4- CHEMICAL KINETICS


Here we are going through the ncert solutions for class 12 chemistry chapter 4- chemical kinetics so before going through the ncert solutions make sure to go through the textbook that helps you to understand the solutions more easily

 ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​​​ Chapter 4 - Chemical Kinetics

 

 ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​​​ INTEXT 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_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



Question 4.3:

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

ANSWER:

The order of the reactionhttps://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.



 

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° 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=eEa/RT⇒In x=−EaRT⇒log x=−Ea2.303RT⇒log x=−209500 J mol−12.303×8.314 JK−1mol−1×581=−18.8323Nowx =Antilog (−18.8323)              = 1.471×10−19x=e-Ea/RTIn x=-EaRTlog x=-Ea2.303RTlog x=-209500 J mol-12.303×8.314 JK-1mol-1×581=-18.8323Now, x =Antilog -18.8323              = 1.471×10-19



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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) + 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 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 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 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 = 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_1bd21519.gif

(iv) Given rate = [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_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 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= 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 = [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 = 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 = [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 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.

(i) Concentration of reactants (pressure in case of gases)

(ii) Temperature

(iii) Presence of a catalyst



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]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. 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, 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 tohttps://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_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.gifhttps://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:

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, 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_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_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

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:

  •  

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, 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 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, 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 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, 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 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 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 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:

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

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)

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

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

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

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= 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(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


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.

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 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