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NEET • CHEMISTRY • CHEMICAL KINETICS

Chemical kinetics practice · Mixed difficulty

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31 Medium
31 of 207
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
Option C (Halved on reducing the concentration of RClRCl to half) is correct because the rate is directly proportional to the concentration of RClRCl in a first-order reaction.
32 Hard
32 of 207
Application +4 | -1
In gaseous reactions important for the understanding of the upper atmosphere H2O{\mathrm{H}_2}\mathrm{O} and O\mathrm{O} react bimolecularly to form two OH\mathrm{OH} radicals. ΔH\Delta H for this reaction is 72  kJ  at  500  K72\;\mathrm{kJ}\;\mathrm{ at}\;500\;\mathrm{K} and Ea{E_a} is 77  kJ  mol−177\;\mathrm{kJ}\;\mathrm{mo}{\mathrm{l}^{ - 1}}, then Ea{E_a} for the bimolecular recombination of two OH radicals to form H2O{\mathrm{H}_2}\mathrm{O} and O\mathrm{O} is:
View answer and explanation
Correct option: C
5  kJ  mol−15\;\mathrm{kJ}\;\mathrm{mo}{\mathrm{l}^{ - 1}}
Option C (5 kJ/mol) is correct because it is calculated by using the relationship between the activation energies of the forward and reverse reactions and the enthalpy change of the reaction.
33 Medium
33 of 207
Understanding +4 | -1
The half life for a reaction … of temperature.
View answer and explanation
Correct option: C
Decreased with increase
Option D (Dependent) is correct because the half-life of a reaction is temperature-dependent, typically decreasing with increasing temperature due to the increased reaction rate.
34 Hard
34 of 207
Understanding +4 | -1
For irreversible elementary reactions in parallel A→K1R{A}\xrightarrow{K_1} {R} and A→K2S,{A} \xrightarrow{K_2} {S}, the rate of disappearance of reactant 'A{A}' is
View answer and explanation
Correct option: B
(K1+K2)CA\left(K_1+K_2\right) C_{{A}}
Option B ((K1+K2)CA(K_1+K_2) C_{{A}}) is correct because the rate of disappearance of A is the sum of the rates of the two parallel reactions.
35 Easy
35 of 207
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 it represents the fraction of molecules with sufficient energy to overcome the activation energy, as defined by the Arrhenius equation.
36 Hard
36 of 207
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
The correct answer is D (B only) because mechanism B's slow step matches the given rate law, while mechanism A does not.
37 Easy
37 of 207
Understanding +4 | -1
Effective collisions are those in which molecules must:
View answer and explanation
Correct option: D
All of the above
Option D (All of the above) is correct because effective collisions require molecules to have sufficient energy (equal to or greater than the threshold energy), proper orientation, and acquire the activation energy to overcome the potential barrier.
38 Easy
38 of 207
Remembering +4 | -1
The inversion of cane sugar into glucose and fructose is:
View answer and explanation
Correct option: A
I order
The correct answer is A (I order) because the inversion of cane sugar is a pseudo-first-order reaction, where the concentration of water is in large excess.
39 Hard
39 of 207
Application +4 | -1
What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20∘C20^{\circ} \mathrm{C} to 35∘C35^{\circ} \mathrm{C}? (R=8.314 J mol−1 K−1)\left(\mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)
View answer and explanation
Correct option: B
34.5 kJ mol−134.5 \mathrm{~kJ} \mathrm{~mol}^{-1}
Option B (34.5 kJ mol−134.5 \mathrm{~kJ} \mathrm{~mol}^{-1}) is correct because it is calculated using the Arrhenius equation relating rate constants and temperatures.
40 Medium
40 of 207
Application +4 | -1
A first order reaction is 20% complete in 10 min. What is the rate constant of the reaction?
View answer and explanation
Correct option: B
0.02230.0223
Option B (0.0223) is correct because it is the rate constant calculated using the integrated rate law for a first-order reaction with 20% completion in 10 minutes.
41 Medium
41 of 207
Understanding +4 | -1
At room temperature, the reaction between NO\mathrm{NO} and O2{\mathrm{O}_2} to give NO2\mathrm{N}{\mathrm{O}_2} is fast, while that between CO\mathrm{CO} and O2{\mathrm{O}_2} is slow. It is due to:
