NCERT Solutions for Class 12 Chemistry Chapter 4 – Chemical Kinetics
Page No 98:
Question 4.1:
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.
Answer:
Average rate of reaction ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6313/NCERT(INTEXT)_13-11-08_Utpal_12_Chemistry_4_9_html_7c1ba064.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6313/NCERT(INTEXT)_13-11-08_Utpal_12_Chemistry_4_9_html_7c1ba064.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6313/NCERT(INTEXT)_13-11-08_Utpal_12_Chemistry_4_9_html_70c2142c.gif)
= 6.67 × 10−6 M s−1
Question 4.2:
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?
Answer:
Average rate ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6314/NCERT(INTEXT)_13-11-08_Utpal_12_Chemistry_4_9_html_m2a103f0e.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6314/NCERT(INTEXT)_13-11-08_Utpal_12_Chemistry_4_9_html_m2a103f0e.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6314/NCERT(INTEXT)_13-11-08_Utpal_12_Chemistry_4_9_html_5e7dae04.gif)
= 0.005 mol L−1 min−1
= 5 × 10−3 M min−1
Page No 103:
Question 4.3:
For a reaction, A + B → Product; the rate law is given by,
. What is the order of the reaction?
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6316/NCERT(INTEXT)_13-11-08_Utpal_12_Chemistry_4_9_html_72be5927.gif)
Answer:
The order of the reaction![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6316/NCERT(INTEXT)_13-11-08_Utpal_12_Chemistry_4_9_html_3d801889.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6316/NCERT(INTEXT)_13-11-08_Utpal_12_Chemistry_4_9_html_3d801889.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6316/NCERT(INTEXT)_13-11-08_Utpal_12_Chemistry_4_9_html_29e9d4f5.gif)
= 2.5
Question 4.4:
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?
Answer:
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:
Rate1 = k .(a)2
= ka2
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(ka2)
Hence, the rate of formation will increase by 9 times.
Page No 111:
Question 4.5:
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?
Answer:
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 1st order reaction,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6321/NCERT(INTEXT)_13-11-08_Utpal_12_Chemistry_4_9_html_m634877af.gif)
= 444.38 s
= 444 s (approx)
Question 4.6:
Time required to decompose SO2Cl2 to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.
Answer:
We know that for a 1st order reaction,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6322/NCERT(INTEXT)_13-11-08_Utpal_12_Chemistry_4_9_html_470b8c21.gif)
It is given that t1/2 = 60 min
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6322/NCERT(INTEXT)_13-11-08_Utpal_12_Chemistry_4_9_html_m2301ba1b.gif)
Page No 116:
Question 4.7:
What will be the effect of temperature on rate constant?
Answer:
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,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6323/NCERT(INTEXT)_13-11-08_Utpal_12_Chemistry_4_9_html_m70f0906.gif)
Where,
A is the Arrhenius factor or the frequency factor
T is the temperature
R is the gas constant
Ea is the activation energy
Question 4.8:
The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate Ea.
Answer:
It is given that T1 = 298 K
∴T2 = (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 k1 = k and that of k2 = 2k
Also, R = 8.314 J K−1 mol−1
Now, substituting these values in the equation:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6325/NCERT(INTEXT)_13-11-08_Utpal_12_Chemistry_4_9_html_75a3d42e.gif)
We get:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6325/NCERT(INTEXT)_13-11-08_Utpal_12_Chemistry_4_9_html_41219bce.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6325/NCERT(INTEXT)_13-11-08_Utpal_12_Chemistry_4_9_html_782090af.gif)
= 52897.78 J mol−1
= 52.9 kJ mol−1
Note: There is a slight variation in this answer and the one given in the NCERT textbook.
Question 4.9:
The activation energy for the reaction
2HI(g) → H2 + I2(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?
Answer:
In the given case:
Ea = 209.5 kJ mol−1 = 209500 J mol−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⇒In x=-EaRT⇒log x=-Ea2.303RT⇒log x=-209500 J mol-12.303×8.314 JK-1mol-1×581=-18.8323Now, x =Antilog -18.8323 = 1.471×10-19
Page No 117:
Question 4.1:
From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.
