Question Set A

CBSE 12th Chemistry Chemical Kinetics Important Questions And Answers Study Material - QB365

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Questions:

Generic 5

3 x 5 = 15

For the reaction 2 A +B \(\longrightarrow\) A₂B, rate = k [A] [B]²with k = 2.0 \(\times 10^{-6}\) mol^-2 s^-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.

The half-life for decay of radioactive ¹⁴C is 5730 years. An archaeological artefact containing wood had only 80% of ¹⁴ C activity as found in a living tree. Calculate the age of the artefact

Identify the reaction order from each of the following rate constants.
(i) k = 2.3 \(\times\) 10^-5 L mol ^-1 s^-1
(ii) k = 3 10^-4 \(\times\) s ^-1

Generic 3

7 x 3 = 21

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?

The decomposition of NH₃ on platinum surface is zero order. What are the rates of production of N₂ and H₂ it k = 2.5 x 10^-4mol Ls^-1?

The first order rate constant for the decomposition of ethyl iodide by the reaction: C₂H₅I(g)\(\longrightarrow\)C₂H₄ (g) + HI(g) at 600K is 1.60\(\times{10}^{-5}sec^{-1}\). Its energy of activation is 209 kJ/mol. Calculate the rate constant of the reaction at 700K.

The rate constant for the decomposition of a hydrocarbon is 2.418 \(\times 10^{-5}\) s^-1at 546 K. If the energy of activation is 179.9 kJ/mol, what will be value of pre-exponential factor?

The activation energy for the reaction, 2 HI (g) \(\longrightarrow\) H₂ + I₂ (g), is 209.5 kJ mol^-1 at 581 K. Calculate the fraction of molecules of reactions having energy equal to or grater than activation energy.

The rate of the chemical reaction doubles for an Increase of 10 K from 298 K. Calculate E_a.

10.

For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction

Generic 2

13 x 2 = 26

11.

Time required to decompose SO₂Cl₂to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.

12.

When reaction is completed 99.9%, [R] _n = [R]₀– 0.999[R]₀

13.

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

14.

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 the Ea for this reaction and what temperature will its half-period be 256 minutes?

15.

A first order reaction is found to have a rate constant k = 5.5 x 10^-14s^-1. Find the half life of this reaction.

16.

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?

17.

Consider a certain reaction A \(\longrightarrow\) Produces with k = 2.0\(\times 10 ^{-2} s^{-1}\) . Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L^-1.

18.

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 year^-1

19.

In a reaction, 2 A \(\longrightarrow\) products, the concentration of A decreases from 0.5 mol L^-1 to 0.4 mol L^-1in 10 minutes. Calculate the rate during this interval.

20.

For the reaction R \(\longrightarrow\) 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.

21.

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?

22.

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. [R = 8.314 JK^-1 mol^-1][log 4 = 0.6021]

23.

The decomposition of hydrocarbon follows the equation k = (4.5 \(\times\) 10¹¹s^-1) e^-28000K/T Calculate E_a.

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Answers Key:

Generic 5

3 x 5 = 15

\(Rate \ = \ k{ \left[ A \right] }^{ -1 }{ \left[ B \right] }^{ 2 }\)
\(Rate \ =2.0\times { 10 }^{ -6 }{ mol }^{ -2 }{ L }^{ 2 }{ s }^{ -1 }\times 0.1\times { \left( 0.2 \right) }^{ 2 }\)
\(=8\times { 10 }^{ -9 }{ mol }{ \ L }^{ -1 }{ s }^{ -1 }\)
After is reduced to \(0.06\ { mol }{ \ L }^{ -1 }\) ,
When 2 moles of 'A' react then 1 mole of 'B' reacts.
When \(0.1-0.06=0.04\) moles of A react then 0.02 moles of B will react.
\(\therefore \ \left[ B \right] =0.20-0.02=0.18 \ { mol }{ \ L }^{ -1 }\)
\(rate \ =2.0\times { 10 }^{ -6 }\times (0.06){ (0.18) }^{ 2 }\)
\(=2\times { 10 }^{ -6 }\times 6\times { 10 }^{ -2 }\times 3.24\times { 10 }^{ -2 }\)
\(=38.88\times { 10 }^{ -10 }\)
\(=3.888\times { 10 }^{ -9 }{ mol }{ \ L }^{ -1 }{ s }^{ -1 }\)
\(=3.89\times { 10 }^{ -9 }{ mol }{ \ L }^{ -1 }{ s }^{ -1 }\)

