Important Questions for Class 12 Chemistry Chapter 3 Chemical Kinetics

Chemical Kinetics

Question 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 = \(-\frac{\Delta[\mathrm{R}]}{\Delta \mathrm{t}}\)

⇒ \(-\frac{[R]_2-[R]_1}{t_2-t_1}=-\frac{0.02-0.03}{25} \mathrm{Mmin}^{-1}\)

⇒ \(-\frac{0.02-0.03}{25} \mathrm{Mmin}^{-1}\)

⇒ \(4 \times 10^{-4} \mathrm{M} \mathrm{Min}^{-1}\)

⇒ \(\frac{4 \times 10^{-4}}{60} \mathrm{Ms}^{-1}\)

⇒ \(6.67 \times 10^{-5} \mathrm{M} \mathrm{s}^{-1}\)

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:

⇒ \(-\frac{1}{2} \frac{\Delta[\mathrm{A}]}{\Delta t}=-\frac{1}{2} \frac{[\mathrm{A}]_2-[\mathrm{A}]_1}{\mathrm{t}_2-\mathrm{t}_1}\)

⇒ \(-\frac{1}{2} \frac{[0.4-0.5]}{10}=-\frac{1}{2} \frac{-0.1}{10}\)

⇒ 0.005 mol L^-1 min^-1 = 5 × 10^-3 M min^-1

Question 3. 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, \(\frac{1}{2}+2=2 \frac{1}{2}=2.5\)

Question 4. The conversion of molecules X to Y follows second-order kinetics. If the concentration of X is increased to three times how will it affect the rate of formation of Y?
Answer:

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

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

Rale = 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,

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Question 5. A first-order reaction has a rate constant of 1.15 × 10^-3 S^-1. How long will 5g of this reactant take to reduce to 3g?
Answer:

We know that for a 1^st order reaction

⇒ \(t=\frac{2.303}{k} \log \frac{[R]_0}{[R]}=\frac{2.303}{1.15 \times 10^{-3}} \log \frac{5}{3}\)

⇒ \(\frac{2.303}{1.15 \times 10^{-3}} \times 0.2219=444.38 \mathrm{~s}=444 \mathrm{~s}\)

Question 6. The time required to decompose SO₂Cl₂ to half of its initial amount is 60 minutes. If decomposition is a first-order reaction, calculate the rate constant of the reaction.
Answer:

We know that fora 1^st order reaction, \(t_{1 / 2}=\frac{0.693}{k}\)

It is given that t_1/2 = 60 min

⇒ \(k=\frac{0.693}{t_{1 / 2}}=\frac{0.693}{60}=0.01155 \mathrm{~min}^{-1}=1.155 \times 10^{-2} \mathrm{~min}^{-1}\)

Question 7. What will be the effect of temperature on the 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 the Arrhenius equation, k =Ae^-Ea/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

Question 8. The rate of the chemical reaction doubles for an increase of 10 K in an absolute temperature from 298 K. Calculate E_a.
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 k₂ = 2k

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

Now, substituting these values in the equation:

⇒ \(\log \frac{k_2}{k_1}=\frac{E_a}{2.303 R}\left[\frac{T_2-T_1}{T_1 T_2}\right]\)

We get:

⇒ \(\log \frac{2 k}{k}=\frac{E_a}{2.303 \times 8.314}\left[\frac{10}{298 \times 308}\right] \Rightarrow \log 2=\frac{E_2}{2.303 \times 8.314}\left[\frac{10}{298 \times 308}\right]\)

⇒ \(E_a=\frac{2.303 \times 8.314 \times 298 \times 308 \times \log 2}{10}\)

⇒ 52897.78 J mol^-1 = 52.9 kJ mol^-1

Question 9. The activation energy for the reaction 2HI_g → H₂+ I_2(g) is 209.5 kJ mol^-1 at 581 K. Calculate the fraction of molecules of reactant having energy equal to or greater than the activation energy.
Answer:

The fraction of molecules of reactants having energy equal to or greater than activation energy is given as:

⇒ \(\mathrm{x}=\mathrm{e}^{-\mathrm{Ea} / \mathrm{RT}}\)

⇒ In x = -Ea/RT

⇒ \(\log x=\frac{-E_a}{2.303 R T}\)

⇒ \(\log x=\frac{209500 \mathrm{~J} \mathrm{~mol}^{-1}}{2.303 \times 8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \times 581}=-18.8323\)

Now, x = Anti log (-18.8323)

= 1.471 × 10^-19

Question 10. From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.
Answer:

1. 3NO(g) → N₂O(g) ; Rate = k[NO]²
2. H₂O₂(aq) + 31 (aq.) + H⁺ → 2H₂O(I) + I^–₃ ; Rate = k[H₂O)][I^–]
3. CH₃CHO(g) → CH₄(g) + CO(g) : Rate = k[CH₃CHO]^3/2
4. C₂H₅Cl(g) → C₂H₂(g) + HCI(g) : Rate = k[C₂H₅Cl]

Answer:

