Class 12 Chemistry Chapter 3 Chemical Kinetics ncert solutions

Chemical Kinetics Class 12 ncert solutions: Chemistry Class 12 Chapter 3 exercise solutions

Textbook	NCERT
Class	Class 12
Subject	Chemistry
Chapter	Chapter 3
Chapter Name	Chemical Kinetics ncert solutions
Category	Ncert Solutions
Medium	English

Are you looking for Class 12 Chemistry Chapter 3 Chemical Kinetics ncert solutions? Now you can download Ncert Chemistry Class 12 Chapter 3 exercise solutions pdf from here.

Question 3.1: From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.

• (a) 3NO(g) → N₂O (g) Rate = k[NO]²
• (b) H₂O₂ (aq) + 3I⁻ (aq) + 2H⁺ → 2H₂O (I) + I⁻₃ Rate = k[H₂O₂][I⁻]
• (c) CH₃CHO (g) → CH₄ (g) + CO(g) Rate = k[CH₃CHO]^3/2
• (d) C₂H₅Cl (g) → C₂H₄ (g) + HCl (g) Rate = k[C₂H₅Cl]

Solution 3.1: (a) Given rate = \(k\left [ NO \right ]^{2}\)

Therefore, the order of the reaction = 2

Dimensions of

\(k = \frac{Rate}{\left [ NO \right ]^{2}}\)

\(\ = \frac{mol \; L^{-1} s^{-1}}{\left ( mol \; L^{-1} \right )^{2}} \ \)

\(= \frac{mol \; L^{-1} s^{-1}}{mol^{2}\; L^{-2}} \ \)

\(= L \; mol^{-1} s^{-1}\)

(b) Given rate = \(k[ H_{2}O_{2} ][ I ^{-}]\)

Therefore, the order of the reaction = 2

Dimensions of  

\(k = \frac{Rate}{\left [ H_{2}O_{2} \right ]\left [ I^{-} \right ]}\)

\(\ = \frac{mol \; L^{-1} S^{-1}}{\left ( mol \; L^{-1} \right ) \left ( mol \; L^{-1} \right )} \ \)

\(= L \; mol^{-1} s^{-1}\)

(c) Given rate =

\(= k \left [ CH_{3} CHO \right ]^{\frac{3}{2}}\)

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

Dimensions of

\(k = \frac{Rate}{\left [ CH_{3} CHO \right ]^{\frac{3}{2}}} \ \)

\(= \frac{mol \; L^{-1}s^{-1}}{\left (mol \; L^{-1} \right )^{\frac{3}{2}}} \ \)

\(= \frac {mol\; L^{-1} s^{-1}}{mol^{\frac{3}{2}}\; L^{\frac{3}{2}}} \ \)

\(L^{\frac{1}{2}}\; mol^{ -\frac{1}{2}} \; s^{-1}\)

(d) Given rate = \(k = \left [ C_{2}H_{5}Cl \right ]\)

Therefore, the order of the reaction = 1

Dimension of

\(k = \frac{Rate}{\left [ C_{2}H_{5}Cl \right ]} \ \)

\(= \frac{mol\; L^{-1} s^{-1}}{mol \; L^{-1}} \ \)

\(= s^{-1}\)

Question 3.2: 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

Solution 3.2: The initial rate of reaction is

Rate = \(k\left [ A \right ]\left [ B \right ]^{2} \ \)

\(= \left ( 2.0 \times 10^{-6} mol^{-2} L^{2} s^{-1} \right ) \)

\(\left( 0.1 \; mol \; L^{-1} \right )\left ( 0.2 \; mol \; L^{-1} \right )^{2} \ \)

\(= 8.0 \times 10^{-9} mol^{-2}L^{2} s^{-1}\)

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

Therefore, the concentration of B reacted

\( = \frac {1}{2} \times 0.04 \; mol \; L^{-1} = 0.02 \; mol \; L^{-1}\)

Then, the concentration of B available,

\(\left [ B \right ] = \left ( 0.2 – 0.02 \right ) mol \; L^{-1} \)

\(= 0.18\; mol \; L^{-1}\)

After [A] is reduced to \(0.06 \; mol \; L^{-1}\) , the rate of the reaction is given by, Rate =

