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NEET • CHEMISTRY • CHEMICAL KINETICS

Chemical kinetics practice · Mixed difficulty

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1 Easy
1 of 207
Understanding +4 | -1
If the reaction rate at a given temperature becomes slower then
View answer and explanation
Correct option: A
The free energy of activation is higher
Option A (The free energy of activation is higher) is correct because a slower reaction rate implies a higher activation energy barrier.
2 Easy
2 of 207
Understanding +4 | -1
1[A]2\frac{1}{{{{\left[ A \right]}^2}}} vs times are a straight line. Order of reaction is
View answer and explanation
Correct option: D
Third
The correct answer is Third because the integrated rate law for a third-order reaction has the form 1[A]2=kt+constant\frac{1}{[A]^2} = kt + constant, which gives a straight line when 1[A]2\frac{1}{[A]^2} is plotted against time.
3 Medium
3 of 207
Application +4 | -1
The rate constant (k′)\left( {k'} \right) of one of the reaction is found to be double that of the rate constant (k′′)\left( {k''} \right) of another reaction. Then the relationship between the corresponding activation energies of the two reaction (E′a  and  E′′a)\left( {{{E'}_a}\;\mathrm{and}\;{{E''}_a}} \right) can be represented as
View answer and explanation
Correct option: D
E′a<E′′a{{E'}_a} < {{E''}_a}
Option D (E′a<E′′a{{E'}_a} < {{E''}_a}) is correct because a higher rate constant implies a lower activation energy, according to the Arrhenius equation.
4 Easy
4 of 207
Understanding +4 | -1
Acid hydrolysis of sucrose is a
View answer and explanation
Correct option: A
Pseudo first order reaction
The correct answer is A (Pseudo first order reaction) because the acid hydrolysis of sucrose involves water in large excess, making it a pseudo-first-order reaction.
5 Easy
5 of 207
Remembering +4 | -1
Order of radioactive disintegration reaction is
View answer and explanation
Correct option: B
First
The correct answer is B (First) because radioactive decay follows first-order kinetics.
6 Medium
6 of 207
Application +4 | -1
Temperature coefficient of a reaction is 2. When temperature is increased from 30∘C  to  100∘C30^\circ C\;to\;100^\circ C, rate of the reaction increases by
View answer and explanation
Correct option: A
128 times
Option A (128 times) is correct because the temperature increases by 70°C, which is 7 increments of 10°C, so the rate increases by 27=1282^7 = 128 times.
7 Easy
7 of 207
Understanding +4 | -1
On addition of AgNO3  to  NaCl,\mathrm{AgN}{\mathrm{O}_3}\;to\;\mathrm{NaCl}, white ppt. occurs:
View answer and explanation
Correct option: A
Instantaneously
Option A (Instantaneously) is correct because the reaction between Ag+ and Cl- ions is a very fast ionic reaction.
8 Medium
8 of 207
Understanding +4 | -1
Which is correct relation in between dcdt,dndt  and  dPdt,\frac{{dc}}{{dt}},\frac{{dn}}{{dt}}\;and\;\frac{{dP}}{{dt}}, where c,n  and  pc,n\;and\;p, represent concentration, mole and pressure terms for gaseous phase reactant A(g)→productsA\left( \mathrm{g} \right) \to products?