(i) 3 NO(g) → N2O (g) Rate = k[NO]2
(ii) H2O2 (aq) + 3 I− (aq) + 2 H+ → 2 H2O (l) +
Rate = k[H2O2][I−]
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6329/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m4c7b31ab.gif)
(iii) CH3CHO(g) → CH4(g) + CO(g) Rate = k [CH3CHO]3/2
(iv) C2H5Cl(g) → C2H4(g) + HCl(g) Rate = k [C2H5Cl]
Answer:
(i) Given rate = k [NO]2
Therefore, order of the reaction = 2
Dimension of ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6329/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m3b22fd6b.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6329/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m3b22fd6b.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6329/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m721084d0.gif)
(ii) Given rate = k [H2O2] [I−]
Therefore, order of the reaction = 2
Dimension of ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6329/NS_14-11-08_Utpal_12_Chemistry_4_30_html_795ae02.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6329/NS_14-11-08_Utpal_12_Chemistry_4_30_html_795ae02.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6329/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m10a6b4b2.gif)
(iii) Given rate = k [CH3CHO]3/2
Therefore, order of reaction = ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6329/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m47d40970.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6329/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m47d40970.gif)
Dimension of ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6329/NS_14-11-08_Utpal_12_Chemistry_4_30_html_71de0a04.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6329/NS_14-11-08_Utpal_12_Chemistry_4_30_html_71de0a04.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6329/NS_14-11-08_Utpal_12_Chemistry_4_30_html_1bd21519.gif)
(iv) Given rate = k [C2H5Cl]
Therefore, order of the reaction = 1
Dimension of ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6329/NS_14-11-08_Utpal_12_Chemistry_4_30_html_6629c3a8.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6329/NS_14-11-08_Utpal_12_Chemistry_4_30_html_6629c3a8.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6329/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m4d709495.gif)
Question 4.2:
For the reaction:
2A + B → A2B
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.
Answer:
The initial rate of the reaction is
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
= 8.0 × 10−9 mol−2 L2 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
= 0.02 mol L−1
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6331/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m4019195.gif)
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 L2 s−1) (0.06 mol L−1) (0.18 mol L−1)2
= 3.89 mol L−1 s−1
Question 4.3:
The decomposition of NH3 on platinum surface is zero order reaction. What are the rates of production of N2 and H2 if k = 2.5 × 10−4 mol−1 L s−1?
Answer:
The decomposition of NH3 on platinum surface is represented by the following equation.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6332/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m60473d8e.gif)
Therefore,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6332/NS_14-11-08_Utpal_12_Chemistry_4_30_html_350a4e76.gif)
However, it is given that the reaction is of zero order.
Therefore,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6332/NS_14-11-08_Utpal_12_Chemistry_4_30_html_1efbb932.gif)
Therefore, the rate of production of N2 is
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6332/NS_14-11-08_Utpal_12_Chemistry_4_30_html_483a3837.gif)
And, the rate of production of H2 is
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6332/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m52c63472.gif)
= 7.5 × 10−4 mol L−1 s−1
Question 4.4:
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.,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6333/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m2469a1fb.gif)
If the pressure is measured in bar andtime in minutes, then what are the units of rate and rate constants?
Answer:
If pressure is measured in bar and time in minutes, then
Unit of rate = bar min−1
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6333/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m2469a1fb.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6333/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m3784d290.gif)
Therefore, unit of rate constants![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6333/NS_14-11-08_Utpal_12_Chemistry_4_30_html_499cfced.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6333/NS_14-11-08_Utpal_12_Chemistry_4_30_html_499cfced.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6333/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m2f900d95.gif)
Question 4.5:
Mention the factors that affect the rate of a chemical reaction.
Answer:
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
Page No 118:
Question 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 half?
