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JEE • CHEMISTRY • CHEMICAL KINETICS

Chemical kinetics practice · Mixed difficulty

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211 Medium
211 of 301
Application +4 | -1
The half life of a first order reaction is 2000
years. If the concentration after 8000 years is
0.02 M, then the initial concentration was:
View answer and explanation
Correct option: C
0.32 M
The correct answer is 0.32 M because after 4 half-lives, the initial concentration is reduced to 0.02 M, so the initial concentration is 0.02 * 24.
212 Easy
212 of 301
Remembering +4 | -1
Rate constant of a reaction depends upon
View answer and explanation
Correct option: D
Temperature
Option D (Temperature) is correct because the rate constant (k) of a reaction is temperature-dependent, as described by the Arrhenius equation: k=Ae−EaRTk = A e^{-\frac{{E_a}}{{RT}}}, where T is the temperature.
213 Medium
213 of 301
Application +4 | -1
Consider the given reaction

N2O4→2NO2\mathrm{N}_2 \mathrm{O}_4 \rightarrow 2 \mathrm{NO}_2

Initial conc. of N2O4=3M\mathrm{N}_2 \mathrm{O}_4=3 \mathrm{M}

Concentration of N2O4\mathrm{N}_2 \mathrm{O}_4 is 2.75M2.75 \mathrm{M}

after 30 sec30 \mathrm{~sec}., find out rate of formation of NO2\mathrm{NO}_2 during this interval (in mol lit −1 min−1{ }^{-1} \mathrm{~min}^{-1} ) (Nearest integer)
View answer and explanation
Correct option: A
1
Option A (1) is correct because the rate of formation of NO2 is twice the rate of consumption of N2O4, and the calculation yields approximately 1 mol lit⁻¹ min⁻¹.
214 Medium
214 of 301
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 (All of these) is correct because the relative rates depend on the concentration of the reactant [A]. If [A] > 1, then r3 > r2 > r1; if [A] < 1, then r1 > r2 > r3; and if [A] = 1, then r1 = r2 = r3.
215 Medium
215 of 301
Application +4 | -1
99%99 \% of a first order reaction was completed in 32 min32 {~min}. When will 99.9%99.9 \% of the reaction complete
View answer and explanation
Correct option: D
48 min48 {~min}
Option D (48 {~min}) is correct because for a first-order reaction, the time required for a certain percentage completion is proportional to the logarithm of the remaining reactant.
216 Easy
216 of 301
Understanding +4 | -1
The half-life period for a zero order reaction is equal to
View answer and explanation
Correct option: B
[A]02k\frac{{{{\left[ A \right]}_0}}}{{2k}}
Option B ([A]02k\frac{{{{\left[ A \right]}_0}}}{{2k}}) is correct because the half-life of a zero-order reaction is directly proportional to the initial concentration and inversely proportional to the rate constant.
217 Easy
217 of 301
Understanding +4 | -1
For an elementary reaction X+2Y→kZ\mathrm{X}+2 \mathrm{Y} \xrightarrow{\mathrm{k}} \mathrm{Z}
Rate law can be expressed as:
View answer and explanation
Correct option: A
k[X]1[Y]2\mathrm{k}[\mathrm{X}]^1[\mathrm{Y}]^2
Option A (k[X]1[Y]2\mathrm{k}[\mathrm{X}]^1[\mathrm{Y}]^2) is correct because for an elementary reaction, the rate law is determined directly from the stoichiometry of the reaction.
218 Easy
218 of 301
Understanding +4 | -1
The reason for almost doubling the rate of reaction on increasing the temperature of the reaction system by 10∘C10^\circ C is
View answer and explanation
Correct option: B
Collision frequency increases
Option B is correct because increasing the temperature increases the kinetic energy of molecules, leading to more frequent and effective collisions, thus increasing the reaction rate.
219 Medium
219 of 301
Application +4 | -1
The energies of activation for forward and reverse reactions for A2+B2  ⇋  2AB{A_2} + {B_2}\;\leftrightharpoons\;2AB are 180  kJ  mol−1180\;kJ\;\mathrm{mo}{\mathrm{l}^{ - 1}} and 200  kJ  mol−1200\;kJ\;\mathrm{mo}{\mathrm{l}^{ - 1}} respectively. The presence of a catalyst lowers the activation energy of both (forward and reverse) reactions by 100  kJ  mol−1100\;kJ\;\mathrm{mo}{\mathrm{l}^{ - 1}}. The enthalpy change of the reaction (A2+B2→2AB{A_2} + {B_2} \to 2AB) in the presence of catalyst will be (in kJ  mol−1kJ\;\mathrm{mo}{\mathrm{l}^{ - 1}})
View answer and explanation
Correct option: D
-20
Option D (-20) is correct because the enthalpy change is the difference between the activation energies of the forward and reverse reactions, and the catalyst lowers both by the same amount.
220 Easy
220 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 absolute value of the enthalpy change.
221 Hard
221 of 301
Application +4 | -1
Naturally occurring potassium consists of 0.01%0.01 \% of 40 K40 \mathrm{~K} which has a half-life of 1.28×109yr1.28 \times 10^9 \mathrm{yr}. Activity of 1.00 g1.00 \mathrm{~g} sample of KCl\mathrm{KCl} is
View answer and explanation
Correct option: C
4.37×108 yr−14.37 \times 10^8 \mathrm{~yr}^{-1}
Option C (4.37×108 yr−14.37 \times 10^8 \mathrm{~yr}^{-1}) is correct because it is the activity of the KCl sample calculated using the half-life and the number of K-40 atoms.
222 Medium
222 of 301
Application +4 | -1
The rate of a reaction A doubles on increasing the temperature from 300 to 310 K. By how much, the temperature of reaction B should be increased from 300 K so that rate doubles if activation energy of the reaction B is twice to that of reaction A.
View answer and explanation
Correct option: B
4.92 K
Option B (4.92 K) is correct because it accurately calculates the temperature increase required for reaction B based on the given activation energies and the Arrhenius equation.
223 Easy
223 of 301
Application +4 | -1
The rate constant of a first order reaction is 4×10−3  sec⁡−1.4 \times {10^{ - 3\;}}{\sec ^{ - 1}}. At a reactant concentration of 0.02  M0.02\;M, the rate of reaction would be:
View answer and explanation
Correct option: A
8×10−5  M  sec⁡−18 \times {10^{ - 5}}\;M\;{\sec ^{ - 1}}
Option A (8×10−5  M  sec⁡−18 \times {10^{ - 5}}\;M\;{\sec ^{ - 1}}) is correct because the rate of a first-order reaction is directly proportional to the reactant concentration.
224 Medium
224 of 301
Analyzing +4 | -1
Assertion (A):All catalytic reactions show zero order for catalyst used.