\(k \left [ A \right ]\left [ B \right ]^{2} \ \)

\( = \left ( 2.0 \times 10^{-6} mol^{-2}L^{2}s^{-1} \right )\)

\(\left ( 0.06\; mol L^{-1} \right )\left ( 0.18 \; mol \; L^{-1} \right )^{2} \ \)

\(= 3.89\; \times 10^{-9} mol \; L^{-1} s^{-1}\)

Question 3.3: The decomposition of NH₃ on the platinum surface is zero order reaction. What are the rates of production of N₂ and H₂ if k = 2.5 × 10^–4 mol^–1 L s ^–1?

Solution 3.3: The decomposition of NH3 on the platinum surface is represented by the following equation:

\(2NH^{3(g)} \overset{Pt}{\rightarrow} N_{2(g)} + 3H_{2(g)}\)

Therefore, Rate =

\(-\frac{1}{2}\frac{d\left [ NH_{3} \right ]}{dt}\)

\(= \frac{d\left [ N_{2} \right ]}{dt} \)

\(= \frac{1}{3}\frac{d\left [ H_{2} \right ]}{dt}\)

However, it is given that the reaction is of zero order.

Therefore,

\(-\frac{1}{2}\frac{d\left [ NH_{3} \right ]}{dt} \)

\(= \frac{d\left [ N_{2} \right ]}{dt} \)

\(= \frac{1}{3}\frac{d\left [ H_{2} \right ]}{dt}= k \ \)

\( \ = 2.5 \times 10^{-4}\; mol L^{-1} s^{-1}\)

Therefore, the rate of production of N₂ is \(\frac{d\left [N_{2} \right ]}{dt} = 2.5 \times 10^{-4} mol\;L^{-1}s^{-1}\) And, the rate of production of H₂ is

\(\frac{d\left [H_{2} \right ]}{dt} = 3 \times 2.5 \times 10^{-4} mol\;L^{-1}s^{-1} \ \)

\( = 7.5 \times 10^{-4} \; mol \;L^{-1}s^{-1}\)

Question 3.4: The decomposition of dimethyl ether leads to the formation of \(CH_{4}, H_{2}, \; and \; CO\) and the reaction rate is given by \(Rate = k\left [ CH_{3} O\, C! H_{3} \right ]^{\frac{3}{2}}\) 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., \(Rate = k\left ( P_{CH_{3} O\, C! H_{3}} \right )^{\frac{3}{2}}\) If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants?

Solution 3.4: If pressure is measured in bar and time in minutes, then

Unit of rate = \(bar \; min^{-1}\)

\(Rate = k\left ( P_{CH_{3} O\, C! H_{3}} \right )^{\frac{3}{2}} \ \)

\( \Rightarrow k = \frac{Rate}{k\left ( P_{CH_{3} O\, C! H_{3}} \right )^{\frac{3}{2}}}\)

Therefore, the unit of rate constants

\((k) = \frac{bar\; min^{-1}}{bar^{\frac{3}{2}}} \ \)

\(= bar^{\frac{-1}{2}}min^{-1}\)

Question 3.5: Mention the factors that affect the rate of a chemical reaction.

Solution 3.5: The factors which are responsible for the effect on the chemical reaction’s rate are

• (a) Reaction temperature
• (b) Presence of a catalyst
• (c) The concentration of reactants (pressure in case of gases)
• (d) Nature of the products and reactants
• (e) Radiation exposure
• (f) Surface area

Question 3.6: 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?

Solution 3.6: Let the concentration of the reactant be [A] = a

Rate of reaction, \(R = k [A]^{2}\ \ = ka^{2}\)

(a) If the concentration of the reactant is doubled, i.e., [A] = 2a, then the rate of the reaction would be \(R’ = k\left ( A \right )^{2} \ \ = 4 ka^{2} \ \ = 4 \; R\)

Therefore, the rate of the reaction now will be 4 times the original rate.