View answer and explanation
Correct option: C
The activation energy for the reaction, 2NO+O2→2NO22\mathrm{NO} + {\mathrm{O}_2} \to 2\mathrm{N}{\mathrm{O}_2} is less than 2CO+O2→2CO22\mathrm{CO} + {\mathrm{O}_2} \to 2\mathrm{C}{\mathrm{O}_2}
Option C (The activation energy for the reaction, 2NO+O2→2NO22\mathrm{NO} + {\mathrm{O}_2} \to 2\mathrm{N}{\mathrm{O}_2} is less than 2CO+O2→2CO22\mathrm{CO} + {\mathrm{O}_2} \to 2\mathrm{C}{\mathrm{O}_2}) is correct because reactions with lower activation energies proceed faster.
42 Hard
42 of 207
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 it represents the condition where B is consumed much faster than it is produced, leading to [C] = [A0](1-e(-K1t)).
43 Hard
43 of 207
Understanding +4 | -1
For a first order reaction the plot of log⁡[A]t\log [A]_t vs tt is linear with a
View answer and explanation
Correct option: D
negative slope and non-zero intercept
Option D (negative slope and non-zero intercept) is correct because the integrated rate law for a first-order reaction is linear with a negative slope and a non-zero intercept.
44 Medium
44 of 207
Understanding +4 | -1
For a reaction of order nn, the unit of the rate constant is
View answer and explanation
Correct option: C
mol1−nLn−1s−1\mathrm{mo}{\mathrm{l}^{1 - n}}{\mathrm{L}^{n - 1}}{\mathrm{s}^{ - 1}}
Option C (mol1−nLn−1s−1\mathrm{mo}{\mathrm{l}^{1 - n}}{\mathrm{L}^{n - 1}}{\mathrm{s}^{ - 1}}) is correct because this is the general formula for the units of the rate constant for a reaction of order n.
45 Medium
45 of 207
Understanding +4 | -1
What is the formula to find value of t1/2{t_{1/2}} for a zero order reaction?
View answer and explanation
Correct option: C
[R]02k\frac{{{{\left[ R \right]}_0}}}{{2k}}
Option C ([R]02k\frac{{{{\left[ R \right]}_0}}}{{2k}}) is correct because this is the standard formula for the half-life of a zero-order reaction.
46 Medium
46 of 207
Remembering +4 | -1
Which of the following does not affect the rate of reaction?
View answer and explanation
Correct option: C
ΔH\Delta H of reaction
Option C is correct because the enthalpy change (ΔH\Delta H) of a reaction is a thermodynamic property and does not directly influence the reaction rate.
47 Hard
47 of 207
Application +4 | -1
The reaction 2A+B+C→D+E2A + B + C \to D + E is found to be first order in AA, second in BB and zero order in CC. What is the effect on the rate of increasing concentration of AA, BB and CC two times?
View answer and explanation
Correct option: B
8 times
Option B (8 times) is correct because the rate is first order in A and second order in B, so doubling A doubles the rate, and doubling B quadruples the rate, resulting in a total increase of 2 * 4 = 8 times.
48 Easy
48 of 207
Remembering +4 | -1
Which of the following theory is not related to chemical kinetics?
View answer and explanation
Correct option: D
VSPER theory
Option D (VSPER theory) is correct because VSEPR theory deals with molecular geometry, not reaction rates or mechanisms, which are the focus of chemical kinetics.
49 Hard
49 of 207
Application +4 | -1
What is the activation energy (KJ/mole) for a reaction if it is rate constant doubles when the temperature is raised from 300K to 400K? [R=314J mol.K−1]\left[ {R = 314J\,mol.{K^{ - 1}}} \right]
View answer and explanation
Correct option: B
6.92
Option B (6.92) is correct because it is calculated using the Arrhenius equation relating rate constants and temperatures.
50 Medium
50 of 207
Understanding +4 | -1
Which statement is correct?
View answer and explanation
Correct option: D
Increase in the total pressure of a gas phase reaction increase the fraction of collisions effective in producing reactions
Option D (Increase in the total pressure of a gas phase reaction increase the fraction of collisions effective in producing reactions) is correct because increasing pressure increases the collision frequency and the likelihood of effective collisions.
51 Hard
51 of 207
Understanding +4 | -1
If aa is the initial concentration of the reactant and (a−(a- x)x) is the concentration at time tt for the first-order reaction (rate constant kk). Then which of the following statements are correct.