View answer and explanation
Correct option: A
−dcdt=−1Vdndt=−1RTdPdt- \frac{{dc}}{{dt}} = - \frac{1}{V}\frac{{dn}}{{dt}} = - \frac{1}{{RT}}\frac{{dP}}{{dt}}
Option A (−dcdt=−1Vdndt=−1RTdPdt- \frac{{dc}}{{dt}} = - \frac{1}{V}\frac{{dn}}{{dt}} = - \frac{1}{{RT}}\frac{{dP}}{{dt}}) is correct because it correctly relates the rate of change of concentration, moles, and pressure using the ideal gas law.
9 Hard
9 of 207
Understanding +4 | -1
In the reaction A⟶P\mathrm{A} \longrightarrow \mathrm{P}, concentration of A\mathrm{A} was registered after equal intervals of time. If these values followed A.P., then order of reaction will be:
View answer and explanation
Correct option: A
0
Option A (0) is correct because if the concentration of A follows an arithmetic progression (A.P.) with equal time intervals, it indicates a zero-order reaction.
10 Easy
10 of 207
Analyzing +4 | -1
The concentration of R in the reaction R→PR \to P was measured as a function of time and the following data is obtained
[R](molar)      1.00            0.75          0.40          0.10t  (min)                  0.00          0.05          0.12          0.18  \begin{array}{llllllllllllllllllll} {\left[ R \right]\left( {molar} \right)\;\;\;1.00\;\;\;\;\;\;0.75\;\;\;\;\;0.40\;\;\;\;\;0.10}\\ {t\;\left( {min} \right)\;\;\;\;\;\;\;\;\;0.00\;\;\;\;\;0.05\;\;\;\;\;0.12\;\;\;\;\;0.18} \end{array}\;
The order of the reaction is
View answer and explanation
Correct option: A
Zero
Option A (Zero) is correct because the concentration decreases linearly with time, indicating a zero-order reaction.
11 Hard
11 of 207
Analyzing +4 | -1
For a homogeneous gaseous reaction A⟶3B\mathrm{A} \longrightarrow 3 \mathrm{B}, if pressure after time t\mathrm{t} was Pt\mathrm{P}_{\mathrm{t}} and after completion of reaction, pressure was P∞\mathrm{P}_{\infty} then select correct relation
View answer and explanation
Correct option: D
k=1tln⁡(2P∞3(P∞−Pt))\mathrm{k}=\frac{1}{\mathrm{t}} \ln \left(\frac{2 \mathrm{P}_{\infty}}{3\left(\mathrm{P}_{\infty}-\mathrm{P}_\mathrm{t}\right)}\right)
The correct answer is D (k=1tln⁡(2P∞3(P∞−Pt))\mathrm{k}=\frac{1}{\mathrm{t}} \ln \left(\frac{2 \mathrm{P}_{\infty}}{3\left(\mathrm{P}_{\infty}-\mathrm{P}_\mathrm{t}\right)}\right)) because it correctly relates the rate constant to the pressures at time t and at completion for a first-order reaction.
12 Medium
12 of 207
Analyzing +4 | -1
For a reaction of the type A+B→xA + B \to x products, it is observed that doubling the concentration of AA causes the reaction rate (k1)({k_1}) to be four times as great but doubling the amount of BB does not affect the rate (k2)({k_2}). The rate equation is
View answer and explanation
Correct option: B
k=k1k2k = {k_1}{k_2}
Option B (k=k1k2k = {k_1}{k_2}) is correct because the rate constant is affected by both A and B, and their effects are multiplicative.
13 Medium
13 of 207
Analyzing +4 | -1
A hypothetical reaction A2+B2→2AB{A_2} + {B_2} \to 2AB follows the mechanism as given below,

