Chemical Kinetics Class 12th Chemistry Part I CBSE Solution

Class 12^th Chemistry Part I CBSE Solution

Intext Questions Pg-98

1. For the reaction R → P, the concentration of a reactant changes from 0.03M to 0.02M in 25…
2. In a reaction, 2A → Products, the concentration of A decreases from 0.5 mol L-1 to 0.4 mol…

Intext Questions Pg-103

1. For a reaction, A + B → Product; the rate law is given by, r = k [A]1/2 [B]^2 . What is…
2. The conversion of molecules X to Y follows second order kinetics. If concentration of X is…

Intext Questions Pg-111

1. A first order reaction has a rate constant 1.15 × 10-3 s-1. How long will 5 g of this…
2. Time required to decompose SO2Cl2 to half of its initial amount is 60 minutes. If the…

Intext Questions Pg-116

1. What will be the effect of temperature on rate constant ?
2. The rate of the chemical reaction doubles for an increase of 10K in absolute temperature…
3. The activation energy for the reaction 2 HI(g) → H2 + I2 (g) is 209.5 kJ mol-1 at…

Exercises

1. From the rate expression for the following reactions, determine their order of reaction…
2. For the reaction : 2A + B → A2B the Rate = k[A][B]^2 with k = 2.0 × 10-6 mol-2 L^2 sec-1.…
3. The decomposition of NH3 on platinum surface is zero order reaction. What are the rates of…
4. The decomposition of dimethyl ether leads to formation of CH4, H2 and CO and the reaction…
5. Mention the factors that affect the rate of chemical reaction?
6. A reaction is second order with respect to a reactant. How is the rate of reaction…
7. What is the effect of temperature on the rate constant of a reaction? How can this effect…
8. In a pseudo first order hydrolysis of ester in water, the following results were obtained…
9. A reaction is First Order in A and second order in B. (i) Write the differential rate…
10. In a reaction between A and B, the initial rate of reaction(r0)was measured for different…
11. The following results have been obtained during the kinetic studies of the reaction: 2A +…
12. The reaction between A and B is first order with respect to A and zero order with respect…
13. Calculate the half life of a first order reaction from their rate constants given below:…
14. The half-life for radioactive decay of 14C is 5730 years. An archeological artifact…
15. The experiemental data for decomposition of N2O5 [2N2O5→ 4NO2 + O2] in gas phase at 318 k…
16. The rate constant for a first order reaction is 60 s-1. How much time will it take to…
17. During nuclear explosion, one of the products is^90 Sr with half-life of 28.1 years. If…
18. For a first order reaction, show that time required for 99% completion is twice the time…
19. A first order reaction takes 40 min for 30% decomposition. Calculate t1/2.…
20. For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data…
21. The following data were obtained during the first order thermal decomposition of SO2Cl2 at…
22. The rate constant for the decomposition of N2O5 at various temperatures is given below:…
23. The rate constant for the decomposition of hydrocarbons is 2.418 × 10-5s-1 at 546 K. If…
24. Consider a certain reaction A → Products with k = 2.0 × 10-2s-1. Calculate the…
25. Sucrose decomposes in acid solution into glucose and fructose according to the first order…
26. The decomposition of hydrocarbon follows the equation k = (4.5 × 10^11 s-1) e-28000K/T.…
27. The rate constant for the first order decomposition of H2O2 is given by the following…
28. The decomposition of A into product has value of k as 4.5 × 10^3 s-1 at 10°C and energy of…
29. The time required for 10% completion of a first order reaction at 298K is equal to that…
30. The rate of a reaction quadruples when the temperature changes from 293 K to 313 K.…

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Intext Questions Pg-98

Question 1.

For the reaction R → P, the concentration of a reactant changes from 0.03M to 0.02M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.

Answer:

Given-

Initial concentration(R₁) = 0.03M

Final concentration(R₂) = 0.02M

Time = 25 mins.