(I) x=a(1−ek1t)x=a\left(1-e^{k_1 t}\right)

(II) (a−x)=a(12)y(a-x)=a\left(\frac{1}{2}\right)^y where y=tt50y=\frac{t}{t_{50}}

(III) k1=2.303tlog⁡(aa−x)k_1=\frac{2.303}{t} \log \left(\frac{a}{a-x}\right)
View answer and explanation
Correct option: D
I, II, III
Option D (I, II, III) is correct because all three statements are correct representations of first-order kinetics.
52 Medium
52 of 207
Understanding +4 | -1
Which one is not correct?
View answer and explanation
Correct option: A
Rate of zero order reaction depends upon initial concentration of reactant
Option A (Rate of zero order reaction depends upon initial concentration of reactant) is correct because the rate of a zero-order reaction is independent of the concentration of the reactant.
53 Hard
53 of 207
Understanding +4 | -1
Read the following statements

(i) Order of reaction can be fractional or zero.

(ii) Molecularity of a reaction can be fractional but cannot be zero.

(iii) Slowest step in the complex reaction is considered as a rate determining step.

(iv) Units of rate constant for second order reaction are mol L s−1{mol}\, {L} \,{s}{ }^{-1}.

(v) Order is applicable to elementary as well as complex reactions whereas molecularity is applicable only for elementary reactions.

Which of the following is the correct code for the statements above ?
View answer and explanation
Correct option: B
TFTFT
Option B (TFTFT) is correct because statements (i), (iii), and (v) are true, while statements (ii) and (iv) are false.
54 Hard
54 of 207
Application +4 | -1
The rate law for a reaction between the substances AA and BB is given by rate=k[A]n[B]mrate = k{\left[ A \right]^n}{\left[ B \right]^m}. On doubling the concentration of AA and halving the concentration of BB, the ratio of the new rate to the earlier rate of the reaction will be as
View answer and explanation
Correct option: D
2(n−m){2^{\left( {n - m} \right)}}
Option D (2(n−m){2^{\left( {n - m} \right)}}) is correct because doubling [A] raises the rate by 2n2^n and halving [B] reduces the rate by (1/2)m(1/2)^m, resulting in a ratio of 2n/(1/2)m=2(n+m)2^n / (1/2)^m = 2^(n+m). However, the question asks for ratio of new rate to the earlier rate, which is 2n∗(1/2)m=2(n−m)2^n * (1/2)^m = 2^(n-m)
55 Medium
55 of 207
Remembering +4 | -1
Rate constant for a reaction is λ\lambda. Average life is represent by
View answer and explanation
Correct option: A
1λ\frac{1}{\lambda }
Option A (1λ\frac{1}{\lambda }) is correct because the average life is the reciprocal of the rate constant for a first-order reaction.
56 Easy
56 of 207
Analyzing +4 | -1
2N2O5  ⇋  4NO2+O22{N_2}{O_5}\;\leftrightharpoons\;4N{O_2} + {O_2}

