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Welcome to Swastik Classes’ NCERT Solutions for Class 12 Chemistry Chapter 4 – Chemical Kinetics. This chapter deals with the study of the rates of chemical reactions, the factors affecting reaction rates, and the mechanisms of reactions.

Our NCERT Solutions for Class 12 Chemistry Chapter 4 aim to provide students with a comprehensive understanding of the concepts covered in the chapter. Our solutions are designed by subject matter experts with vast experience in the field of chemistry. We have provided step-by-step solutions to all the questions in the chapter, making it easy for students to understand and follow the concepts.

Our solutions also include illustrations and diagrams to help students visualize the processes and remember them better. Additionally, we have included sample questions and answers to help students prepare for their exams.

With our NCERT Solutions for Class 12 Chemistry Chapter 4, students can gain a deeper understanding of the factors affecting reaction rates, the various methods of determining reaction rates, and the concept of order and molecularity of a reaction. They can also score well in their exams and build a strong foundation in chemical kinetics.

NCERT Solutions For Class 12 Chemistry Chapter 4 – PDF Download

Answers for chemistry class 12 chapter 4 Chemical Kinetics

Chemistry Class 12 NCERT solutions chapter 4 intext questions and answers

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_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_5e7dae04.gif

= 0.005 mol L−1 min−1

= 5 × 10−3 M min−1

Question 4.3:

For a reaction, A + B → Product; the rate law is given by, What is the order of the reaction?

Answer:

The order of the reaction

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  — (i)

Let [X] = a mol L−1, then equation (i) 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.

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

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°C 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.gifwhere,

A is the Arrhenius factor 

T is the temperature

R is the Universal 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°C (or 10K ).

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

Question 4.9:

The activation energy for the reaction2HI(g) → H2(g) + 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:

7 2Cd73zF9ws 0HusM079kkLUTRwmKLrvRbPN6969844fXp6ooPFBM2adyYU8bqQ1jQc0yTVWnOe2LzRqy4XAhszRh8HdwV9fQkNEwr1G5odCivDne47h3Tkd sH4WOnTUg2c g

 

NCERT Class 12 Chemistry Chapter 4 Exercise Solutions

*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)  H2O(aq) + 3 I− (aq) + 2 H+ → 2 H2O (l) +  Rate = k[H2O2][I]

(iii)  CH3CHO(g) → CH4(g) + CO(g) Rate = [CH3CHO]3/2

(iv)  C2H5Cl(g) → C2H4(g) + HCl(g) Rate = [C2H5Cl]

Answer:

(i)  Given rate = [NO]2

Therefore, order of the reaction = 2

Dimension of k

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 = [H2O2] [I]

Therefore, order of the reaction = 2

Dimension of k

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 = [CH3CHO]3/2

Therefore, order of reaction = 3/2

Dimension of k

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 = [C2H5Cl]

Therefore, order of the reaction = 1

Dimension of k

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 = [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 

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 = [A][B]2= (2.0 × 10−6 mol−2 L2 s−1) (0.06 mol L−1) (0.18 mol L−1)2

= 3.89× 10−9 mol L−1 s−1

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 = 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 byRate = [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 min, then unit of rate = barmin−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 constantshttps://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.

  1. Concentration of reactants (pressure in case of gases)
  2. Temperature
  3. Presence of a catalyst
  4. Surface Area

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 = [A]2ka2

(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/editlive_ncert/62/2013_06_26_17_38_46/SA.png

Therefore, the rate of the reaction would be reduced to one-fourth. 

*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°C 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_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

= 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

= 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

= 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_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−1 0.20 0.20 0.40
B/ mol L−1 0.30 0.10 0.05
r0/ mol L−1 s−1 5.07 × 10−5 5.07 × 10−5 1.43 × 10−4

What is the order of the reaction with respect to A and B?

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

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.

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 = [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 mol−2L2min−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:

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

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 = [A]1 [B]0

⇒ Rate = [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

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 is given by

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 is given by

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 is given by

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 year (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:

All radioactive decay are first order reactions. 

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
https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6372/NS_14-11-08_Utpal_12_Chemistry_4_30_html_5cc8818c.gif 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 and compare it with (ii).

Answer:

(i)

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,

CAkgFHMobPUO1afjM1AQJj S QIj FoMd8MzsXLEddgc9gwuo6Vzl9cR2VEX7nGuovdt8kzZDilce 8o4emywxtgKaxa

is the half-life. From the graph, the half-life is obtained as 1450 s.

(iii)

t(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_27d7e51c.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_6d9ac63a.gif
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
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, log[N2O5] 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_40b6aaf6.gif

(v)  From the plot, log[N2O5] 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_720ab5a7.gif

Again, slope of the line of the plot log[N2O5] 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_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:

Relation between k and half-life is given by

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
HhDKWYKXgruA5Qd13FQvzzyZSHci7QMSmPbaxWbnbxRMQ9z3iyhMa4T63xP5nFhVRBo iB1G gRX0H7pSbzuPhhPF8i3j2vCWd7SQvP 5 BHIfkEIyoT 3w3zjHUM4Wwj5ogaLU

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
TFa4wKMhJn 9ctFl73T8ER8PFbO5NfQPptThaFNfxBHCUSDbV ame5ZnTCC Mq 886maJHlCRycy95kYZTC8B9JN8

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)
0 35.0
360 54.0
720 63.0

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_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

= 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

= 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

*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
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.