Reason (R): The rate of reaction does not depend on concentration of catalyst.
View answer and explanation
Correct option: C
Both A and R are correct and R is correct explanation of A
Option A (A is correct but R is wrong) is incorrect. Both A and R are incorrect. Catalytic reactions do not always show zero order for the catalyst, and the rate of reaction *does* depend on the concentration of the catalyst.
225 Hard
225 of 301
Application +4 | -1
The rate of a reaction double when its temperature changes from 300 K to 310 K. Activation energy of such a reaction will be (R=8.314   JK−1mol−1    and   log⁡2=0.301)(R = 8.314\,\,\,\mathrm{J}{\mathrm{K}^{ - \mathrm{1}}}\mathrm{mo}{\mathrm{l}^{ - \mathrm{1}}}\,\,\,\,\mathrm{and}\,\,\,\log 2 = 0.301)
View answer and explanation
Correct option: A
53.6  kJ  mol−153.6\,\,\mathrm{kJ}\,\,\mathrm{mo}{\mathrm{l}^{ - \mathrm{1}}}
Option A (53.6  kJ mol−153.6\,\,\mathrm{kJ}\,\mathrm{mo}{\mathrm{l}^{ - \mathrm{1}}}) is correct because it is the activation energy calculated using the Arrhenius equation, given the rate doubles when the temperature changes from 300 K to 310 K.
226 Easy
226 of 301
Application +4 | -1
By increase in temperature by 10  K10\;K, the rate of reaction becomes double. How many times the rate of reaction will be if the temperature is increased from 303  K  to  353  K303\;K\;to\;353\;K?
View answer and explanation
Correct option: D
32
Option D (32) is correct because with every 10K increase, the rate doubles, and increasing from 303K to 353K (50K increase) results in the rate increasing by a factor of 25 = 32.
227 Medium
227 of 301
Application +4 | -1
If rate constant for a first order reaction is 60 s−160 \mathrm{~s}^{-1}, then how much time will take to reduce the initial concentration of the reactant to its 116\frac{1}{16} th value?
View answer and explanation
Correct option: B
4.6×10−2 s4.6 \times 10^{-2} \mathrm{~s}
Option B (4.6×10−2 s4.6 \times 10^{-2} \mathrm{~s}) is correct because it's calculated using the first-order kinetics formula with the given rate constant and the fraction of reactant remaining.
228 Easy
228 of 301
Understanding +4 | -1
Which expression is wrong for fist order reaction?
View answer and explanation
Correct option: B
k=t2.303log⁡(A0At)k = \frac{t}{{2.303}}\log \left( {\frac{{{A_0}}}{{{A_t}}}} \right)
Option B (k=t2.303log⁡(A0At)k = \frac{t}{{2.303}}\log \left( {\frac{{{A_0}}}{{{A_t}}}} \right)) is wrong because the correct integrated rate law for a first-order reaction is k=2.303tlog⁡(A0At)k = \frac{{2.303}}{t}\log \left( {\frac{{{A_0}}}{{{A_t}}}} \right).
229 Easy
229 of 301
Application +4 | -1
A first order reaction is formed to have a rate constant, K=5.5×10−14S−1K = 5.5 \times {10^{ - 14}}{S^{ - 1}} , then the half life of the reactions is
View answer and explanation
Correct option: A
1.26×1013S1.26 \times {10^{13}}S
The correct answer is A (1.26×1013S1.26 \times {10^{13}}S) because the half-life of a first-order reaction is calculated using the formula t1/2 = 0.693/K.
230 Medium
230 of 301
Analyzing +4 | -1
As the initial concentration increases from 0.750.75 to 1.55 M1.55 \mathrm{~M} in a reaction, t1/2t_{1 / 2} decreases from 6060 to 29 s.29 \mathrm{~s}. The order of the reaction is
View answer and explanation
Correct option: C
Second
Option C is correct because the half-life is inversely proportional to the initial concentration, indicating a second-order reaction.
231 Easy
231 of 301
Understanding +4 | -1
Following mechanism has been proposed for a reaction,

