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JEE • CHEMISTRY • CHEMICAL KINETICS

Chemical kinetics practice · Medium

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121 Medium
121 of 144
Application +4 | -1
For the reaction 2 N2O5( g)→4NO2( g)+O2( g)2 \mathrm{~N}_2 \mathrm{O}_5(\mathrm{~g}) \rightarrow 4 \mathrm{NO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) If the concentration of O2\mathrm{O}_2 increases by 2×10−3M2 \times 10^{-3} \mathrm{M} in 100 seconds then the rate of disappearance of N2O5\mathrm{N}_2 \mathrm{O}_5 is :-
View answer and explanation
Correct option: C
4×10−5Ms−14 \times 10^{-5} \mathrm{Ms}^{-1}
Option C (4×10−5Ms−14 \times 10^{-5} \mathrm{Ms}^{-1}) is correct because it represents the rate of disappearance of N2O5\mathrm{N}_2 \mathrm{O}_5 calculated from the given rate of increase of O2\mathrm{O}_2 using the stoichiometry of the reaction.
122 Medium
122 of 144
Application +4 | -1
For a first order reaction it takes 15 second for 75% completion. The half life of the reaction is
View answer and explanation
Correct option: B
7.5 sec
Option B (7.5 sec) is correct because it accurately calculates the half-life of a first-order reaction given the time for 75% completion.
123 Medium
123 of 144
Application +4 | -1
A given sample of milk turns sour at room temperature (27∘C)\left(27^{\circ} \mathrm{C}\right) in 16 h.16 \mathrm{~h}. In a refrigerator at 7∘C,7^{\circ} \mathrm{C}, milk can be stored three times as long before it sours. How long (in hours) it takes for milk to sour at 57∘C?57^{\circ} \mathrm{C} ?(ln⁡2=0.7,ln⁡3=1.1)(\ln 2=0.7, \ln 3=1.1)
View answer and explanation
Correct option: A
4
Option A (4) is correct because it uses the Arrhenius equation to relate the rate constant to temperature and calculates the time taken for milk to sour at 57°C.
124 Medium
124 of 144
Understanding +4 | -1
In general, the rate of a reaction can be increased by all the factors except
View answer and explanation
Correct option: C
Increasing the activation energy
Option C (Increasing the activation energy) is correct because increasing activation energy decreases the rate of reaction, as it requires more energy for the reaction to proceed.
125 Medium
125 of 144
Application +4 | -1
Hydrolysis of ethyl acetate is catalysed by acid.

CH3COOC2H5+H2O→H+CH3COOH+C2H5OH\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5+\mathrm{H}_2 \mathrm{O} \xrightarrow{\mathrm{H}^{+}} \mathrm{CH}_3 \mathrm{COOH}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}

Rate constant of the reaction is 5.56×10−6 s−15.56 \times 10^{-6} \mathrm{~s}^{-1}. What is its rate constant when it is second-order reaction (pseudo unimolecular)?
View answer and explanation
Correct option: C
1.0×10−7 L mol−1 s−11.0 \times 10^{-7} \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~s}^{-1}
The correct answer is 1.0×10−7 L mol−1 s−11.0 \times 10^{-7} \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~s}^{-1} because it is calculated by dividing the pseudo-first-order rate constant by the concentration of water.
126 Medium
126 of 144
Application +4 | -1
For the reaction A→B,A \to B, the rate expression is r=k[A]n.r = k{\left[ A \right]^n}. When the concentration of AA is doubled, the rate of reaction is quadrupled. The value of nn is
View answer and explanation
Correct option: D
2
Option D (2) is correct because when the concentration of A is doubled, and the rate quadruples, the order n is 2, indicating a second-order reaction.
127 Medium
127 of 144
Analyzing +4 | -1
For the second-order reaction A+B→Products,{A}+{B} \rightarrow \text{Products}, the rate constant kk, is given as
k=2.303(a−b)tlog⁡b(a−x)a(b−x),k=\frac{2.303}{(a-b) t} \log \frac{b(a-x)}{a(b-x)}, where, aa and bb are the initial concentrations of 'AA' and 'BB' and xx is the change in concentration after time t.t. If b≫a,b \gg a, the reaction reduces to
View answer and explanation
Correct option: A
First-order with respect to AA
Option A (First-order with respect to A) is correct because when b >> a, the concentration of B remains essentially constant, making the reaction pseudo-first-order with respect to A.
128 Medium
128 of 144
Application +4 | -1
The rate of the reaction 3A+2B→Products3 \mathrm{A}+2 \mathrm{B} \rightarrow \text{Products} is given by the rate expression rate =K[A][B]2.=K[\mathrm{A}][\mathrm{B}]^2. If A\mathrm{A} is taken in large excess, then the reaction start obeying
View answer and explanation
Correct option: B
Second-order kinetics
Option B (Second-order kinetics) is correct because when A is in large excess, its concentration becomes constant, and the rate law simplifies to depend only on the concentration of B squared, making it second order.
129 Medium
129 of 144
Application +4 | -1
In a reaction for every 10∘C10^{\circ} \mathrm{C} rise of temperature, the rate is tripled. If the temperature is increased from 10∘C10^{\circ} \mathrm{C} to 50∘C50^{\circ} \mathrm{C}, the rate of the reaction will become
View answer and explanation
Correct option: A
81 times
Option A (81 times) is correct because the rate increases by a factor of 3 for every 10°C rise, and the temperature increases by 40°C.
130 Medium
130 of 144
Analyzing +4 | -1
Assertion (A):A plot of log⁡dXdt vs. log⁡(a−x)\log \frac{d X}{d t} \text{ vs. }\log (a-x) leads to value of KK as antilog (intercept) and slope equal to order of reaction.

