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NEET • CHEMISTRY • CHEMICAL KINETICS

Chemical kinetics practice · Mixed difficulty

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181 Hard
181 of 207
Application +4 | -1
Find the two third life (t1/2)\left( {{t_{1/2}}} \right) of a first order reaction in which k=5.48×10−14k = 5.48 \times {10^{ - 14}} per second
View answer and explanation
Correct option: B
2.01×1013  s2.01 \times {10^{13}}\;s
Option B (2.01×1013  s2.01 \times {10^{13}}\;s) is correct because it is the calculated two-thirds life of a first-order reaction using the given rate constant.
182 Hard
182 of 207
Analyzing +4 | -1
The activation energies of two reactions are E1  and  E2(E1>E2){E_1}\;and\;{E_2}({E_1} > {E_2}). If the temperature of the system is increased from T1  and  T2{T_1}\;and\;{T_2}, the rate constant of the reactions changes from k1  to  k2{k_1}\;to\;{k_2} in the first reaction and k1  and  k2{k_1}\;and\;{k_2} in the second reaction. Predict which of the following expression is correct?
View answer and explanation
Correct option: B
k1′k1′>k2′k2′\frac{{{k_1}^\prime }}{{{k_1}^\prime }} > \frac{{{k_2}^\prime }}{{{k_2}^\prime }}
Option B is correct because the rate constant is more sensitive to temperature changes for reactions with higher activation energies.
183 Medium
183 of 207
Analyzing +4 | -1
Which of the following statement is true for the reaction, H2+Br2→2HBr{H_2} + B{r_2} \to 2HBr. The rate law is dxdt=K[H2][Br2]1/2\frac{{dx}}{{dt}} = K\left[ {{H_2}} \right]{\left[ {B{r_2}} \right]^{1/2}}
View answer and explanation
Correct option: D
All the above are correct
Option D (All the above are correct) is correct because the rate law indicates a 1.5 order, molecularity is 2, and quadrupling [Br2] doubles the rate.
184 Easy
184 of 207
Analyzing +4 | -1
For the reaction system

2NO(g)+O2(g)→2NO2(g)2\mathrm{NO}\left( \mathrm{g} \right) + {\mathrm{O}_2}\left( \mathrm{g} \right) \to 2\mathrm{N}{\mathrm{O}_2}\left( \mathrm{g} \right) if the volume of the reaction vessel is reduced to one-third of its original volume, what will be the order of the reaction?
View answer and explanation
Correct option: C
Increase to twenty seven times of its initial value
Option C (Increase to twenty seven times of its initial value) is correct because the rate is proportional to the product of the reactant concentrations, and reducing the volume increases the concentrations.
185 Easy
185 of 207
Application +4 | -1
A first order reaction is 10% complete in 20 min. The time taken for 19% completion is
View answer and explanation
Correct option: B
40 Min
Option B (40 Min) is correct because using the first-order kinetics equation, we can find the rate constant and then calculate the time for 19% completion.
186 Hard
186 of 207
Analyzing +4 | -1
The following statement(s) is (are) correct :

(i) A plot of log⁡kp\log k_p versus 1/T1 / T is linear

(ii) A plot of log⁡[X]\log [\mathrm{X}] versus time is linear for a first order reaction, X→P{X} \rightarrow {P}