\({ t }_{ { 1 }/{ 2 } }=5730 \ years\)
\({ t }_{ { 1 }/{ 2 } }=\frac { 0.693 }{ k } \)
\(k=\frac { 0.693 }{ 5730 } { years }^{ -1 }\)
\(k=\frac { 2.303 }{ t } \log { \frac { { \left[ N \right] }_{ 0 } }{ \left[ N \right] } }\)
\(t=\frac { 2.303 }{ k } \log { \frac { 100 }{ 80 } }\)
\(=\frac { 2.303 }{ 0.693 } \times 5730\left[ \log { 5 } -\log { 4 } \right] \)
\(=\frac { 5730 }{ 0.3010 } \times \left[ 0.6990-0.6021 \right] \)
\(=\frac { 5730 }{ 0.3010 } \times 0.0969=\frac { 55.237 }{ 0.3010 } \)
\(=1844.64 \ years\approx 1845 \ years\)

(i) The unit of second order rate constant is L mol^–1 s^–1, therefore k = 2.3 x 10^–5 L mol^–1 s^–1 represents a second order reaction.
(ii) The unit of a first order rate constant is s^–1 therefore k = 3 x 10^–4 s^–1 represents a first order reaction.

Generic 3

7 x 3 = 21

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, k₁ = 4.5 × 10³ s⁻¹
T₁ = 273 + 10 = 283 K
k₂ = 1.5 × 10⁴ s⁻¹
_Ea = 60 kJ mol⁻¹ = 6.0 × 10⁴ J mol⁻¹
Then,
\(\log \frac{1.5 \times 10^{4}}{4.5 \times 10^{3}}=\frac{6.0 \times 10^{4} \mathrm{Jmol}^{-1}}{2.303 \times 8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}}\left(\frac{T_{2}-283}{283 T_{2}}\right)\)
\(\Rightarrow 0.5229=3133.627\left(\frac{T_{2}-283}{283 T_{2}}\right)\)
\(\Rightarrow 0.0427 T_{2}=T_{2}-283\)
\(\Rightarrow 0.9528 T_{2}=283\)
\(\Rightarrow T_{2}=297.019 K \text { (approximately) }\)
= 297K
= 24º C
Hence, k would be 1.5 × 10⁴ s⁻¹ at 24°C.

\(2NH_x\rightarrow N_2+3H_2\)
\(Rate=-\frac{s[NH_3]}{dt}=\frac{d[N_2]}{dt}=\frac{1}{3}\frac{d[H_2]}{dt}\)
For zero order reaction, rate=k
\(\therefore\ -\frac{1}{2}\frac{s[NH_3]}{dt}=\frac{d[N_2]}{dt}=\frac{1}{3}\frac{d[H_2]}{dt}\)
\(2.5 \times 10^{-4}\ mol L^{-1}\ s^{-1}\)
Rate of production of N₂=\(\frac{d[N_2]}{dt}\)
\(2.5 \times 10^{-4}\ mol L^{-1}\ s^{-1}\)
Rate of production of N₂=\(\frac{d[N_2]}{dt}=3\times2.5\times10^{-4}\)
\(=7.5 \times 10^{-4}\ mol\ L^{-1}\ s^{-1}\)

We know that
\(\log k_{2}-\log k_{1}=\frac{E_{\mathrm{a}}}{2.303 R}\left[\frac{1}{T_{1}}-\frac{1}{T_{2}}\right]\)
\(\log k_{2}=\log k_{1}+\frac{E_{\mathrm{a}}}{2.303 R}\left[\frac{1}{T_{1}}-\frac{1}{T_{2}}\right]\)
\(=\log \left(1.60 \times 10^{-5}\right)+\frac{209000 \mathrm{~J} \mathrm{~mol} \mathrm{~L}^{-1}}{2.303 \times 8.314 \mathrm{Jmol} \mathrm{L}^{-1} \mathrm{~K}^{-1}}\left[\frac{1}{600 \mathrm{~K}}-\frac{1}{700 \mathrm{~K}}\right]\)
log k₂ = – 4.796 + 2.599 = – 2.197
k₂ = 6.36 x 10^–3 s^–1