Given rate = k[NO]²

Therefore, the order of the reaction = 2

Dimension of k = \(\frac{\text { Rate }}{[\mathrm{NO}]^2}=\frac{\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}}{\left(\mathrm{~mol} \mathrm{~L}^{-1}\right)^2}=\frac{\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}}{\mathrm{~mol}^{-2} \mathrm{~L}^{-2}}=\mathrm{L} \mathrm{mol}^{-1} \mathrm{~s}^{-1}\)

Give rate = k[H₂O][I^–]

Therefore, the order of the reaction = 2

Dimension of k = \(\frac{\text { Rate }}{\left[\mathrm{H}_2 \mathrm{O}_2\right]\left[\mathrm{I}^{-}\right]}=\frac{\mathrm{molL}^{-1} \mathrm{~s}^{-1}}{\left(\mathrm{molL}^{-1}\right)\left(\mathrm{molL}^{-1}\right)}=\mathrm{L} \mathrm{mol}^{-1} \mathrm{~s}^{-1}\)

Given rate = k[CHCHO]^3/2

Therefore, order of reaction = \(\frac{3}{2}\)

⇒ \(\text { Dimension of } \mathrm{k}=\frac{\text { Rate }}{\left[\mathrm{CH}_3 \mathrm{CHO}\right]{\frac{3}{2}}}=\frac{\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}}{\left(\mathrm{~mol} \mathrm{~L}^{-1}\right)^{\frac{3}{2}}}=\frac{\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}}{\mathrm{~mol}^{\frac{3}{2}} \mathrm{~L}^{-\frac{3}{2}}}=\mathrm{L}^{\frac{1}{2}} \mathrm{~mol}^{-\frac{1}{2}} \mathrm{~s}^{-1}\)

Given rate- k[C₂H₅Cl]

Therefore, the order of the reaction = 1

⇒ \(\text { Dimension of } \mathrm{k}=\frac{\text { Rate }}{\left[\mathrm{C}_2 \mathrm{H}_5 \mathrm{Cl}\right]}=\frac{\mathrm{molL}^{-1} \mathrm{~s}^{-1}}{\mathrm{~mol} \mathrm{~L}^{-1}}=\mathrm{s}^{-1}\)

Question 11. For the reaction: 2A + B → A₂B the rate = k [A][B]² with k = 2.0 × 10^-6 mol^-2 L²S^-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 fAJ is reduced to 0.06 mol L^-1.
Answer:

The initial rate of the reaction is Rate = k[A][B]² = (2.0 × 10^-6 mol^-2 L² S^-1) (0.1 mol L^-1) (0.2 mol L^-1)² = 8.0 × 10^-9 mol ^-2 L² s^-1 when [A] is reduced from 0.1 mol L^-1 to 0.06 mol^-1, the concentration of A reacted = (0.1 -0.06) mol L^-1 =0.04 mol L^-1.

Therefore, concentration of B reacted \(=\frac{1}{2} \times 0.04 \mathrm{~mol} \mathrm{~L}^{-1}=0.02 \mathrm{~mol} \mathrm{~L}^{-1}\)

Then, a concentration of B is available. |B] = (0.2 – 0.02) mol L^-1 = 0. 1 8 mol L^-1 After [A] is reduced to 0.06 mol L^-1. the rate of the reaction is given by.

Rate = k(A][B]² = (2.0 × 10^-6 mol^-2 L² s^-1) (0.06 mol L^-1) (0.18 mol L^-1)² = 3.88 × 10 mol L^-1s^-1.

Question 12. The decomposition of NH₃, on the platinum surface is an order reaction. What is the rate of production of N₂, and H₂ if k = 2.5 × 10^-4 mol liter s^-1?
Answer:

2NH₃ → N₂ + 3H₂

⇒ \(\text { Rate of reaction }(\mathrm{r})=-\frac{1}{2} \frac{\mathrm{d}\left[\mathrm{NH}_3\right]}{\mathrm{dt}}=\frac{\mathrm{d}\left[\mathrm{N}_2\right]}{\mathrm{dt}}=\frac{1}{3} \frac{\mathrm{d}\left[\mathrm{H}_2\right]}{\mathrm{dt}}\)

Rate (r) = k[NH₃]° = k

(∵ zero order reaction)

= 2.5 × 10^-4

⇒ \(\frac{\mathrm{d}\left[\mathrm{N}_2\right]}{\mathrm{dt}}=\mathrm{r}=2.5 \times 10^{-4} \mathrm{~mol} \mathrm{lit}^{-1} \mathrm{~s}^{-1}\)

⇒ \(\frac{\mathrm{d}\left[\mathrm{H}_2\right]}{\mathrm{dt}}=3 \mathrm{r}=3 \times 2.5 \times 10^{-4}=7.5 \times 10^{-4} \mathrm{~mol} \mathrm{lit}^{-1} \mathrm{sec}^{-1}\)

Question 13. The decomposition of dimethyl ether leads to the formation of CH_4, H_2, 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.c., Rate = k[P_CH3OCH3],^3/2 If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants?
Answer:

⇒ \(\mathrm{CH}_3 \mathrm{OCH}_3 \longrightarrow \mathrm{CH}_4+\mathrm{CO}+\mathrm{H}_2\)

