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Chemical Kinetics — NEET Chemistry MCQs with Solutions
Free NEET Chemistry Chemical Kinetics MCQs with step-by-step solutions covering Rate of a Chemical Reaction, Rate Law & Rate Constant, Order & Molecularity of a Reaction, Integrated Rate Equations (Zero & First Order), Half-Life & Pseudo First Order Reactions, Temperature Dependence, Arrhenius Equation & Collision Theory. Practise online on Prepizo — no login needed.
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Sample questions with solutions
Q1 — Rate of a Chemical Reaction · easy · numerical
In a reaction the concentration of a reactant falls from $0.50\,M$ to $0.40\,M$ in $10\,s$. The average rate of disappearance of the reactant is:
A. $1 \times 10^{-2}\,mol\,L^{-1}s^{-1}$ ✓ Correct
B. $5 \times 10^{-2}\,mol\,L^{-1}s^{-1}$
C. $1 \times 10^{-1}\,mol\,L^{-1}s^{-1}$
D. $1 \times 10^{-3}\,mol\,L^{-1}s^{-1}$
Solution: Average rate $= \dfrac{\Delta[R]}{\Delta t} = \dfrac{0.50-0.40}{10} = \dfrac{0.10}{10} = 1 \times 10^{-2}\,mol\,L^{-1}s^{-1}$.
Q2 — Rate of a Chemical Reaction · easy · numerical
The rate of a chemical reaction (change of concentration with time) is expressed in the units:
A. $mol\,L^{-1}$
B. $mol\,L^{-1}s^{-1}$ ✓ Correct
C. $L\,mol^{-1}s^{-1}$
D. $mol\,L\,s^{-1}$
Solution: Rate = concentration per unit time $= \dfrac{mol\,L^{-1}}{s} = mol\,L^{-1}s^{-1}$.
Q3 — Rate of a Chemical Reaction · easy · numerical
For $2N_2O_5 \rightarrow 4NO_2 + O_2$, if oxygen is formed at a rate of $2 \times 10^{-3}\,mol\,L^{-1}s^{-1}$, the rate of formation of $NO_2$ is:
A. $5 \times 10^{-4}\,mol\,L^{-1}s^{-1}$
B. $4 \times 10^{-3}\,mol\,L^{-1}s^{-1}$
C. $8 \times 10^{-3}\,mol\,L^{-1}s^{-1}$ ✓ Correct
D. $2 \times 10^{-3}\,mol\,L^{-1}s^{-1}$
Solution: Rate $= \dfrac{1}{4}\dfrac{d[NO_2]}{dt} = \dfrac{d[O_2]}{dt}$, so $\dfrac{d[NO_2]}{dt} = 4 \times 2 \times 10^{-3} = 8 \times 10^{-3}\,mol\,L^{-1}s^{-1}$.
Q4 — Rate of a Chemical Reaction · easy · numerical
The concentration of a reactant decreases from $2.0\,M$ to $1.6\,M$ in $20\,s$. The average rate of disappearance is:
A. $4 \times 10^{-1}\,mol\,L^{-1}s^{-1}$
B. $2 \times 10^{-2}\,mol\,L^{-1}s^{-1}$ ✓ Correct
C. $8 \times 10^{-2}\,mol\,L^{-1}s^{-1}$
D. $2 \times 10^{-3}\,mol\,L^{-1}s^{-1}$
Solution: Average rate $= \dfrac{2.0-1.6}{20} = \dfrac{0.4}{20} = 2 \times 10^{-2}\,mol\,L^{-1}s^{-1}$.
Q5 — Rate of a Chemical Reaction · easy · numerical
For $N_2 + 3H_2 \rightarrow 2NH_3$, if nitrogen disappears at $1 \times 10^{-2}\,mol\,L^{-1}s^{-1}$, the rate of disappearance of hydrogen is:
A. $1 \times 10^{-2}\,mol\,L^{-1}s^{-1}$
B. $3 \times 10^{-2}\,mol\,L^{-1}s^{-1}$ ✓ Correct
C. $3.3 \times 10^{-3}\,mol\,L^{-1}s^{-1}$
D. $6 \times 10^{-2}\,mol\,L^{-1}s^{-1}$
Solution: Hydrogen is consumed three times as fast as nitrogen: $-\dfrac{d[H_2]}{dt} = 3\left(-\dfrac{d[N_2]}{dt}\right) = 3 \times 10^{-2}\,mol\,L^{-1}s^{-1}$.