Here, \(k=2.418\times { 10 }^{ -5 }{ s }^{ -1 },{ E }_{ a }=179.9\quad kJ\quad { mol }^{ -1 },\quad T=546K.\)
\(or \ logA=logk+\frac { { E }_{ a } }{ 2.303RT } =log(2.418\times { 10 }^{ -5 }{ s }^{ -1 })+\frac { 179.9 \ kJ \ { mol }^{ -1 } \ }{ 2.303\times 8.314\times { 10 }^{ -3 }kJ \ { K }^{ -1 } \ { mol }^{ -1 }\times 546K } \)
\(\\=(-5+0.3834){ s }^{ -1 }+17.2081=12.5924{ s }^{ -1 }\)
\(A=Antilog(12.5924){ s }^{ -1 }=3.912\times { 10 }^{ 12 }{ s }^{ -1 }\)
According to Arrhenius equation,
\(k=A{ e }^{ { -{ E }_{ a } }/{ RT } }or \ ln \ k=lnA-\frac { { E }_{ a } }{ RT } or \ logk=logA-\frac { { E }_{ a } }{ 2.303RT }\)

The fraction of molecules having energy equal to or greater than activation energy,
Fraction of molecules \(={ e }^{ { -{ E }_{ a } }/{ RT } }\)
\(={ e }^{ \frac { -209.5\times 1000 }{ 8.314\times 581 } }\)
\(\\ ={ e }^{ -43.37 }=1.461\times { 10 }^{ -19 }\).

\(\log { \frac { { k }_{ 2 } }{ { k }_{ 1 } } } =\frac { { E }_{ a } }{ 2.303R } \left( \frac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 1 }{ T }_{ 2 } } \right)\)
\(\frac { { k }_{ 2 } }{ { k }_{ 1 } } =2,\quad { T }_{ 1 }=298K,\quad { T }_{ 2 }=308K,\quad R=8.314\quad J{ K }^{ -1 }{ mol }^{ -1 }\)
\(\log { 2 } =\frac { { E }_{ a } }{ 2.303\times 8.314 } \times \left[ \frac { 308-298 }{ 308\times 298 } \right] \)
\(0.3010=\frac { { E }_{ a } }{ 2.303\times 8.314 } \times \frac { 10 }{ 298\times 308 } \)
\(\therefore \ { E }_{ a }=\frac { 0.3010\times 2.303\times 8.314\times 298\times 308 }{ 10 } \)
\(52.898 \ kJ\)

10.

For first order reaction, \(t=\frac { 2.303 }{ k } \log { \frac { a }{ a-x } } \)
99% completion means that x = 99% of a = 0.99 a
\({ t }_{ 99 }\%\)\(=\frac { 2.303 }{ k } \log { \frac { a }{ a-0.99a } } =\frac { 2.303 }{ k } \log { { 10 }^{ 2 } } =2\times \frac { 2.303 }{ k } \)
90% completion means that x = 90% of a = 0.90 a
\(\therefore \quad { t }_{ 90 }\%\)\(=\frac { 2.303 }{ k } \log { \frac { a }{ a-0.99a } } =\frac { 2.303 }{ k } \log { { 10 } } =2\times \frac { 2.303 }{ k } \)
\(\therefore \quad \frac { { t }_{ 99 } }{ { t }_{ 90 } } ={ \left( \frac { 2\times 2.303 }{ k } \right) }/{ \left( \frac { 2.303 }{ k } \right) }=2\quad \)\({ t }_{ 99 }\%\)\(=2\times { t }_{ 90 }\%\)

Generic 2

13 x 2 = 26

11.

We know that for a 1^st order reaction,
\(t_{1 / 2}=\frac{0.693}{k}\)
It is given that t_1/2 = 60 min
\(\therefore k=\frac{0.693}{t_{1 / 2}}\)
= 0.693/60
= 0.01155 min^-1
= 1.155 min^-1
Or k = 1.925 x 10^-4 s^-1

12.

\(k=\frac{2.303}{t} \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]}\)
\(=\frac{2.303}{t} \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]_{0}-0.999[\mathrm{R}]_{0}}=\frac{2.303}{t} \log 10^{3}\)
t = 6.909/k
For half-life of the reaction
t _1/2= 0.693/k
\(\frac{t}{t_{1 / 2}}=\frac{6.909}{k} \times \frac{k}{0.693}=10\)

13.