Rate = k(CH₃OCH₃]^3/2; = k[P(CH₃OCH₃)]^3/2

unit of rate = bar min^-1

⇒ \(\text { Unit of } \mathrm{K}=\frac{\text { Rate }}{\left[\mathrm{P}_{\mathrm{CH}_3 \mathrm{OCH}_3}\right]^{3 / 2}}=\frac{\text { bar } \mathrm{min}^{-1}}{\text { bar }^{3 / 2}}=\mathrm{bar}^{-1 / 2} \mathrm{~min}^{-1}\)

Question 14. Mention the factors that affect the rate of a chemical reaction.
Answer:

The important factors on which the rate of a chemical reaction depends are

1. Nature of the reacting species.
2. Concentration of the reacting species.
3. The temperature at which a reaction proceeds.
4. The surface area of the reactants,
5. Presence of a catalyst

Question 15. A reaction is of second order concerning a reactant. How is its rate affected if the concentration of the reactant is (1) doubled and (2) reduced to half?
Answer:

Given rate (r₀ ) = K [A]²

If [A] is doubled: \(r_1=k[2 A]^2\)

r₁ = 4r₀

If [A] is reduced to half: \(\mathrm{r}_2=\mathrm{k}\left[\frac{\mathrm{A}}{2}\right]^2\)

⇒ \(r_2=\frac{1}{4} r_0\)

Question 16. What is the effect of temperature on the rate constant, of a reaction? How can this temperature effect on the rate constant be represented quantitatively?
Answer:

The rate constant is nearly doubled with a rise in temperature by 100C for a chemical reaction.

The temperature effect on the rate constant can be represented quantitatively by Arrhenius equation, k = Ac^-Ea/RT

⇒ \(2.303 \log \frac{\mathrm{K}_2}{\mathrm{~K}_1}=\frac{\mathrm{E}_a}{\mathrm{R}}\left[\frac{\mathrm{T}_2-\mathrm{T}_1}{\mathrm{~T}_1 \mathrm{~T}_2}\right]\)

where K is the rate constant at the temperature

A is the Arrhenius parameter,

R is the gas constant,

T is the temperature,

and E_a is the energy of activation which is always positive.

Question 17. In a pseudo-first-order hydrolysis of ester in water, the following results were obtained:

1. Calculate the average rate of reaction between the time interval 30 to 60 seconds.
2. Calculate the pseudo-first-order rate constant for the hydrolysis of ester.

Answer:

The average rate of reaction between the time interval 30 to 60 seconds

⇒ \(=\frac{\mathrm{d}[\text { Ester }]}{\mathrm{dt}}=-\frac{(0.17-0.31)}{60-30}=\frac{-(-0.14)}{30}=4.67 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)

For a pseudo-first-order reaction.

⇒ \(\text { For } t=30 s, \quad k_1=\frac{2.303}{30} \log \frac{0.55}{0.31}=1.911 \times 10^{-2} \mathrm{~s}^{-1}\)

⇒ \(\text { For } \mathrm{t}=60 \mathrm{~s}, \quad \mathrm{k}_2=\frac{2.303}{60} \log \frac{0.55}{0.17}=1.957 \times 10^{-2} \mathrm{~s}^{-1}\)

⇒ \(\text { For } \mathrm{t}=90 \mathrm{~s}, \quad \mathrm{k}_3=\frac{2.303}{90} \log \frac{0.55}{0.85}=2.075 \times 10^{-2} \mathrm{~s}^{-1}\)

Then, average rate constant, k = \(\frac{k_1+k_2+k_3}{3}\)

⇒ \(=\frac{\left(1.911 \times 10^{-2}\right)+\left(1.957 \times 10^{-2}\right)+\left(2.075 \times 10^{-2}\right)}{3}=1.98 \times 10^{-2} \mathrm{~s}^{-1}\)

Question 18. A reaction is first order in A and second order in B:

1. Write the differential rate equation.
2. How is the rate affected when the concentration of B is tripled?
3. How is the rate affected when the concentration of both A and B are doubled?

Answer:

Rate = k [A]¹[B]²

r₀= K[A]¹[B]²

r₁,= k [A]¹[3 B]²

r₁, = 9 × r0

r₀ = K [A]¹[B]²

r₀ = k [2A]¹[2B]²

r₂ = – 8 × r₀

Question 19. In a reaction between A and B the initial rate of reaction (r₀) was measured for different initial concentrations A and B as given below:

What is the order of the reaction concerning A and B?

Answer:

Let the order of the reaction concerning A be x and B be y.