The formula for average rate of the reaction is,

 → Equation 1

∵ {R} = (R₂)-( R₁), the equation 1 is written as,

= 4 × 10^-4 mol L^-1 min^-1

The average rate of reaction in seconds is given by,

= 

(dividing by 60 to convert minutes to seconds)

= 6.6 × 10^-6 mol L^-1 s^-1

The average rate of the reaction in minutes is4 × 10^-4 mol L^-1 min^-1 and in seconds is 6.6 × 10^-6 mol L^-1 s^-1

Question 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:

Given-

Initial concentration(A₁) = 0.5M

Final concentration(A₂) = 0.4M

Time = 10 mins.

The formula for average rate of the reaction is,

 → Equation 1

∵ {A} = (A₂)-( A₁), the equation 1 is written as,

= 

= 0.005 mol L^-1 min^-1

= 5 × 10^-3 mol L^-1 min^-1

The average rate of the reaction is5 × 10^-3 mol L^-1 min^-1

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Intext Questions Pg-103

Question 1.

For a reaction, A + B → Product; the rate law is given by, r = k [ A]^1/2 [B]². What is the order of the reaction?

Answer:

The order of the reaction is sum of the powers of concentration of reactants in the rate law. According to this,

The order of the reaction = 1/2 + 2

= 

The order of the reaction is 2.5

Question 2.

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:

Let the reaction be X→Y

As, this reaction follows second order kinetics, the rate of the reaction will be,

Rate = k(X)² → Equation 1

Since the concentration of X is increased three times, Equation 1

will become,

Rate = k(3X)²

= k × 9(X)²

∴ The rate of formation of Y will become 9 times.

Thus, the rate of formation of Y will become 9 times

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Intext Questions Pg-111

Question 1.

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:

Given-

Rate constant, k = 1.15 × 10^-3 s^-1

Initial quantity, R₀ = 5 g

Final quantity, R = 3 g

According to the formula of first order reaction,

t = 

t = 443.8 s

t = 4.438 × 10² s

The time taken for 5g of the reactant to reduce to 3 g is 4.438 × 10² s

Question 2.

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:

Given-

60 mins

Using the formula for half life, , we get,

∴ k = 1.155 × 10^-2 min^-1

The rate constant of the reaction, k is1.155 × 10^-2 min^-1

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Intext Questions Pg-116

Question 1.

What will be the effect of temperature on rate constant ?

Answer:

The rate constant of a chemical reaction normally increases with increase in temperature. It is observed that for a chemical reaction with rise in temperature by 10°C, the rate constant is nearly doubled and this temperature dependence of the rate of a chemical reaction can

be accurately explained by Arrhenius equation,

k = A 

Thus, the rate constant of a chemical reaction normally increases with increase in temperature.

Question 2.

The rate of the chemical reaction doubles for an increase of 10K in absolute temperature from 298K. Calculate E_a.

Answer:

Given-

Initial temperature, T₁ = 298 K

Final temperature, T₂ = 298 K + 10 K = 308 K

Knowing that the rate constant of a chemical reaction normally increases with increase in temperature, we assume that,

Initial value of rate constant, k₁ = k

Final value of rate constant, k₂ = 2k

Using Arrhenius equation,

 → Equation 1

where, R = 8.314 J K^-1 mol^-1 (gas constant).

Substituting all the values in equation 1, we get,

log 2 = 

E_a = 

E_a = 

E_a = 52897 J mol^-1

E_a = 52.897 kJ mol^-1

The energy of activation, E_a is 52.897 kJ mol^-1

Question 3.

The activation energy for the reaction 2 HI(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:

Given-

Energy of activation, E_a= 209.5 kJ mol^–1

E_a in joules = 209.5 × 1000 = 209500 J mol^–1

Temperature, T = 581K

Gas constant, R = 8.314 J K^-1 mol^-1

Using Arrhenius equation,

k = A 

In this equation, the term  represents the number of molecules which have kinetic energy greater than the activation energy, E_a.

∴ The number of molecules =  → Equation 1

Substituting all the known values in equation 1, we get,

⇒ 

= e^-43.3708

Finding the value of anti ln(43.3708), we get, 1.47 × 10^-19

The fraction of molecules of reactants having energy equal to or greater than activation energy is1.47 × 10^-19

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Exercises

Question 1.

