Chemical Kinetics Question Answers: NCERT Class 12 Chemistry

The order of the reaction = 1/2 + 2

= 2.5

The order of reaction is defined as the sum of the powers of concentrations In the rate law.

The rate of second order reaction can be expressed as     rate = k [A]²

The reaction X → Y follows second order kinetics.

Therefore, the rate equation for this reaction will be:

Rate = k [X]^{2                                                    ________________________}   (1)

Let [X] = a mol L^-1 , then equation (1) can be written as:

Rate = 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.

From the question, we can write down the following information:

Initial amount = 5 g

Final concentration = 3 g

Rate constant = 1.15 10 ^{- 3}s ^{- 1}

We know that for a 1^st order reaction,

= 444.38 s

= 444 s (approx)

We know that for a 1^st order reaction,

t½  = 0.693 / k

It is given that t_1/2= 60 min

k = 0.693 / t½ 

  = 0.693 / 60

  = 0.01155 min^-1

   = 1.155 min^-1

Or

k = 1.925 x 10^-2 s^-1

The rate constant of a reaction is nearly doubled with a 10° rise in temperature. However, the exact dependence of the rate of a chemical reaction on temperature is given by Arrhenius equation,

k = Ae ^{- E_a / RT}

Where,

A is the Arrhenius factor or the frequency factor

T is the temperature

R is the gas constant

E_a is the activation energy

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:

= 52897.78 J mol ^{- 1}

= 52.9 kJ mol ^{- 1}

In the given case:

E_a = 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

(i) Given rate = k [NO]²

Therefore, order of the reaction = 2

Dimension of k = Rate / [NO]²

= mol L^-1 s^-1 / (mol L^-1)²

= mol L^-1 s^-1 / mol² L^-2

= L mol^-1 s^-1

(ii) Given rate = k [H₂O₂][I ^- ]

Therefore, order of the reaction = 2

Dimension of k = Rate  / [H₂O₂][I ^- ]

= mol L^-1 s^-1  / (mol L^-1) (mol L^-1)

= L mol^-1 s^-1

(iii) Given rate =k [CH₃CHO]^3/2

Therefore, order of reaction = 3/2

Dimension of k = Rate / [CH₃CHO]^3/2

=  mol L^-1 s^-1   / (mol L^-1)^3/2

= mol L^-1 s^-1   / mol^3/2  L^-3/2

= L^½ mol^-½  s^-1

(iv) Given rate = k [C₂H₅Cl]

Therefore,order of the reaction = 1

Dimension of k = Rate /  [C₂H₅Cl]

= mol L^-1 s^-1   / mol L^-1

= s^-1

Let the order of the reaction with respect to A be xand with respect to B be y.

Therefore,

= 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.

The given reaction is of the first order with respect to A and of zero order with respect to B.

Therefore, the rate of the reaction is given by,

Rate = k [A]¹[B]⁰

⇒ Rate = k [A]

From experiment I, we obtain

2.0 x 10^-2mol L^-1min^-1= k (0.1 mol L^-1)

⇒ k = 0.2 min^-1

From experiment II, we obtain

4.0 x 10^-2mol L^-1min^-1= 0.2 min^-1[A]

⇒ [A] = 0.2 mol L^-1

From experiment III, we obtain

Rate = 0.2 min^-1 x 0.4 mol L^-1

= 0.08 mol L^-1min^-1

From experiment IV, we obtain

2.0 x 10^-2mol L^-1min^-1= 0.2 min^-1[A]

⇒ [A] = 0.1 mol L^-1

(i) Half life, t_½ = 0.693 / k

= 0.693 / 200 s^-1

= 3.47 ×10 ^-3 s (approximately)

(ii) Half life, t_½ = 0.693 / k

= 0.693 / 2 min^-1

= 0.35 min (approximately)

(iii) Half life,t_½ = 0.693 / k

= 0.693 / 4 years^-1

= 0.173 years (approximately)

Here,

K = 0.693 / t_½

= 0.693 / 5730 years^-1

It is known that,

= 1845 years (approximately)

Hence, the age of the sample is 1845 years.

(i)

(ii) Time corresponding to the concentration, 1630x10² / 2 mol L^-1 = 81.5 mol L^-1 is the half life. From the graph, the half life is obtained as 1450 s.

(iii)

(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

Rate = k [N₂O₅]

(v) From the plot,  log[N₂O₅]

v/s t, we obtain

Slope = -2.46 -(1.79) / 3200-0

= -0.67 / 3200

Again, slope of the line of the plot log[N₂O₅] v/s t is given by

- k / 2.303

.Therefore, we obtain,

- k / 2.303  = - 0.67 / 3200

⇒ k = 4.82 x 10^-4 s^-1

(vi) half life is given by,

t_½ = 0.693 / k

= 0.639 / 4.82x10^-4 s

=1.438 x 10³ s

Or we can say

1438 S

Which is very near to what we obtain from graph.

Therefore, 0.7814 μg of ⁹⁰Sr will remain after 10 years.

