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JEE • CHEMISTRY • CHEMICAL KINETICS

Chemical kinetics practice · Medium

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31 Medium
31 of 144
Analyzing +4 | -1
Assertion (A):The acid hydrolysis of ester takes place more rapidly in D2O\mathrm{D}_2 \mathrm{O} than H2O\mathrm{H}_2 \mathrm{O}

Reason (R): D3O+\mathrm{D}_3 \mathrm{O}^{+} is stronger acid than H3O+\mathrm{H}_3 \mathrm{O}^{+} or H2D+O\mathrm{H}_2 \mathrm{D}^{+} \mathrm{O}
View answer and explanation
Correct option: C
Both A and R are correct and R is correct explanation of A
Option C is correct because both the assertion and reason are correct, and the reason correctly explains the assertion. D3_3O+^+ is a stronger acid than H3_3O+^+, leading to faster acid hydrolysis in D2_2O.
32 Medium
32 of 144
Analyzing +4 | -1
A kinetic study of the reaction A→products\mathrm{A} \rightarrow \text{products} provides the data t=0 s,[A]=2.00 M;t=0 \mathrm{~s},[\mathrm{A}]=2.00 \mathrm{~M} ; 500 s,500 \mathrm{~s}, 1.00 M;1.00 \text{ M}; 1500 s,1500 \mathrm{~s}, 0.50M;0.50 \mathrm{M} ; 3500 s,3500 \mathrm{~s}, 0.25 M.0.25 \mathrm{~M}. In the simplest possible way determine whether this reaction is of
View answer and explanation
Correct option: C
Second-order
Option C (Second-order) is correct. The half-life doubles with each successive interval, indicating a second-order reaction.
33 Medium
33 of 144
Analyzing +4 | -1
Assertion: Two different reaction can never have same rate of reaction

Reason: Rate of reaction always depends only on frequency of collision and Arrhenius factor
View answer and explanation
Correct option: D
If both Assertion & Reason are false
Option D (If both Assertion & Reason are false) is correct because two different reactions can have the same rate, and the rate of reaction depends on more than just collision frequency and the Arrhenius factor.
34 Medium
34 of 144
Application +4 | -1
We are given with following information about concentration of reactant with initial rate of reaction.

Initial concentrationInitial rate0.005 M7.5×10−40.02 M3×10−3\begin{array}{cccccccccccccccccccc} {{\text{Initial concentration}}}&{{\text{Initial rate}}} \\ {0.005{\text{ M}}}&{7.5 \times {{10}^{ - 4}}} \\ {0.02{\text{ M}}}&{3 \times {{10}^{ - 3}}} \end{array}

Find out order of reaction with respect to that reactant.
View answer and explanation
Correct option: A
1
The correct answer is 1 because the order of the reaction is determined by comparing the change in rate with the change in concentration.
35 Medium
35 of 144
Application +4 | -1
T50\mathrm{T}_{50} of a first order reaction is 10 min10 \mathrm{~min}. Starting with 10 mol L−110 \mathrm{~mol} \mathrm{~L}^{-1}, rate after 20 min20 \mathrm{~min} is
View answer and explanation
Correct option: B
0.0693×2.5 mol L−1 min−10.0693 \times 2.5 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}
Option B (0.0693×2.5 mol L−1 min−10.0693 \times 2.5 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}) is correct because it calculates the rate of the first-order reaction after 20 minutes (two half-lives) using the rate constant and the concentration of the reactant at that time.
36 Medium
36 of 144
Analyzing +4 | -1
Select the incorrect statement among the following
View answer and explanation
Correct option: C
For an endothermic reaction Activation energy is always less than heat of reaction
Option C is correct because for an endothermic reaction, the activation energy is always *greater* than the heat of reaction (enthalpy change).
37 Medium
37 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.
38 Medium
38 of 144
Understanding +4 | -1
What is the role of a catalyst in a catalysed
reaction?
View answer and explanation
Correct option: A
Lowers the activation energy
Option A (Lowers the activation energy) is correct because a catalyst provides an alternative reaction pathway with a lower activation energy, thus increasing the reaction rate.
39 Medium
39 of 144
Application +4 | -1
The rate of first order reaction are 0.8 M min−10.8 \mathrm{~M} \mathrm{~min}^{-1} and 0.08 M min−10.08\mathrm{~M} \mathrm{~min}^{-1} at 40 min and 60 min respectively. Then half life is:
View answer and explanation
Correct option: C
6 min
Option A (43 g) is correct because it represents the lowest possible molecular mass calculated from the empirical formula.
40 Medium
40 of 144
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.
41 Medium
41 of 144
Analyzing +4 | -1
STATEMENT-1 : If the activation energy of reaction is zero, temperature will have no effect on the rate constant.