(iii) A plot of log p versus 1/T1 / {T} is linear at constant volume

(iv) A plot of p{p} versus 1/V1 / {V} is linear at constant temperature
View answer and explanation
Correct option: D
(i), (ii) and (iv)
The correct answer is D ((i), (ii) and (iv)) because these statements accurately reflect the relationships described by the Arrhenius equation, first-order kinetics, and Boyle's law.
187 Medium
187 of 207
Application +4 | -1
For an elementary reaction 2A+3B2C2A + 3B2C . The rate of disappearance of B is
View answer and explanation
Correct option: B
3K1[A]2[B]3−3k2[C]23{K_1}{\left[ A \right]^2}{\left[ B \right]^3} - 3{k_2}{\left[ C \right]^2}
Option B (3K1[A]2[B]3−3k2[C]23{K_1}{\left[ A \right]^2}{\left[ B \right]^3} - 3{k_2}{\left[ C \right]^2}) is correct because it represents the rate of disappearance of B considering both forward and reverse reactions, and the stoichiometry.
188 Easy
188 of 207
Remembering +4 | -1
The temperature coefficient of most of the reactions lies between
View answer and explanation
Correct option: B
2 and 3
Option B is correct because the temperature coefficient of most reactions typically falls between 2 and 3.
189 Easy
189 of 207
Application +4 | -1
The rate of a reaction is doubled when temperature increases by 10∘C10^\circ C. If temperature is increased by 100∘C100^\circ C, then rate of reaction will become
View answer and explanation
Correct option: D
1024 times
Option D (1024 times) is correct because the rate doubles for every 10°C increase, and with a 100°C increase, it doubles 10 times, resulting in 2102^{10} times the original rate.
190 Hard
190 of 207
Application +4 | -1
In a second order reaction when the concentration of both reactant are equal, the reaction is completed in 500 s. How long will it take for the reaction to go to 60% completion?
View answer and explanation
Correct option: C
3000 s
Option C (3000 s) is correct because it is the time calculated for 60% completion of a second-order reaction with equal initial concentrations.
191 Medium
191 of 207
Remembering +4 | -1
Which of the following is the correct statement?
View answer and explanation
Correct option: D
Order of reaction is predicted even without knowing the rate law
The correct answer is D (Order of reaction is predicted even without knowing the rate law) because the order of a reaction can only be determined experimentally and cannot be predicted without experimental data or the rate law.
192 Easy
192 of 207
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 D (All of the above) is correct because it accurately describes how the rate constants and activation energies affect the reaction rates at different temperatures.
193 Easy
193 of 207
Remembering +4 | -1
Effect of temperature on reaction rate is given by
View answer and explanation
Correct option: B
Arrhenius equation
Option B is correct because the Arrhenius equation describes the relationship between temperature and the reaction rate constant.
194 Hard
194 of 207
Analyzing +4 | -1
The inversion of cane sugar proceeds with half-life of 500 minute at pH=5\mathrm{pH}=5 for any concentration of sugar. However, if pH=6,\mathrm{pH}=6, the half-life changes to 50 minute. The rate law expression for the sugar inversion can be written as
View answer and explanation
Correct option: D
r=K[sugar][H+]−1r=K[\text{sugar}]\left[\mathrm{H}^{+}\right]^{-1}
The correct answer is D (r=K[sugar][H+]−1r=K[\text {sugar}][\mathrm{H}^{+}]^{-1}) because the half-life is inversely proportional to the hydrogen ion concentration, indicating a negative first-order dependence on [H+].
195 Easy
195 of 207
Remembering +4 | -1
The slope in Arrhenius plot, is equal to:
View answer and explanation
Correct option: A
−Ea2.303R-\frac{E_a}{2.303 R}
Option A (−Ea2.303R-\frac{E_a}{2.303 R}) is correct because it represents the slope of the Arrhenius plot (ln k vs 1/T) when the Arrhenius equation is expressed in logarithmic form.
196 Hard
196 of 207
Application +4 | -1
t1/4  {t_{1/4}}\; can be taken as the time taken for the concentration of reactant to drop to 34{\frac{3}{4}} of its initial value. If the rate constant for a first order reaction is kk the t1/4  {t_{1/4}}\; can be written as
View answer and explanation
Correct option: C
0.29/k0.29\mathrm{ }/k
Option C (0.29/k0.29\mathrm{ }/k) is correct because it is the correct expression for t1/4t_{1/4} in terms of the rate constant k for a first-order reaction.
197 Easy
197 of 207
Analyzing +4 | -1
Consider the reaction 2A+B→product2A + B \to product
When concentration of B alone was doubled, the half-life did not change. When the concentration of A alone was doubled, the rate increased by two times. The unit of rate constant for this reaction is
View answer and explanation
Correct option: A
L.  mol−1  s−1L.\mathrm{ }\;mo{l^{ - 1}}\;{s^{ - 1}}
Option A (L.  mol−1  s−1L.\mathrm{ }\;mo{l^{ - 1}}\;{s^{ - 1}}) is correct because the rate law is rate = k[A][B]0 = k[A], making the overall order 1, and rate = k[A]2, making the overall order 2. Therefore, the rate law is rate = k[A]2[B]0, overall order 2. The unit of k is L mol-1 s-1.
198 Hard
198 of 207
Application +4 | -1
For a first reaction; 2A(g)→2B(g)+C(g)2A\left( g \right) \to 2B\left( g \right) + C\left( g \right), the initial pressure is P0{P_0} and after time ′t′'t', total pressure of reaction mixture becomes Pt{P_t}, the rate constant is
View answer and explanation
Correct option: B
K=1tln⁡P03P0−2PtK = \frac{1}{t}\ln \frac{{{P_0}}}{{3{P_0} - 2{P_t}}}
Option D (K=1tln⁡P03P0−2PtK = \frac{1}{t}\ln \frac{{{P_0}}}{{3{P_0} - 2{P_t}}}) is correct because it correctly applies the integrated rate law for a first-order reaction in terms of partial pressures.
199 Hard
199 of 207
Application +4 | -1
The rate constants K1{K_1} and K2{K_2} for two different reactions are 1016e−2000/T{10^{16}}{e^{ - 2000/T}} and 1015e−1000/T{10^{15}}{e^{ - 1000/T}}, respectively. The temperature at whichK1=K2  {K_1} = {K_{2\;}} is:
View answer and explanation
Correct option: C
10002.303  K\frac{{1000}}{{2.303}}\;\mathrm{K}
Option C (10002.303  K\frac{{1000}}{{2.303}}\;\mathrm{K}) is correct because it's the temperature at which the two rate constants are equal, calculated using the Arrhenius equation.
200 Hard
200 of 207
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
The correct answer is C (Second, 2) because the half-life is constant at 2 days, indicating a second-order reaction.
201 Easy
201 of 207
Analyzing +4 | -1
In the reaction; A+2B→3C+DA + 2B \to 3C + D which of the following expressions does not describe changes in the concentration of various species as a function of time?
View answer and explanation
Correct option: D
2d[B]dt=d[A]dt\frac{{2d\left[ B \right]}}{{dt}} = \frac{{d\left[ A \right]}}{{dt}}
Option D (2d[B]dt=d[A]dt\frac{{2d\left[ B \right]}}{{dt}} = \frac{{d\left[ A \right]}}{{dt}}) is correct because the rate of change of B should be twice the rate of change of A, but with opposite signs due to B being a reactant.
202 Easy
202 of 207
Understanding +4 | -1
Activation energy of a reaction
View answer and explanation
Correct option: D
Decreases with temperature
Option A is correct because the activation energy of a reaction is a constant value that is independent of temperature, according to the Arrhenius equation.
203 Easy
203 of 207
Understanding +4 | -1
The differential rate expression for the reaction H2+I2→2HI{\mathrm{H}_2} + {\mathrm{I}_2} \to 2\mathrm{HI} is:
View answer and explanation
Correct option: D
−2d[H2]dt=−2d[I2]dt=d[HI]dt- 2\frac{{d\left[ {{\mathrm{H}_2}} \right]}}{{dt}} = - 2\frac{{d\left[ {{\mathrm{I}_2}} \right]}}{{dt}} = \frac{{d\left[ {\mathrm{HI}} \right]}}{{dt}}
Option D (−2d[H2]dt=−2d[I2]dt=d[HI]dt- 2\frac{{d\left[ {{\mathrm{H}_2}} \right]}}{{dt}} = - 2\frac{{d\left[ {{\mathrm{I}_2}} \right]}}{{dt}} = \frac{{d\left[ {\mathrm{HI}} \right]}}{{dt}}) is correct because it correctly represents the rate of disappearance of reactants and the rate of appearance of products, considering the stoichiometric coefficients.
204 Medium
204 of 207
Analyzing +4 | -1
A reaction, A2+B2→2AB{A_2} + {B_2} \to 2AB occurs by the following mechanism;
A2→A+A{A_2} \to A + A ……(Slow)
A+B2→AB+BA + {B_2} \to AB + B …..(Fast)
A+B→ABA + B \to AB ......(Fast)
Its order would be:
View answer and explanation
Correct option: B
1
Option B (1) is correct because the rate-determining step is the slow step, which involves only A2, making the reaction first order overall.
205 Medium
205 of 207
Application +4 | -1
For the non-equilibrium process, A+B→productsA + B \to products, the rate is first order with respect to AA and second order respect to BB. If 1.0  mole1.0\;mole each of AA and BB are introduced into a 1 litre vessel and the initial rate was 1.0×10−2  mol/litre    sec.1.0 \times {10^{ - 2}}\;\mathrm{mol}/\mathrm{litre}\;\;\mathrm{sec}\mathrm{.} The rate (in mol  litre−1sec⁡−1\mathrm{mol}\;\mathrm{litr}{\mathrm{e}^{ - 1}}{\sec ^{ - 1}} ) when half of the reactants have been used:
View answer and explanation
Correct option: A
1.2×10−31.2 \times {10^{ - 3}}
Option A (1.2×10−31.2 \times {10^{ - 3}}) is correct because it is the calculated rate when half of the reactants have been used, considering the rate law and initial conditions.
206 Easy
206 of 207
Application +4 | -1
The rate of the reaction