(b) If the concentration of the reactant is reduced to half, i.e., \(\left [ A \right ] = \frac{1}{2}a\) , then the rate of the reaction would be

\(R” = k\left ( \frac{1}{2} a \right )^{2} \ \)

\(= \frac{1}{4} k a \ \ = \frac{1}{4} R\)

Therefore, the rate of the reaction will be reduced to \(\frac{1}{4} ^{th}\)

Question 3.7: 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?

Solution 3.7: When a temperature of 10° rises for a chemical reaction, then the rate constant increases and becomes near to double its original value.

The temperature effect on the rate constant can be represented quantitatively by the Arrhenius equation.

\(k = Ae^{ -E_{a} / RT}\)

Where,

• k = rate constant
• A = Frequency factor / Arrhenius factor
• R = gas constant
• T = temperature
• E_a = activation energy for the reaction.

Question 3.8: 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

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

Solution 3.8: The avg rate of reaction between the time intervals, 30 to 60 seconds,

\(= \frac{d\left [ Ester \right ]}{dt} \ \)

\( = \frac{0.31 – 0.17}{60 – 30} \ \)

\(= \frac{0.14}{30} \ \)

\(= 4.67 \times 10^{-3}\; mol \; l^{-1}\; s^{-1}\)

Question 3.9: A reaction is first order in A and second order in B. (i) Write the differential rate equation. (ii) How is the rate affected by increasing the concentration of B three times? (iii) How is the rate affected when the concentrations of both A and B are doubled?

Solution 3.9: (a) The differential rate equation will be

\(-\frac{d\left [ R \right ]}{dt} = k\left [ A \right ]\left [ B \right ]^{2}\)

(b) If the concentration of B is increased three times, then

\(-\frac{d\left [ R \right ]}{dt} = k\left [ A \right ]\left [ 3B \right ]^{2} \ \)

\(= 9. k\left [ A \right ]\left [ B \right ]^{2}\)

Therefore, the reaction rate will be increased by 9 times.

(c) When the concentrations of both A and B are doubled,

\(-\frac{d\left [ R \right ]}{dt} = k\left [ 2A \right ]\left [ 2B \right ]^{2} \ \)

\(= 8. k\left [ A \right ]\left [ B \right ]^{2}\)

Therefore, the rate of reaction will increase 8 times.

Question 3.10: 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?

Solution 3.10: Let the order of the reaction with respect to A be x and with respect to B be y.

Then,

\(r_{0} = k\left [ A \right ]^{x} \left [ B \right ]^{y} \ \)

\(5.07 \times 10^{-5} = k\left [ 0.20 \right ]^{x}\left [ 0.30 \right ]^{y} \;\;\;\; (i) \ \)

\( 5.07 \times 10^{-5} = k\left [ 0.20 \right ]^{x}\left [ 0.10 \right ]^{y} \;\;\;\; (ii) \ \)

\( 1.43 \times 10^{-4} = k\left [ 0.40 \right ]^{x}\left [ 0.05 \right ]^{y} \;\;\;\; (iii)\)

Dividing equation (i) by (ii), we get

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

\( \Rightarrow 1 = \frac{\left [ 0.30 \right ]^{y}}{\left [ 0.10 \right ]^{y}} \ \)

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

\(\Rightarrow y = 0\)

Dividing equation (iii) by (ii), we get

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

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

\( \begin{bmatrix} Since\; y = 0,\ \left [ 0.05 \right ]^{y} = \left [ 0.30 \right ]^{y} = 1 \end{bmatrix} \ \)

\( \Rightarrow 2.821 = 2^{x} \ \)

\(\Rightarrow \log 2.821 = x \log 2 \) (taking log on both sides)

\( \Rightarrow x = \frac{\log 2.821}{\log 2} \ \)

= 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 3.11: 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.

Solution 3.11: Let the order of the reaction with respect to A be x and with respect to B be y.