Q6 — Rate of a Chemical Reaction · easy · numerical
Which statement correctly distinguishes average rate from instantaneous rate?
A. Average rate is the change in concentration over a finite time interval, whereas instantaneous rate is the rate at a particular moment. ✓ Correct
B. Average rate and instantaneous rate are always numerically equal for every reaction.
C. Instantaneous rate is measured over a long time interval, while average rate is measured at a single instant.
D. The average rate of a reaction stays constant throughout the entire reaction.
Solution: Average rate uses a finite interval $\Delta t$; instantaneous rate is the limit as $\Delta t \to 0$ (the tangent slope at one instant).
Q7 — Rate of a Chemical Reaction · easy · numerical
In the reaction $A \rightarrow B$, the concentration of $B$ rises from $0$ to $0.6\,M$ in $30\,s$. The average rate of formation of $B$ is:
A. $2 \times 10^{-1}\,mol\,L^{-1}s^{-1}$
B. $1.8 \times 10^{1}\,mol\,L^{-1}s^{-1}$
C. $5 \times 10^{-2}\,mol\,L^{-1}s^{-1}$
D. $2 \times 10^{-2}\,mol\,L^{-1}s^{-1}$ ✓ Correct
Solution: Average rate $= \dfrac{\Delta[B]}{\Delta t} = \dfrac{0.6}{30} = 2 \times 10^{-2}\,mol\,L^{-1}s^{-1}$.
Q8 — Rate of a Chemical Reaction · easy · numerical
For the reaction $2A \rightarrow B$, if $B$ is formed at $0.05\,mol\,L^{-1}s^{-1}$, the rate of disappearance of $A$ is:
A. $0.20\,mol\,L^{-1}s^{-1}$
B. $0.025\,mol\,L^{-1}s^{-1}$
C. $0.10\,mol\,L^{-1}s^{-1}$ ✓ Correct
D. $0.05\,mol\,L^{-1}s^{-1}$
Solution: $A$ is consumed twice as fast as $B$ is formed: $-\dfrac{d[A]}{dt} = 2\dfrac{d[B]}{dt} = 2 \times 0.05 = 0.10\,mol\,L^{-1}s^{-1}$.
Q9 — Rate of a Chemical Reaction · easy · numerical
The instantaneous rate of a reaction is best described as:
A. The slope of the tangent to the concentration versus time curve at that instant. ✓ Correct
B. The same as the average rate taken over the whole reaction.
C. The slope of the straight line joining the initial and final points of the curve.
D. Always zero at the very start of a reaction.
Solution: The instantaneous rate is $\lim_{\Delta t \to 0}\dfrac{\Delta[X]}{\Delta t}$, i.e. the slope of the tangent to the concentration-time curve at that moment.
Q10 — Rate Law & Rate Constant · easy · numerical
The rate constant of a first-order reaction has the units:
A. $mol\,L^{-1}s^{-1}$
B. $s^{-1}$ ✓ Correct
C. $L\,mol^{-1}s^{-1}$
D. $mol^{-2}L^2\,s^{-1}$
Solution: For order $n$, units of $k = mol^{1-n}L^{n-1}s^{-1}$; for $n=1$ this gives $mol^{0}L^{0}s^{-1} = s^{-1}$.
Q11 — Rate Law & Rate Constant · easy · numerical
The rate constant of a zero-order reaction has the units:
A. $s^{-1}$
B. $L\,mol^{-1}s^{-1}$
C. $mol^{-2}L^2\,s^{-1}$
D. $mol\,L^{-1}s^{-1}$ ✓ Correct
Solution: For $n=0$, units of $k = mol^{1-0}L^{0-1}s^{-1} = mol\,L^{-1}s^{-1}$ (same as the rate itself).
