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JEE • CHEMISTRY • CHEMICAL KINETICS

Chemical kinetics practice · Mixed difficulty

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91 Medium
91 of 301
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.
92 Easy
92 of 301
Application +4 | -1
The t12{t_{\frac{1}{2}}} of a reaction is halved as the initial concentration of the reactant is doubled. What is order of reaction?
View answer and explanation
Correct option: C
2
Option C (2) is correct because the half-life of a second-order reaction is inversely proportional to the initial concentration of the reactant.
93 Easy
93 of 301
Understanding +4 | -1
For the elementary step,

(CH3)3⋅CBr(aq)→(CH3)3C+(aq)+Br−(aq){\left( {\mathrm{C}{\mathrm{H}_3}} \right)_3} \cdot \mathrm{CBr}\left( {aq} \right) \to {\left( {\mathrm{C}{\mathrm{H}_3}} \right)_3}{\mathrm{C}^ + }\left( {aq} \right) + \mathrm{B}{\mathrm{r}^ - }\left( {aq} \right) the molecularity is:
View answer and explanation
Correct option: B
1
Option B (1) is correct because molecularity is the number of reactant molecules involved in an elementary step, which is one in this case.
94 Medium
94 of 301
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.
95 Easy
95 of 301
Understanding +4 | -1
In the reaction A+B→productsA + B \to products, if B is taken in excess, then it is an example of
View answer and explanation
Correct option: C
Pseudo unimolecular reaction
The correct answer is C (Pseudo unimolecular reaction) because when one reactant is in excess, the reaction order appears to be determined only by the other reactant.
96 Easy
96 of 301
Understanding +4 | -1
Which of the following statement is incorrect about the molecularity of a reaction?
View answer and explanation
Correct option: D
There is no difference between order and molecularity of a reaction
Option D (There is no difference between order and molecularity of a reaction) is correct because order and molecularity are different concepts.
97 Easy
97 of 301
Remembering +4 | -1
The unit of rate constant for a zero order reaction is
View answer and explanation
Correct option: A
mol.L−1.S−1mol.{L^{ - 1}}.{S^{ - 1}}
Option A (mol.L-1.S-1) is correct because the rate constant for a zero-order reaction has units of concentration per time.
98 Medium
98 of 301
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.
99 Medium
99 of 301
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.
100 Hard
100 of 301
Analyzing +4 | -1
For the reaction 2NH3→N2+3H22 \mathrm{NH}_3 \rightarrow \mathrm{N}_2+3 \mathrm{H}_2,

−d[NH3]dt=k1[NH3],−d[N2]dt=k2[NH3],d[H2]dt=k3[NH3]-\frac{\mathrm{d}\left[\mathrm{NH}_3\right]}{\mathrm{d} t}=k_1\left[\mathrm{NH}_3\right],-\frac{\mathrm{d}\left[\mathrm{N}_2\right]}{\mathrm{d} t}=k_2\left[\mathrm{NH}_3\right], \frac{\mathrm{d}\left[\mathrm{H}_2\right]}{\mathrm{d} t}=k_3\left[\mathrm{NH}_3\right]

Then relation between k1,k2k_1, k_2 and k3k_3 is
View answer and explanation
Correct option: A
1.5k1=3k2=k31.5 k_1=3 k_2=k_3
Option A (1.5k1=3k2=k31.5 k_1=3 k_2=k_3) is correct because it correctly relates the rate constants based on the stoichiometry of the reaction.
101 Medium
101 of 301
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.
102 Medium
102 of 301
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.
103 Easy
103 of 301
Application +4 | -1
For A+B→C+D,  ΔH=−20  kJ  mol−1A + B \to C + D,\;\Delta H = - 20\;\mathrm{kJ}\;\mathrm{mo}{\mathrm{l}^{ - 1}} the activation energy of the forward reaction is 85  kJ  mol−1.85\;\mathrm{kJ}\;\mathrm{mo}{\mathrm{l}^{ - 1}}. The activation energy for backward reaction is …kJ  mol−1\mathrm{kJ}\;\mathrm{mo}{\mathrm{l}^{ - 1}}.
View answer and explanation
Correct option: B
105
Option B (105) is correct because the activation energy for the backward reaction is the sum of the activation energy for the forward reaction and the enthalpy change.
104 Hard
104 of 301
Application +4 | -1
After how many second will the concentration of the reactant in a first order reaction be halved if the rate constant is 1.155×10−3  s−1?1.155 \times {10^{ - 3}}\;{s^{ - 1}}?
View answer and explanation
Correct option: A
600
Option A (600) is correct because the half-life of a first-order reaction is calculated using the formula t1/2=0.693kt_{1/2} = \frac{{0.693}}{k}, and substituting the given rate constant yields approximately 600 seconds.
105 Easy
105 of 301
Analyzing +4 | -1
The rate constant K1{K_1} of a reaction is found to be double that of rate constant K2{K_2} of another reaction. The relationship between corresponding activation energies of the two reactions at same temperature (E1  and  E2)({E_1}\;and\;{E_2}) can be represented as:
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because without knowing the pre-exponential factors (A1 and A2), we cannot definitively relate E1 and E2 even with the given rate constant relationship.
106 Medium
106 of 301
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.
107 Medium
107 of 301
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.
108 Medium
108 of 301
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.
109 Medium
109 of 301
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.
110 Medium
110 of 301
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.
111 Easy
111 of 301
Application +4 | -1
Calculate the overall order of a reaction which has the rate expression