Reason (R):dXdt=K(a−x)n\frac{d X}{d t}=K(a-x)^n
View answer and explanation
Correct option: C
Both A and R are correct and R is correct explanation of A
Option D (Both A and R are correct but R is not correct explanation of A) is incorrect. Option C (Both A and R are correct and R is correct explanation of A) is correct because the integrated rate law can be expressed in logarithmic form, where the intercept corresponds to the rate constant and the slope corresponds to the order of the reaction.
131 Medium
131 of 144
Analyzing +4 | -1
Statement I Kinetics of study of decomposition of N2O5(g)\mathrm{N}_2 \mathrm{O}_5(g) to NO2(g)\mathrm{NO}_2(g) and O2(g)\mathrm{O}_2(g) can be studied by measuring pressure of O2\mathrm{O}_2 gas in CCl4\mathrm{CCl}_4 solvent.

2 N2O5(g)⟶4NO2(g)+O2(g)2 \mathrm{~N}_2 \mathrm{O}_5(g) \longrightarrow 4 \mathrm{NO}_2(g)+\mathrm{O}_2(g)

Statement II At the start, volume of O2(g)\mathrm{O}_2(g) is zero.
View answer and explanation
Correct option: B
Both Statement I and Statement II are correct but
Statement II is not the correct explanation of Statement I
Option B (Both Statement I and Statement II are correct but Statement II is not the correct explanation of Statement I) is correct because the kinetics can be studied by measuring the pressure of O2{}_2 in CCl4{}_4, and the initial volume of O2{}_2 is zero, but the second statement is not the reason for the first.
132 Medium
132 of 144
Application +4 | -1
The decomposition of N2O5{N}_2 {O}_5 is a first order reacton represented by N2O5⟶N2O4+12O2{N}_2 {O}_5 \longrightarrow {N}_2 {O}_4+\frac{1}{2} {O}_2. After 15 minutes the volume of O2{O}_2 produced is 9 ml9 \, \mathrm{ml} and at the end of the reaction 35 mL35 \mathrm{~mL}. The rate constant it equal to
View answer and explanation
Correct option: C
115In⁡4435\frac{1}{15} \operatorname{In} \frac{44}{35}
Option C (115In⁡4435\frac{1}{15} \operatorname{In} \frac{44}{35}) is correct because it is derived from the integrated rate law for a first-order reaction, considering the volume of oxygen produced at different times.
133 Medium
133 of 144
Application +4 | -1
Which of the following represents the expression for 3/43 / 4th the life of a first-order reaction?
View answer and explanation
Correct option: C
2.303klog⁡4\frac{2.303}{k} \log 4
Option C (2.303klog⁡4\frac{2.303}{k} \log 4) is correct because it represents the time taken for 75% completion of a first-order reaction.
134 Medium
134 of 144
Application +4 | -1
Timetaken for 12.8 g12.8 {~g} of a radioactive substance to decay to 0.4 g0.4 {~g} is: (half life is 138 s138 {~s} )
View answer and explanation
Correct option: B
690 s690 {~s}
Option B (690 s690 {~s}) is correct because it is calculated by determining the number of half-lives required for the decay and multiplying by the half-life duration.
135 Medium
135 of 144
Application +4 | -1
Which of the following rate law has an overall order of 0.5 for reaction involving substances X,Y\mathrm{X}, \mathrm{Y} and Z\mathrm{Z} ?
View answer and explanation
Correct option: C
rate =k(CX)1.5(CY)−1(CZ)0=k\left(C_{\mathrm{X}}\right)^{1.5}\left(C_{\mathrm{Y}}\right)^{-1}\left(C_{\mathrm{Z}}\right)^0
The correct answer is C (rate =k(CX)1.5(CY)−1(CZ)0=k(C_{\mathrm{X}})^{1.5}(C_{\mathrm{Y}})^{-1}(C_{\mathrm{Z}})^0) because the sum of the exponents in the rate law equals 0.5.
136 Medium
136 of 144
Application +4 | -1
t99.9%{t_{99.9\% }} of a first order reaction is 100 minutes. Rate constant of the reaction is
View answer and explanation
Correct option: C
0.0693 minutes
Option C (0.0693 minutes) is correct because the rate constant is calculated using the relationship between t99.9%t_{99.9\%} and half-life for a first-order reaction.
137 Medium
137 of 144
Analyzing +4 | -1
Consider the following statements,