STATEMENT-2 : Lower the activation energy fasten is the reaction.
View answer and explanation
Correct option: B
If both the statements are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT-1
Option B (If both the statements are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT-1) is correct because both statements are true, but the second statement doesn't explain the first.
42 Medium
42 of 144
Application +4 | -1
If the rate constant of the reaction, X→YX \rightarrow Y is 2.5×10−4 mol L−1 s−12.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} and initial concentration of XX is 3 M3 \mathrm{~M} then the concentration of Y{Y} after 100 minutes will be
View answer and explanation
Correct option: C
1.5 M1.5 \mathrm{~M}
Option C (1.5 M1.5 \mathrm{~M}) is correct because it's calculated by multiplying the rate constant by the time to find the concentration of Y formed in a zero-order reaction.
43 Medium
43 of 144
Application +4 | -1
After 20%20 \% completion, the rate of reaction A→products\mathrm{A} \rightarrow \text{products} is 10 unit and after 80%80 \% completion, the rate is 0.625 unit. The order of the reaction is
View answer and explanation
Correct option: C
Second
Option C (Second) is correct because the rate is proportional to the square of the reactant concentration, indicating a second-order reaction.
44 Medium
44 of 144
Analyzing +4 | -1
Statement I: Rate constants determined from Arrhenius equation are fairly accurate for simple as well as complex molecules

Statement II: Reactant molecules undergo chemical change irrespective of their orientation during collision
View answer and explanation
Correct option: C
If statement I is true but statement II is false
Option C is correct because Statement I is false, as Arrhenius equation is not accurate for complex molecules, and Statement II is false, as reactant molecules' orientation matters during collisions.
45 Medium
45 of 144
Application +4 | -1
Following reaction takes places by mechanism

NO2+CO⟶NO+CO2\mathrm{NO}_2+\mathrm{CO} \longrightarrow \mathrm{NO}+\mathrm{CO}_2
Step I NO2+NO2→k1. Slow NO3+NO\mathrm{NO}_2+\mathrm{NO}_2 \xrightarrow{k_1 \text {. Slow }} \mathrm{NO}_3+\mathrm{NO}

Step II NO3+CO→k2. Fast NO2+CO2\mathrm{NO}_3+\mathrm{CO} \xrightarrow{k_2 \text {. Fast }} \mathrm{NO}_2+\mathrm{CO}_2

