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JEE • CHEMISTRY • CHEMICAL KINETICS

Chemical kinetics practice · Mixed difficulty

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241 Easy
241 of 301
Understanding +4 | -1
The integrated rate equations can be determined for
View answer and explanation
Correct option: D
Both (1) and (2)
Option D (Both (1) and (2)) is correct because integrated rate equations can be derived for both zero-order and first-order reactions.
242 Hard
242 of 301
Understanding +4 | -1
For the first-order reaction with C0C_0 as the initial concentration and CC at time t,(C0−C)=t,\left(C_0-C\right)=
View answer and explanation
Correct option: A
C0[1−e−kt]C_0\left[1-e^{-k t}\right]
Option A is correct because for a first-order reaction, the concentration at time t, C, is related to the initial concentration, C_0, by the equation C=C0e−ktC = C_0e^{-kt}, thus (C0−C)=C0(1−e−kt)(C_0 - C) = C_0(1 - e^{-kt}).
243 Medium
243 of 301
Application +4 | -1
When concentration of the reactant AA in the reaction A→BA \to B increases to 8 times, then rate increases to 2 times only. The order of reaction is
View answer and explanation
Correct option: B
13\frac{1}{3}
Option B (1/3) is correct because the rate is proportional to the concentration raised to the power of the order of the reaction, and solving for the order gives 1/3.
244 Hard
244 of 301
Application +4 | -1
The half-life for the first-order decomposition of nitramide NH4NO2(aq)→N2O(g)+H2O(l)\mathrm{NH}_4 \mathrm{NO}_2(\mathrm{aq}) \rightarrow \mathrm{N}_2 \mathrm{O}(\mathrm{g})+\mathrm{H}_2 \mathrm{O}({l}) is 123 minutes at 27∘C27^{\circ} \mathrm{C}. If 200 ml200 \mathrm{~ml} of 0.02 M0.02 \mathrm{~M} NH4NO2\mathrm{NH}_4 \mathrm{NO}_2 solution is allowed to decompose, then how long must the reaction proceed to produce 49.26 ml49.26 \mathrm{~ml} of 'wet' N2O(g)\mathrm{N}_2 \mathrm{O}(\mathrm{g}) measured at 27∘C27^{\circ} \mathrm{C} and 785 mm Hg?785 \mathrm{~mm} \mathrm{~Hg}? The vapour pressure of water at 27∘C27^{\circ} \mathrm{C} is 25 mm Hg25 \mathrm{~mm} \mathrm{~Hg}.
View answer and explanation
Correct option: A
123 min123 \mathrm{~min}
Option A (123 min) is correct because the reaction proceeds for one half-life to produce the given amount of N₂O.
245 Easy
245 of 301
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 rate of change of concentration with respect to time is constant, indicating a zero-order reaction.
246 Medium
246 of 301
Application +4 | -1
The energy of activation of reaction, if its rate double when the temperature is raised from 290 K290 \mathrm{~K} to 300 K300 \mathrm{~K} is x\mathrm{x} kcal then the value of x/3\mathrm{x} / 3 is (Take R=2cal/k/mol\mathrm{R}=2 \mathrm{cal} / \mathrm{k} / \mathrm{mol} )
View answer and explanation
Correct option: C
12
Option C (12) is correct because the activation energy is calculated using the Arrhenius equation.
247 Easy
247 of 301
Understanding +4 | -1
For the reaction 2A+B→3 C+D2A + B \to 3\,C + D, which of the following does not express the reaction rate
View answer and explanation
Correct option: D
−13d[C]dt- \frac{1}{3}\frac{{d\left[ C \right]}}{{dt}}
Option D (−13d[C]dt- \frac{1}{3}\frac{{d\left[ C \right]}}{{dt}}) is the correct answer because the rate expression should be +13d[C]dt+\frac{1}{3}\frac{d[C]}{dt} since C is a product.
248 Hard
248 of 301
Analyzing +4 | -1
A reaction was observed for 15 days and the percentage of the reactant remaining after the days indicated was recorded in the following table.