The rate law for the acid catalysed hydrolysis of an ester being given as

Rate=k  [H+]  [ester]=k′  [ester]Rate = k\;\left[ {{H^ + }} \right]\;\left[ {ester} \right] = k'\;\left[ {ester} \right]

If the acid concentration is doubled at constant ester concentration

1. The second order rate constant, kk is doubled.

2. The pseudo first order rate constant, kk is double.

3. The rate of the reaction is doubled.

Which of the above statements are correct?
View answer and explanation
Correct option: B
2 and 3
Option B (2 and 3) is correct because doubling the acid concentration doubles both the pseudo-first-order rate constant and the reaction rate, while the second-order rate constant remains unchanged.
138 Medium
138 of 144
Analyzing +4 | -1
For the reaction A→K1B→K2C→K3D,\mathrm{A} \xrightarrow{K_1} \mathrm{B} \xrightarrow{K_2} \mathrm{C} \xrightarrow{K_3} \mathrm{D}, where K3>K2>K1;K_3>K_2>K_1; the rate of formation of D\mathrm{D} is determined by
View answer and explanation
Correct option: A
The rate of reaction: A→B\mathrm{A} \rightarrow \mathrm{B}
The rate of formation of D is determined by the rate of reaction A → B because K1 is the smallest rate constant, making it the slowest and rate-determining step.
139 Medium
139 of 144
Application +4 | -1
For the reaction 2N2O5→4NO2+O22{N_2}O5 \to 4N{O_2} + {O_2} rate and rate constant are 1.02×10−4  mol.L−1.sec⁡−11.02 \times {10^{ - 4}}\,\,mol.{L^{ - 1}}.{\sec ^{ - 1}} and 3.4×10−5sec⁡−13.4 \times {10^{ - 5}}{\sec ^{ - 1}} respectively then concentration of N2O5{N_2}{O_5} at that time will be
View answer and explanation
Correct option: B
3 M3\,M
Option B (3 M) is correct because the concentration of N2O5 is calculated using the rate law for the reaction.
140 Medium
140 of 144
Application +4 | -1
Half life of a first order reaction is 4 s4 {~s} and the initial concentration of the reactant is 0.12M0.12 {M}. The concentration of the reactant left after 16 s16 {~s} is
View answer and explanation
Correct option: A
0.0075M0.0075 {M}
Option D (-x) is correct because the energy of an electron depends on the atomic number squared divided by the principal quantum number squared.
141 Medium
141 of 144
Analyzing +4 | -1
Consider the following statements

a) Increase in concentration of reactant increases the rate of zero order reaction

b) Rate constant K is equal to collision frequency A if Ea=0{E_a} = 0

c) Rate constant K is equal to collision frequency A if Ea= ∝{E_a} = \, \propto

d) ln K Vs T is a straight line

e) ln K Vs 1/T is a straight line

Correct statements are
View answer and explanation
Correct option: B
b and e
Option B (b and e) is correct because the rate constant equals the collision frequency when activation energy is zero, and ln K vs 1/T yields a straight line according to the Arrhenius equation.
142 Medium
142 of 144
Application +4 | -1
If for a radio-isotope, at the end of 4.2 days, N=0.798 N0{N}=0.798 {~N}_0. What will be half life period of its nucleus?
View answer and explanation
Correct option: B
12.83 days
Option B (12.83 days) is correct because it is the half-life period calculated using the radioactive decay formula.
143 Medium
143 of 144
Application +4 | -1
The ratio of 14C12C\frac{{ }^{14} \mathrm{C}}{{ }^{12} \mathrm{C}} in a piece of wood is 18\frac{1}{8} part that of atmosphere. If half life of 14C{ }^{14} \mathrm{C} is 5730 years, the age of wood sample is …..years.
View answer and explanation
Correct option: A
17190
Option A (17190) is correct because it accurately calculates the age of the wood sample using the half-life of Carbon-14 and the given ratio.
144 Medium
144 of 144
Analyzing +4 | -1
Statement I A high activation energy corresponds to a
reaction rate that is very sensitive to temperature.
Statement II In following gaseous phase reaction CH3+CH3⟶CH3−CH3\mathrm{CH}_3+\mathrm{CH}_3 \longrightarrow \mathrm{CH}_3-\mathrm{CH}_3 rate is independent of temperature.
View answer and explanation
Correct option: B
Both Statement I and Statement II are correct but
Statement II is not the correct explanation of Statement I
Option B (Both Statement I and Statement II are correct but Statement II is not the correct explanation of Statement I) is correct because both statements are factually true, but the second statement doesn't explain the first.
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