Let \( a=100, a-x=100-30=70, t=40 \mathrm{~min} \)
\(k=\frac{2.303}{t} \log \frac{a}{(a-x)}=\frac{2.303}{40 \mathrm{~min}} \log \frac{100}{70}\)
\(=\frac{2.303}{40} \log \frac{10}{7}=\frac{2.303}{40} \log 1.428=\frac{2.303}{40} \times 0.1545\)
\(k=8.91 \times 10^{-3} \min ^{-1} \)
\(t_{1 / 2}=\frac{0.693}{\mathrm{k}}=\frac{0.693}{8.91 \times 10^{-3} \min ^{-1}}=77.78 \mathrm{~min} \)
\(t_{1 / 2}=77.78 \mathrm{~min}\)

14.

\(\log { k } =14.34-\frac { 1.25\times { 10 }^{ 4 }K }{ T }\)
\(\log { k } =logA-\frac { { E }_{ a } }{ 2.303RT } \)
\(-\frac { { E }_{ a } }{ 2.303RT } =\frac { 1.25\times { 10 }^{ 4 }K }{ T } \)
\({ E }_{ a }=1.25\times { 10 }^{ 4 }\times 8.314\times 2.303\)
\(=239.34 \ KJ \ { mol }^{ -1 }\)
\(k=\frac { 0.693 }{ 256 } { min }^{ -1 }\)
\(k=\frac { 0.693 }{ 256 } \times \frac { 1 }{ 60 }\)
\(=4.51\times { 10 }^{ -5 }{ s }^{ -1 }\)
\(\log { k } =14.34-\frac { 1.25\times { 10 }^{ 4 }K }{ T } \)
\( -4.35=14.34-\frac { 1.25\times { 10 }^{ 4 } }{ T }\)
\(T=\frac { 1.25\times { 10 }^{ 4 } }{ 18.69 }\)
\( =\frac { 12500 }{ 18.69 } =668.80 \ K\)

15.

Half-life for a first order reaction is
\(t_{1 / 2}=\frac{0.693}{k}\)
\(t_{1 / 2}=\frac{0.693}{5.5 \times 10^{-14} \mathrm{~s}^{-1}}=1.26 \times 10^{13} \mathrm{~s}\)
Show that in a first order reaction, time required for completion of 99.9% is 10 times of half-life (t_1/2) of the reaction.

16.

As sucrose decomposes according to first order rate law, \(k=\frac { 2.303 }{ t } \log { \frac { { \left[ A \right] }_{ 0 } }{ { \left[ A \right] } } } \)
The aim is to find \({ \left[ A \right] }/{ { \left[ A \right] }_{ 0 } }\)
As \({ t }_{ { 1 }/{ 2 } }=3.0\quad hour,\quad \therefore k=\frac { 0.693 }{ { t }_{ { 1 }/{ 2 } } } =\frac { 0.693 }{ 3\quad hr } =0.231\quad { hr }^{ -1 }\)
Hence, \(0.231 \ { hr }^{ -1 }\)\(=\frac { 2.303 }{ 8\quad hr } \log { \frac { { \left[ A \right] }_{ 0 } }{ { \left[ A \right] } } } \) or \(\frac { { \left[ A \right] }_{ 0 } }{ { \left[ A \right] } } =0.8024\quad \)or \(\frac { { \left[ A \right] }_{ 0 } }{ { \left[ A \right] } } =Antilog\quad (0.8024)=6.345\)or \(\frac { { \left[ A \right] } }{ { { \left[ A \right] }_{ 0 } } } =\frac { 1 }{ 6.345 } =0.158\)

17.

The units of k show that the reaction is of first order.Hence, \(k=\frac { 2.303 }{ t } \log { \frac { { \left[ A \right] }_{ 0 } }{ { \left[ A \right] } } } \)
\(\therefore \ 2.0\times { 10 }^{ -2 }{ s }^{ -1 }=\frac { 2.303 }{ 100s } \log { \frac { 1.0 \ mol \ { L }^{ -1 } }{ \left[ A \right] } } or \ \log { \left[ A \right] } =-0.8684\)
\(\ \therefore \ \left[ A \right] =Antilog(-0.8684)=Antilog(1.1316)=0.1354 \ mol \ { L }^{ -1 }\)

18.