Therefore, the rate of the reaction is given by,

Rate = \begin{array}{l}k\left [ A \right ]^{x} \left [ B \right ]^{y}\end{array}

According to the question,

\(6.0 \times 10^{-3} = k\left [ 0.1 \right ]^{x} \left [ 0.1 \right ]^{y}\) ….1

\(7.2 \times 10^{-2} = k\left [ 0.3 \right ]^{x} \left [ 0.2 \right ]^{y}\) …..2

\(2.88 \times 10^{-1} = k\left [ 0.3 \right ]^{x} \left [ 0.4 \right ]^{y}\) ….3

\(2.4 \times 10^{-2} = k\left [ 0.4 \right ]^{x} \left [ 0.1 \right ]^{y}\) ….4

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

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

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

\(4 = \left (\frac{0.4}{0.1 } \right )^{x}\)

\(\left (4 \right )^{1} = \left (4 \right )^{x}\)

x = 1

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

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

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

4 = 2^y

2² = 2^y

y = 2

Hence, the rate law is

Rate = \(k \left [ A \right ] \left [ B \right ]^{2}\)

\(k = \frac{Rate}{\left [ A \right ] \left [ B \right ]^{2}}\)

From experiment 1, we get

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

= 6.0 \(L^{2} \;mol^{-2} \;min^{-1}\)

From experiment 2, we get

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

= 6.0 \(L^{2} \;mol^{-2} \;min^{-1}\)

From experiment 3, we get

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

= 6.0 \(L^{2} \;mol^{-2} \;min^{-1}\)

From experiment 4, we get

\(k = \frac{2.4 \times 10^{-2} mol\; L^{-1} \;min^{-1}}{\left (0.4 \;mol \;L^{-1} \right ) \left (0.1 \;mol \;L^{-1} \right )^{2}}\)

= 6.0 \(L^{2} \;mol^{-2} \;min^{-1}\)

Thus, rate constant, k = 6.0 \(L^{2} \;mol^{-2} \;min^{-1}\)

Question 3.12: 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

Solution 3.12: The given reaction is of the first order with respect to A and of zero order with respect to B.

Thus, the rate of the reaction is given by,

Rate = \(k \left [ A \right ]^{1} \left [ B \right ]^{0}\)

Rate = \(k \left [ A \right ]\)

From experiment 1, we get

\(2.0 \times 10^{-2}\, mol\; L^{-1} min^{-1} \)

\(= k\left ( 0.1 \; mol \; L^{-1} \right ) \ \)

\(\Rightarrow k = 0.2\; min^{-1}\)

From experiment 2, we get

\(4.0 \times 10^{-2}\, mol\; L^{-1} min^{-1} \)

\(= 0.2 min^{-1} \left [ A \right ] \ \)

\( \Rightarrow \left [ A \right ] = 0.2\; mol\; L^{-1}\)

From experiment 3, we get

Rate =

\(0.2 \; min^{-1} \times 0.4 \; mol \; L^{-1}\ \)

\(= 0.08\; mol\; L^{-1}min^{-1}\)

From experiment 4, we get

\(2.0 \times 10^{-2}\; mol \; L^{-1} min^{-1} \)

\(= 0.2\; min^{-1} \left [ A \right ] \ \)

\(\Rightarrow \left [ A \right ] = 0.1\; mol \; L^{-1}\)

Question 3.13: Calculate the half-life of a first-order reaction from their rate constants given below. (a) 200 s⁻¹ (b) 2 min⁻¹ (c) 4 years⁻¹

Solution 3.13: (a) Half life,

\(t_{ \frac{1}{2}} = \frac{0.693}{k} \ \)

\( = \frac {0.693}{200\; s^{-1}} \ \)

\( = 3.47\;\times 10^{-3} s\) (Approximately)

(b) \(t_{ \frac{1}{2}} = \frac{0.693}{k} \ \)

\( = \frac {0.693}{2\; min^{-1}} \ \)

\(= 0.35\; min\) (Approximately)

(c) \(t_{ \frac{1}{2}} = \frac{0.693}{k} \ \)

\(= \frac {0.693}{4\; years^{-1}} \ \)

\(= 0.173\; years\) (Approximately)

Question 3.14: The half-life for the radioactive decay of ¹⁴C is 5730 years. An archaeological artefact containing wood had only 80% of the ¹⁴C found in a living tree. Estimate the age of the sample.

Solution 3.14: Here,

\(k = \frac{0.693}{t_{\frac{1}{2}}} \ \)

\(= \frac{0.693}{5730}years^{-1}\)

It is known that,

\(t = \frac{2.303}{k} \log \frac{\left [ R \right ]_{0}}{\left [ R \right ]} \ \)

\( = \frac{2.303}{0.693/5730} \log \frac{100}{80} \ \)

\(= 1845\; years\) (approximately)

Hence, the age of the sample is 1845 years.