Therefore, r₀ = k [A]^x[B]^y 

5.07 x 10^-5 =k [0.20]^x[10.30]^y _______ 1

5.07 x 10^-5 = k [0.20]^x[0.10]^y _______ 2

1.43 x 10^-4 = k [0.40]^x[0.05]^y _______ 3

Dividing equation (1) by (2), we obtain

⇒ \(\frac{5.07 \times 10^{-5}}{5.07 \times 10^{-5}}=\frac{\mathrm{k}[0.20]^{\mathrm{x}}[0.30]^y}{\mathrm{k}[0.20]^{\mathrm{x}}[0.10]^y}\)

⇒ \(I=\frac{[0.30]^y}{[0.10]^y}\)

⇒ \(\left(\frac{0.30}{0.10}\right)^0=\left(\frac{0.30}{0.10}\right)^y\)

⇒ y = 0

Dividing equation (3) by (1), we obtain

⇒ \(\frac{1.43 \times 10^{-4}}{5.07 \times 10^{-5}}=\frac{k[0.40]^x[0.05]^y}{k[0.20]^x[0.30]^y}\)

⇒ \(\frac{1.43 \times 10^{-4}}{5.07 \times 10^{-5}}=\frac{[0.40]^x}{[0.20]^x}\)

[Since y = 0 [0.05]y =[0.30] = 1]

⇒ 2.821 =2^x

⇒ log 2.821 = x log 2 (Taking log on both sides)

⇒ \(x=\frac{\log 2.821}{\log 2}=1.496=1.5\)

Hence, the order of the reaction concerning A is 1.5, and concerning B is zero.

Question 20. 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:

Let the order of the reaction concerning A be x and for B be y.

Therefore, the rate of the reaction is given by,

Rate = k[A]^x[B]^y

According to the question.

6.0 × 10^-3 = k[0.1]^x [0.1]^y

7.2 × 10^-2 = k[0.3]^x[0.2]^y

2.88 × 10^-1 = k[0.3]^x [0.4]^y

2.40 × 10^-2 = k[0.4]^x [0.1]^y

Dividing equation (4) by (1), we obtain

⇒ \(\frac{2.4 \times 10^{-2}}{6.0 \times 10^{-3}}=\frac{k[0.4]^x[0.1]^y}{k[0.1]^x[0.1]^y}\)

⇒ \(4=\frac{[0.4]^x}{[0.1]^x} \Rightarrow 4=\left(\frac{0.4}{0.1}\right)^x \Rightarrow(4)^1=4^x \Rightarrow x=1\)

Dividing equation (3) by (2), we obtain

⇒ \(\frac{2.88 \times 10^{-1}}{7.2 \times 10^{-2}}=\frac{k[0.3]^x[0.4]^y}{k[0.3]^x[0.2]^y}\)

⇒ \(4=\left(\frac{0.4}{0.2}\right)^y \Rightarrow 4=2^y \Rightarrow 2^2=2^y \Rightarrow y=2\)

Therefore, the rate law is

⇒ \(\text { Rate }=k[A][B]^2 \Rightarrow k=\frac{\text { Rate }}{[A][B]^2}\)

From experiment I, we obtain

⇒ \(\mathrm{k}=\frac{6.0 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}}{\left(0.1 \mathrm{~mol} \mathrm{~L}^{-1}\right)\left(0.1 \mathrm{~mol} \mathrm{~L}^{-1}\right)^2}=6.0 \mathrm{~L}^2 \mathrm{~mol}^{-2} \mathrm{~min}^{-1}\)

From experiment 2, we obtain

⇒ \(\mathrm{k}=\frac{7.2 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}}{\left(0.3 \mathrm{~mol} \mathrm{~L}^{-1}\right)\left(0.2 \mathrm{~mol} \mathrm{~L}^{-1}\right)^2}=6.0 \mathrm{~L}^2 \mathrm{~mol}^{-2} \mathrm{~min}^{-1}\)

From exp. 3, we obtain,

⇒ \(\mathrm{k}=\frac{2.88 \times 10^{-1} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}}{\left(0.3 \mathrm{~mol} \mathrm{~L}^{-1}\right)\left(0.4 \mathrm{~mol} \mathrm{~L}^{-1}\right)^2}=6.0 \mathrm{~L}^2 \mathrm{~mol}^{-2} \mathrm{~min}^{-1}\)

From exp. 4, we obtain,

⇒ \(\mathrm{k}=\frac{2.40 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}}{\left(0.4 \mathrm{~mol} \mathrm{~L}^{-1}\right)\left(0.1 \mathrm{~mol} \mathrm{~L}^{-1}\right)^2}=6.0 \mathrm{~L}^2 \mathrm{~mol}^{-2} \mathrm{~min}^{-1}\)

Therefore, the rate is constant.

k = 6.0 L² mol^-2 min^-1.

Question 21. The reaction between A and B is first order concerning A and zero order concerning B. Fill in the blanks in the following table:

Answer:

The given reaction is of the first order concerning A and zero order concerning B.

Therefore, the rate of the reaction is given by.

Rate = k [A]¹[B]⁰

Rate = k [A]

From experiment 1. we obtain

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

k = 0.2 min^-1

From experiment 2. we obtain

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

[A] = [0.2] mol L^-1

From experiment 3, we obtain

Rate = 0.2 min^-1 x 0.4 mol L^-1 = 0.08 mol L^-1 min^-1

From experiment 4, we obtain

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

[A] = 0. 1 mol L^-1

Question 22. Calculate the half-life of a first-order reaction from their rate constants given below:

1. 200 s^-1
2. 2 min^-1
3. 4 year^-1

⇒ \(t_{1 / 2}=\frac{0.693}{k}=\frac{0.693}{200 \mathrm{~min}^{-1}}=\mathbf{3 . 4} \times 10^{-3} \sec \text { (approx.) }\)