A2  ⇋  A+A(fast){A_2}\;\leftrightharpoons\;A + A\left( {fast} \right)

A+B2→AB+B(slow)A + {B_2} \to AB + B(slow)

A+B→AB(fast)A + B \to AB\left( {fast} \right)

The order of the overall reaction is
View answer and explanation
Correct option: C
1121\frac{1}{2}
Option C (1121\frac{1}{2}) is correct because the rate law is determined by the slow step and the equilibrium of the fast step.
14 Easy
14 of 207
Understanding +4 | -1
If aa is the initial concentration then time required to decompose half of the substance for nth{n^{\mathrm{th}}} order is inversely proportional to:
View answer and explanation
Correct option: C
a1−n{a^{1 - n}}
Option B (an−1{a^{n - 1}}) is correct because the half-life for an nth order reaction is inversely proportional to an−1{a^{n-1}}.
15 Hard
15 of 207
Understanding +4 | -1
Which of the following statement(s) is/are correct?

(i) Rate of reaction decreases with passage of time as the concentration of reactants decrease.

(ii) For a reaction
pP+qQ⟶rR+sS Rate =k[P]x[Q]y where x=p and y=q\begin{aligned} & {pP}+{qQ} \longrightarrow {rR}+{sS} \\ & \text { Rate }={k}[{P}]^{{x}}[{Q}]^{{y}} \text { where } {x}={p} \text { and } {y}={q} \end{aligned}

(iii) Rate law is the expression in which reaction rate is given in terms of molar concentration of reactants with each term raised to some power, which may or may not be same as the stichiometric coefficient of the reacting species in a balanced chemical equation.
View answer and explanation
Correct option: A
(i) and (iii)
Option A ((i) and (iii)) is correct because the rate of reaction decreases with time and the rate law is determined experimentally, not directly from stoichiometry.

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CHEMICAL BONDING & MOLECULAR STRUCTURE 595 MCQs THERMODYNAMICS 583 MCQs STRUCTURE OF ATOM 499 MCQs CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES 429 MCQs Equilibrium 412 MCQs STATES OF MATTER 410 MCQs
16 Medium
16 of 207
Application +4 | -1
The activation energy of exothermic reaction A→BA \to B is 80  kJ  mol−180\;kJ\;\mathrm{mo}{\mathrm{l}^{ - 1}}. The heat of reaction is 200  kJ    mol−1200\;kJ\;\;\mathrm{mo}{\mathrm{l}^{ - 1}}. The activation energy for the reaction B→AB \to A (in kJ    mol−1)kJ\;\;\mathrm{mo}{\mathrm{l}^{ - 1}}) will be
View answer and explanation
Correct option: D
280
Option D (280) is correct because the activation energy for the reverse reaction is the sum of the activation energy for the forward reaction and the heat of reaction.
17 Medium
17 of 207
Application +4 | -1
For a reaction A+B →A + B\, \to Products, the rate of the reaction was doubled when the concentration of AA was doubled. When the concentration of AA and BB were doubled, the rate was again doubled, the order of the reaction with respect to AA and BB are:
View answer and explanation
Correct option: C
1,01,0
Option C (1,0) is correct because doubling the concentration of A doubles the rate, indicating a first-order dependence on A, while changing B has no effect.
18 Medium
18 of 207
Remembering +4 | -1
A catalyst alters, which of the following in a chemical reaction?
View answer and explanation
Correct option: D
Activation energy
Option D is correct because a catalyst lowers the activation energy of a reaction, providing an alternative pathway with a lower energy barrier.
19 Medium
19 of 207
Analyzing +4 | -1
Consider the reaction,

Cl2(aq)+H2S(aq)→S(s)+2H+(aq)+2Cl−(aq)C{l_2}\left( {aq} \right) + {H_2}S\left( {aq} \right) \to S\left( s \right) + 2{H^ + }\left( {aq} \right) + 2C{l^ - }\left( {aq} \right) the rate equation for this reaction is , rate=k[Cl2][H2S]rate = k\left[ {C{l_2}} \right]\left[ {{H_2}S} \right]

Which of these mechanisms is/are consistent with this rate equation?

A.    Cl2+H2S(aq)→H+Cl−+Cl++HS−(slow)          Cl++HS−→H+Cl−+S(fast)B.    H2S↔H++HS−(fast  equilibrium)        Cl2+HS−→2Cl−+H+S(slow)\begin{array}{l} A.\;\;C{l_2} + {H_2}S\left( {aq} \right) \to {H^ + }C{l^ - } + C{l^ + } + H{S^ - }\left( {slow} \right)\\ \;\;\;\;\;C{l^ + } + H{S^ - } \to {H^ + }C{l^ - } + S\left( {fast} \right)\\ B.\;\;{H_2}S \leftrightarrow {H^ + } + H{S^ - }\left( {fast\;equilibrium} \right)\\ \;\;\;\;C{l_2} + H{S^ - } \to 2C{l^ - } + {H^ + }S\left( {slow} \right) \end{array}
View answer and explanation
Correct option: D
(A)Only
Option D (All of the above) is correct because all the statements about the rate constants and equilibrium constant are derived correctly from the given net rate equation.
20 Medium
20 of 207
Analyzing +4 | -1
The reaction obey I order with respect to H2{\mathrm{H}_2} and ICl\mathrm{ICl} both