Question 3.15: The experimental data for decomposition of N₂O₅ [2N₂O₅ → 4NO₂ + O₂]  in gas phase at 318K are given below:

t(s)	0	400	800	1200	1600	2000	2400	2800	3200
102 × [N2O5]  mol L-1	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 k and compare it with (ii).

Solution 3.15: (a)

(ii) Time corresponding to the concentration, 1630×10² / 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)

t(s)	102 × [N2O5]  mol L-1	Log[N2O5]
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

(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

– k /2.303

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

– k / 2.303. = -0.67 / 3200

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 / 4.82×10^-4 s

=1.438

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

Question 3.16: 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 1/16^th value?

Solution 3.16: It is known that,

\(t = \frac{2.303}{k} \log \frac{\left [ R \right ]_{0}}{\left [ R \right ]} \ \)

\( = \frac{2.303}{60\; s^{-1}} \log \frac{1}{1/16} \ \)

\( = \frac{2.303}{60\; s^{-1}} \log 16 \ \)

\(= 4.6 \times 10^{-2} \left ( approximately \right )\)

Hence, the required time is \(4.6 \times 10^{-2}\; s\).

Question 3.17: During the nuclear explosion, one of the products is ⁹⁰Sr, with a 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?

Solution 3.17:

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

Here,

It is known that

\(t = \frac{2.303}{k} \log \frac{\left [ R \right ]_{0}}{\left [ R \right ]} \ \)

\( \Rightarrow 10 = \frac{2.303}{\frac{0.693}{28.1}} \log \frac{1}{\left [ R \right ]} \ \)

\( \Rightarrow 10 = \frac{2.303}{\frac{0.693}{28.1}}\left ( – \log \left [ R \right ] \right ) \ \)

\( \Rightarrow \log \left [ R \right ] = – \frac{10 \times 0.693}{2.303 \times 28.1}\ \)

\( \Rightarrow \left [ R \right ] = antilog \left ( – 0.1071 \right )\ \)

\( = antiog \left ( 1.8929 \right ) \ \)

\( = 0.7814 \mu g\)

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

Again,

\(t = \frac{2.303}{k} \;log \frac{\left [ R \right ]_{0}}{\left [ R \right ]} \ \)

\( \Rightarrow 60 = \frac{2.303}{\frac{0.693}{28.1}} \log \frac{1}{\left [ R \right ]}\ \)

\( \Rightarrow log \left [ R \right ] = – \frac{60 \times 0.693}{2.303 \times 28.1}\ \)

\( \left [ R \right ] = antilog \; \left ( – 0.6425 \right ) \ \)

\( = antilog \left ( 1.3575 \right )\ \)

\( = 0.2278 \mu g\)

Therefore, 0.2278 μg of \( ^{90}\text{Sr} \) will remain after 60 years.

Question 3.18:  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.

Solution 3.18: 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

\(t_{1} = \frac{2.303}{k} \log \frac{100}{100 – 90} \ \)

\( = \frac{2.303}{k} \log 10 \ \)

\( = \frac{2.303}{k}\)

Therefore, t₁ = 2 t₂ 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 3.19: A first-order reaction takes 40 min for 30% decomposition. Calculate t_1/2.

Solution 3.19: For a first-order reaction,

\(t = \frac{2.303}{k} \log \frac{\left [ R \right ]_{0}}{\left [ R \right ]} \ \)

\( k = \frac{2.303}{40\; min} \log \frac{100}{100 – 30} \ \)

\(= \frac{2.303}{40 \; min} \log \frac{10}{7} \ \)

\( = 8.918 \times 10^{-3} \; min^{-1}\)

Therefore, \(t _{\frac{1}{2}}\) of the decomposition reaction is \(t _{\frac{1}{2}} = \frac{0.693}{k} \ \)

\( = \frac{0.693}{8.918 \times 10^{-3}} min \ \)

= 77.7 min ( approximately )

Question 3.20: 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.