2A+B→D+E2A + B \to D + E

A+B→C+D                  ...(slow)A + B \to C + D\;\;\;\;\;\;\;\;\;...(slow)

A+C→E              ....(fast)A + C \to E\;\;\;\;\;\;\;....(fast)

The rate law expression for the reaction is:
View answer and explanation
Correct option: B
r=K[A][B]r = K\left[ A \right]\left[ B \right]
Option B (r=K[A][B]r = K[A][B]) is correct because the rate law is determined by the slowest step in the reaction mechanism, which is A+B→C+DA + B \to C + D.
232 Medium
232 of 301
Analyzing +4 | -1
A reaction is 50% complete in 2 hours and 75% complete in 4 hours. The order of reaction is
View answer and explanation
Correct option: A
1
Option A (1) is correct because the reaction being 50% complete in 2 hours and 75% complete in 4 hours indicates a first-order reaction, where t75%=2×t50%t_{75\%} = 2 \times t_{50\%}.
233 Medium
233 of 301
Understanding +4 | -1
Rate constant (k)(k) of a reaction has least value at
View answer and explanation
Correct option: D
low TT and high Ea E_{\text {a }}
Option D (low TT and high Ea E_{\text {a }}) is correct because the rate constant kk decreases exponentially with increasing activation energy (EaE_a) and decreasing temperature (TT) according to the Arrhenius equation.
234 Hard
234 of 301
Application +4 | -1
The order of the reaction A→products\mathrm{A} \rightarrow \text{products} is 0.5.0.5 . The time for complete reaction TT and time for 50%50 \% reaction, t1/2t_{1 / 2} will be related as
View answer and explanation
Correct option: C
t1/2=0.3Tt_{1 / 2}=0.3 T
Option D (t1/2=0.7Tt_{1 / 2}=0.7 T) is incorrect. Option C (t1/2=0.3Tt_{1 / 2}=0.3 T) is correct because it relates the half-life to the time for complete reaction for a reaction of order 0.5.
235 Hard
235 of 301
Analyzing +4 | -1
What is order with respect to A, B, C, respectively