⇒ \(t_{1 / 2}=\frac{0.693}{k}=\frac{0.693}{2 \mathrm{~min}^{-1}}=\mathbf{0 . 3 5} \mathbf{~ m i n} \text { (approx.) }\)

⇒ \(t_{1 / 2}=\frac{0.693}{k}=\frac{0.693}{4 \text { year }^{-1}}=1.733 \times 10^{-1} \text { years (approx.) }\)

Question 23. 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, \(k=\frac{0.693}{t_2}=\frac{0.693}{5730} \text { years }^{-1}\)

⇒ \(t=\frac{2.303}{k} \log \frac{a}{(a-x)}\)

⇒ \(\mathrm{t}_{1 / 2} \text { of }{ }^{14} \mathrm{C}=5730 \mathrm{yr} \text {; }\)

Also, a = 1 00, (a – x) = 80

⇒ \(\mathrm{t}=\frac{2.303 \times 5730}{0.693} \log \frac{100}{80}=1845 \mathrm{yr} \)

Question 24. The experimental data for the decomposition of N₂O₃

⇒ \(\left[2 \mathrm{~N}_2 \mathrm{O}_5 \longrightarrow 4 \mathrm{NO}_2+\mathrm{O}_2\right]\) in the gas phase at 3 18 K are given below:

1. Plot [N₂O₅] against t.
2. Find the half-life period for the reaction.
3. Draw a graph between log [N₂O₅] and t.
4. What is the rate law?
5. Calculate the rate constant.
6. Calculate the half-life period from k and compare it with (2).

Answer:

Time corresponding to the concentration, \(\frac{1.630 \times 10^2}{2} \mathrm{molL}^{-1}=81.5 \mathrm{~mol} \mathrm{~L}^{-1}\) is the half life. From the graph, the half-life is obtained as 1450 s.

The given reaction 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]

From the plot, log [N₂O₅] v/st, we obtain

⇒ \(\text { Slope }=\frac{-2.46-(-1.79)}{3200-0}=\frac{-0.67}{3200}\)

Again .slope of the line of the plot log [N,05] v/s t is given by \(-\frac{k}{2.303}\)

Therefore, we obtain,

⇒ \(-\frac{k}{2.303}=-\frac{0.67}{3200}\)

k = 4.82 × 10^-4,s^-1

Half-life is given by,

⇒ \(t_{1 / 2}=\frac{0.639}{k}=\frac{0.693}{4.82 \times 10^{-4}} \mathrm{~s}\)

⇒ 1.438 x 10%- 1438s

This value, 1438 s. is very close to the value that was obtained from the graph.

Question 25. 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 1/16^th value?
Answer:

It is known that,

⇒ \(t=\frac{2.303}{k} \log \frac{a}{(a-x)}\)

⇒ \(\text { If } a=1 \text { then }(a-x)=\frac{1}{16}\)

⇒ \(\mathrm{t}=\frac{2.303}{60 \mathrm{~s}^{-1}} \log \frac{1}{1 / 16}=\mathbf{0 . 0 4 6 2} \mathrm{s}\)

Question 26. During a nuclear explosion, one of the products is ⁹⁰Sr with a half-life of 28.1 years. If I pg of 90Sr was absorbed in the bones of a newborn baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically?
Answer:

Here, \(k=\frac{0.693}{t_{1 / 2}}=\frac{0.693}{28.1} y^{-1}\)

It is known that,

⇒ \(t=\frac{2.303}{k} \log \frac{[R]_0}{[R]} \quad \Rightarrow 10=\frac{2.303}{\frac{0.693}{28.1}} \log \frac{1}{[R]} \Rightarrow 10=\frac{2.303}{\frac{0.693}{28.1}}(-\log [R])\)

⇒ \(\log [R]=-\frac{10 \times 0.693}{2.303 \times 28.1} \Rightarrow[R]=\text { anti } \log (-0.1071)\)

= anti log(T.8929) = 0.74814 μg

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

Again,

⇒ \(t=\frac{2.303}{k} \log \frac{[R]_0}{[R]} \Rightarrow 60=\frac{2.303}{\frac{0.693}{28.1}} \log \frac{1}{[R]} \Rightarrow \log [R]=-\frac{60 \times 0.693}{2.303 \times 28.1}\)

[R] = antilog (-0.6425) = 0.2278 μg

Therefore, 0.2278 μg of ⁹⁰Sr will remain after 60 years.

Question 27. For a first-order reaction, show that the time required for 99% completion is twice the time required for the completion of 90% of the reaction.
Answer:

For a first-order reaction, the time required for 99% completion is

⇒ \(t_1=\frac{2.303}{k} \log \frac{100}{100-99}=\frac{2.303}{k} \log 100=2 \times \frac{2.303}{k}\)

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

⇒ \(t_2=\frac{2.303}{k} \log \frac{100}{100-90}=\frac{2.303}{k} \log 10=\frac{2.303}{k}\)

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.