Q12 — Rate Law & Rate Constant · easy · numerical
The rate constant of a second-order reaction has the units:
A. $L\,mol^{-1}s^{-1}$ ✓ Correct
B. $s^{-1}$
C. $mol\,L^{-1}s^{-1}$
D. $L^2\,mol^{-2}s^{-1}$
Solution: For $n=2$, units of $k = mol^{1-2}L^{2-1}s^{-1} = mol^{-1}L\,s^{-1} = L\,mol^{-1}s^{-1}$.
Q13 — Rate Law & Rate Constant · easy · numerical
For a reaction with rate law $Rate = k[A]$, doubling the concentration of $A$ makes the rate:
A. become 4 times the original
B. become 2 times the original ✓ Correct
C. become half the original
D. remain the same
Solution: The reaction is first order in $A$, so rate $\propto [A]$; doubling $[A]$ doubles the rate.
Q14 — Rate Law & Rate Constant · easy · numerical
For the rate law $Rate = k[A][B]$, the overall order of the reaction is:
A. 1
B. 0
C. 2 ✓ Correct
D. 3
Solution: Overall order is the sum of the exponents: $1 + 1 = 2$.
Q15 — Rate Law & Rate Constant · easy · numerical
Which statement correctly describes the difference between the rate and the rate constant of a reaction?
A. The rate changes as reactant concentrations change, but the rate constant stays the same at a fixed temperature. ✓ Correct
B. The rate and the rate constant always have the same numerical value.
C. Both the rate and the rate constant decrease as reactant concentration decreases.
D. The rate constant depends on concentration, whereas the rate does not.
Solution: Rate $= k \times$ (concentration terms), so it falls as reactants are used up; $k$ is constant at a given temperature and is independent of concentration.
Q16 — Rate Law & Rate Constant · easy · numerical
On which of the following does the rate constant of a reaction depend?
A. Only on the initial concentration of the reactants.
B. On nothing; it has the same value for every reaction.
C. Temperature and the presence of a catalyst, but not on reactant concentrations. ✓ Correct
D. The concentration of the reactants but not on temperature.
Solution: The rate constant is fixed by temperature (via the Arrhenius equation) and by a catalyst; it does not depend on reactant concentrations.
Q17 — Rate Law & Rate Constant · easy · numerical
For the rate law $Rate = k[A]^2[B]$, the overall order of the reaction is:
A. 3 ✓ Correct
B. 2
C. 1
D. 4
Solution: Overall order $= 2 + 1 = 3$.
Q18 — Rate Law & Rate Constant · easy · numerical
For a reaction with rate law $Rate = k[A][B]$, doubling both $[A]$ and $[B]$ changes the rate to:
A. 2 times the original
B. 8 times the original
C. 4 times the original ✓ Correct
D. the same as the original
Solution: Rate $\propto [A][B]$, so the factor is $2 \times 2 = 4$.
Q19 — Order & Molecularity of a Reaction · easy · numerical
For a reaction the rate law is Rate $= k[A][B]^2$. The overall order of the reaction is:
A. 2
B. 3 ✓ Correct
C. 4
D. 1
Solution: Overall order is the sum of the exponents in the rate law: $1 + 2 = 3$.
Q20 — Order & Molecularity of a Reaction · easy · numerical
For a reaction, Rate $= k[A]^2[B]$. The order with respect to $B$ is:
A. 1 ✓ Correct
B. 0
C. 3
D. 2
Solution: The exponent of $[B]$ in the rate law is 1, so the order with respect to $B$ is 1.
Q21 — Order & Molecularity of a Reaction · easy · numerical
The rate of a reaction is found to be independent of the concentration of $A$, i.e. Rate $= k[A]^0$. The order with respect to $A$ is:
A. fractional
B. 2
C. 0 ✓ Correct
D. 1
Solution: Since the exponent of $[A]$ is 0, the reaction is zero order with respect to $A$; the rate does not change with $[A]$.