From the rate expression for the following reactions, determine their order of reaction and dimensions of the rate constants.

(i) 3NO(g) → N₂O (g) Rate = k[NO]²

(ii) H₂O₂ (aq) + 3I^–(aq) + 2H⁺→ 2H₂O(l) + I^–₃ Rate = k[H₂O₂][I^–]

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

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

Order is sum of all powers of the concentrations of the reactant in “rate law expression”

Answer:

(i) 3NO(g) → N₂O(g)

Rate = k[NO]²

Here rate law is dependent on concentration of NO only.And power raised to concentration of NO is 2. Hence order of reaction is 2.

To find dimensions of rate constant (K)- We know rate of chemical reaction is measured in . So, unit of K will be

 i.e. [concentration]^-1[ time]^-1 . If concentration is in mol L^-1 and time is in sec. then dimensions of k will be : mol^-1 L sec^-1

(ii) H₂O₂(aq) + 3I^-(aq) + 2H⁺ → 2 H₂O(l) + I₃^-

Rate = k[H₂O₂][I^-]

Here rate law is dependent on concentration of H₂O₂ and I^-. And power raised to concentration of H₂O₂ is 1, and power raised to concentration of I^- is 1. order is sum of powers hence, order of given reaction is 2.

To find dimensions of rate constant (K)- We know rate of chemical reaction is measured in [concentration/time]. So, unit of K will be [Concentration]/[time][concentration]² i.e. [concentration]^-1[ time]^-1. If concentration is in mol L^-1 and time is in sec then dimensions of k will bemol^-1 L sec^-1

(iii) CH₃CHO(g) → CH₄(g) + CO(g)

Rate = k[CH₃CHO]^3/2

Here rate law is dependent on concentration of CH₃CHO.And power raised to CH₃CHO is 3/2. Hence order of reaction is 3/2.

To find dimensions of rate constant (K)- We know rate of chemical reaction is measured in [concentration/time]. So, unit of K will be [Concentration]/[time][concentration]^3/2 i.e. [concentration]^-1/2[time]^-1.

If concentration is in mol L^-1 and time is in sec. Then dimensions of K will be mol^-1/2L^1/2 sec^-1

(iv) C₂H₅Cl (g) → C₂H₄(g) + HCl(g)

Rate = k[C₂H₅Cl]

Here rate law is dependent on concentration of C₂H₅Cl. And power raised to C₂H₅Cl is 1. Hence order of reaction is 1.

To find dimensions of rate constant (K)- We know rate of chemical reaction is measured in [concentration/time]. So, unit of K will be [Concentration]/[time][concentration]¹. i.e. [time]^-1

If concentration is in mol L^-1 and time is in sec. Then dimensions of K will be sec^-1.

Question 2.

For the reaction :

2A + B → A₂B

the Rate = k[A][B]² with k = 2.0 × 10^-6 mol^-2 L² sec^-1. Calculate the initial rate of the reaction when [A] = 0.1 mol L^-1 and [B] = 0.2 mol L^-1. Calculate the rate of reaction after [A] is reduced to 0.06 mol L^-1.

Answer:

a) Rate = k[A][B]², Rate = 2.0 × 10^-6[0.1][0.2]²

Rate = 8 × 10^-9 mol L^-1sec^-1

b) 

Situation when A is remained 0.06 mol L^-1

Now, According to rate law, Rate = k[A][B]²

Rate = 2 × 10^-6[0.06][0.18]²

i.e. Rate = 3.888 × 10^-9 mol L^-1sec^-1

Initial rate of reaction is 8 × 10^-9 mol L^-1sec^-1. and rate when concentration of A is 0.06 mol L^-1 is 3.888 × 10^-9 mol L^-1sec^-1.

Question 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 L^-1 sec^-1?

Answer:

2NH₃(g) → N₂(g) + 3H₂(g)

Rate of zero order reaction is equal to rate constant. i.e. Rate = 2.5 × 10^-4mol L^-1sec^-1.

