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Chemical Kinetics

JEE Chemistry · 84 questions · Page 3 of 9 · Click an option or "Show Solution" to reveal answer

Q21
For a reaction 12A→2B{1 \over 2}A \to 2B rate of disappearance of ‘A’ is related to the rate of appearance of ‘B’ by the expression
A −d[A]dt - {{d[A]} \over {dt}} = 12d[B]dt{1 \over 2}{{d[B]} \over {dt}}
B −d[A]dt - {{d[A]} \over {dt}} = 14d[B]dt{1 \over 4}{{d[B]} \over {dt}}
C −d[A]dt - {{d[A]} \over {dt}} = d[B]dt{{d[B]} \over {dt}}
D −d[A]dt - {{d[A]} \over {dt}} = 4d[B]dt4{{d[B]} \over {dt}}
Correct Answer
Option B
Solution

The rates of reactions for the reaction

12A→2B{1 \over 2}A \to 2B

can be written either as

−2ddt[A]  - 2{d \over {dt}}\left[ A \right]\,\,

with respect to

′A′'A'

or

     \,\,\,\,\,
12ddt[B]    {1 \over 2}{d \over {dt}}\left[ B \right]\,\,\,\,

with respect to

′B′'B'

From the above, we have

−2ddt[A]=12ddt[B]- 2{d \over {dt}}\left[ A \right] = {1 \over 2}{d \over {dt}}\left[ B \right]

or

     \,\,\,\,\,
−ddt[A]=14ddt[B]- {d \over {dt}}\left[ A \right] = {1 \over 4}{d \over {dt}}\left[ B \right]

i.e., correct answer is (b)

Q22
The time for half life period of a certain reaction A →\to products is 1 hour. When the initial concentration of the reactant ‘A’, is 2.0 mol L–1, how much time does it take for its concentration to come from 0.50 to 0.25 mol L–1 if it is a zero order reaction ?
A 4 h
B 0.5 h
C 0.25 h
D 1 h
Correct Answer
Option C
Solution

For the reaction

A  →  A\,\, \to \,\,

Product ; given

t1/2=1  {t_{1/2}} = 1\,\,

hour for a zero order reaction

tcompletion  =[A0]k=initial  conc.rate  constan⁡t{t_{completion}}\,\, = {{\left[ {{A_0}} \right]} \over k} = {{initial\,\,conc.} \over {rate\,\,cons\tan t}}

∴\therefore

     t1/2=[A0]2K\,\,\,\,\,{t_{1/2}} = {{\left[ {{A_0}} \right]} \over {2K}}

or

     k=[A0]2t1/2=22×1=1  mol lit−1 hr−1\,\,\,\,\,k = {{\left[ {{A_0}} \right]} \over {2{t_{1/2}}}} = {2 \over {2 \times 1}} = 1\,\,mol\,li{t^{ - 1}}\,h{r^{ - 1}}

Further for a zero order reaction

k=dxdt=change   in   concentrationtimek = {{dx} \over {dt}} = {{change\,\,\,in\,\,\,concentration} \over {time}}
I=0.50−0.25time   I = {{0.50 - 0.25} \over {time}}\,\,\,

∴\therefore

     \,\,\,\,\,

time

=0.25  hr.= 0.25\,\,hr.
Q23
For the reaction 2A + 3B + 32{3 \over 2}C → \to 3P, which statement is correct ?
A dnAdt=dnBdt=dnCdt{{d{n_A}} \over {dt}} = {{d{n_B}} \over {dt}} = {{d{n_C}} \over {dt}}
B dnAdt=23dnBdt=34dnCdt{{d{n_A}} \over {dt}} = {2 \over 3}{{d{n_B}} \over {dt}} = {3 \over 4}{{d{n_C}} \over {dt}}
C dnAdt=32dnBdt=34dnCdt{{d{n_A}} \over {dt}} = {3 \over 2}{{d{n_B}} \over {dt}} = {3 \over 4}{{d{n_C}} \over {dt}}
D dnAdt=23dnBdt=43dnCdt{{d{n_A}} \over {dt}} = {2 \over 3}{{d{n_B}} \over {dt}} = {4 \over 3}{{d{n_C}} \over {dt}}
Correct Answer
Option D
Solution

2A + 3B +

32{3 \over 2}

C →\to 3P rate =

−12d[A]dt=−13d[B]dt=−23d[C]dt=13d[P]dt- {1 \over 2}{{d\left[ A \right]} \over {dt}} = - {1 \over 3}{{d\left[ B \right]} \over {dt}} = - {2 \over 3}{{d\left[ C \right]} \over {dt}} = {1 \over 3}{{d\left[ P \right]} \over {dt}}