Hence, (dxdt)\left(\frac{d x}{d t}\right) for the given reaction is
View answer and explanation
Correct option: C
k1[NO2]2k_1\left[\mathrm{NO}_2\right]^2
Option C (k1[NO2]2k_1[NO_2]^2) is correct because the rate of the reaction is determined by the slow step, which involves the reaction of two NO2NO_2 molecules.
46 Medium
46 of 144
Application +4 | -1
The half-life for decay of radioactive 14C{ }^{14} {C} is 5730 years. An archaeological artefact containing wood has only 80%80 \% of the 14C{ }^{14} {C} activity as found in living trees. The age of the artefact would be [Given : log⁡1.25=0.0969\log 1.25=0.0969]
View answer and explanation
Correct option: A
1845 years
The correct answer is A (1845 years) because it is calculated using the first-order kinetics equation for radioactive decay, considering the half-life of 14C^{14}C and the remaining activity in the artefact.
47 Medium
47 of 144
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 is correct because the rate is proportional to [A]2 and independent of [B], so the rate law is k[A]2, which can be represented as k = k1*k2.
48 Medium
48 of 144
Understanding +4 | -1
The inversion of cane sugar into glucose and fructose according to the equation C12H22O11+H2O→C6H12O6+C6H12O6{\mathrm{C}_{12}}{\mathrm{H}_{22}}{\mathrm{O}_{11}} + {\mathrm{H}_2}\mathrm{O} \to {\mathrm{C}_6}{\mathrm{H}_{12}}{\mathrm{O}_6} + {\mathrm{C}_6}{\mathrm{H}_{12}}{\mathrm{O}_6} is an example of
View answer and explanation
Correct option: C
Pseudo unimolecular
Option C (Pseudo unimolecular) is correct because the reaction appears to be first order with respect to cane sugar, as the concentration of water is in large excess and remains nearly constant.
49 Medium
49 of 144
Application +4 | -1
The half-life period of a first order reaction is 1 min 40 s. Calculate its rate constant.
View answer and explanation
Correct option: B
6.93×10−3  s−16.93 \times {10^{ - 3}}\;{s^{ - 1}}
Option B (6.93×10−3  s−16.93 \times {10^{ - 3}}\;{s^{ - 1}}) is correct because it is the rate constant calculated using the half-life formula for a first-order reaction, with time in seconds.
50 Medium
50 of 144
Analyzing +4 | -1
Rate of a reaction A+B⟶C\mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C} is given below as a function of differential concentrations of A\mathrm{A} and B\mathrm{B}.

Exp.[A][B]Ratemol L−1mol L−1mol L−1 time−11.0.010.010.0052.0.020.010.0103.0.010.020.005\begin{array}{cccccccccccccccccccc} {{\text{Exp}}{\text{.}}}&{\left[ {\text{A}} \right]}&{\left[ {\text{B}} \right]}&{{\text{Rate}}} \\ {}&{{\text{mol }}{{\text{L}}^{ - 1}}}&{{\text{mol }}{{\text{L}}^{ - 1}}}&{{\text{mol }}{{\text{L}}^{ - 1}}\,{\text{tim}}{{\text{e}}^{ - 1}}} \\ {1.}&{0.01}&{0.01}&{0.005} \\ {2.}&{0.02}&{0.01}&{0.010} \\ {3.}&{0.01}&{0.02}&{0.005} \end{array}

Determine the order of reaction with respect to A\mathrm{A} and and also calculate the half-life of A\mathrm{A} in the reaction.
View answer and explanation
Correct option: A
1.3861.386
The correct answer is A (1.3861.386) because it is the calculated half-life of A based on the determined rate law and rate constant from the experimental data.
51 Medium
51 of 144
Analyzing +4 | -1
Rate constant (K)(\mathrm{K}) varies with temperature as given by equation log⁡K(min−1)=5−2000T\log \mathrm{K}\left(\mathrm{min}^{-1}\right)=5-\frac{2000}{\mathrm{T}}
Consider the following about this equation
(I) Pre exponential factor is 10510^5.
(II) Eα\mathrm{E}_\alpha is 9.212 kcal9.212 \mathrm{~kcal}.
(III) Variation of log⁡K\log \mathrm{K} with 1/T1 / \mathrm{T} is linear.
Select the correct statement.
View answer and explanation
Correct option: A
I, II and III
Option A (I, II and III) is correct because all three statements are true based on the Arrhenius equation and its logarithmic form.
52 Medium
52 of 144
Application +4 | -1
If the initial concentration of reactants are doubled half life becomes half. Then order of the reaction will be
View answer and explanation
Correct option: B
2
Option B (2) is correct because the half-life is inversely proportional to the initial concentration raised to the power of (n-1), where n is the order of the reaction.
53 Medium
53 of 144
Application +4 | -1
For the elementary reaction 2NO+O2→2NO2,2 \mathrm{NO}+\mathrm{O}_2 \rightarrow 2 \mathrm{NO}_2, the volume of a container is tripled. The rate becomes
View answer and explanation
Correct option: C
127\frac{1}{27} times
Option C (127\frac{1}{27} times) is correct because the rate of the elementary reaction is inversely proportional to the cube of the volume, given the rate law r = k[NO]²[O₂].
54 Medium
54 of 144
Understanding +4 | -1
The rate for the reaction, RCl+NaOH(aq)→ROH+NaClRCl + NaOH\left( {aq} \right) \to ROH + NaCl is given by rate=K1[RCl].rate\mathrm{ } = K_1\left[ {RCl} \right]. The rate of the reaction is
View answer and explanation
Correct option: C
Halved on reducing the concentration of RClRCl to half
The correct answer is C (Halved on reducing the concentration of RClRCl to half) because the rate law is first order with respect to RClRCl, so halving its concentration will halve the rate.
55 Medium
55 of 144
Understanding +4 | -1
When two reactants AA and BB are mixed to give products CC and DD, the reaction quotient QQ at the initial stage of the reaction is best described by-
View answer and explanation
Correct option: B
Increase with time
Option D (Decrease the degree of ionisation of group reagent) is correct because adding NH4ClNH_4Cl (containing NH4+NH_4^+) decreases the ionization of NH4OHNH_4OH (a weak base) due to the common ion effect, thus controlling the hydroxide ion concentration.
56 Medium
56 of 144
Understanding +4 | -1
Assertion: Hydrolysis of methyl ethanoate is a pseudo first order reaction.