For the above reaction which of the following is not correct above rates of reaction?
View answer and explanation
Correct option: D
−2d[N2O5]dt=4d[NO2]dt=d[O2]dt\frac{{ - 2d\left[ {{N_2}{O_5}} \right]}}{{dt}} = 4\frac{{d\left[ {N{O_2}} \right]}}{{dt}} = \frac{{d\left[ {{O_2}} \right]}}{{dt}}
Option D (−2d[N2O5]dt=4d[NO2]dt=d[O2]dt\frac{{ - 2d\left[ {{N_2}{O_5}} \right]}}{{dt}} = 4\frac{{d\left[ {N{O_2}} \right]}}{{dt}} = \frac{{d\left[ {{O_2}} \right]}}{{dt}}) is correct because the rates are not correctly related according to the stoichiometry of the reaction.
57 Medium
57 of 207
Application +4 | -1
The rate constant of a reaction at temperature 200  K200\;K is 10 times less than the rate constant at 400  K400\;K. What is the activation energy (Ea{E_a} ) of the reaction?
View answer and explanation
Correct option: B
921.2  R921.2\;R
Option B (921.2  R921.2\;R) is correct because it is the activation energy calculated using the Arrhenius equation, given the rate constants at two different temperatures.
58 Hard
58 of 207
Understanding +4 | -1
For the chemical reaction A→products,{A} \rightarrow \text{products}, the rate of disappearance of A{A} is a given by

rA=−dCAdt=K1⋅CA1+K2⋅CAr_{{A}}=-\frac{{d} C_{{A}}}{{d} t}=\frac{K_1 \cdot C_{{A}}}{1+K_2 \cdot C_{{A}}}

At low concentration of A,{A}, expressed as CA,C_{{A}}, the reaction is of the first-order with the rate constant
View answer and explanation
Correct option: B
K1K_1
Option B (K1K_1) is correct because at low concentrations of A, the term K2⋅CAK_2 \cdot C_A becomes negligible, simplifying the rate expression to rA=K1⋅CAr_A = K_1 \cdot C_A, which is first order.
59 Easy
59 of 207
Application +4 | -1
For the reaction N2O5(g)→2NO2(g)  +  12O2(g),{N_2}{O_5}(g) \to 2N{O_2}(g)\; + \;\frac{1}{2}{O_2}(g), the rate of disappearance of N2O5{N_2}{O_5} is 6.25×10−3  mol  L−1  s−1.6.25 \times {10^{ - 3}}\;\mathrm{mol}\;{\mathrm{L}^{ - 1}}\;{\mathrm{s}^{ - 1}}. The rate of formation of NO2N{O_2} and O2{O_2} will be respectively.
View answer and explanation
Correct option: B
1.25×10−2  mol  L−1  s−1  and  3.125×10−3  mol  L−1  s−11.25 \times {10^{ - 2}}\;\mathrm{mol}\;{\mathrm{L}^{ - 1}}\;{\mathrm{s}^{ - 1}}\;and\;3.125 \times {10^{ - 3}}\;\mathrm{mol}\;{\mathrm{L}^{ - 1}}\;{\mathrm{s}^{ - 1}}
Option B (1.25×10−2  mol  L−1  s−1  and  3.125×10−3  mol  L−1  s−11.25 \times {10^{ - 2}}\;\mathrm{mol}\;{\mathrm{L}^{ - 1}}\;{\mathrm{s}^{ - 1}}\;and\;3.125 \times {10^{ - 3}}\;\mathrm{mol}\;{\mathrm{L}^{ - 1}}\;{\mathrm{s}^{ - 1}}) is correct because it correctly calculates the rates of formation of NO2NO_2 and O2O_2 based on the stoichiometry of the reaction.
60 Easy
60 of 207
Analyzing +4 | -1
The following homogeneous gaseous reactions were experimentally found to be second order overall.

1.  2NO→N2+O21.\;2\mathrm{NO} \to {\mathrm{N}_2} + {\mathrm{O}_2}

2.  3O2→2O32.\;3{\mathrm{O}_2} \to 2{\mathrm{O}_3}

3.  N2O3→NO+NO2\mathrm{3}\mathrm{.}\;{\mathrm{N}_2}{\mathrm{O}_3} \to \mathrm{NO} + \mathrm{N}{\mathrm{O}_2}

4.  H2+I2→2HI\mathrm{4}\mathrm{.}\;{\mathrm{H}_2} + {\mathrm{I}_2} \to 2\mathrm{HI}

Which of these are most likely to be elementary reaction that occur in one step?
View answer and explanation
Correct option: C
1 and 4
Option C (1 and 4) is correct because bimolecular elementary reactions are more common and likely to occur in one step.
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