\;\;\begin{array}{llllllllllllllllllll} {\mathbf{Time}\mathrm{ }\left( {\mathbf{days}} \right)\;\;\;\;\;\% \mathrm{ }\mathbf{Reactant}\mathrm{ }\mathbf{remaining}}\\ {\;\;\;\;\;\;0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;100}\\ {\;\;\;\;\;2\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;50}\\ {\;\;\;\;\;4\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;39}\\ {\;\;\;\;\;6\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;25}\\ {\;\;\;\;\;8\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;21}\\ {\;\;\;\;10\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;18}\\ {\;\;\;\;12\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;15}\\ {\;\;\;\;14\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;12.5}\\ {\;\;\;\;15\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;10} \end{array}\;\;\begin{array}{llllllllllllllllllll} {\mathbf{Time}\mathrm{ }\left( {\mathbf{days}} \right)\;\;\;\;\;\% \mathrm{ }\mathbf{Reactant}\mathrm{ }\mathbf{remaining}}\\ {\;\;\;\;\;\;0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;100}\\ {\;\;\;\;\;2\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;50}\\ {\;\;\;\;\;4\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;39}\\ {\;\;\;\;\;6\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;25}\\ {\;\;\;\;\;8\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;21}\\ {\;\;\;\;10\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;18}\\ {\;\;\;\;12\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;15}\\ {\;\;\;\;14\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;12.5}\\ {\;\;\;\;15\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;10} \end{array}
Which one of following best describes the order and the half-life of the reaction?

Reaction order Half-life (days)
View answer and explanation
Correct option: C
Second 2
Option C (Second, 2) is correct. The half-life doubles as the initial concentration halves, indicating a second-order reaction. The half-life is calculated to be 2 days.
249 Hard
249 of 301
Application +4 | -1
For the consecutive first-order reactions A→K1B→K2C,B\mathrm{A} \xrightarrow{K_{\mathrm{1}}}\mathrm{B} \xrightarrow{K_2} \mathrm{C}, \mathrm{B} is in steady state. What will be the concentration of C\mathrm{C} at this stage? At t=0,[A]=[A]0t=0,[\mathrm{A}]=[\mathrm{A}]_0 and [B]=[C]=0[\mathrm{B}]=[\mathrm{C}]=0.
View answer and explanation
Correct option: C
[A]0⋅(1−e−K1t)[\mathrm{A}]_0 \cdot\left(1-e^{-K_1 t}\right)
Option C ([A]0⋅(1−e−K1t)[\mathrm{A}]_0 \cdot\left(1-e^{-K_1 t}\right)) is correct because, under steady-state approximation, the concentration of C is determined by the decay of A with rate constant K1.
250 Easy
250 of 301
Analyzing +4 | -1
Assertion (A):The instantaneous rate of reaction can be studied in chemical kinetics.

Reason (R): Ionic reactions occur instantaneously.
View answer and explanation
Correct option: B
A is wrong but R is correct
Option B is correct because ionic reactions are indeed very fast and occur almost instantaneously, but they are generally not studied using standard chemical kinetics methods due to their speed.
251 Easy
251 of 301
Application +4 | -1
For the reaction,

2N2O5(g)→4NO2(g)+O2(g)2{N_2}{O_5}\left( g \right) \to 4N{O_2}\left( g \right) + {O_2}\left( g \right)