Answer:
Letthe concentration of the reactant be [A] = a
Rate of reaction, R = k [A]2
= ka2
(i)If the concentration of the reactant is doubled, i.e. [A] = 2a, then the rate of the reaction would be
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6337/NS_14-11-08_Utpal_12_Chemistry_4_30_html_590225c8.gif)
= 4ka2
= 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.
, then the rate of the reaction would be
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6337/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m1669ca00.gif)
![](https://img-nm.mnimgs.com/img/study_content/editlive_ncert/62/2013_06_26_17_38_46/SA.png)
Therefore, the rate of the reaction would be reduced to![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6337/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m77622ec.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6337/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m77622ec.gif)
Question 4.7:
What is the effect of temperature on the rate constant of a reaction? How can this temperature effect on rate constant be represented quantitatively?
Answer:
The rate constant is nearly doubled with a rise in temperature by 10° for a chemical reaction.
The temperature effect on the rate constant can be represented quantitatively by Arrhenius equation,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6338/NS_14-11-08_Utpal_12_Chemistry_4_30_html_69b9e530.gif)
where, k is the rate constant,
A is the Arrhenius factor or the frequency factor,
R is the gas constant,
T is the temperature, and
Ea is the energy of activation for the reaction
Question 4.8:
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.
Answer:
(i) Average rate of reaction between the time interval, 30 to 60 seconds, ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6340/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m4f229d69.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6340/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m4f229d69.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6340/NS_14-11-08_Utpal_12_Chemistry_4_30_html_38ed6d91.gif)
= 4.67 × 10−3 mol L−1 s−1
(ii) For a pseudo first order reaction,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6340/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m7f521f40.gif)
For t = 30 s, ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6340/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m17404f4d.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6340/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m17404f4d.gif)
= 1.911 × 10−2 s−1
For t = 60 s, ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6340/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m5221cd51.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6340/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m5221cd51.gif)
= 1.957 × 10−2 s−1
For t = 90 s, ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6340/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m42b9eee6.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6340/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m42b9eee6.gif)
= 2.075 × 10−2 s−1
Then, average rate constant, ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6340/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m14005130.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6340/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m14005130.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6340/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m7eed4be9.gif)
Question 4.9:
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 are doubled?
Answer:
(i) The differential rate equation will be
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6342/NS_14-11-08_Utpal_12_Chemistry_4_30_html_2e1296ab.gif)
(ii) If the concentration of B is increased three times, then
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6342/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m114b68f8.gif)
Therefore, the rate of reaction will increase 9 times.
(iii) When the concentrations of both A and B are doubled,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6342/NS_14-11-08_Utpal_12_Chemistry_4_30_html_2e57ae95.gif)
Therefore, the rate of reaction will increase 8 times.
Question 4.10:
In a reaction between A and B, the initial rate of reaction (r0) was measured for different initial concentrations of A and B as given below:
- A/ mol L−10.200.200.40B/ mol L−10.300.100.05r0/ mol L−1 s−15.07 × 10−55.07 × 10−51.43 × 10−4
What is the order of the reaction with respect to A and B?
Answer:
Let the order of the reaction with respect to A be x and with respect to B be y.
Therefore,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6343/NS_14-11-08_Utpal_12_Chemistry_4_30_html_77e512e4.gif)
Dividing equation (i) by (ii), we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6343/NS_14-11-08_Utpal_12_Chemistry_4_30_html_33efb25d.gif)
Dividing equation (iii) by (ii), we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6343/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m61c7192d.gif)
= 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.
Question 4.11:
The following results have been obtained during the kinetic studies of the reaction:
2A + B → C + D
- ExperimentA/ mol L−1B/ mol L−1Initial rate of formation of D/mol L−1 min−1I0.10.16.0 × 10−3II0.30.27.2 × 10−2III0.30.42.88 × 10−1IV0.40.12.40 × 10−2
Determine the rate law and the rate constant for the reaction.