According to rate law, 

2.5 × 10^-4mol L^-1sec^-1 = 

i.e. the rate of production of N₂ is 2.5 × 10^-4mol L^-1 sec^-1.

According to rate law,

i.e. rate of formation of H₂ is 3 times rate of reaction = 3 × 2.5 × 10^-4mol L^-1sec^-1

= 7.5 × 10^-4mol L^-1sec^-1

Rate of formation of N₂ and H₂ is 2.5 × 10^-4 mol L^-1sec^-1 and 7.5 × 10^-4 mol L^-1sec^-1 respectively.

Question 4.

The decomposition of dimethyl ether leads to 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 an increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether i.e. Rate = k[pCH₃OCH₃]^3/2. If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants?

Answer:

Rate of given chemical reaction will be represented as

.

Hence units of rate is bar min^-1

To find units of K, K = rate/[pCH₃OCH₃]^3/2

The unit of k = bar ^-1/2min^-1.

Units of Rate- bar min^-1. and Units of Rate constant K : bar ^-1/2min ^-1

Question 5.

Mention the factors that affect the rate of chemical reaction?

Answer:

Factors affecting rate of reaction-

1) Temperature (increasing temperature increases rate of reaction)

2) Concentration or pressure of reactants. (Increasing concentration or pressure increases rate of reaction)

3) Presence or absence of a catalyst. (Adding catalyst mostly increases the rate of reaction)

4) The surface area of solid reactant. (If reaction is processing over solid reactant then increasing its surface area increases rate of reaction)

5) Nature of reactants

Question 6.

A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration is

(i) doubled (ii) reduced to half

Answer:

Let suppose reaction

A ⇒ B;

having rate law, Rate = K [A]²

(i) If the concentration of A is doubled then the rate will affect by the square of concentration i.e. rate will become 2² = 4 times.

(ii) If the concentration of A is halved then the rate will affect the square of concentration i.e (1/2)² = 1/4 times.

(i) Rate becomes 4 times of initial Rate (ii) Rate becomes 1/4 times of initial Rate

Question 7.

What is the effect of temperature on the rate constant of a reaction? How can this effect of temperature on rate constant be represented quantitatively?

Answer:

Increase in temperature increases the rate constant of a reaction. as we know increase in temperature increases the rate of reaction to satisfy the equation Rate = k [concentration]ⁿ where n can be any real number. k have to increase as concentration is almost not changing over small temperature change. Increasing temperature by 10⁰C almost double the rate constant.

This can be represented quantitively by the help of arrehinus equation-

K = Ae^-Ea/RT, where k is rate constant, E_a is activation energy, R is universal gas constant, T is absolute temperature.

Question 8.

In a pseudo first order hydrolysis of ester in water, the following results were obtained


(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 the ester.

Answer:

(i) Average rate of reaction over interval is [change in concentration]/[time taken] i.e.

(ii) the pseudo first-order rate constant can be calculated by

K = (2.303/t) log(C_i/C_t) where K is Rate constant,

t is time taken,

C_i is initial concentration

C_t is Concentration at time t.

K = (2.303/30) log (0.55/0.31)

⇒ K = 1.9 × 10^-2 sec^-1

(i) Average rate between 30 to 60 sec is 0.00467 mol L^-1sec^-1

(ii) Pseudo first order rate constant is 1. × 10^-2sec^-1

Question 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) Order is power raised to reactant in rate law, hence,

Rate = k[A][B]²

(ii) When the concentration of B is increased three times then the rate is affected by the square of reactant. The rate is increased by 9 Times.

(iii) When concentration of reactant both A and B is doubled then the rate will have affected as square of reactant B and Two times of Reactant A. Overall increase in rate is 8 times

(i) When the concentration of B is increased by three times, the rate is increased by nine times.

(ii) When conc. of both reactants doubled then Rate increased 8 times.

Question 10.

In a reaction between A and B, the initial rate of reaction(r₀)was measured for different initial concentrations of A and B are given below:


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

Answer:

When concentration of B is changed then rate of reaction doesn't change that means order with respect to B is 0. But when the concentration of A is doubled rate increased by 2.82 times i.e.2^1.5 = 2.82.Hence order with respect to A is 1.5.