Question 28. A first-order reaction takes 40 min for 30% decomposition. Calculate t_1/2.
Answer:

For a first-order reaction,

⇒ \(t=\frac{2.303}{k} \log \frac{[R]_0}{[R]}\)

⇒ \(\mathrm{k}=\frac{2.303}{40 \mathrm{~min}} \log \frac{100}{100-30}=\frac{2.303}{40 \mathrm{~min}} \log \frac{10}{7}=8.918 \times 10^{-3} \mathrm{~min}^{-1}\)

Therefore. t_1/2 of the decomposition reaction is

⇒ \(t_{1 / 2}=\frac{0.693}{k}=\frac{0.693}{8.918 \times 10^{-3}} \min =77.7 \mathrm{~min} \text { (approx.) }\)

Question 29. For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained.

Calculate the rate constant.

Answer:

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

(CH₃)₂ CHN = NCH (CH₃)₂(g → N₂(g) + C₆H₁₄(g)

After time, t, total pressure, P_t = (P₀– p) + p + p

P_t = P₀+p

⇒  p = P_t-P₀

Therefore, P₀ – p = P₀ (P_t – P0) = 2P₀ – P₁.

For a first-order reaction,

⇒ \(k=\frac{2.303}{t} \log \frac{P_0}{P_0-x}=\frac{2.303}{t} \log \frac{P_0}{2 P_0-P_t}\)

When t = 360 s,

⇒ \(\mathrm{k}=\frac{2.303}{360 \mathrm{~s}} \log \frac{35.0}{(2 \times 35.0-54.0)}=2.175 \times 10^{-3} \mathrm{~s}^{-1}\)

When t = 720 s,

⇒ \(\mathrm{k}=\frac{2.303}{720 \mathrm{~s}} \log \frac{35.0}{2 \times 35.0-63.0}=2.235 \times 10^{-3} \mathrm{~s}^{-1}\)

Hence, the average value of the rate constant is

⇒ \(\mathrm{k}=\frac{\left(2.175 \times 10^{-3}\right)+\left(2.235 \times 10^{-3}\right)}{2} \mathrm{~s}^{-1}=2.205 \times 10^{-3} \mathrm{~s}^{-1}\)

Question 30. The following data were obtained during the first-order thermal decomposition of SO₂Cl₂, at a constant volume.

⇒ \(\mathrm{SO}_2 \mathrm{Cl}_2(\mathrm{~g}) \longrightarrow \mathrm{SO}_2(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g})\)

Calculate the rate of the reaction when the total pressure is 0.65 atm.
Answer:

The thermal decomposition of SO₂Cl₂, at a constant volume, is represented by the following equation.

⇒\(\mathrm{SO}_2 \mathrm{Cl}_2(\mathrm{~g}) \longrightarrow \mathrm{SO}_2(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g})\)

After time, t, total pressure, P, = (P0- p) + p + p

⇒ P₁ = P₀+P

⇒ P = P₁-P₀

Therefore, P₀– p = P₀– (P_t – P₂) = 2P₀– P₁,

For a first-order reaction,

⇒ \(\mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{\mathrm{P}_0}{\mathrm{P}_0-\mathrm{x}}=\frac{2.303}{\mathrm{t}} \log \frac{\mathrm{P}_0}{2 \mathrm{P}_0-\mathrm{P}_{\mathrm{t}}}\)

When t = 100 s, \(\mathrm{k}=\frac{2.303}{100 \mathrm{~s}} \log \frac{0.5}{2 \times 0.5-0.6}=2.231 \times 10^{-3} \mathrm{~s}^{-1}\)

When P_t = 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, the Pressure of SOCl₂

P_SO2Cl2= 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_SO2Cl2) = (2.23 × 10^-3 s^-1 ) (0.35 atm) = 7.8 × 10^-4 atm s^-1

Question 31. The rate constant for the decomposition of N₂O₅ at various temperatures is given below:

Draw a graph between In k and 1/T and calculate the values of A and E_a. Predict the rate constant at 30° and 50°C.
Answer:

For the given data, we obtain

The slope of the line.

⇒ \(\frac{y_2-y_1}{x_2-x_1}=12.301 K\)

According to Arrhenius’s equations. Slope = \(-\frac{E_a}{R}\)

⇒ \(\mathrm{E}_{\mathrm{a}}=- \text { Slope } \times \mathrm{R}=-(-12.301 \mathrm{~K}) \times\left(8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)=102.27 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Again, In k = In \(A-\frac{E_a}{R T}\)

In A = In \(k+\frac{E_a}{R T}\)

When T = 273 K,

In k =-7,147

Then, In A = \(-7.147+\frac{102.27 \times 10^3}{8.314 \times 273}=37.911\)

Therefore, A = 2.91 x 1016

When T = 30 + 273 K = 303 K

⇒ \(\frac{1}{\mathrm{~T}}=0.0033 \mathrm{~K}=3.3 \times 10^{-3} \mathrm{~K}\)

Then, at \(\frac{1}{\mathrm{~T}}=3.3 \times 10^{-3} \mathrm{~K} \text {,}\)

In k = -2.8

Therefore, k = 6.08 x 10~2 s-1

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

⇒ \(\frac{1}{\mathrm{~T}}=0.0031 \mathrm{~K}=3.1 \times 10^{-3} \mathrm{~K}\)

Then, at \(\frac{1}{\mathrm{~T}}=3.1 \times 10^{-3} \mathrm{~K} \text {, }\)