Answer:
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,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6345/NS_14-11-08_Utpal_12_Chemistry_4_30_html_mc37c1bf.gif)
According to the question,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6345/NS_14-11-08_Utpal_12_Chemistry_4_30_html_5994d239.gif)
Dividing equation (iv) by (i), we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6345/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m35ffc9b0.gif)
Dividing equation (iii) by (ii), we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6345/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m59fb76c4.gif)
Therefore, the rate law is
Rate = k [A] [B]2
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6345/NS_14-11-08_Utpal_12_Chemistry_4_30_html_74849e29.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6345/NS_14-11-08_Utpal_12_Chemistry_4_30_html_46a305db.gif)
From experiment I, we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6345/NS_14-11-08_Utpal_12_Chemistry_4_30_html_25904772.gif)
= 6.0 L2 mol−2 min−1
From experiment II, we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6345/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m4358181f.gif)
= 6.0 L2 mol−2 min−1
From experiment III, we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6345/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m4df1f4b9.gif)
= 6.0 L2 mol−2 min−1
From experiment IV, we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6345/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m44b3dd36.gif)
= 6.0 L2 mol−2 min−1
Therefore, rate constant, k = 6.0 L2 mol−2 min−1
Question 4.12:
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:
- ExperimentA/ mol L−1B/ mol L−1Initial rate/mol L−1 min−1I0.10.12.0 × 10−2II—0.24.0 × 10−2III0.40.4—IV—0.22.0 × 10−2
Answer:
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
Page No 119:
Question 4.13:
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
Answer:
(i) Half life, ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6348/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m7f8f9078.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6348/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m7f8f9078.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6348/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m7ad0f420.gif)
= 3.47
×10 -3 s (approximately)
(ii) Half life, ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6348/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m7f8f9078.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6348/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m7f8f9078.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6348/NS_14-11-08_Utpal_12_Chemistry_4_30_html_753ea805.gif)
= 0.35 min (approximately)
(iii) Half life, ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6348/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m7f8f9078.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6348/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m7f8f9078.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6348/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m4451f518.gif)
= 0.173 years (approximately)
Question 4.14:
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.
Answer:
Here,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6350/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m37f6ea30.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6350/NS_14-11-08_Utpal_12_Chemistry_4_30_html_2c431e67.gif)
It is known that,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6350/NS_14-11-08_Utpal_12_Chemistry_4_30_html_26185dc1.gif)
= 1845 years (approximately)
Hence, the age of the sample is 1845 years.
Question 4.15:
The experimental data for decomposition of N2O5
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6372/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m7508b7ed.gif)
in gas phase at 318K are given below:
- t(s)
0 400 800 1200 1600 2000 2400 2800 3200 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).
Answer:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6372/NS_14-11-08_Utpal_12_Chemistry_4_30_html_me506e51.jpg)
(ii) Time corresponding to the concentration,
is the half life. From the graph, the half life is obtained as 1450 s.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6372/NS_14-11-08_Utpal_12_Chemistry_4_30_html_7c1699df.gif)
(iii)
- t(s)01.63− 1.794001.36− 1.878001.14− 1.9412000.93− 2.0316000.78− 2.1120000.64− 2.1924000.53− 2.2828000.43− 2.3732000.35− 2.46
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6372/NS_14-11-08_Utpal_12_Chemistry_4_30_html_47ff23ff.jpg)
(iv) The given reaction is of the first order as the plot,
v/s t, is a straight line. Therefore, the rate law of the reaction is
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6372/NS_14-11-08_Utpal_12_Chemistry_4_30_html_2c7892c1.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6372/NS_14-11-08_Utpal_12_Chemistry_4_30_html_40b6aaf6.gif)
(v) From the plot,
v/s t, we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6372/NS_14-11-08_Utpal_12_Chemistry_4_30_html_2c7892c1.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6372/NS_14-11-08_Utpal_12_Chemistry_4_30_html_720ab5a7.gif)
Again, slope of the line of the plot
v/s t is given by
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6372/NS_14-11-08_Utpal_12_Chemistry_4_30_html_2c7892c1.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6372/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m30f2689c.gif)
Therefore, we obtain,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6372/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m29d84705.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6372/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m2841e4e4.gif)
(vi) Half-life is given by,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6372/NS_14-11-08_Utpal_12_Chemistry_4_30_html_3ee340dc.gif)
This value, 1438 s, is very close to the value that was obtained from the graph.
