.175 × 10^{−3 }s^{−1}
When t = 720 s, k = 2.303 / 720s log 35.0 / 2x35.0 - 63.0
= 2.235 × 10^{−3} s^{−1}
Hence, the average value of rate constant is
k = (2.175 × 10^{ - 3} + 2.235 × 10 ^{- 3} ) / 2 s ^{- 1}
= 2.21 × 10^{−3 }s^{−1}
Question30. The following data were obtained during the first order thermal decomposition of SO_{2}Cl_{2} at a constant volume.
SO_{2}Cl_{2}(g) → SO_{2}(g) + Cl_{2}(g)
Experiment |
Time/s^{−1} |
Total pressure/atm |
1 |
0 |
0.5 |
2 |
100 |
0.6 |
Calculate the rate of the reaction when total pressure is 0.65 atm.
Solution :
The thermal decomposition of SO_{2}Cl_{2} at a constant volume is represented by the following equation.
After time, t, total pressure, P_{t} = (P_{º} - p) + p + p
⇒ P_{t} = (P_{º} + p)
⇒ p = Pt - Pº
therefore, Pº - p = Pº - Pt - Pº
= 2 P_{º }- P_{t}
For a first order reaction,
k = 2.303/t Log P_{º }/ P_{º} - p
= 2.303/t Log P_{º} / 2 P_{º} - Pt
When t= 100 s,
k = 2.303 / 100s log 0.5 / 2x0.5 - 0.6
= 2.231 × 10 - 3s - 1
When Pt= 0.65 atm,
P0+ p= 0.65
⇒ p= 0.65 - P0
= 0.65 - 0.5
= 0.15 atm
Therefore, when the total pressure is 0.65 atm, pressure of SOCl2 is
_{PSOCL2} = P_{0} - p
= 0.5 - 0.15
= 0.35 atm
Therefore, the rate of equation, when total pressure is 0.65 atm, is given by,
Rate = k(p_{SOCL2})
= (2.23 × 10 - 3s - 1) (0.35 atm)
= 7.8 × 10^{ - 4} atm s^{ - 1}
Question31. The rate constant for the decomposition of N_{2}O_{5} at various temperatures is given below:
T/°C |
0 |
20 |
40 |
60 |
80 |
10^{5} X K /S^{-1} |
0.0787 |
1.70 |
25.7 |
178 |
2140 |
Draw a graph between ln k and 1/T and calculate the values of A and E_{a}.
Predict the rate constant at 30º and 50ºC.
Solution :
From the given data, we obtain
T/°C |
0 |
20 |
40 |
60 |
80 |
T/K |
273 |
293 |
313 |
333 |
353 |
1/T / k^{-1} |
3.66×10^{−3} |
3.41×10^{−3} |
3.19×10^{−3} |
3.0×10^{−3} |
2.83 ×10^{−3} |
10^{5} X K /S^{-1} |
0.0787 |
1.70 |
25.7 |
178 |
2140 |
ln k |
−7.147 |
− 4.075 |
−1.359 |
−0.577 |
3.063 |
Slope of the line,
In k= - 2.8
Therefore, k = 6.08x10-2s-1
Again when T = 50 + 273K = 323K,
1/T = 3.1 x 10-3 K
In k = - 0.5
Therefore, k = 0.607 s-1
Question32. The rate constant for the decomposition of hydrocarbons 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.
Solution :
k = 2.418 × 10^{−5} s^{−1}
T = 546 K
E_{a} = 179.9 kJ mol^{−1} = 179.9 × 10^{3 }J mol^{−1}
According to the Arrhenius equation,
= (0.3835 − 5) + 17.2082
= 12.5917
Therefore, A = antilog (12.5917)
= 3.9 × 10^{12} s^{−1} (approximately)
Question33. 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}.
Solution :
k = 2.0 × 10^{−2} s^{−1}
T = 100 s
[A]_{o} = 1.0 moL^{−1}
Since the unit of k is s^{−1}, the given reaction is a first order reaction.
Therefore, k = 2.303/t Log [A]º / [A]
⇒2.0 × 110-2 s-1 = 2.303/100s Log 1.0 / [A]
⇒2.0 × 110-2 s-1 = 2.303/100s ( - Log [A] )
⇒ - Log [A] = - (2.0 x 10-2 x 100) / 2.303
⇒ [A] = antilog [- (2.0 x 10-2 x 100) / 2.303]
= 0.135 mol L^{−1 (approximately)}
Hence, the remaining concentration of A is 0.135 mol L^{−1.}
Question34. Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t_{1/2 }= 3.00 hours. What fraction of sample of sucrose remains after 8 hours?
