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NEET • CHEMISTRY • CHEMICAL KINETICS

Chemical kinetics practice · Mixed difficulty

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121 Medium
121 of 207
Remembering +4 | -1
A catalyst is used in a reaction to
View answer and explanation
Correct option: D
decrease the time required for the reaction
Option D is correct because a catalyst speeds up a reaction, thus decreasing the time required for the reaction to reach completion.
122 Easy
122 of 207
Application +4 | -1
The time taken for the completion of 3/43/4 of a first order reaction is
View answer and explanation
Correct option: B
(2.303/k)log⁡4\left( {2.303/k} \right)\log 4
Option B ((2.303/k)log⁡4\left( {2.303/k} \right)\log 4) is correct because it represents the time taken for 3/4 completion of a first-order reaction based on the integrated rate law.
123 Medium
123 of 207
Application +4 | -1
The reaction 2SO2(g)+O2(g)  ⇋  2SO3(g)2S{O_2}\left( g \right) + {O_2}\left( g \right)\;\leftrightharpoons\;2S{O_3}\left( g \right) is carried out in a 1  dm31\;d{m^3} vessel and 2  dm32\;d{m^3} vessel separately. The ratio of the reaction velocities will be
View answer and explanation
Correct option: D
8:18:1
Option D (8:18:1) is correct because the reaction velocity is proportional to the concentration of reactants, and changing the volume affects the concentrations.
124 Medium
124 of 207
Application +4 | -1
If rate constant of a reaction is 1.155×10−3s−11.155 \times {10^{ - 3}}{s^{ - 1}} then the half life period for the reaction will be
View answer and explanation
Correct option: D
10 min
Option D (10 min) is correct because it is the half-life calculated from the given rate constant for a first-order reaction.
125 Easy
125 of 207
Analyzing +4 | -1
The following equilibrium constants are given
N2+3H2⇄2NH3;K1{N_2} + 3{H_2} \rightleftarrows 2N{H_3};{K_1}
N2+O2⇄2NO;K2{N_2} + {O_2} \rightleftarrows 2NO;{K_2}
H2+12O2⇄H2O;K3{H_2} + \frac{1}{2}{O_2} \rightleftarrows {H_2}O;{K_3}
The equilibrium constant for the oxidation of 2 mol NH32\,{\text{mol}}\,N{H_3} by oxygen to NONO is
View answer and explanation
Correct option: A
K2K33K1\frac{{{K_2}K_3^3}}{{{K_1}}}
Option A (K2K33K1\frac{{{K_2}K_3^3}}{{{K_1}}}) is correct because it represents the equilibrium constant for the oxidation of ammonia to NO, derived from the given equilibrium constants.
126 Medium
126 of 207
Application +4 | -1
The rate constant for a first order reaction which has half-life 231 s is
View answer and explanation
Correct option: B
3×10−3 s−13 \times {10^{ - 3}}\,{s^{ - 1}}
Option B (3×10−3 s−13 \times {10^{ - 3}}\,{s^{ - 1}}) is correct because it is the rate constant calculated from the given half-life for a first-order reaction.
127 Medium
127 of 207
Analyzing +4 | -1
The rate constant is numerically the same for three reactions of first, second and third order respectively. Which one is true for rate of three reaction?
View answer and explanation
Correct option: D
All of these
Option D ((A)Only) is correct because mechanism A is consistent with the rate equation, as the slow step determines the rate, and its reactants match the rate law.
128 Medium
128 of 207
Application +4 | -1
The rate of first order reaction, A→  productsA \to \;products, is 7.5×10−4  mol  litre−1  sec⁡−1.7.5 \times {10^{ - 4}}\;\mathrm{mol}\;\mathrm{litr}{\mathrm{e}^{ - 1}}\;{\sec ^{ - 1}}. If the concentration of A  is  0.5  mol  litre−1A\;is\;0.5\;\mathrm{mol}\;\mathrm{litr}{\mathrm{e}^{ - 1}} the rate constant is:
View answer and explanation
Correct option: C
1.5×10−3  sec⁡−11.5 \times {10^{ - 3}}\;{\sec ^{ - 1}}
Option C (1.5×10−3  sec⁡−11.5 \times {10^{ - 3}}\;{\sec ^{ - 1}}) is correct because for a first-order reaction, k = rate/[A].
129 Hard
129 of 207
Understanding +4 | -1
Identify correct statements.