If the concentration of NO2N{O_2} increase by 5.2×10−3  M5.2 \times {10^{ - 3}}\;M in 100 s then the rate of the reactions
View answer and explanation
Correct option: A
1.3×10−5  M  s−11.3 \times {10^{ - 5}}\;M\;{s^{ - 1}}
Option A (1.3×10−5  M  s−11.3 \times {10^{ - 5}}\;M\;{s^{ - 1}}) is correct because the rate of reaction is calculated based on the rate of change of NO2 concentration.
252 Medium
252 of 301
Application +4 | -1
The unit of rate constant obeying the rate expression r=k[A]1[B]2/3r = k{\left[ A \right]^1}{\left[ B \right]^{2/3}} is
View answer and explanation
Correct option: A
mole−2/3lit2/3time−1{\text{mol}}{{\text{e}}^{ - 2/3}}{\text{li}}{{\text{t}}^{2/3}}{\text{tim}}{{\text{e}}^{ - 1}}
Option A (mole−2/3lit2/3time−1{\text{mol}}{{\text{e}}^{ - 2/3}}{\text{li}}{{\text{t}}^{2/3}}{\text{tim}}{{\text{e}}^{ - 1}}) is correct because it is derived from the rate law expression and the overall order of the reaction.
253 Easy
253 of 301
Understanding +4 | -1
Which of the following expression is correct for second order reaction. (C0{C_0} refers to initial concentration of reactant)?
View answer and explanation
Correct option: C
t1/2∝C0−1{t_{1/2}} \propto C_0^{ - 1}
Option C (t1/2∝C0−1{t_{1/2}} \propto C_0^{ - 1}) is correct because for a second-order reaction, the half-life is inversely proportional to the initial concentration.
254 Medium
254 of 301
Analyzing +4 | -1
Assertion (A):Larger is the activation energy, lesser is the effect of a given temperature rise on rate constant K.K.

Reason (R): K=Ae−Ea/RTK=A e^{-E_a / R T}
View answer and explanation
Correct option: B
A is wrong but R is correct
Option B is correct because the Arrhenius equation shows that a larger activation energy results in a greater change in the rate constant for a given temperature change.
255 Hard
255 of 301
Analyzing +4 | -1
For the consecutive first-order reactions: A→K1B→K2C,{A} \xrightarrow{K_1} {B} \xrightarrow{K_2} {C},
in what condition, [C]=[A0].(1−e−K1t)?[{C}]=\left[{A}_0\right] .\left(1-e^{-K_1 t}\right)?
View answer and explanation
Correct option: B
K1≪K2K_1 \ll K_2
Option B (K1≪K2K_1 \ll K_2) is correct because when K1K_1 is much smaller than K2K_2, the concentration of C simplifies to the given expression.
256 Easy
256 of 301
Application +4 | -1
The rate of a chemical reaction doubled for every 10∘C10^\circ C rise in temperature. If the temperature is increased by 60∘C60^\circ C the rate of reaction increase by
View answer and explanation
Correct option: C
64 times
Option C (64 times) is correct because the rate doubles for every 10°C increase, and with a 60°C increase, the rate increases by 2 raised to the power of 6.
257 Easy
257 of 301
Understanding +4 | -1
Activation energy of a chemical reaction can be determined by
View answer and explanation
Correct option: B
Determining the rate constant at two temperatures
Option B (Determining the rate constant at two temperatures) is correct because the Arrhenius equation relates activation energy to the rate constant at different temperatures.
258 Medium
258 of 301
Application +4 | -1
The rate of reaction 2X+Y→2 \mathrm{X}+\mathrm{Y} \rightarrow Products is given by d[Y]dt=k[X]2[Y]\frac{\mathrm{d}[\mathrm{Y}]}{\mathrm{d} t}=k[\mathrm{X}]^2[\mathrm{Y}], if X\mathrm{X} is present in large excess, the order of the reaction is
View answer and explanation
Correct option: C
One
Option C (One) is correct because when X is in large excess, its concentration remains nearly constant, making the reaction's rate dependent only on the concentration of Y, thus first order.
259 Easy
259 of 301
Analyzing +4 | -1
With the help of following reactions, arrange metals A, B, D and E in decreasing order of their reactivity ?