CCl3CHO+NO→CHCl3+NO+CO\mathrm{CC}{\mathrm{l}_3}\mathrm{CHO} + \mathrm{NO} \to \mathrm{CHC}{\mathrm{l}_3} + \mathrm{NO} + \mathrm{CO} is equal to rate k[CCl3CHO][NO]k\left[ {\mathrm{CC}{\mathrm{l}_3}\mathrm{CHO}} \right]\left[ {\mathrm{NO}} \right].

If concentration is expressed in mol/L. The unit of k is
View answer and explanation
Correct option: A
L  mol−1  s−1\mathrm{L}\;\mathrm{mo}{\mathrm{l}^{ - 1}}\;{\mathrm{s}^{ - 1}}
Option A (L  mol−1  s−1\mathrm{L}\;\mathrm{mo}{\mathrm{l}^{ - 1}}\;{\mathrm{s}^{ - 1}}) is correct because it is the unit of the rate constant for a second-order reaction.
207 Hard
207 of 207
Analyzing +4 | -1
It takes 40 minutes for the completion of 50%50 \% of the reaction A→Products.\mathrm{A} \rightarrow \text{Products}. For the completion of next 50%50 \% of the reaction, the time taken is 80 minutes. The order of the reaction is
View answer and explanation
Correct option: A
2
Option A (2) is correct because the half-life increases with time, indicating a second-order reaction.
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