As the temperature increases.

A) Rate of reaction increases

B) Number of activated molecules increases

C) Collision frequency increases
View answer and explanation
Correct option: C
A, B, and C
Option C (A, B, and C) is correct because increasing the temperature generally increases the rate of reaction, the number of activated molecules, and the collision frequency.
130 Easy
130 of 207
Application +4 | -1
In a first order reaction the a/(a−x)a/\left( {a - x} \right) was found to be 8 after 10 minute. The rate constant is:
View answer and explanation
Correct option: A
(2.303×3log⁡2)/10\left( {2.303 \times 3\log 2} \right)/10
Option A ((2.303×3log⁡2)/10\left( {2.303 \times 3\log 2} \right)/10) is correct because it correctly applies the integrated rate law for a first-order reaction to find the rate constant.
131 Easy
131 of 207
Remembering +4 | -1
The temperature coefficient of a reaction is:
View answer and explanation
Correct option: C
The ratio of rate constant differing by 10∘C10^\circ C, preferably 35∘C  and  25∘C35^\circ C\;and\;25^\circ C
Option C is correct because the temperature coefficient is defined as the ratio of rate constants at two temperatures differing by 10°C.
132 Easy
132 of 207
Remembering +4 | -1
In the respect of the equation k=Ae−Ea/RTk = A{e^{ - {E_a}/RT}} in chemical kinetics, which one of the following statements is correct?
View answer and explanation
Correct option: C
Ea{E_a} is energy of activation
The correct answer is C (Ea{E_a} is energy of activation) because EaE_a represents the activation energy in the Arrhenius equation.
133 Medium
133 of 207
Understanding +4 | -1
K350K340>1\frac{{{K_{{{35}^0}}}}}{{{K_{{{34}^0}}}}} > 1, this means that
View answer and explanation
Correct option: A
Rate increases with rise in temperature
Option A (Rate increases with rise in temperature) is correct because a rate constant increasing with temperature indicates a faster reaction rate at higher temperatures.
134 Hard
134 of 207
Analyzing +4 | -1
The two reactions (I) A→Products,\text{(I) } \mathrm{A} \rightarrow \text{Products}, and (II) B→Products\text{(II) } \mathrm{B}\rightarrow \text{Products} follow first-order kinetics. The rate of reaction-I is doubled when temperature is raised from 300300 to 310 K.310 \mathrm{~K}. The half-life for this reaction at 310 K310 \mathrm{~K} is 30 minutes. At the same temperature, B decomposes twice as fast as A. If the energy of activation for the reaction-II is half that of reaction-I, the rate constant of reaction-II at 300 K is
View answer and explanation
Correct option: B
0.0327 min−10.0327 \mathrm{~min}^{-1}
Option B (0.0327 min⁻¹) is correct because it is calculated by using the Arrhenius equation and the given information about the rate constants and activation energies of the two reactions.
135 Easy
135 of 207
Understanding +4 | -1
The activation energy of a reaction is zero. The rate constant for the reaction
View answer and explanation
Correct option: D
Is nearly independent of temp
Option D (Is nearly independent of temp) is correct because when the activation energy is zero, the rate constant is independent of temperature according to the Arrhenius equation.
136 Medium
136 of 207
Analyzing +4 | -1
During the kinetic study of the reaction 2A+B→C+D2A + B \to C + D following results were obtained.