In k = -0.968

Therefore, k = 0.38s-1

Question 32. The rate constant for the decomposition of hydrocarbons is 2.418 x 10^-5 s^-1 at 546 K. If the energy of activation is 179.9 KJ/mol, what will be the value of the pre-exponential factor?
Answer:

According to the Arrhenius equation,

⇒ \(\mathrm{k}=\mathrm{Ae}^{-\mathrm{E}_{\mathrm{a}} / \mathrm{RT}} \Rightarrow{In} \mathrm{k}={In} \mathrm{A}-\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}}\)

⇒ \(\log \mathrm{k}=\log \mathrm{A}-\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R} T}\)

⇒ \(\log \mathrm{A}=\log \mathrm{k}+\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}}\)

⇒ \(=\log \left(2.418 \times 10^{-5} \mathrm{~s}^{-1}\right)+\frac{179.9 \times 10^3 \mathrm{~J} \mathrm{~mol}^{-1}}{2.303 \times 8.314 \mathrm{Jk}^{-1} \mathrm{~mol}^{-1} \times 546 \mathrm{~K}}\)

= (0.3855 – 5) + 1 7.2082 = 12,59 17

Therefore, A = antilog ( 12.5917) = 3.9 x 10¹² s^-1

Question 33. Consider a certain reaction A → Products with k = 2.0 x 10^-2 s ^-1. Calculate the concentration of A remaining after 100s if the initial concentration of A is 1.0 mol L^-1.
Answer:

Since the unit of k is s^-1. the given reaction is a first-order reaction.

Therefore, \(k=\frac{2.303}{t} \log \frac{[\mathrm{A}]_0}{[\mathrm{~A}]}\)

⇒ \(2.0 \times 10^{-2} \mathrm{~s}^{-1}=\frac{2.303}{100 \mathrm{~s}} \log \frac{1.0}{[\mathrm{~A}]}\)

⇒ \(2.0 \times 10^{-2} \mathrm{~s}^{-1}=\frac{2.303}{100 \mathrm{~s}}(-\log [\mathrm{A}])\)

⇒ \(\log [A]=\frac{-2.0 \times 10^{-2} \times 100}{2.303}\)

⇒ \([A]={anti} \log \left(\frac{-2.0 \times 10^{-2} \times 100}{2.303}\right)=0.135 \mathrm{~mol} \mathrm{~L}^{-1}\)

Hence, the remaining concentration of A is 0,135 mol L^-1.

Question 34. 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 the sample of sucrose remains after 8 hours?
Answer:

For a first order reaction, k = \(\frac{2.303}{t} \log \frac{[R]_0}{[R]}\)

It is given that, t_1/2 = 3.00 hours

Therefore, k = \(\frac{0.693}{t_{1 / 2}}=\frac{0.693}{3} h^{-1}=0.231 h^{-1}\)

Then, \(0.231 \mathrm{~h}^{-1}=\frac{2.303}{8 \mathrm{~h}} \log \frac{[\mathrm{R}]_0}{[\mathrm{R}]}\)

⇒ \(\log \frac{[\mathrm{R}]_0}{[\mathrm{R}]}=\frac{0.23 \mathrm{Ih}^{-1} \times 8 \mathrm{~h}}{2.303}\)

⇒ \(\frac{[R]_0}{|R|}={anti} \log (0.8024)\)

⇒ \(\frac{[\mathrm{R}]_0}{[\mathrm{R}]}=6.3445\)

⇒ \(\frac{[\mathrm{R}]}{[\mathrm{R}]_0}=0.1576 \approx 0.158\)

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

Question 35. The decomposition of hydrocarbon follows the equation k = (4.5 ¹ 10¹¹ s^-1) e^{-28000 K/T.} Calculate E,
Answer:

The given equation is k = (4.5 × 10¹¹ s^-1) e^{-28000 K/T} ______ 1

Arrhenius equation is given by, k – Ae^-Ea/RT ______ 2

From equation (1) and (2). we obtain

⇒ \(\frac{E_a}{R T}=\frac{28000}{T}\)

E₂ = R × 28000 K = 8.3141 K^-1 mol^-1 × 28000 K = 2327921 J mol^-1 = 232.792 kJ mol^-1

Question 36. The following equation gives the rate constant for the first-order decomposition of H2O2: log k = 14.34- 1.25 × 10⁴ K/T. Calculate E for this reaction and at what temperature will its half-period be 256 minutes?
Answer:

Arrhenius equation is given by.