Order with respect to A and B is 1.5 and 0 respectively.

Question 11.

The following results have been obtained during the kinetic studies of the reaction:

2A + B → C + D


Determine the Rate law and the rate constant for the reaction.

Answer:

By comparing Experiment I and IV if we increase the concentration of A by 4 times then Rate also increased by 4 times. That means order with respect to A is 1.

By comparing Experiment II and III if we double the concentration of B Rate increases by 4 times that means order with respect to B is 2.

Rate law of reaction will be, Rate = k [A][B]²

To find K, K = rate/[A][B]² i.e. K = 6.0 × 10^-3/[0.1][0.1]²

K = 6 mol^-2L²sec^-1

Order with respect to A and B is 1 and 2 Respectively. And value of K(rate constant) is = 6 mol^-2L²sec^-1

Question 12.

The reaction between A and B is first order with respect to A and zero order with respect to B.Phil. in the blanks in the following table:

Answer:

As reaction is first order with respect to A and zero Order with respect to B. Then changing the concentration of B won’t affect the rate of reaction and increasing concentration of A ‘n’ times will increase the rate by ‘n’ times. By this logic lets fill the table- In first blank space concentration of A will be 0.2 mol L^-1 because the rate is doubled. In second blank space, Rate will be 8 × 10^-2mol L^-1min^-1 because the concentration of A is increased 4 Times. In third blank space concentration of A will be 0.1 mol L^-1 because the rate is same as in experiment I.

Question 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:

Half life of first order reaction is, t_1/2 = ln2/K where t_1/2 is half life of first order reaction, K is rate constant of First order reaction.

(i) t_1/2 = ln2/200 s^-1

⇒ t_1/2 = 0.693/200 s^-1 (∵ ln2 = 0.693)

⇒ t_1/2 = 0.003465 sec.

(ii) t_1/2 = ln2/2 min^-1

⇒ t_1/2 = 0.693/2 min^-1 (∵ ln2 = 0.693)

⇒ t_1/2 = 0.3465 min

(iii) t_1/2 = ln2/4 year^-1

⇒ t_1/2 = 0.693/4 years ^-1 (∵ ln2 = 0.693)

⇒ t_1/2 = 0.17325 year.

Half life of 3 reactions are 0.003465 sec, 0.3465 min, 0.17325 year respectively.

Question 14.

The half-life for radioactive decay of 14C is 5730 years. An archeological artifact containing wood had only 80%of the 14C found in a living tree. Estimate the age of the sample.

Answer:

Radio active decay occurs via first order rate law,

t_1/2 = 5730 years. rate constant(k) of given decay is 0.693/t_1/2

⇒ 0.693/5730 = 1.2 × 10^-4 year^-1

By first order integrated rate law age of sample will be,

where T is the age of sample A₀ is the initial activity of the sample. and A_t is the activity of sample at any time t.

T = 0.18 × 10⁴ years.

Age of given sample is 0.18 × 10⁴ years.

Question 15.

The experiemental data for decomposition of N₂O₅

[2N₂O₅→ 4NO₂ + O₂]

in gas phase at 318 k are given below:


(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: 

(i) 

(ii) initial conc.is 1.63 × 10^-2 so half conc. is 0.815 × 10^-2M From given information and graph t_1/2 is 1440 sec.

(iii) 

(iv) As log [N₂O₅] vs time is straight line given reaction is first order. Hence its rate law will be, Rate = k[N₂O₅]

(v) The slope of above graph is slope = 0.000209

K = 2.303 × slope

⇒ 4.82 × 10^-4sec^-1

Now, t_1/2 = 0.693/K.