⇒\Rightarrow

d[A]dt=23d[B]dt=43d[C]dt{{d\left[ A \right]} \over {dt}} = {2 \over 3}{{d\left[ B \right]} \over {dt}} = {4 \over 3}{{d\left[ C \right]} \over {dt}}
=−23d[P]dt= - {2 \over 3}{{d\left[ P \right]} \over {dt}}
Q24
Consider the reaction : Cl2(aq) + H2S(aq) → S(s) + 2H+ (aq) + 2Cl– (aq) The rate equation for this reaction is rate = k [Cl2] [H2S] Which of these mechanisms is/are consistent with this rate equation? (A) Cl2 + H2S →\to H+ + Cl– + Cl+ + HS– (slow) Cl+ + HS– →\to H+ + Cl– + S (fast) (B) H2S ⇔ \Leftrightarrow H+ + HS– (fast equilibrium) Cl2 + HS– →\to 2Cl– + H+ + S (slow)
A B only
B Both A and B
C Neither A nor B
D A only
Correct Answer
Option D
Solution

Since the slow step is the rate determining step hence- if we consider option

(1)(1)

we find Rate

=k[Cl2][H2S]= k\left[ {C{l_2}} \right]\left[ {{H_2}S} \right]

Now if we consider option

(2)(2)

we find Rate

=k[Cl2][HS−]     ...(1)= k\left[ {C{l_2}} \right]\left[ {H{S^ - }} \right]\,\,\,\,\,...\left( 1 \right)

From equation

(i)(i)
k=[H+][HS−]H2Sk = {{\left[ {{H^ + }} \right]\left[ {H{S^ - }} \right]} \over {{H_2}S}}
     \,\,\,\,\,

or

[HS−]=k[H2S]H+\left[ {H{S^ - }} \right] = {{k\left[ {{H_2}S} \right]} \over {{H^ + }}}

Substituting this value in equation

(1)(1)

we find Rate

=k[Cl2]K[H2S]H+= k\left[ {C{l_2}} \right]K{{\left[ {{H_2}S} \right]} \over {{H^ + }}}
=k′[Cl2][H2S][H+]= k'{{\left[ {C{l_2}} \right]\left[ {{H_2}S} \right]} \over {\left[ {{H^ + }} \right]}}

hence only, mechanism

(1)(1)

is consistent with the given rate equation.

Q25
For a first order reaction, (A) →\to products, the concentration of A changes from 0.1 M to 0.025 M in 40 minutes. The rate of reaction when the concentration of A is 0.01 M is :
A 1.73 x 10–5 M/ min
B 3.47 x 10–4 M/min
C 3.47 x 10–5 M/min
D 1.73 x 10–4 M/min
Correct Answer
Option B
Solution

For a first order reaction

k=2.0303t log⁡ aa−xk = {{2.0303} \over t}\,\log \,{a \over {a - x}}
=2.30340log⁡0.10.025= {{2.303} \over {40}}\log {{0.1} \over {0.025}}
=2.30340log⁡4= {{2.303} \over {40}}\log 4
=2.303×0.602040= {{2.303 \times 0.6020} \over {40}}
=3.47×10−2= 3.47 \times {10^{ - 2}}
R=K(A)1=3.47×10−2×0.01R = K{\left( A \right)^1} = 3.47 \times {10^{ - 2}} \times 0.01
=3.47×10−4= 3.47 \times {10^{ - 4}}
Q26
The rate of a reaction doubles when its temperature changes from 300K to 310K. Activation energy of such a reaction will be:(R = 8.314 JK–1 mol–1 and log 2 = 0.301)
A 48.6 kJ mol–1
B 58.5 kJ mol–1
C 60.5 kJ mol–1
D 53.6 kJ mol–1
Correct Answer
Option D
Solution

Activation energy can be calculated from the equation

log⁡ k2log⁡ k1=Ea2.303R(1T1−1T2){{\log \,{k_2}} \over {\log \,{k_1}}} = {{{E_a}} \over {2.303R}}\left( {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right)

given

k2k1=2  T2=310 K  T1=300 K{{{k_2}} \over {{k_1}}} = 2\,\,{T_2} = 310\,K\,\,{T_1} = 300\,K
=log⁡2=−Ea2.303×8.314(1310−1300)= \log 2 = {{ - {E_a}} \over {2.303 \times 8.314}}\left( {{1 \over {310}} - {1 \over {300}}} \right)
Ea=53598.6J/mol{E_a} = 53598.6J/mol
=53.6kJ/mol.= 53.6kJ/mol.
Q27
Higher order (>3) reactions are rare due to
A increase in entropy and activation energy as more molecules are involved
B shifting of equilibrium towards reactants due to elastic collisions
C loss of active species on collision
D low probability of simultaneous collision of all the reacting species
Correct Answer
Option D
Solution

Reactions of higher order

(>3)\left( { > 3} \right)

are very rate due to very less chances of many molecules to undergo effective collisions.