Question 4.16:
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?
Answer:
It is known that,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6351/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m774930f5.gif)
Hence, the required time is 4.6 × 10−2 s.
Question 4.17:
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.
Answer:
Here, ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6353/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m4f924778.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6353/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m4f924778.gif)
It is known that,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6353/NS_14-11-08_Utpal_12_Chemistry_4_30_html_5cd18479.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6353/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m25b3f4d2.gif)
Therefore, 0.7814 μg of 90Sr will remain after 10 years.
Again,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6353/NS_14-11-08_Utpal_12_Chemistry_4_30_html_5cd18479.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6353/NS_14-11-08_Utpal_12_Chemistry_4_30_html_16b7b6cb.gif)
Therefore, 0.2278 μg of 90Sr will remain after 60 years.
Question 4.18:
For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
Answer:
For a first order reaction, the time required for 99% completion is
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6354/NS_14-11-08_Utpal_12_Chemistry_4_30_html_b7c8ada.gif)
For a first order reaction, the time required for 90% completion is
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6354/NS_14-11-08_Utpal_12_Chemistry_4_30_html_1367300f.gif)
Therefore, t1 = 2t2
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.
Question 4.19:
A first order reaction takes 40 min for 30% decomposition. Calculate t1/2.
Answer:
For a first order reaction,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6356/NS_14-11-08_Utpal_12_Chemistry_4_30_html_5cd18479.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6356/NS_14-11-08_Utpal_12_Chemistry_4_30_html_5fff9d0b.gif)
Therefore, t1/2 of the decomposition reaction is
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6356/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m624d0e08.gif)
= 77.7 min (approximately)
Question 4.20:
For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained.
- t (sec)P(mm of Hg)035.036054.072063.0
Calculate the rate constant.
Answer:
The decomposition of azoisopropane to hexane and nitrogen at 543 K is represented by the following equation.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6358/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m5e306863.gif)
After time, t, total pressure, ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6358/NS_14-11-08_Utpal_12_Chemistry_4_30_html_35bdefa5.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6358/NS_14-11-08_Utpal_12_Chemistry_4_30_html_35bdefa5.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6358/NS_14-11-08_Utpal_12_Chemistry_4_30_html_670c95fe.gif)
= 2P0 − Pt
For a first order reaction,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6358/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m510c488d.gif)
When t = 360 s, ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6358/NS_14-11-08_Utpal_12_Chemistry_4_30_html_5f71fbd9.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6358/NS_14-11-08_Utpal_12_Chemistry_4_30_html_5f71fbd9.gif)
= 2.175 × 10−3 s−1
When t = 720 s, ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6358/NS_14-11-08_Utpal_12_Chemistry_4_30_html_9f83889.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6358/NS_14-11-08_Utpal_12_Chemistry_4_30_html_9f83889.gif)
= 2.235 × 10−3 s−1
Hence, the average value of rate constant is
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6358/NS_14-11-08_Utpal_12_Chemistry_4_30_html_3e8a9bbb.gif)
= 2.21 × 10−3 s−1
Note: There is a slight variation in this answer and the one given in the NCERT textbook.
Question 4.21:
The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6359/NS_14-11-08_Utpal_12_Chemistry_4_30_html_3572ff23.gif)
- ExperimentTime/s−1Total pressure/atm100.521000.6
Calculate the rate of the reaction when total pressure is 0.65 atm.