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_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

= 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

= 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

= P0 − 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 (pSOCl2)

= (2.23 × 10−3 s−1) (0.35 atm)

= 7.8 × 10−4 atm s−1

Question 4.22:

The rate constant for the decomposition of N2O5 at various temperatures is given below:

T/°C 0 20 40 60 80
https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6373/NS_14-11-08_Utpal_12_Chemistry_4_30_html_2066c989.gif 0.0787 1.70 25.7 178 2140

Draw a graph between ln and 1/and calculate the values of and Ea.

Predict the rate constant at 30º and 50ºC.

Answer:

From the given data, we obtain

T/°C 0 20 40 60 80
T/K 273 293 313 333 353
https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6373/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m20ff4268.gif 3.66×10−3 3.41×10−3 3.19×10−3 3.0×10−3 2.83 ×10−3
https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6373/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m33822587.gif 0.0787 1.70 25.7 178 2140
ln k −7.147 − 4.075 −1.359 −0.577 3.063
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_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 = 2.0 × 10−2 s−1. Calculate the concentration of remaining after 100 s if the initial concentration of 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_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_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_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

= (4.5 × 1011 s−1) e−28000 K/T

Calculate Ea.

Answer:

The given equation is

= (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  (ii)

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 = 14.34 − 1.25 × 10K/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 mol1 (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 as 4.5 × 103 s−1 at 10°C and energy of activation 60 kJ mol−1. At what temperature would 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.

*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 is

4 × 1010 s−1. Calculate 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_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_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 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

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.

 

Topics to study in chemistry class 12 chapter 4

Section number

Topics

4.1

The rate of a Chemical Reaction

4.2

Factors Influencing the Rate of a Reaction

4.2.1

Dependence of Rate on Concentration

4.2.2

Rate Expression and Rate Constant

4.2.3

Order of a Reaction

4.2.4

Molecularity of a Reaction

4.3

Integrated Rate Equations

4.3.1

Zero Order Reactions

4.3.2

First-Order Reaction

4.3.3

Half-Life of a Reaction

4.4

Pseudo First Order Reaction

4.5

Temperature Dependence of the Rate of a Reaction

 

Weightage of chemistry class 12 chapter 4

By preparing for chemical kinetics, you will secure your 4 marks in the CBSE board Term II exam. 

Chapters

Marks

Chemical kinetics

4 marks

Conclusion

Swastik Classes’ NCERT Solutions for Class 12 Chemistry Chapter 4 – Chemical Kinetics, provide students with a comprehensive understanding of the concepts of chemical kinetics. Our solutions cover the factors affecting reaction rates, the different methods of determining reaction rates, and the concept of order and molecularity of a reaction.

We have provided step-by-step solutions to all the questions in the chapter, making it easy for students to understand and follow the concepts. Our solutions also include illustrations and diagrams to help students visualize the processes and remember them better.

Our NCERT Solutions for Class 12 Chemistry Chapter 4 are designed to help students prepare for their exams and score well. By using our solutions, students can build a strong foundation in chemical kinetics and gain a deeper understanding of the subject.

Overall, our NCERT Solutions for Class 12 Chemistry Chapter 4 are an effective resource for students to learn and master the concepts of chemical kinetics. Students can use our solutions to prepare for their exams, revise the concepts covered in the chapter, and gain confidence in their understanding of chemical kinetics.

Related Links

NCERT Solution for Class XIIth Maths Chapter 13 Probability

NCERT Solution for Class XIIth Chemistry Chapter 6 General Principles and Processes of Isolation of Elements

NCERT Solution for Class XIIth Chemistry Chapter 1 The Solid State

NCERT Solution for Class XIIth Chemistry Chapter 5 Surface Chemistry

NCERT Solution for Class XIIth Chemistry Chapter 10 Haloalkanes and Haloarenes

NCERT Solution for Class XIIth Chemistry Chapter 11 Alcohols, Phenols and Ethers

 

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Videos on chemistry class 12 NCERT solutions chapter 4 – Chemical Kinetics

FAQs on chemistry class 12 chapter 4

What is a first-order reaction?

In a first-order chemical reaction, the reaction rate varies according to the concentration of only one of the reactants. In simple words, it is a chemical reaction in which the reaction rate is directly proportional to the concentration of reactants. 

 

How do you find the unit of rate law?

To find the units of rate law, divide the units of rate by the units of molarity in the concentration term of the rate law.

What are the factors affecting the Rate of Reaction?

Following are the factors which affect the rate of reaction:

  • Concentration of reactants
  • Effect of temperature
  • Catalysts
  • Effect of radiations
  • Nature of reacting species.

Factors affecting rate of reaction class 12 chemistry chapter 3 electrochemistry
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2021 Result Highlight of Swastik Classes

NCERT Solutions Class 12 Chemistry Chapters

  • Chapter 1 The Solid State
  • Chapter 2 Solutions
  • Chapter 3 Electrochemistry
  • Chapter 4 Chemical Kinetics
  • Chapter 5 Surface Chemistry
  • Chapter 6 General Principles and Processes of Isolation of Elements
  • Chapter 7 The p-Block Elements
  • Chapter 8 The d-and f-Block Elements
  • Chapter 9 Coordination Compounds
  • Chapter 10 Haloalkanes and Haloarenes
  • Chapter 11 Alcohols, Phenols and Ethers
  • Chapter 12 Aldehydes, Ketones and Carboxylic Acids
  • Chapter 13 Amines
  • Chapter 14 Biomolecules
  • Chapter 15 Polymers
  • Chapter 16 Chemistry in Everyday Life

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