Run        [A]    in  M        [B]  in  M        Initial  rate  of  formation  of  D  in  M  s−1I                        0.1                              0.1                                      6.0×10−3  II                      0.3                            0.2                                      7.2×10−2    III                    0.3                            0.4                                      2.88×10−1        IV                    0.4                            0.1                                      2.40×10−2        \begin{array}{l} Run\;\;\;\;[A]\;\;in\;M\;\;\;\;[B]\;in\;M\;\;\;\;Initial\;rate\;of\;formation\;of\;D\;in\;M\;{s^{ - 1}}\\ I\;\;\;\;\;\;\;\;\;\;\;\;0.1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;0.1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;6.0 \times {10^{ - 3}}\;\\ II\;\;\;\;\;\;\;\;\;\;\;0.3\;\;\;\;\;\;\;\;\;\;\;\;\;\;0.2\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;7.2 \times {10^{ - 2}}\;\;\\ III\;\;\;\;\;\;\;\;\;\;0.3\;\;\;\;\;\;\;\;\;\;\;\;\;\;0.4\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;2.88 \times {10^{ - 1}}\;\;\;\;\\ IV\;\;\;\;\;\;\;\;\;\;0.4\;\;\;\;\;\;\;\;\;\;\;\;\;\;0.1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;2.40 \times {10^{ - 2}}\;\;\;\; \end{array}

On the basis of above data which one is correct:
View answer and explanation
Correct option: D
r=k[A][B]2\mathrm{r} = k\left[ A \right]{\left[ B \right]^2}
Option D (r=k[A][B]2\mathrm{r} = k[A][B]^2) is correct because it is the rate law that fits the experimental data provided.
137 Medium
137 of 207
Analyzing +4 | -1
Consider the following statements

a) Increase in concentration of reactant increases the rate of zero order reaction

b) Rate constant K is equal to collision frequency A if Ea=0{E_a} = 0

c) Rate constant K is equal to collision frequency A if Ea= ∝{E_a} = \, \propto

d) ln K Vs T is a straight line

e) ln K Vs 1/T is a straight line

Correct statements are
View answer and explanation
Correct option: B
b and e
Option B (b and e) is correct because statement b is true when activation energy is zero, and statement e is the correct form of the Arrhenius equation.
138 Medium
138 of 207
Understanding +4 | -1
A first order reaction is carried out with an initial concentration of 10 mole per litre and 80%80\% of the reactant changes into the product. Now if the same reaction is carried out with an initial concentration of 5 mol per litre the percentage of the reactant changing to the product is:
View answer and explanation
Correct option: B
80
Option B (80) is correct because the percentage of reactant changing into product in a first-order reaction is independent of the initial concentration.
139 Medium
139 of 207
Analyzing +4 | -1
Two reactions A→productsA \to products and B→productsB \to products have rate constants KA{K_A} and KB{K_B} at temperature TT and activation energies EA{E_A} and EB{E_B} respectively. If KA>KB{K_A} > {K_B}&nbsp; and EA<EB{E_A} < {E_B} and assuming that AA for both the reactions is same, then:
View answer and explanation
Correct option: D
All of the above
Option D (All of these) is correct because the rates of the reactions depend on the concentrations of the reactants, and without knowing the concentrations, we cannot definitively compare the rates.
140 Easy
140 of 207
Remembering +4 | -1
For the first order reaction with the rate constant kk, which expression gives the rate half-life period? (Initialconc.=a)(Initial\mathrm{ }conc.\mathrm{ } = a)
View answer and explanation
Correct option: D
0.693k\frac{{0.693}}{k}
Option D (0.693k\frac{{0.693}}{k}) is correct because it is the standard formula for the half-life of a first-order reaction.
141 Hard
141 of 207
Application +4 | -1
A+B→productsA + B \to products
If concentration of AA is doubled, rate increases 4 times. If concentration of AA and BB are doubled, rate increases 8 times. The differential rate equation of the reaction will be
View answer and explanation
Correct option: C
dCdt=kCA2×CB\frac{{dC}}{{dt}} = kC_A^2 \times {C_B}
Option C (dCdt=kCA2×CB\frac{{dC}}{{dt}} = kC_A^2 \times {C_B}) is correct because it matches the rate law derived from the given experimental data.
142 Hard
142 of 207
Understanding +4 | -1
Which one is not correct, according to Arrhenius equation:
View answer and explanation
Correct option: A
A high activation energy usually implies a fast reaction
Option A (A high activation energy usually implies a fast reaction) is correct because high activation energy implies a slow reaction, not a fast one.
143 Medium
143 of 207
Understanding +4 | -1
The correct difference between first and second – order reactions is that
View answer and explanation
Correct option: D
The half-life of a first order reaction does not depend on [A]0{\left[ \mathrm{A} \right]_\mathrm{0}}; the half-life of a second order reaction does
depend on [A]0{\left[ \mathrm{A} \right]_\mathrm{0}}
Option D (The half-life of a first order reaction does not depend on [A]0{\left[ \mathrm{A} \right]_\mathrm{0}}; the half-life of a second order reaction does depend on [A]0{\left[ \mathrm{A} \right]_\mathrm{0}}) is correct because this is the key difference in how half-life relates to initial concentration for first and second-order reactions.
144 Easy
144 of 207
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 it is the rate of the reaction calculated from the rate of formation of NO2NO_2 considering the stoichiometry.
145 Easy
145 of 207
Analyzing +4 | -1
The rate equation for the reaction 2A+B→C2A + B \to C is found to be
Rate=k[A][B]Rate = k\left[ A \right]\left[ B \right]
The correct statement in relation to this reaction that the
View answer and explanation
Correct option: C
Rate of formation of CC is twice the rate of disappearance of AA
Option D (Value of kk is independent of the initial concentration of AA and BB) is correct because the rate constant k is independent of the initial concentrations of reactants.
146 Easy
146 of 207
Application +4 | -1
For the reaction, 2A+B→products2A + B \to products, the active mass of B is kept constant, and that of A is doubled. The rate of reaction will be then
View answer and explanation
Correct option: C
Increase 4 times
Option C (Increase 4 times) is correct because if the rate law is rate = k[A]²[B], doubling [A] while keeping [B] constant will quadruple the rate.
147 Medium
147 of 207
Understanding +4 | -1
Activation energy is given by
View answer and explanation
Correct option: A
The energy difference between activated complex and the energy reactant molecules
Option A (The energy difference between activated complex and the energy reactant molecules) is correct because activation energy is the energy required to transform reactants into the activated complex, representing the energy difference between these two states.
148 Hard
148 of 207
Application +4 | -1
In the reaction,