Solution 3.20: The decomposition of azoisopropane to hexane and nitrogen at 543 K is represented by the following equation.

After time, t, total pressure, Pt = (P_º – p) + p + p

⇒ Pt = (P_º + p)

⇒ p = P_t  – P₀

therefore, P_º – p = P₀  – Pt  – P₀

= 2P₀ − P_t

For a first order reaction,

k = 2.303/t  Log P₀ /P₀  – p

=   2.303/t  Log P₀ / 2 P₀  –  P_t

When t = 360 s, k = 2.303 / 360s log 35.0 / 2×35.0 – 54.0

= 2.175 × 10⁻³ s⁻¹

When t = 720 s, k = 2.303 / 720s log 35.0 / 2×35.0 – 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

Question 3.21: The following data were obtained during the first order thermal decomposition of SO₂Cl₂ at a constant volume.

SO₂Cl₂(g)  →  SO₂(g) + Cl₂(g)

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.

Solution 3.21: The thermal decomposition of SO₂Cl₂ at 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 = Pt  – Pº

therefore, Pº – p = Pº  – Pt  – Pº

= 2 P_º –  P_t

For a first order reaction,

\(k = \frac{2.303}{t} \log \frac{P_0}{P_0 – p} \ \)

\( = \frac{2.303}{t} \log \frac{P_0}{2P_0 – P_t}\)

When t= 100 s,

\(k = \frac{2.303}{100 \; s} \log \frac{0.5}{2 \times 0.5 – 0.6} \ \)

\( = 2.231 \times 10^{-3} \; s^{-1}\)

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}

Question 3.22: The rate constant for the decomposition of N₂O₅ at various temperatures is given below:

T/°C	0	20	40	60	80
105 X K /S-1	0.0787	1.70	25.7	178	2140

Draw a graph between ln k and 1/T and calculate the values of A and E_a. Predict the rate constant at 30º and 50ºC.

Solution 3.22: From the given data, we obtain

T/°C	0	20	40	60	80
T/K	273	293	313	333	353
1/T / k-1	3.66×10−3	3.41×10−3	3.19×10−3	3.0×10−3	2.83 ×10−3
105 X K /S-1	0.0787	1.70	25.7	178	2140
ln k	−7.147	− 4.075	−1.359	−0.577	3.063

Slope of the line,

In k= – 2.8

Therefore, k = 6.08×10-2s-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

Question 3.23: 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 the pre-exponential factor?

Solution 3.23:

\( k = 2.418 \times 10^{-5} s^{-1}\)

T = 546 K

\(E_a = 179.9 kJ\; mol^{-1} \)

\(= 179.9 \times 10^3\; J\; mol^{-1}\)

According to the Arrhenius equation,

\(k =Ae^{-E_a /RT}\ \)

\( \Rightarrow I!n \;k = I!n A – \frac{E_a}{RT} \ \)

\( \Rightarrow \log k = \log A – \frac{E_a}{2.303 \; RT} \ \)

\( \Rightarrow \log A = \log k + \frac{E_a}{2.303 \; RT} \ \)

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

\( = \left ( 0.3835 – 5 \right ) + 17.2082 \ \ = 12.5917\)

Therefore, A = antilog (12.5917)

\(= 3.9 \times 10^{12 } \; s^{-1}\) (approximately)

Question 3.24: Consider a certain reaction A → Products with k = 2.0 × 10^–2 s^–1. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L^–1.

Solution 3.24:

\(k = 2.0 \times 10^{-2} \; s^{-1} \; T = 100 \; s\)

\(\left [ A \right ]_0 = 1.0 \; mol^{-1}\)

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

Therefore,

\(k = \frac{2.303}{t} \log \frac{\left [ A \right ]_0}{\left [ A \right ]} \ \)

\( \Rightarrow 2.0 \times 10^{-2}\; s^{-1} = \frac{2.303}{100 \; s} \log \frac{1.0}{\left [ A \right ]} \ \)

\( \Rightarrow 2.0 \times 10^{-2} \; s^{-1} = \frac{2.303}{100 \; s}\left ( – \log \left [ A \right ] \right ) \ \)

\( \Rightarrow – \log \left [ A \right ] = \frac{2.0 \times 10 ^{-2} \times 100}{2.303} \ \)

\( \Rightarrow \left [ A \right ] = antilog \left ( – \frac{2.0 \times 10 ^{-2} \times 100}{2.303} \right ) \ \)

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

Hence, the remaining concentration of A is \(0.135 \; mol \; L^{-1}\)

Question 3.25:  Sucrose decomposes in an acid solution into glucose and fructose according to the first-order rate law, with t_1/2 = 3.00 hours. What fraction of a sample of sucrose remains after 8 hours?