\((a) \ \ { t }_{ { 1 }/{ 2 } }=\frac { 0.693 }{ k } =\frac { 0.693 }{ 200 \ { s }^{ -1 } } \)
\(=3.465\times { 10 }^{ -3 }s\)
\(=3.47\times { 10 }^{ -3 }s\)
\((b) \ { t }_{ { 1 }/{ 2 } }=\frac { 0.693 }{ 2{ min }^{ -1 } } =0.3465 \ min\)
\(=0.35 \ min\)
\(\ (c) \ { t }_{ { 1 }/{ 2 } }=\frac { 0.693 }{ 4year^{ -1 } } =0.173 \ years\)

19.

Average rate = \(-\frac{1}{2} \frac{\Delta[A]}{\Delta t}\)
\(=-\frac{1}{2} \frac{[A]_{2}-[A]_{1}}{t_{2}-t_{1}}\)
\(=-\frac{1}{2} \frac{0.4-0.5}{10}\)
\(=-\frac{1}{2} \frac{-0.1}{10}\)
= 0.005 mol L⁻¹min⁻¹
= 5 x 10⁻³ M min⁻¹

20.

\(Rate=-\frac { \Delta \left[ R \right] }{ \Delta t } \)
\(=-\frac { \left[ 0.02M-0.03M \right] }{ 25 } \)
\(=\frac { 0.01 }{ 25 } mol\quad { L }^{ -1 }\quad { min }^{ -1 }\)
\(Rate=0.4\times { 10 }^{ -3 }mol\quad { L }^{ -1 }\quad { min }^{ -1 }\)
\(=4\times { 10 }^{ -4 }mol\quad { L }^{ -1 }\quad { min }^{ -1 }\)
\(=\frac { 4\times { 10 }^{ -4 }mol\quad { L }^{ -1 } }{ 60\quad s } \)
\(=0.066\times { 10 }^{ -4 }\)
\(=6.6\times { 10 }^{ -6 }mol\quad { L }^{ -1 }\quad { s }^{ -1 }\)

21.

For first order reaction.
\(t =\frac{2.303}{k} \log \frac{a}{(a-x)} \)
\(\text { Given, }(a-x) =\frac{a}{16} ; k=60 \mathrm{~s}^{-1}\)
On putting the values in Eq. (i)
\(t =\frac{2.303}{60 \mathrm{~s}^{-1}} \log \frac{a \times 16}{a}=\frac{2.303}{60 \mathrm{~s}^{-1}} \log 16 \)
\(=\frac{2.303}{60 \mathrm{~s}^{-1}} \log 2^{4}\)
\(=\frac{2.303}{60 \mathrm{~s}^{-1}} \times 4 \log 2\)
\(=\frac{2.303}{60 \mathrm{~s}^{-1}} \times 4 \times 0.3010 \)
\( \text {Time }(t)= 4.62 \times 10^{-2} \mathrm{~s}\)

22.

\(\log { \frac { { k }_{ 2 } }{ { k }_{ 1 } } } =\frac { { E }_{ a } }{ 2.303R } \left( \frac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 1 }{ T }_{ 2 } } \right) \)
\({ E }_{ a }=\frac { 2.303R\times { T }_{ 1 }{ T }_{ 2 } }{ { T }_{ 2 }-{ T }_{ 1 } } \log { \frac { { k }_{ 2 } }{ { k }_{ 1 } } } \)
Quadruples means rate becomes four time \({ k }_{ 2 }=4{ k }_{ 1 }\)
\({ E }_{ a }=\frac { 2.303\times 8.314\times 313\times 293 }{ 20 } \times \log { 4 }\)
\({ E }_{ a }=\frac { 19.147\times 313\times 293 }{ 20 } \times 0.6021\)
\(=52.86\ kJ\ { mol }^{ -1 }\)

23.

\(k=(4.5\times { 10 }^{ 11 }{ s }^{ -1 }){ e }^{ \frac { -28000K }{ T } }\)
\(k=A{ e }^{ { { -E }_{ a } }/{ RT } }\)
\(\frac { { -E }_{ a } }{ RT } =\frac { -28000K }{ T } \)
\({ E }_{ a }=28000\quad K\times 8.314\ J{ K }^{ -1 }{ mol }^{ -1 }\)
\( { E }_{ a }=232.792\ kJ{ { \ mol }^{ -1 } }\)