Again,

Therefore, 0.2278 μg of ⁹⁰Sr will remain after 60 Years

For a first order reaction, the time required for 99% completionis

t₁ = 2.303/k Log 100/100-99

= 2.303/k Log 100

= 2x 2.303/k

For a first order reaction, the time required for 90% completion is

t₂ = 2.303/k Log 100/100-90

= 2.303/k Log 10

= 2.303/k

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.

For a first order reaction,

t = 2.303/k Log [R] _º / [R]

k = 2.303/40min  Log 100 / 100-30

= 2.303/40min  Log 10 / 7

= 8.918 x 10^-3 min^-1

Therefore, t_1/2 of the decomposition reaction is

t_1/2 = 0.693/k

=  0.693 / 8.918 x 10^-3  min

= 77.7 min (approximately)

The initial rate of the reaction is

Rate = k [A][B]²

= (2.0 × 10 ^{- 6}mol ^{- 2}L²s ^{- 1}) (0.1 mol L ^{- 1}) (0.2 mol L ^{- 1})²

= 8.0 × 10 ^{- 9}mol ^{- 2}L²s ^{- 1}

When [A] is reduced from 0.1 mol L ^{- 1}to 0.06 mol ^{- 1}, the concentration of A reacted = (0.1 - 0.06) mol L ^{- 1} = 0.04 mol L ^{- 1}

Therefore, concentration of B reacted= 1/2 x 0.04 mol L^-1 = 0.02 mol L ^{- 1}

Then, concentration of B available, [B] = (0.2 - 0.02) mol L ^{- 1}

= 0.18 mol L ^{- 1}

After [A] is reduced to 0.06 mol L ^{- 1}, the rate of the reaction is given by,

Rate = k [A][B]²

= (2.0 × 10 ^{- 6}mol ^{- 2}L²s ^{- 1}) (0.06 mol L ^{- 1}) (0.18 mol L ^{- 1})²

= 3.89 mol L ^{- 1}s ^{- 1}

The decomposition of azoisopropane to hexane and nitrogen at 543 K is represented by the following equation.

After time, t, total pressure, P_t = (P_º - p) + p + p

⇒ P_t = (P_º + p)

⇒ p = P_t  - P_º

therefore, P_º - p = P_º  - P_t  - P_º

= 2 P_º -  P_t

For a first order reaction,

k = 2.303/t  Log  P_º / P_º  - p

=   2.303/t  Log  P_º / 2 P_º  -  P_t

When t = 360 s, k = 2.303 / 360s log 35.0 / 2x35.0 - 54.0

= 2.175 × 10 ^{- 3} s ^{- 1}

When t = 720 s, k = 2.303 / 720s log 35.0 / 2x35.0 - 63.0

= 2.235 × 10 ^{- 3} s ^{- 1}

Hence, the average value of rate constant is

k = (2.175 × 10 ^{- 3}  + 2.235 × 10 ^{- 3} ) / 2   s ^{- 1}

= 2.21 × 10 ^{- 3} s ^{- 1}

The thermal decomposition of SO₂Cl₂at a constant volume is represented by the following equation.

After time, t, total pressure,P_t = (P_º - p) + p + p

⇒ P_t = (P_º + p)

⇒ p = P_t  - P_º

therefore, P_º - p = P_º  - P_t  - P_º

= 2 P_º -  P_t

For a first order reaction,

k = 2.303/t  Log  P_º / P_º  - p

=   2.303/t  Log  P_º / 2 P_º  -  P_t

When t= 100 s,

k = 2.303 / 100s log 0.5 / 2x0.5 - 0.6

= 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

p_SOCL2 = 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(p_SOCL2)

= (2.23 × 10 ^{- 3}s ^{- 1}) (0.35 atm)

= 7.8 × 10 ^{- 4}atm s ^{- 1}

From the given data, we obtain

Slope of the line,

In k= - 2.8

Therefore, k = 6.08x10^-2s^-1

Again when T = 50 + 273K = 323K,

1/T = 3.1 x 10^-3 K

In k = - 0.5

Therefore, k = 0.607 s^-1

k= 2.418 × 10^-5 s^-1 

T= 546 K

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

According to the Arrhenius equation,

= (0.3835 - 5) + 17.2082

= 12.5917

Therefore, A = antilog (12.5917)

= 3.9 × 10¹² s ^{- 1}(approximately)

k= 2.0 × 110^-2 s^-1

T= 100 s

[A]_o= 1.0 moL ^{- 1}

Since the unit of kis s ^{- 1}, the given reaction is a first order reaction.

Therefore, k = 2.303/t  Log  [A]_º / [A]

⇒2.0 × 110^-2 s^-1  = 2.303/100s  Log  1.0 / [A]

⇒2.0 × 110^-2 s^-1  = 2.303/100s  ( - Log [A] )

⇒ - Log [A] = -  (2.0 x 10^-2 x 100) /   2.303

⇒ [A] = antilog [-  (2.0 x 10^-2 x 100)  /  2.303]

= 0.135 mol L ^{- 1} (approximately)

Hence, the remaining concentration of A is 0.135 mol L ^{- 1}.