⇒ 0.693/4.82 × 10^-4

⇒ t_1/2 = 1438 sec. which is almost equal to (ii)

Question 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:

Given:

Order of the reaction = 1

Let, Initial concentration [R]^°= x

Final concentration [R] = x/16

Rate constant k = 60 s^-1

We know, time 

⇒ 

⇒ 

Solving, we get t = 4.6 × 10^-2s

Question 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:

Initial concentration, [R]_° = 1μg

Final concentration, [R]

Half-life t_1/2 = 28.1 yrs

Solution:

We know, t_1/2 = 0.693/k

Where, k – rate constant

⇒ k = 0.693/ t_1/2

⇒ k = 0.693/(28.1 yrs)

⇒ k = 0.0246 yrs^-1

Also, 

If t = 10yrs, then, using the formula, we get,

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

⇒  = 0.7824μg

If t = 60yrs, then again, we get,

⇒ 

⇒ 

⇒ 

⇒ 

⇒ [R]₆₀ = 0.228 μg

∴ 0.2278μg of ⁹⁰Sr will be left after 60 years.

Question 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:

Let, initial concentration be [R]_°

Concentration at 90% completion be ((100-90)/100)×[R]_°

∴ Concentration at 90% be 0.1[R]_°

Concentration at 99% completion be ((100-99)/100)× [R]_°∴ Concentration at 99% be 0.01[R]_°

We know, time 

Time taken for 90% completion is 

⇒ 

⇒ 

Time taken for 99% completion is 

⇒ 

⇒ 

⇒ t₉₉ = 2t₉₀

Hence, the time taken to complete 9% of the first order reaction is twice the time required for the completion of 90% of the reaction.

Question 19.

A first order reaction takes 40 min for 30% decomposition. Calculate t_1/2.

Answer:

Given:

Time t = 40 min

When 30% decomposition is undergone, 70% is the concentration.

We know, time taken 

Where, k- rate constant

[R]_° -Initial concentration

[R]-Concentration at time ‘t’

⇒ 

⇒ 

⇒ 

⇒ 258.23 = (2.303/k)

We know, Half-life t_1/2 = 0.693/k

Which can be written as, t_1/2 = 0.3010 × (2.303/k)

⇒ t_1/2 = 0.3010 × 258.23

⇒ t_1/2 = 77.72 min

Question 20.

For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained.


Calculate the rate constant.

Answer:

When t = 0, the total partial pressure is P₀ = 35.0 mm of Hg

When time t = t, the total partial pressure is P_t = P₀ + p

P₀-p = P_t-2p, but by the above equation, we know p = P_t-P₀

Hence, P₀-p = P_t-2( P_t-P₀)

Thus, P₀-p = 2P₀ – P_t

We know that time 

Where, k- rate constant

[R]_° -Initial concentration of reactant

[R]-Concentration of reactant at time ‘t’

Here concentration can be replaced by the corresponding partial pressures.

Hence, the equation becomes, 

 → equation 1

At time t = 360 s, P_t = 54 mm of Hg and P₀ = 30 mm of Hg,

Substituting in equation 1,

⇒ 

⇒ k = 2.175 × 10^-3 s^-1

At time t = 720 s, P_t = 63 mm of Hg and P₀ = 30 mm of Hg,

Substituting in equation 1,

⇒ 

Thus, k = 2.235 × 10^-3 s^–1

Taking average, k = (2.235 × 10^-3 s^–1 + 2.175 × 10^-3 s^–1)/2

∴ k = 2.21 × 10^-3 s^–1.

Question 21.

The following data were obtained during the first order thermal decomposition of SO₂Cl₂ at a constant volume.
SO₂Cl_2(g)→ SO_{2 (g)} + Cl_{2 (g)}
Calculate the rate of the reaction when total pressure is 0.65 atm

Answer:

When t = 0, the total partial pressure is P₀ = 0.5 atm

When time t = t, the total partial pressure is P_t = P₀ + p

P₀-p = P_t-2p, but by the above equation, we know p = P_t-P₀

Hence, P₀-p = P_t-2(P_t-P₀)

Thus, P₀-p = 2P₀ – P_t

We know that time 

Where, k- rate constant

[R]_° -Initial concentration of reactant

[R]-Concentration of reactant at time ‘t’

Here concentration can be replaced by the corresponding partial pressures.