Solution :^{For a first order reaction, k = 2.303/t Log [R]º / [R] It is given that, t1/2 = 3.00 hours Therefore, k = 0.693 / t1/2 = 0.693 / 3 h-1 = 0.231 h - 1 Then, 0.231 h - 1 = 2.303 / 8h Log [R]º / [R]}
Hence, the fraction of sample of sucrose that remains after 8 hours is 0.158.
Question35. The decomposition of hydrocarbon follows the equation k = (4.5 × 10^{11} s^{−1}) e^{−28000} K/T Calculate E_{a}.
Solution :
The given equation is
k = (4.5 × 10^{11 }s^{−1}) e^{−28000} K/T (i)
Arrhenius equation is given by,
k= Ae^{ -Ea/RT} (ii)
From equation (i) and (ii), we obtain
Ea / RT = 28000K / T
⇒ Ea = R x 28000K
= 8.314 J K^{−1} mol^{−1} × 28000 K
= 232792 J mol^{−1}
= 232.792 kJ mol^{−1}
Question36. The rate constant for the first order decomposition of H_{2}O_{2} is given by the following equation:
log k = 14.34 − 1.25 × 10^{4 K/T}
Calculate E_{a} for this reaction and at what temperature will its half-period be 256 minutes?
Solution :
Arrhenius equation is given by,
k= Ae -Ea/RT
⇒In k = In A - Ea/RT
⇒In k = Log A - Ea/RT
⇒ Log k = Log A - Ea/2.303RT (i)
The given equation is
Log k = 14.34 - 1.25 104 K/T (ii)
From equation (i) and (ii), we obtain
Ea/2.303RT = 1.25 104 K/T
= 1.25 × 10^{4} K × 2.303 × 8.314 J K^{−1} mol^{−1}
= 239339.3 J mol^{−1 }(approximately)
= 239.34 kJ mol^{−1}
Also, when t_{1/2} = 256 minutes,
k = 0.693 / t^{1/2}
= 0.693 / 256
= 2.707 × 10^{ - 3 }min^{ - 1}
= 4.51 × 10 - 5s^{ - 1}
= 2.707 × 10^{−3} min^{−1}
= 4.51 × 10^{−5} s^{−1}
It is also given that, log k = 14.34 − 1.25 × 10^{4 }K/T
= 668.95 K
= 669 K (approximately)
Question37. The decomposition of A into product has value of k as 4.5 × 10^{3 }s^{−1} at 10°C and energy of activation 60 kJ mol^{−1}. At what temperature would k be 1.5 × 10^{4} s^{−1}?
Solution :
From Arrhenius equation, we obtain
log k2/k1 = Ea / 2.303 R (T_{2} - T_{1}) / T_{1}T_{2}
Also, k_{1} = 4.5 × 10^{3 s−1}
T_{1} = 273 + 10 = 283 K
k_{2} = 1.5 × 10^{4} s^{−1}
E_{a} = 60 kJ mol^{−1} = 6.0 × 10^{4} J mol^{−1}
Then,
= 297 K
= 24°C
Hence, k would be 1.5 × 10^{4 }s^{−1} at 24°C.
Question38. 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 × 10^{10 }s^{−1}. Calculate k at 318 K and E_{a}.
Solution :
For a first order reaction,
t = 2.303 / k log a / a - x
At 298 K, t = 2.303 / k log 100 / 90
= 0.1054 / k
At 308 K, t' = 2.303 / k' log 100 / 75
= 2.2877 / k'
According to the question,
t = t'
⇒ 0.1054 / k = 2.2877 / k'
⇒ k' / k = 2.7296
From Arrhenius equation, we obtain
To calculate k at 318 K,
It is given that, A = 4 x 1010 s-1, T = 318K
Again, from Arrhenius equation, we obtain
Therefore, k = Antilog (-1.9855)
= 1.034 x 10-2 s -1
Question39. The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.
Solution :
From Arrhenius equation, we obtain
Hence, the required energy of activation is 52.86 kJmol^{−1.}