(a) B+ANO3→BNO3+AB +{ANO}_3 \rightarrow {BNO}_3+ A

(b) A+HCl→ACl+12H2A +{HCl} \rightarrow {ACl}+\frac{1}{2} {H}_2

(c) D+ECl→DCl+ED +{ECl} \rightarrow {DCl}+ E

(d) D+HNO3→D +{HNO}_3 \rightarrow No reaction
View answer and explanation
Correct option: B
B >> A >> D >> E
Option B (B > A > D > E) is correct because it correctly orders the metals based on their reactivity as determined by the given reactions.
260 Medium
260 of 301
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.
261 Medium
261 of 301
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.
262 Medium
262 of 301
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.
263 Hard
263 of 301
Application +4 | -1
Rate constant of the following reaction is 1.0×10−5 s−11.0 \times 10^{-5} \mathrm{~s}^{-1}.

2 N2O5⟶2 N2O4+O22 \mathrm{~N}_2 \mathrm{O}_5 \longrightarrow 2 \mathrm{~N}_2 \mathrm{O}_4+\mathrm{O}_2

At a time when [N2O5]=0.2 mol L−1\left[\mathrm{N}_2 \mathrm{O}_5\right]=0.2 \mathrm{~mol} \mathrm{~L}^{-1}, average rate of the reaction
View answer and explanation
Correct option: B
1.0×10−6 M L−1 s−11.0 \times 10^{-6} \mathrm{~M} \mathrm{~L}^{-1} \mathrm{~s}^{-1}
Option B (1.0×10−6 M L−1 s−11.0 \times 10^{-6} \text{ M L}^{-1} \text{ s}^{-1}) is correct because the average rate of the reaction is half the rate of disappearance of N2O5, given it's a first-order reaction.
264 Medium
264 of 301
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.
265 Hard
265 of 301
Analyzing +4 | -1
The reaction of hydrogen and iodine monochloride is given as:

H2( g)+2ICl(g)⟶2HCl(g)+I2( g){H}_2({~g})+2 {ICl}({g}) \longrightarrow 2 {HCl}({g})+{I}_2({~g})

The reaction is of first order with respect to H2( g){H}_2({~g}) and ICI(g){ICI}({g}), following mechanisms were proposed.

Mechanism A:

H2( g)+2ICl(g)⟶2HCl(g)+I2( g){H}_2({~g})+2 {ICl}({g}) \longrightarrow 2 {HCl}({g})+{I}_2({~g})

Mechanism B:

H2( g)+ICl(g)⟶HI(g); slow HI(g)+ICl(g)⟶HCl(g)+I2( g); fast \begin{aligned} & {H}_2({~g})+{ICl}({g}) \longrightarrow {HI}({g}) ; \text { slow } \\ & {HI}({g})+{ICl}({g}) \longrightarrow {HCl}({g})+{I}_2({~g}) ; \text { fast } \end{aligned}

Which of the above mechanism(s) can be consistent with the given information about the reaction?
View answer and explanation
Correct option: D
B only
Option D (B only) is correct because mechanism B's slow step matches the given rate law, while mechanism A is a single-step reaction that doesn't align with the rate law.
266 Easy
266 of 301
Application +4 | -1
In a reaction A+B→CA + B \to C, the rate expression is R=k[A][B]2R = k\left[ A \right]{\left[ B \right]^2} If the concentration of both the reaction is doubled at constant volume then the rate of the reaction will be
View answer and explanation
Correct option: A
Eight time
Option A (Eight time) is correct because the rate is proportional to [A][B]2, so doubling both concentrations increases the rate by a factor of 2 * 22 = 8.
267 Medium
267 of 301
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.
268 Medium
268 of 301
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.
269 Easy
269 of 301
Understanding +4 | -1
Which of the following statements is not correct?
View answer and explanation
Correct option: C
Catalyst lowers the energy of activation of the forward direction without affecting the energy of activation of the backward direction
Option C (Catalyst lowers the energy of activation of the forward direction without affecting the energy of activation of the backward direction) is correct because a catalyst lowers the activation energy for both forward and reverse reactions equally.
270 Medium
270 of 301
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.
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