Hence, the equation becomes, 

⇒  → equation 1

At time t = 100 s, P_t = 0.6 atm and P₀ = 0.5 atm,

Substituting in equation 1,

⇒ 

Thus, k = 2.231 × 10^-3 s^-1

The rate of reaction R = k × P_S02Cl2

When total pressure P_t = 0.65 atm and P₀ = 0.5 atm, then

P_S02Cl2 = 2P₀-P_t

Thus, substituting the values, P_S02Cl2 = 2(0.5)-0.6 = 0.35 atm

R = k × P_S02Cl2 = 2.231 × 10^-3 s^–1 × 0.35

Rate of the reaction R = 7.8 × 10^-4atm s^–1

Question 22.

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


Draw a graph between ln k and 1/T and calculate the values of A and Ea. Predict the rate constant at 30° and 50°C.

Answer:

To convert the temperature in °C to °K we add 273 K.

The graph is given as:

The Arrhenius equation is given by k = Ae^-Ea/RT

Where, k- Rate constant

A- Constant

E_a-Activation Energy

R- Gas constant

T-Temperature

Taking natural log on both sides,

ln k = ln A-(E_a/RT)

→ equation 1

By plotting a graph, ln K Vs 1/T, we get y-intercept as ln A and Slope is –E_a/R.

Slope = (y₂-y₁)/(x₂-x₁)

By substituting the values, slope = -12.301

⇒ –E_a/R = -12.301

But, R = 8.314 JK^-1mol^-1

⇒ E_a = 8.314 JK^-1mol^-1 × 12.301 K

⇒ E_a = 102.27 kJ mol^-1

Substituting the values in equation 1 for data at T = 273K

⇒ 

⇒ 

(∵ At T = 273K, ln k = -7.147)

On solving, we get ln A = 37.911

∴ A = 2.91×10⁶

When T = 30⁰C, hence T = 30 + 273 = 303K

⇒ 

⇒ 

⇒ ln k = -2.8

⇒ k = 6.08x10^-2s^-1.

When T = 50⁰C, hence T = 50 + 273 = 323K

Substituting in equation 1,

A = 2.91 × 10⁶, E_a = 102.27 kJ mol^-1

Ln K = ln(2.91 × 10⁶)- 102.27/(8.314 × 323)

Thus, ln k = -0.5 and k = 0.607s^-1.

Question 23.

The rate constant for the decomposition of hydrocarbons is 2.418 × 10^–5s^–1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.

Answer:

Given,

k = 2.418 × 10^-5 s^-1

T = 546 K

E_a = 179.9 kJ mol^-1 = 179.9 × 10³J mol^-1

The Arrhenius equation is given by k = Ae^-Ea/RT

Taking natural log on both sides,

Ln k = ln A-(E_a/RT)

Substituting the values,

ln(2.418 × 10^-5 ) = ln A-179.9/(8.314 × 546)

ln A = 12.5917

A = 3.9 × 10¹² s^-1(approximately)

Question 24.

Consider a certain reaction A → Products with k = 2.0 × 10^–2s^-1. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L^–1.

Answer:

Given,

k = 2.0 × 10^–2s^-1

time t = 100s

Concentration [A₀] = 1.0 mol L^-1

We know, 

On substituting the values, 

Log(1/[A]) = 2.303/2

Log[A] = -2.303/2

[A] = 0.135 mol L^–1

Question 25.

Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t_1/2 = 3.00 hours. What fraction of sample of sucrose remains after 8 hours?

Answer:

t_1/2 = 3.00 hours

We know, t_1/2 = 0.693/k

∴ k = 0.693/3

k = 0.231hrs^-1

We know, time 

Where, k- rate constant

[R]_° -Initial concentration

[R]-Concentration at time ‘t’

Thus, substituting the values, 

log([R]₀/[R]) = 0.8

log([R]/[R]₀) = -0.8

[R]/[R]₀ = 0.158

Hence, 0.158 fraction of sucrose remains.

Question 26.

The decomposition of hydrocarbon follows the equation

k = (4.5 × 10¹¹s^–1) e^-28000K/T. Calculate E_a.