H2(g)+2ICl(g)→2HCl(g)+I2(g){\mathrm{H}_2}\left( \mathrm{g} \right) + 2\mathrm{ICl}\left( \mathrm{g} \right) \to 2\mathrm{HCl}\left( \mathrm{g} \right) + {\mathrm{I}_2}\left( \mathrm{g} \right)

Which of the following mechanism is in consistent with the given fact?

MechanismA:H2(g)+2ICl→2HCl(g)+I2(g)A:{\mathrm{H}_2}\left( \mathrm{g} \right) + 2\mathrm{ICl} \to 2\mathrm{HCl}\left( \mathrm{g} \right) + {\mathrm{I}_2}\left( \mathrm{g} \right)

MechanismB:(i)H2(g)+ICl(g)→SlowHCl(g)+HI(g)B:\left( \mathrm{i} \right){\mathrm{H}_2}\left( \mathrm{g} \right) + \mathrm{ICl}\left( \mathrm{g} \right)\xrightarrow{{Slow}}\mathrm{HCl}\left( \mathrm{g} \right) + \mathrm{HI}\left( \mathrm{g} \right)

(ii)HI(g)+ICl(g)→HCl(g)+I2\left( {\mathrm{ii}} \right)\mathrm{HI}\left( \mathrm{g} \right) + \mathrm{ICl}\left( \mathrm{g} \right) \to \mathrm{HCl}\left( \mathrm{g} \right) + {\mathrm{I}_2}
View answer and explanation
Correct option: D
BB only
Option D (BB only) is correct because Mechanism B is consistent with the rate law, while Mechanism A is not.
21 Easy
21 of 207
Application +4 | -1
In the following reaction A→B+C,A \to B + C, rate constant is 0.001  Ms−1.0.001\;\mathrm{M}{\mathrm{s}^{ - 1}}\mathrm{.} Half-life and completion time of the given reaction are
View answer and explanation
Correct option: A
500 s, 1000 s
Option A (500 s, 1000 s) is correct because for a zero-order reaction, the half-life is a/2k and the completion time is a/k, where a is the initial concentration and k is the rate constant. Assuming initial concentration is 1M, half-life is 500s and completion time is 1000s.
22 Medium
22 of 207
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 the pseudo-first-order rate constant and the overall reaction rate, while the second-order rate constant remains unchanged.
23 Hard
23 of 207
Application +4 | -1
The velocity constant of a reaction at 290  K290\;K was found to be 3.2×10−3  at  300  K3.2 \times {10^{ - 3}}\;at\;300\;K, it will be
View answer and explanation
Correct option: A
6.4×10−36.4 \times {10^{ - 3}}
Option A (6.4×10−36.4 \times {10^{ - 3}}) is correct because the rate constant increases with temperature, and it can be estimated using the Arrhenius equation or a simplified approach.
24 Hard
24 of 207
Analyzing +4 | -1
The rate of the reaction 2 N2O5→4NO2+O22 \mathrm{~N}_2 \mathrm{O}_5 \rightarrow 4 \mathrm{NO}_2+\mathrm{O}_2 can be written in three ways:

−d[N2O5]dt=k[ N2O5];d[NO2]dt=k′[N2O5]d[O2]dt=k′′[N2O5]\begin{aligned} & \frac{-\mathrm{d}\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\mathrm{dt}}={k}\left[\mathrm{~N}_2 \mathrm{O}_5\right] ; \frac{\mathrm{d}\left[\mathrm{NO}_2\right]}{\mathrm{dt}}={k}^{\prime}\left[\mathrm{N}_2 \mathrm{O}_5\right] \\ & \frac{\mathrm{d}\left[\mathrm{O}_2\right]}{\mathrm{dt}}={k}^{\prime \prime}\left[\mathrm{N}_2 \mathrm{O}_5\right] \end{aligned}