Solution 3.25: For the first-order reaction,

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

It is given that

\(t_{\frac{1}{2}} = 3.00 \; hours\)

Therefore,

\(k = \frac{0.693}{t_{\frac{1}{2}}}\)

\(\frac{0.693}{3} h^{-1} \ \ = 0.231 \; h^{-1}\)

Then,

\(0.231 \; h^{-1} = \frac{2.303}{8h} \log \frac{\left [ R \right ]_0}{\left [ R \right ]} \ \)

\( \Rightarrow \log \frac{\left [ R \right ]_0}{\left [ R \right ]} = \frac{0.231\; h^{-1} \times 8\; h}{2.303} \ \)

\( \Rightarrow \frac{\left [ R \right ]_0}{\left [ R \right ]} = antilog \left ( 0.8024 \right ) \ \)

\( \Rightarrow \frac{\left [ R \right ]_0}{\left [ R \right ]} = 6.3445 \ \)

\( \Rightarrow \frac{\left [ R \right ]}{\left [ R \right ]_0} \)

\(= 0.1576\) (approx)

= 0.158

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

Question 3.26: The decomposition of hydrocarbon follows the equation k = (4.5 × 10¹¹ s⁻¹) e^−28000K/T Calculate E_a.

Solution 3.26: The given equation is

\(k = \left ( 4.5 \times 10_{11} S – 1 \right ) e_{-28000} K/T\) ….1

The Arrhenius equation is given by

\(k = Ae^{-E_a /RT}\) ….2

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

\(\frac{E_a }{RT} = \frac{28000\; K}{T} \ \)

\( \Rightarrow E_a = R \times 28000 \; K \ \)

\( = 8.314\; J \; K^{-1} mol^{-1} \times 28000 \; K \ \)

= 232.79 kJ mol⁻¹

Question 3.27: The rate constant for the first-order decomposition of H₂O₂ is given by the following equation: \(log \; k = 14.34 – 1.25 \times 10^4 \; K/T\) Calculate Ea for this reaction, and at what temperature will its half-period be 256 minutes?

Solution 3.27: Arrhenius equation is given by

\(k = Ae^{-E_a/RT} \ \)

\( \Rightarrow I!n\; k = I!n\; A – \frac{E_a}{RT} \ \)

\( \Rightarrow \log k = \log A – \frac{E_a}{2.303\; RT}\) ….1

The given equation is

\(\log k = 14.34 – 1.25 \times 10^{4}\; K/T\) ….2

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

\(\frac{E_a}{2.303\; RT} = \frac{1.25 \times 10^4\; K}{T}\ \)

\( \Rightarrow E_a = 1.25 \times 10^4\; K \times 2.303 \times R \ \)

= 1.25 × 10⁴ K × 2.303 × 8.314 J K⁻¹ mol⁻¹

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

\(= 239.34\; kJ\; mol^{-1}\)

Also, when \(t_{\frac{1}{2}} = 256\) minutes,

\(k = \frac{0.693}{t_{\frac{1}{2}}} \ \)

\( = \frac{0.693}{256} \ \)

\( = 2.707 \times 10^{-3} \; min^{-1}\ \)

\( = 4.51 \times 10^{-5}\; s^{-1}\)

It is also given that,

\(log\; k = 14.34 – 1.25 \times 10^{4}\; K/T \ \)

\( \Rightarrow log \left ( 4.51 \times 10^{-5} \right ) \)

\(= 14.34 – \frac{1.25 \times 10^4 \; K}{T}\ \)

\( \Rightarrow log \left ( 0.654 – 05 \right ) \)

\(= 14.34 – \frac{1.25 \times 10^4 \; K}{T}\ \)

\( \Rightarrow \frac{1.25 \times 10^4\; K}{T} = 18.686 \ \)

\( = 668.95\; K \ \ = 669\) (approximately)

Question 3.28: The decomposition of A into the product has a 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 × 10⁴ s^–1?