Reason: Water is present in large excess and therefore its concentration remained constant throughout the reaction.
View answer and explanation
Correct option: A
Both assertion and reason are correct statements, and reason is the correct explanation of the assertion.
Option A is correct because both the assertion and reason are true, and the reason accurately explains why the hydrolysis of methyl ethanoate is a pseudo-first-order reaction.
57 Medium
57 of 144
Application +4 | -1
If half life of radioactive bromine ( Br−82)\mathrm{Br}-82) is 36hr36 \mathrm{hr}, find percentage remaining after one day. [nearest integer]
View answer and explanation
Correct option: D
63
Option D (63) is correct because after one day (24 hours), approximately 63% of the radioactive bromine remains, calculated using the half-life formula.
58 Medium
58 of 144
Understanding +4 | -1
What is not correct for a zero order reaction
View answer and explanation
Correct option: D
The graph of concentration of reactant against time is a straight line with positive slope
Option D (The graph of concentration of reactant against time is a straight line with positive slope) is correct because for a zero-order reaction, the graph of concentration vs. time is a straight line with a *negative* slope.
59 Medium
59 of 144
Application +4 | -1
A first order reaction has a specific reaction rate of 10−1 s−110^{-1} \mathrm{~s}^{-1}. How much time will it take for 100 g100 \mathrm{~g} of its reactant to be reduced to 25 g25 \mathrm{~g}?
View answer and explanation
Correct option: A
14 sec
Option A (14 sec) is correct because it is calculated using the integrated rate law for a first-order reaction.
60 Medium
60 of 144
Understanding +4 | -1
For the reaction,

H2(g)+Br2(g)→2HBr(g){\mathrm{H}_2}\left( \mathrm{g} \right) + \mathrm{B}{\mathrm{r}_2}\left( \mathrm{g} \right) \to 2\mathrm{HBr}\left( \mathrm{g} \right)

The experimental data suggest

rate=k[H2][Br2]1/2\mathrm{rate} = k\left[ {{\mathrm{H}_2}} \right]{\left[ {\mathrm{B}{\mathrm{r}_2}} \right]^{1/2}}

the molecularity and order of the reaction are respectively
View answer and explanation
Correct option: D
2,322,\frac{3}{2}
Option D (2,322,\frac{3}{2}) is correct because molecularity is the number of reacting species in an elementary reaction (2 in this case), and the order of the reaction is the sum of the exponents in the rate law (1+12=321 + \frac{1}{2} = \frac{3}{2}).
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