Q22 — Order & Molecularity of a Reaction · easy · numerical
The molecularity of the elementary reaction $2HI \rightarrow H_2 + I_2$ is:
A. 2 (bimolecular) ✓ Correct
B. 1 (unimolecular)
C. 3 (termolecular)
D. 0
Solution: Two $HI$ molecules take part in the single elementary step, so the molecularity is 2 (bimolecular).
Q23 — Order & Molecularity of a Reaction · easy · numerical
Which of the following can never be zero or a fractional value?
A. Rate of a reaction
B. Order of a reaction
C. Rate constant of a reaction
D. Molecularity of a reaction ✓ Correct
Solution: Molecularity counts the species colliding in an elementary step, so it is always a whole number (1, 2 or 3). Order is experimental and may be zero or fractional.
Q24 — Order & Molecularity of a Reaction · easy · numerical
The rate constant of a reaction has the units $mol\,L^{-1}s^{-1}$. The order of the reaction is:
A. 0 ✓ Correct
B. 2
C. 1
D. 3
Solution: Units $mol\,L^{-1}s^{-1}$ correspond to $(mol\,L^{-1})^{1-n}s^{-1}$ with $n = 0$; a zero-order rate constant has these units.
Q25 — Order & Molecularity of a Reaction · easy · numerical
The molecularity of the elementary reaction $PCl_5 \rightarrow PCl_3 + Cl_2$ is:
A. 2 (bimolecular)
B. 3 (termolecular)
C. 0
D. 1 (unimolecular) ✓ Correct
Solution: A single $PCl_5$ molecule decomposes in the elementary step, so the molecularity is 1 (unimolecular).
Q26 — Order & Molecularity of a Reaction · easy · numerical
For a reaction, Rate $= k[A]^2$. If the concentration of $A$ is doubled, the rate becomes:
A. 8 times the original
B. 4 times the original ✓ Correct
C. 2 times the original
D. unchanged
Solution: Rate $\propto [A]^2$, so doubling $[A]$ multiplies the rate by $2^2 = 4$.
Q27 — Order & Molecularity of a Reaction · easy · numerical
Which statement about molecularity is correct?
A. Molecularity is always a whole number and can never be zero or a fraction. ✓ Correct
B. Molecularity is determined experimentally from rate data.
C. Molecularity can be a fraction for complex reactions.
D. Molecularity can be zero for photochemical reactions.
Solution: Molecularity is the number of species in an elementary step, so it must be a small whole number (1, 2 or 3).
Q28 — Integrated Rate Equations (Zero & First Order) · easy · numerical
For a zero-order reaction, $[R] = [R]_0 - kt$. If $[R]_0 = 0.50$ M and $k = 0.05$ mol $L^{-1}s^{-1}$, the concentration of $R$ after 5 s is:
A. 0.30 M
B. 0.05 M
C. 0.20 M
D. 0.25 M ✓ Correct
Solution: $[R] = 0.50 - (0.05)(5) = 0.50 - 0.25 = 0.25$ M.
Q29 — Integrated Rate Equations (Zero & First Order) · easy · numerical
In a zero-order reaction the concentration of the reactant falls from 0.80 M to 0.40 M in 20 s. The rate constant is:
A. 0.01 mol $L^{-1}s^{-1}$
B. 0.02 mol $L^{-1}s^{-1}$ ✓ Correct
C. 0.20 mol $L^{-1}s^{-1}$
D. 0.04 mol $L^{-1}s^{-1}$
Solution: $k = \dfrac{[R]_0 - [R]}{t} = \dfrac{0.80 - 0.40}{20} = 0.02$ mol $L^{-1}s^{-1}$.
Q30 — Integrated Rate Equations (Zero & First Order) · easy · numerical
For a zero-order reaction with $k = 0.02$ mol $L^{-1}s^{-1}$, the concentration of the reactant falls from $0.50$ M to $0.40$ M in:
A. 10 s
B. 5 s ✓ Correct
C. 25 s
D. 2 s
Solution: For zero order, $t = \dfrac{[R]_0 - [R]}{k} = \dfrac{0.50 - 0.40}{0.02} = 5$ s.