Q28
Decomposition of H2O2 follows a first order reaction. In fifty minutes the concentration of H2O2 decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of H2O2 reaches 0.05 M, the rate of formation of O2 will be :
A 6.93 ×\times 10-4 mol min-1
B 6.96 ×\times 10-2 mol min-1
C 1.34 ×\times 10-2 mol min-1
D 2.66 L min–1 at STP
Correct Answer
Option A
Solution
H2O2(aq)→H2O(aq)+12O2(g){H_2}{O_2}\left( {aq} \right) \to {H_2}O\left( {aq} \right) + {1 \over 2}{O_2}\left( g \right)

For a first order reaction

k=2.303tlog⁡a(a−x)k = {{2.303} \over t}\log {a \over {\left( {a - x} \right)}}

Given

a=0.5,(a−x)=0.125, t=50  a = 0.5,\left( {a - x} \right) = 0.125,\,t = 50\,\,

min ∴\therefore

     k=2.30350log⁡0.50.125\,\,\,\,\,k = {{2.303} \over {50}}\log {{0.5} \over {0.125}}
=2.78×10−2 min⁡−1= 2.78 \times {10^{ - 2}}\,{\min ^{ - 1}}
r=k[H2O2]=2.78×10−2×0.05r = k\left[ {{H_2}{O_2}} \right] = 2.78 \times {10^{ - 2}} \times 0.05
=1.386×10−3  mol min⁡−1= 1.386 \times {10^{ - 3}}\,\,mol\,{\min ^{ - 1}}

Now

−d[H2O2]dt=d[H2O]dt=2d[O2]dt- {{d\left[ {{H_2}{O_2}} \right]} \over {dt}} = {{d\left[ {{H_2}O} \right]} \over {dt}} = {{2d\left[ {{O_2}} \right]} \over {dt}}

∴\therefore

     2d[O2]dt−d[H2O2]dt\,\,\,\,\,{{2d\left[ {{O_2}} \right]} \over {dt}} - {{d\left[ {{H_2}{O_2}} \right]} \over {dt}}

∴\therefore

     \,\,\,\,\,
d[O2]dt=12×d[H2O2]dt{{d\left[ {{O_2}} \right]} \over {dt}} = {1 \over 2} \times {{d\left[ {{H_2}{O_2}} \right]} \over {dt}}
=1.386×10−32=6.93×10−4 mol min⁡−1= {{1.386 \times {{10}^{ - 3}}} \over 2} = 6.93 \times {10^{ - 4}}\,mol\,{\min ^{ - 1}}
Q29
The rate law for the reaction below is given by the expression k [A] [B] A + B → \to Product If the concentration of B is increased from 0.1 to 0.3 mole, keeping the value of A at 0.1 mole, the rate constant will be :
A k
B k/3
C 3k
D 9k
Correct Answer
Option A
Solution

Rate constant only depends on temperature only and it is independed of concentration of reactants.

Q30
The reaction of ozone with oxygen atoms in the presence of chlorine atoms can occur by a two step process shown below : O3(g) + Cl∙{^ \bullet } (g) → \to O2(g) + ClO∙{^ \bullet } (g) . . . . . .(i) ki = 5.2 × 109 L mol−1 s−1 ClO∙{^ \bullet }(g) + O∙{^ \bullet }(g) → \to O2(g) + Cl∙{^ \bullet } (g) . . . . . . (ii) kii = 2.6 × 1010 L mol−1 s−1 The closest rate constant for the overall reaction O3(g) + O∙{^ \bullet } (g) → \to 2 O2(g) is :
A 5.2 × 109 L mol−1 s−1
B 2.6 × 1010 L mol−1 s−1
C 3.1 × 1010 L mol−1 s−1
D 1.4 × 1020 L mol−1 s−1
Correct Answer
Option D
Solution

We have

O3(g)+Cl∙(g)→O2(g)+ClO∙(g){O_3}(g) + C{l^ \bullet }(g) \to {O_2}(g) + Cl{O^ \bullet }(g)

...... (1)

ClO∙(g)+O∙(g)→O2(g)+Cl∙(g)Cl{O^ \bullet }(g) + {O^ \bullet }(g) \to {O_2}(g) + C{l^ \bullet }(g)

...... (2) On adding Eq. (1) and Eq. (2), we get the required equation for overall reaction

O3(g)+O∙(g)→2O2(g){O_3}(g) + {O^ \bullet }(g) \to 2{O_2}(g)

Therefore,

koverall=k1×k2{k_{overall}} = {k_1} \times {k_2}
=5.2×109×2.6×1010= 5.2 \times {10^9} \times 2.6 \times {10^{10}}
=1.4×1020= 1.4 \times {10^{20}}

L mol−-1 s−-1

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