The relationship between k{k} and k′{k}^{\prime} and between k{k} and k′′{k}^{\prime \prime} are:
View answer and explanation
Correct option: B
k′=2k;k′′=k/2k^{\prime}=2 k ; k^{\prime \prime}=k / 2
The correct answer is B (k′=2k;k′′=k/2k^{\prime}=2 k ; k^{\prime \prime}=k / 2) because the rate constants are related by the stoichiometry of the reaction.
25 Medium
25 of 207
Application +4 | -1
The activation energies of forward and backward reaction: A2+B2⇋2AB{A_2} + {B_2}\leftrightharpoons2AB are 180kJ mol−1\mathrm{mo}{\mathrm{l}^{ - 1}} and 200 kJ mol−1\mathrm{mo}{\mathrm{l}^{ - 1}} respectively. The presence of a catalyst lowers the activation energy of both (forward and backward) reactions by 100 kJ mol−1\mathrm{mo}{\mathrm{l}^{ - 1}}. The enthalpy change of the reaction in the presence of catalyst will be (in kJ mol−1\mathrm{mo}{\mathrm{l}^{ - 1}}):
View answer and explanation
Correct option: A
−20- 20
Option A (−20- 20) is correct because the enthalpy change is the difference between the activation energies of the forward and backward reactions, and the catalyst lowers both by the same amount, so the enthalpy change remains the same.
26 Easy
26 of 207
Remembering +4 | -1
Temperature dependent equation can be written as
View answer and explanation
Correct option: D
None of these
Option D is correct because the temperature-dependent equation, derived from the Arrhenius equation, is ln⁡k=ln⁡A−EaRT\ln k = \ln A - \frac{{E_a}}{{RT}}.
27 Easy
27 of 207
Remembering +4 | -1
Rate of reaction depends upon
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because the rate of a reaction is influenced by temperature, catalysts, and concentration.
28 Hard
28 of 207
Understanding +4 | -1
For the following SNS_N reaction,

(CH3)3CCl+H2O⟶(CH3)3COH+HCl\left(\mathrm{CH}_3\right)_3 \mathrm{CCl}+\mathrm{H}_2 \mathrm{O} \longrightarrow\left(\mathrm{CH}_3\right)_3 \mathrm{COH}+\mathrm{HCl}

Rate-determining step is
View answer and explanation
Correct option: A
(CH3)3CCl⟶(CH3)3C++Cl−\left(\mathrm{CH}_3\right)_3 \mathrm{CCl} \longrightarrow\left(\mathrm{CH}_3\right)_3 \mathrm{C}^{+}+\mathrm{Cl}^{-}
Option A ((CH3)3CCl⟶(CH3)3C++Cl−\mathrm{CH}_3)_3 \mathrm{CCl} \longrightarrow(\mathrm{CH}_3)_3 \mathrm{C}^{+}+\mathrm{Cl}^{-}) is correct because the rate-determining step in an SN1 reaction is the formation of the carbocation.
29 Easy
29 of 207
Remembering +4 | -1
Half-life period of second order reaction is
View answer and explanation
Correct option: C
Inversely proportional to initial concentration of reactants
Option C (Inversely proportional to initial concentration of reactants) is correct because the half-life of a second-order reaction is inversely proportional to the initial concentration.
30 Easy
30 of 207
Application +4 | -1
The activation energy of a reaction at a given temperature is found to be 2.303  RT  J  mol−12.303\;R\mathrm{T}\;\mathrm{ J}\;\mathrm{ mo}{\mathrm{l}^{ - 1}}. The ratio of rate constant to the Arrhenius factor is
View answer and explanation
Correct option: B
0.10.1
Option B (0.1) is correct because the ratio of the rate constant to the Arrhenius factor is equal to e−EaRTe^{-\frac{E_a}{RT}}, and with the given activation energy, this simplifies to e−2.303=0.1e^{-2.303} = 0.1.
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