Answer:
The thermal decomposition of SO2Cl2 at a constant volume is represented by the following equation.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6359/NS_14-11-08_Utpal_12_Chemistry_4_30_html_6f5f1162.gif)
After time, t, total pressure, ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6359/NS_14-11-08_Utpal_12_Chemistry_4_30_html_35bdefa5.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6359/NS_14-11-08_Utpal_12_Chemistry_4_30_html_35bdefa5.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6359/NS_14-11-08_Utpal_12_Chemistry_4_30_html_73a3ad92.gif)
Therefore, ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6359/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m2b66381a.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6359/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m2b66381a.gif)
= 2 P0 − Pt
For a first order reaction,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6359/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m3f1e8ec2.gif)
When t = 100 s, ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6359/NS_14-11-08_Utpal_12_Chemistry_4_30_html_110029b4.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6359/NS_14-11-08_Utpal_12_Chemistry_4_30_html_110029b4.gif)
= 2.231 × 10−3 s−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
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6359/NS_14-11-08_Utpal_12_Chemistry_4_30_html_68f4cf67.gif)
= 0.5 − 0.15
= 0.35 atm
Therefore, the rate of equation, when total pressure is 0.65 atm, is given by,
Rate = k(
)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6359/NS_14-11-08_Utpal_12_Chemistry_4_30_html_68f4cf67.gif)
= (2.23 × 10−3 s−1) (0.35 atm)
= 7.8 × 10−4 atm s−1
Page No 120:
Question 4.22:
The rate constant for the decomposition of N2O5 at various temperatures is given below:
T/°C 0 20 40 60 80 0.0787 1.70 25.7 178 2140
Draw a graph between ln k and 1/T and calculate the values of A and Ea.
Predict the rate constant at 30º and 50ºC.
Answer:
From the given data, we obtain
- T/°C020406080T/K2732933133333533.66×10−33.41×10−33.19×10−33.0×10−32.83 ×10−30.07871.7025.71782140ln k−7.147− 4.075−1.359−0.5773.063
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6373/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m5d4d0ef3.jpg)
Slope of the line,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6373/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m10e5221b.gif)
According to Arrhenius equation,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6373/NS_14-11-08_Utpal_12_Chemistry_4_30_html_1553283a.gif)
Again,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6373/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m4d3d2169.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6373/NS_14-11-08_Utpal_12_Chemistry_4_30_html_7613e420.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6373/NS_14-11-08_Utpal_12_Chemistry_4_30_html_5a454947.gif)
When
,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6373/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m2e75f310.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6373/NS_14-11-08_Utpal_12_Chemistry_4_30_html_mcce9817.gif)
Then, ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6373/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m21d3621e.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6373/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m21d3621e.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6373/NS_14-11-08_Utpal_12_Chemistry_4_30_html_md01823a.gif)
Again, when
,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6373/NS_14-11-08_Utpal_12_Chemistry_4_30_html_38d00a94.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6373/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m26741d.gif)
Then, at
,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6373/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m194334f6.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6373/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m20d30914.gif)
Question 4.23:
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.
Answer:
k = 2.418 × 10−5 s−1
T = 546 K
Ea = 179.9 kJ mol−1 = 179.9 × 103 J mol−1
According to the Arrhenius equation,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6360/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m6d7bd495.gif)
= (0.3835 − 5) + 17.2082
= 12.5917
Therefore, A = antilog (12.5917)
= 3.9 × 1012 s−1 (approximately)
Question 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.
Answer:
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, ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6362/NS_14-11-08_Utpal_12_Chemistry_4_30_html_42cab5cd.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6362/NS_14-11-08_Utpal_12_Chemistry_4_30_html_42cab5cd.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6362/NS_14-11-08_Utpal_12_Chemistry_4_30_html_7af8db04.gif)
= 0.135 mol L−1 (approximately)
Hence, the remaining concentration of A is 0.135 mol L−1.
Question 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 sample of sucrose remains after 8 hours?