(a) Rate =k[ A]1/2[ B]3/2=k[\mathrm{~A}]^{1 / 2}[{~B}]^{3 / 2}

(b) Rate =k[ A]3/2[ B]−1=k[\mathrm{~A}]^{3 / 2}[{~B}]^{-1}
View answer and explanation
Correct option: A
2, 1/2
Option A (2, 1/2) is correct because the overall order of a reaction is the sum of the exponents in the rate expression.
112 Medium
112 of 301
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].
113 Medium
113 of 301
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.
114 Easy
114 of 301
Understanding +4 | -1
The given reaction,

2FeCl3+SnCl2→SnCl4+2FeCl22\mathrm{FeC}{\mathrm{l}_3} + \mathrm{SnC}{\mathrm{l}_2} \to \mathrm{SnC}{\mathrm{l}_4} + 2\mathrm{FeC}{\mathrm{l}_2} is an example of:
View answer and explanation
Correct option: B
Third order reaction
Option D (None of these) is correct because the order of a reaction cannot be determined solely from the stoichiometry of the balanced chemical equation; it must be determined experimentally.
115 Hard
115 of 301
Analyzing +4 | -1
Consider the two hypothetical reactions given below:

I aA→a A \rightarrow Products, k=x mol−1 L min−1k=x \mathrm{~mol}^{-1} \mathrm{~L} \mathrm{~min}^{-1}
II bB→b B \rightarrow Products, k=y min−1k=y \mathrm{~min}^{-1}

The half-lives of both the reactions are the same, equal to 1hr1 \mathrm{hr} when molar concentration of the reactant is 1.0M1.0 \mathrm{M} in each case. If these reactions are started at the same time taking 1M1 \mathrm{M} of the reactant in each case, the ratio [A]/[B][A]/[B] after 3hr3 \mathrm{hr} will be:
View answer and explanation
Correct option: D
2
Option D is correct. The ratio [A]/[B] after 3 hours is 2, calculated by determining the order of each reaction and then finding the concentration of A and B after 3 hours.
116 Easy
116 of 301
Analyzing +4 | -1
During a particular reaction 20%20 \% of the reactant decomposes in 1 h, 40%1 \mathrm{~h},\,40 \% in 2 h,60%2 \mathrm{~h}, 60 \% in 3 h3 \mathrm{~h} and so on. The unit of rate constant is
View answer and explanation
Correct option: C
mol L−1 h−1\mathrm{mol} \mathrm{~L}^{-1} \mathrm{~h}^{-1}
Option C (mol L−1 h−1\mathrm{mol} \mathrm{~L}^{-1} \mathrm{~h}^{-1}) is correct because the amount of reactant decomposing is directly proportional to time, indicating a zero-order reaction.
117 Easy
117 of 301
Analyzing +4 | -1
Which of the following is an incorrect statement about first-order reaction?
View answer and explanation
Correct option: B
Half-life is the half of the time of complete reaction
Option B is correct because the half-life of a first-order reaction is independent of the initial concentration and is not half of the time for complete reaction (which is theoretically infinite).
118 Easy
118 of 301
Remembering +4 | -1
In Arrhenius equation K=A  e−Ea/RTK = A\;{e^{ - {E_a}/RT}}, the quantity −Ea/RT- {E_a}/RT is referred as:
View answer and explanation
Correct option: A
Boltzmann factor
Option A (Boltzmann factor) is correct because in the Arrhenius equation, the term −EaRT\frac{{ - {E_a}}}{{RT}} is known as the Boltzmann factor, representing the fraction of molecules with sufficient energy to react.
119 Easy
119 of 301
Remembering +4 | -1
In Arrhenius equation, the fraction of effective collisions is given by
View answer and explanation
Correct option: C
e−Ea/RT\mathrm{e}^{-\mathrm{E}_{\mathrm{a}} / \mathrm{RT}}
Option C (e−Ea/RT\mathrm{e}^{-\mathrm{E}_{\mathrm{a}} / \mathrm{RT}}) is correct because this term represents the fraction of molecules with sufficient energy to overcome the activation energy barrier.
120 Medium
120 of 301
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.
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