[A][B][C]Rate(M/sec)0.20.10.028.08×10 - 30.10.20.022.01×10 - 30.11.80.186.03×10 - 30.20.10.086.464×10 - 2\begin{array}{cccccccccccccccccccc} {[{\text{A}}]}&{[{\text{B}}]}&{[{\text{C}}]}&{{\text{Rate(M/sec)}}} \\ {{\text{0}}{\text{.2}}}&{{\text{0}}{\text{.1}}}&{{\text{0}}{\text{.02}}}&{{\text{8}}.{\text{08}} \times {\text{1}}{{\text{0}}^{{\text{ - 3}}}}} \\ {{\text{0}}{\text{.1}}}&{{\text{0}}{\text{.2}}}&{{\text{0}}{\text{.02}}}&{{\text{2}}.{\text{01}} \times {\text{1}}{{\text{0}}^{{\text{ - 3}}}}} \\ {{\text{0}}{\text{.1}}}&{{\text{1}}{\text{.8}}}&{{\text{0}}{\text{.18}}}&{{\text{6}}.{\text{03}} \times {\text{1}}{{\text{0}}^{{\text{ - 3}}}}} \\ {{\text{0}}{\text{.2}}}&{{\text{0}}{\text{.1}}}&{{\text{0}}{\text{.08}}}&{{\text{6}}.{\text{464}} \times {\text{1}}{{\text{0}}^{{\text{ - 2}}}}} \end{array}
View answer and explanation
Correct option: D
1,−1,3/21,-1,3 / 2
Option D (1,−1,3/21,-1,3 / 2) is correct because it represents the orders of the reaction with respect to A, B, and C, respectively, as determined from the given rate data.
236 Medium
236 of 301
Application +4 | -1
The first order rate constant k\mathrm{k} is related to temperature as log⁡k=15.0−(106/T)\log \mathrm{k}=15.0-\left(10^6 / \mathrm{T}\right) Which of the following pair of value is correct?
View answer and explanation
Correct option: A
A=1015\mathrm{A}=10^{15} and Ea=1.9×104 KJ\mathrm{E}_{\mathrm{a}}=1.9 \times 10^4 \mathrm{~KJ}
Option D (0.1  M  Na3PO40.1\;M\;\mathrm{N}{\mathrm{a}_3}\mathrm{P}{\mathrm{O}_4}) is correct because it has the highest van't Hoff factor (i=4), leading to the greatest lowering of vapor pressure.
237 Easy
237 of 301
Understanding +4 | -1
If the half-time for a particular reaction is found to be constant and independent of the initial concentration of the reactants, then the ratio is of
View answer and explanation
Correct option: A
First order
Option A is correct because the half-life of a first-order reaction is independent of the initial concentration of the reactants.
238 Easy
238 of 301
Remembering +4 | -1
Number of mole of a substance present in 1 litre volume is known as:
View answer and explanation
Correct option: B
Molar concentration
Option B (Molar concentration) is correct because molar concentration is defined as the number of moles of a substance present in 1 liter of solution.
239 Medium
239 of 301
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}
The correct answer is 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}) because it correctly represents the rate of disappearance of B in terms of the forward and reverse reaction rates, considering the stoichiometric coefficient of B.
240 Easy
240 of 301
Application +4 | -1
By increase in temperature by 10  K10\;K, the rate of reaction becomes double. How many times the rate of reaction will be if the temperature is increased from 303  K  to  353  K303\;K\;to\;353\;K?
View answer and explanation
Correct option: D
32
Option D (32) is correct because the rate of reaction doubles for every 10K increase, and the temperature increases by 50K, resulting in a 25 = 32-fold increase.
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