2N2O5→4NO2+O22{N_2}{O_5} \to 4N{O_2} + {O_2} initial pressure is 500  atm500\;atm and rate constant kk is
3.38×10−5  s−13.38 \times {10^{ - 5}}\;{s^{ - 1}} after 10 min the final pressure of N2O5{N_2}{O_5} is
View answer and explanation
Correct option: A
490 atm
Option A (490 atm) is correct because it is calculated using the first-order kinetics equation and the given rate constant to determine the pressure change of N2O5 after 10 minutes.
149 Hard
149 of 207
Analyzing +4 | -1
A mixture of 239Pu{ }^{239} \mathrm{Pu} and 240Pu{ }^{240} \mathrm{Pu} has specific activity of 6×6 \times 109(dps)gm−110^9(\mathrm{dps}) \mathrm{gm}^{-1}. The half-life of the isotopes are 2.44×1042.44 \times 10^4 years (for 239Pu{ }^{239} \mathrm{Pu} ) and 6.58×1036.58 \times 10^3 years (for 240Pu{ }^{240} \mathrm{Pu} ) respectively. Determine the isotropic composition of this sample in nearest possible integers.
View answer and explanation
Correct option: A
39
The correct answer is 39 because it represents the approximate percentage of Plutonium-239 in the mixture, calculated using the specific activity and half-lives of the isotopes.
150 Medium
150 of 207
Analyzing +4 | -1
For an elementary reaction 2A→B,2 {A} \rightarrow {B}, consider following statements:

(a) Rate law can be written as rate =k[A]2={k}[{A}]^2

(b) On increasing temperature, rate constant kk decreases.

(c) On doubling the volume, rate becomes one fourth.

(d) On doubling the concentration, rate becomes one fourth.

Choose the correct answer from the options given below:
View answer and explanation
Correct option: A
(a), (c) only
Option A ((a), (c) only) is correct because for an elementary reaction 2A -> B, the rate law is indeed rate = k[A]2, and doubling the volume quarters the rate.
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