Solution 3.28: From Arrhenius equation, we obtain

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

Also,

\(k_1 = 4.5 \times 10^3\; s^{-1}\)

\(T_1 = 273 + 10 = 283\; K \; k_2\ \)

\(= 1.5 \times 10^4\; s^{-1}\ \)

\( E_a = 60\; kJ\; mol^{-1}\)

\(= 6.0 \times 10^{4}\; J \; mol^{-1}\)

Then,

\(\log \frac{1.5 \times 10^4}{4.5 \times 10^3} \)

\(= \frac{6.0 \times 10^4\; J\; mol^{-1}}{2.303 \times 8.314\; J\; K^{-1}\; mol^{-1}}\left ( \frac{T_2 – 283}{283\; T_2} \right )\ \)

\( \Rightarrow 0.5229 = 3133.627 \left ( \frac{T_2 – 283}{283\; T_2} \right ) \ \)

\( \Rightarrow \frac{0.5229 \times 283\; T_2}{3133.627} = T_2 – 283 \ \)

\( \Rightarrow 0.9528 \;T_2 = 283\ \)

\( \Rightarrow T_2 = 297.019\; K\) (approximately)

\(= 297 \; K \ \)

\(= 24^{\circ}\)

Question 3.29: 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 × 10¹⁰ s ^–1. Calculate k at 318K and E_a.

Solution 3.29: For a first-order reaction,

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

\( at \; 298\; K, \; t = \frac{2.303}{k} \log \frac{100}{90}\ \)

\( = \frac{0.1054}{k}\ \)

\( at\; 308\; K, \; t’ = \frac{2.303}{k’} \log \frac{100}{75} \ \)

\(= \frac{2.2877}{k’}\)

According to the question,

\(t = t’ \ \)

\( \Rightarrow \frac{0.1054}{k} = \frac{0.2877}{k’} \ \)

\( \Rightarrow \frac{k’}{k} = 2.7296\)

From the Arrhenius equation, we get

\(\log \frac{k’}{k} = \frac{E_a}{2.303\, R}\left ( \frac{T’ – T}{TT’} \right ) \ \)

\(\Rightarrow \log \left ( 2.7296 \right ) = \frac{E_a}{2.303 \times 8.314}\left ( \frac{308 – 298}{298 \times 308} \right ) \ \)

\( \Rightarrow E_a = \frac{2.303 \times 8.314 \times 298 \times 308 \times \log \left ( 2.7296 \right )}{308 – 298} \ \)

\( = 76640.096\; J \, mol^{-1} \ \)

\( = 76.64\; kJ\; mol^{-1}\)

To calculate k at 318 K,

It is given that,

\(A = 4 \times 10^{10} s^{-1}\)

, T = 318 K

Again, from Arrhenius equation, we get

\(\log k = \log A – \frac{E_a}{2.303\; R\, T} \ \)

\(= \log \left ( 4 \times 10^{10} \right ) – \frac{76.64 \times 10^{3}}{2.303 \times 8.314 \times 318} \ \)

\( = \left ( 0.6021 + 10 \right ) – 12.5876 \ \)

\( = -1.9855\)

Therefore, k = Antilog(-1.9855)

\(= 1.034 \times 10^{-2} \; s^{-1}\)

Question 3.30: 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.

Solution 3.30: From Arrhenius equation, we get

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

From the question we have,

\(K_2 = 4 K_1\)

\(\ T_1 = 293\; K \ \)

\(T_2 = 313\; K\)

Therefore,

\(\log \frac{4 K_1}{K_2} = \frac{E_a}{2.303 \times 8.314}\left ( \frac{313 – 293}{293 \times 313} \right )\ \)

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

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

\( = 52863.33\; J\; mol^{-1}\ \)

\( = 52.86\; kJ\; mol^{-1}\)

Hence, the required energy of activation is \(52.86\; kJ\; mol^{-1}\)