Answer:

The given equation is

k = (4.5 x 10¹¹ s^-1) e^{-28000 K/T} (i)

Comparing, Arrhenius equation

k = Ae ^-E_a^/RT (ii)

We get, E_a / RT = 28000K / T

⇒ E_a = R x 28000K

= 8.314 J K^-1mol^-1 × 28000 K

= 232792 J mol^–1 or 232.792 kJ mol^–1

Question 27.

The rate constant for the first order decomposition of H₂O₂ is given by the following equation: log k = 14.34 – 1.25 × 10⁴K/T

Calculate E_a for this reaction and at what temperature will its half-period be 256 minutes?

Answer:

We know, The Arrhenius equation is given by k = Ae^-Ea/RT

Taking natural log on both sides,

Ln k = ln A-(E_a/RT)

Thus, log k = log A -(E_a/2.303RT) → eqn 1

The given equation is log k = 14.34 – 1.25 × 10⁴K/T → eqn 2

Comparing 2 equations,

E_a/2.303R = 1.25 × 10⁴K

E_a = 1.25 × 10⁴K × 2.303 × 8.314

E_a = 239339.3 J mol^-1 (approximately)

E_a = 239.34 kJ mol^-1

Also, when t_1/2 = 256 minutes,

k = 0.693 / t_1/2

= 0.693 / 256

= 2.707 × 10^-3 min^-1

k = 4.51 × 10^-5s^–1

Substitute k = 4.51 × 10^-5s^–1 in eqn 2,

log 4.51 × 10^-5 s^–1 = 14.34 – 1.25 × 10⁴K/T

log(0.654-5) = 14.34– 1.25 × 10⁴K/T

T = 1.25 × 10⁴/[ 14.34- log(0.654-5)]

T = 668.9K or T = 669 K

Question 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 × 10⁴s^–1?

Answer:

From Arrhenius equation, we obtain

Also, k₁ = 4.5 × 10³ s ^{- 1}

T₁ = 273 + 10 = 283 K

k₂ = 1.5 × 10⁴ s ^{- 1}

E_a = 60 kJ mol ^{- 1} = 6.0 × 10⁴ J mol ^{- 1}

Then,

→ 0.5229 = 3133.627 × (T₂-283)/(283 × T₂)

→ 0.0472T₂ = T₂-283

T₂ = 297K or T₂ = 24⁰ C

Question 29.

The time required for 10% completion of a first order reaction at 298K is equal to that required for its 25% completion at 308K. If the value of A is 4 × 10¹⁰s^–1. Calculate k at 318K and E_a.

Answer:

We know, time t = (2.303/k) × log([R]₀/[R])

Where, k- rate constant

[R]_° -Initial concentration

[R]-Concentration at time ‘t’

At 298K, If 10% is completed, then 90% is remaining.

t = (2.303/k) × log ([R]₀/0.9[R]₀)

t = (2.303/k) × log (1/0.9)

t = 0.1054 / k

At temperature 308K, 25% is completed, 75% is remaining

t’ = (2.303/k’) × log ([R]₀/0.75[R]₀)

t’ = (2.303/k’) × log (1/0.75)

t’ = 2.2877 / k'

But, t = t’

0.1054 / k = 2.2877 / k'

k' / k = 2.7296

From Arrhenius equation, we obtain

log k₂/k₁ = (E_a / 2.303 R) × (T₂ - T₁) / T₁T₂

Substituting the values,

⇒ 

E_a = 76640.09 J mol^-1 or 76.64 kJ mol^-1

We know, log k = log A –E_a/RT

Log k = log(4 × 10¹⁰)-(76.64kJ mol^-1/(8.314 × 318))

Log k = -1.986

∴ k = 1.034 x 10^-2 s ^-1

Question 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:

Given, k₂ = 4k₁, T₁ = 293K and T₂ = 313K

We know, From Arrhenius equation, we obtain

⇒ 

⇒ 

On solving we get,

E_a = 58263.33 J mol^-1 or 58.26 kJ mol^-1