Answer:
For a first order reaction,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6363/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m7f521f40.gif)
It is given that, t1/2 = 3.00 hours
Therefore, ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6363/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m37f6ea30.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6363/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m37f6ea30.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6363/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m5b89395e.gif)
= 0.231 h−1
Then, 0.231 h−1 ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6363/NS_14-11-08_Utpal_12_Chemistry_4_30_html_17d9f90d.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6363/NS_14-11-08_Utpal_12_Chemistry_4_30_html_17d9f90d.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6363/NS_14-11-08_Utpal_12_Chemistry_4_30_html_27c263e8.gif)
Hence, the fraction of sample of sucrose that remains after 8 hours is 0.158.
Question 4.26:
The decomposition of hydrocarbon follows the equation
k = (4.5 × 1011 s−1) e−28000 K/T
Calculate Ea.
Answer:
The given equation is
k = (4.5 × 1011 s−1) e−28000 K/T (i)
Arrhenius equation is given by,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6365/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m14693e7a.gif)
From equation (i) and (ii), we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6365/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m1fcffae4.gif)
= 8.314 J K−1 mol−1 × 28000 K
= 232792 J mol−1
= 232.792 kJ mol−1
Question 4.27:
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?
Answer:
Arrhenius equation is given by,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6366/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m14693e7a.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6366/NS_14-11-08_Utpal_12_Chemistry_4_30_html_1c450463.gif)
The given equation is
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6366/NS_14-11-08_Utpal_12_Chemistry_4_30_html_54fa18f5.gif)
From equation (i) and (ii), we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6366/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m6b2e6e1a.gif)
= 1.25 × 104 K × 2.303 × 8.314 J K−1 mol−1
= 239339.3 J mol−1 (approximately)
= 239.34 kJ mol−1
Also, when t1/2 = 256 minutes,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6366/NS_14-11-08_Utpal_12_Chemistry_4_30_html_6d29612.gif)
= 2.707 × 10−3 min−1
= 4.51 × 10−5 s−1
It is also given that, log k = 14.34 − 1.25 × 104 K/T
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6366/NS_14-11-08_Utpal_12_Chemistry_4_30_html_7cd6c23e.gif)
= 668.95 K
= 669 K (approximately)
Question 4.28:
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?
Answer:
From Arrhenius equation, we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6367/NS_14-11-08_Utpal_12_Chemistry_4_30_html_3dfd91da.gif)
Also, k1 = 4.5 × 103 s−1
T1 = 273 + 10 = 283 K
k2 = 1.5 × 104 s−1
Ea = 60 kJ mol−1 = 6.0 × 104 J mol−1
Then,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6367/NS_14-11-08_Utpal_12_Chemistry_4_30_html_352723eb.gif)
= 297 K
= 24°C
Hence, k would be 1.5 × 104 s−1 at 24°C.
Note: There is a slight variation in this answer and the one given in the NCERT textbook.
Question 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.
Answer:
For a first order reaction,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6368/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m3dc2484c.gif)
At 298 K, ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6368/NS_14-11-08_Utpal_12_Chemistry_4_30_html_me013af8.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6368/NS_14-11-08_Utpal_12_Chemistry_4_30_html_me013af8.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6368/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m40158ae0.gif)
At 308 K, ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6368/NS_14-11-08_Utpal_12_Chemistry_4_30_html_4bd7d91c.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6368/NS_14-11-08_Utpal_12_Chemistry_4_30_html_4bd7d91c.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6368/NS_14-11-08_Utpal_12_Chemistry_4_30_html_5255b362.gif)
According to the question,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6368/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m7629dcdf.gif)
From Arrhenius equation, we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6368/NS_14-11-08_Utpal_12_Chemistry_4_30_html_2c6b4ed9.gif)
To calculate k at 318 K,
It is given that, ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6368/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m104d34ff.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6368/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m104d34ff.gif)
Again, from Arrhenius equation, we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6368/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m1e4ce7c7.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6368/NS_14-11-08_Utpal_12_Chemistry_4_30_html_5e008bec.gif)
Question 4.30:
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.
Answer:
From Arrhenius equation, we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6370/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m1aef9111.gif)
Hence, the required energy of activation is 52.86 kJmol−1.