⇒ \(\mathrm{k}=\mathrm{Ae}^{-\mathrm{E}_a / \mathrm{RT}}\)

⇒ \(\text { In } k={In} A-\frac{E_a}{R T}\)

⇒ \(\log \mathrm{k}=\log \mathrm{A}-\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}}\) ______ 1

The given equation is, log k = 14.34 – 1.23 x 10⁴ K/T _______ 2

From equation (1) and (2), we obtain

⇒ \(\frac{\mathrm{E}_{\mathrm{i}}}{2.303 \mathrm{RT}}=\frac{1.25 \times 10^4 \mathrm{~K}}{\mathrm{~T}}\)

E_a = 1 .25 x 1 0⁴ K × 2.303 xR = 1 .25 × 10⁴ K × 2.303 × 8.314 K^-1 mol^-1

= 239339.3 J mol^-1 = 239.34 kJ mol^-1

Also, when t_1/2= 256 minutes,

⇒ \(\mathrm{k}=\frac{0.693}{\mathrm{t}_{1 / 2}}=\frac{0.693}{256}=2.707 \times 10^{-3} \mathrm{~min}^{-1}=4.51 \times 10^{-5} \mathrm{~s}^{-1}\)

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

⇒ \(\log \left(4.51 \times 10^{-5}\right)=14.34-\frac{1.25 \times 10^4}{T}\)

⇒ \(0.654-5=14.34-\frac{1.25 \times 10^4 \mathrm{~K}}{\mathrm{~T}}\)

⇒ \(\frac{1.25 \times 10^4 \mathrm{~K}}{\mathrm{~T}}=18.686\)

⇒ \(\mathrm{T}=\frac{1.25 \times 10^4 \mathrm{~K}}{18.686}=668.95 \mathrm{~K}=669 \mathrm{~K}\)

Question 37. The decomposition of A into product has a value of k as 4.5 x 103s_1 at 10°C and energy of activation 60 kJ mol^-1. At what temperature would k be 1.5 x 10⁴ s^-1?
Answer:

From Arrhenius equation, we obtain,

⇒ \(\log \frac{k_2}{k_1}=\frac{E_a}{2.303 R}\left(\frac{T_2-T_1}{T_1 T_2}\right)\)

⇒ \(\log \frac{1.5 \times 10^4}{4.5 \times 10^3}=\frac{6.0 \times 10^4 \mathrm{~J} \mathrm{~mol}^{-1}}{2.303 \times 8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}}\left(\frac{\mathrm{T}_2-283}{283 \mathrm{~T}_2}\right)\)

⇒ \(0.5229=3133.627\left(\frac{\mathrm{T}_2-283}{283 \mathrm{~T}_2}\right)\)

⇒ \(\frac{0.5229 \times 283 \mathrm{~T}_2}{3133.627}=\mathrm{T}_2-283\)

⇒ 0.9528 T₂ = 283

⇒ T₂ = 297.019 K = 297 K = 24° C

Hence, k would be 1.5 x 10⁴ s^-1 at 24° C

Question 38. The time required for 10% completion of a first-order reaction at 298K is equal to that required for its 25% completion at 308 K. If the value of A is 4 x 10¹⁰ s^-1. Calculate k at 318 K and F.
Answer:

For a first-order reaction, \(t=\frac{2.303}{k} \log \frac{a}{a-x}\)

⇒ \(\text { At } 298 \mathrm{~K}, \mathrm{t}_1=\frac{2.303}{k_1} \log \frac{100}{90}=\frac{0.1054}{k_1}\)

⇒ \(\text { At } 308 \mathrm{~K}, \mathrm{t}_2=\frac{2.303}{\mathrm{k}_2} \log \frac{100}{75}=\frac{0.2877}{\mathrm{k}_2}\)

According to the question, t = t’

⇒ \(\frac{0.1054}{k_1}=\frac{0.2877}{k_2} \Rightarrow \frac{k_2}{k_1}=2.73\)

From the Arrhenius equation, we obtain

⇒ \(\log \frac{k_2}{k_1}=\frac{E_1}{2.303 R}\left(\frac{T_2-T_1}{T_1 T_2}\right)\)

⇒ \(\log (2.73)=\frac{E_a}{2.303 \times 8.314}\left(\frac{308-298}{298 \times 308}\right)\)

⇒ \(\mathrm{E}_{\mathrm{a}}=\frac{2.303 \times 8.314 \times 298 \times 308 \times \log (2.73)}{308-298}\)

⇒ 76640.26 J mol^-1 = 76.64 Id mol^-1

To calculate k at 318 K,

It is given that, A = 4 x 10^-1 s^-1, T = 318 K

Again, from the Arrhenius equation, we obtain

⇒ \(\log \mathrm{k}=\log \mathrm{A}-\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R} T}=\log \left(4 \times 10^{10}\right)-\frac{76.6 \times 10^3}{2.303 \times 8.3\lceil 4 \times 318}\)

= (0.6021 + 10)- 12.5876 =- 1.9855

Therefore, k = Antilog (- 1.9855) = 1.05 x 10^-1 s^-1

Question 39. 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,

⇒ \(\log \frac{k_2}{k_1}=\frac{E_a}{2.303 R}\left(\frac{T_2-T_1}{T_1 T_2}\right)\)

It is given that, k₂ = 4k₁

T₁ = 293 K: T₂ = 313 K

Therefore, \(\log \frac{4 k_1}{k_1}=\frac{E_a}{2.303 \times 8.314}\left(\frac{313-293}{293 \times 313}\right)\)

⇒ \(0.6021=\frac{20 \times E_a}{2.303 \times 8.314 \times 293 \times 313}\)

⇒ \(\mathrm{E}_{\mathrm{a}}=\frac{0.6021 \times 2.303 \times 8.314 \times 293 \times 313}{20}\)

⇒ 52863.33 J mol^-1 = 52.86 kJ mol^-1

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