For a first order reaction,

k = 2.303/t  Log  [R]_º / [R]

It is given that, t_1/2 = 3.00 hours

Therefore, k = 0.693 / t_1/2

= 0.693 / 3  h^-1

= 0.231 h ^{- 1}

Then, 0.231 h ^{- 1} = 2.303 / 8h  Log  [R]_º / [R]

Hence, the fraction of sample of sucrose that remains after 8 hours is 0.158.

The given equation is

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

Arrhenius equation is given by,

k= Ae ^-E_a/RT                                                (ii)

From equation (i) and (ii), we obtain

E_a  / RT  =  28000K / T

⇒ E_a  = R x 28000K 

= 8.314 J K ^{- 1}mol ^{- 1}× 28000 K

= 232792 J mol ^{- 1}

= 232.792 kJ mol ^{- 1}

Arrhenius equation is given by,

k= Ae ^-E_a/RT  

⇒In k = In A - E_a/RT

⇒In k = Log A - E_a/RT

⇒ Log k = Log A - E_a/2.303RT         (i)

The given equation is

Log k = 14.34 - 1.25 10⁴ K/T             (ii)

From equation (i) and (ii), we obtain

E_a/2.303RT  = 1.25 10⁴ K/T  

⇒ E_a  =1.25 × 10⁴K × 2.303 × R

= 1.25 × 10⁴K × 2.303 × 8.314 J K ^{- 1}mol ^{- 1}

= 239339.3 J mol ^{- 1} (approximately)

= 239.34 kJ mol - 1

Also, when t_1/2= 256 minutes,

k = 0.693 / t_1/2

= 0.693 / 256

= 2.707 × 10 ^{- 3} min ^{- 1}

= 4.51 × 10 ^{- 5}s ^{- 1}

It is also given that, log k= 14.34 - 1.25 × 10⁴K/T

= 668.95 K

= 669 K (approximately)

From Arrhenius equation, we obtain

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

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,

= 297 K = 24°C

Hence, k would be 1.5 × 10⁴ s ^{- 1} at 24°C.

For a first order reaction,

t = 2.303 / k log a  / a - x

At 298 K,

t = 2.303 / k log 100  / 90

= 0.1054 / k

At 308 K,

t' = 2.303 / k' log 100  / 75

= 2.2877 / k'

According to the question,

t = t'

⇒ 0.1054 / k  =  2.2877 / k'

⇒ k' / k  = 2.7296

From Arrhenius equation,we obtain

 To calculate k at 318 K,

It is given that, A = 4 x 10¹⁰ s^-1, T = 318K

Again, from Arrhenius equation, we obtain

Therefore, k = Antilog (-1.9855)

= 1.034 x 10^-2 s ^-1

The decomposition of NH₃on platinum surface is represented by the following equation.

= 7.5 × 10 ^{- 4}mol L ^{- 1}s ^{- 1}

From Arrhenius equation, we obtain

Hence, the required energy of activation is 52.86 kJ mol ^{- 1}.

If pressure is measured in bar and time in minutes, then

Unit of rate = bar min ^{- 1}

Rate = k (P_CH3OCH3)^3/2

⇒ k = Rate / (P_CH3OCH3)^3/2

Therefore, unit of rate constants (k) = bar min^-1  / bar^3/2

= bar^-½ min ^{- 1}

The factors that affect the rate of a reaction areas follows.

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

(ii) Temperature

(iii) Presence of a catalyst

Let the concentration of the reactant be [A] = a

Rate of reaction, R = k [A]² = ka²

(i) If the concentration of the reactant is doubled, i.e. [A] = 2a, then the rate of the reaction would be

R = k(2a)²

= 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. [A] = 1/2 a, then the rate of the reaction would be

R = k(1/2a)²

= 1/4 Ka²

= 1/4 R

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,

K = Ae ^{-E_a / RT}

where, kis the rate constant,

A is the Arrhenius factor or the frequency factor,

R is the gas constant,

T is the temperature, and

E_a is the energy of activation for the reaction

(i) Average rate of reaction between the time interval, 30 to 60 seconds, = d[ester] / dt

= (0.31-0.17) / (60-30)

= 0.14 / 30

= 4.67 × 10 - 3mol L ^{- 1}s ^{- 1}

(ii) For a pseudo first order reaction,

k = 2.303/ t log [R]_º / [R]

For t= 30 s, k₁ = 2.303/ 30 log 0.55 / 0.31

= 1.911 × 10 ^{- 2}s ^{- 1}

For t = 60 s, k₂ = 2.303/ 60 log 0.55 / 0.17

= 1.957 × 10 ^{- 2}s ^{- 1}

For t= 90 s, k₃ = 2.303/ 90 log 0.55 / 0.085

= 2.075 × 10 ^{- 2}s ^{- 1}

Then, average rate constant, k = k₁ + k₂+ k₃  / 3

= 1.911 × 10 ^{- 2}  + 1.957 × 10 ^{- 2} + 2.075 × 10 ^{- 2} / 3

= 1.981 x 10^-2 s ^{- 1}