Rate Law & Rate Constant — NEET Chemistry MCQs with Solutions
Free NEET Chemistry Rate Law & Rate Constant MCQs with step-by-step solutions (30 questions). Part of Chemical Kinetics. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — 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 $mol^{0}L^{0}s^{-1} = s^{-1}$.
Q2 — 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).
Q3 — 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 $k = mol^{1-2}L^{2-1}s^{-1} = mol^{-1}L\,s^{-1} = L\,mol^{-1}s^{-1}$.
Q4 — 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.
Q5 — 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: $1 + 1 = 2$.
Q6 — 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.
Q7 — 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.
Q8 — 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 $= 2 + 1 = 3$.
Q9 — 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 $2 \times 2 = 4$.
Q10 — Rate Law & Rate Constant · medium · numerical
For $A + B \rightarrow$ products the data are: Exp 1 $[A]=0.1, [B]=0.1$, rate $=2 \times 10^{-3}$; Exp 2 $[A]=0.2, [B]=0.1$, rate $=4 \times 10^{-3}$; Exp 3 $[A]=0.1, [B]=0.2$, rate $=8 \times 10^{-3}$ (all in mol, L, s units). The order with respect to $B$ is:
A. 1
B. 3
C. 0
D. 2 ✓ Correct
Solution: From Exp 1 to 3, $[B]$ doubles (with $[A]$ fixed) and the rate rises fourfold ($2^2$), so the order in $B$ is 2.
Q11 — Rate Law & Rate Constant · medium · numerical
For $A + B \rightarrow$ products: Exp 1 $[A]=0.1, [B]=0.1$, rate $=2 \times 10^{-3}$; Exp 2 $[A]=0.2, [B]=0.1$, rate $=4 \times 10^{-3}$; Exp 3 $[A]=0.1, [B]=0.2$, rate $=8 \times 10^{-3}$ (mol, L, s). The overall order of the reaction is:
A. 4
B. 3 ✓ Correct
C. 1
D. 2
Solution: Order in $A$ is 1 (rate doubles when $[A]$ doubles) and order in $B$ is 2 (rate quadruples when $[B]$ doubles); overall order $= 1 + 2 = 3$.
Q12 — Rate Law & Rate Constant · medium · numerical
For $A + B \rightarrow$ products with $Rate = k[A][B]^2$, Exp 1 gives $[A]=0.1\,M, [B]=0.1\,M$, rate $=2 \times 10^{-3}\,mol\,L^{-1}s^{-1}$. The rate constant $k$ is:
A. $2 \times 10^{-3}\,mol^{-2}L^2\,s^{-1}$
B. $0.2\,mol^{-2}L^2\,s^{-1}$
C. $2\,s^{-1}$
D. $2\,mol^{-2}L^2\,s^{-1}$ ✓ Correct
Solution: $k = \dfrac{rate}{[A][B]^2} = \dfrac{2 \times 10^{-3}}{(0.1)(0.1)^2} = \dfrac{2 \times 10^{-3}}{1 \times 10^{-3}} = 2\,mol^{-2}L^2\,s^{-1}$ (third-order units).
Q13 — Rate Law & Rate Constant · medium · numerical
For $2A + B \rightarrow$ products the data are: Exp 1 $[A]=0.1, [B]=0.1$, rate $=6 \times 10^{-3}$; Exp 2 $[A]=0.2, [B]=0.1$, rate $=2.4 \times 10^{-2}$; Exp 3 $[A]=0.2, [B]=0.2$, rate $=2.4 \times 10^{-2}$ (mol, L, s). The order with respect to $A$ is:
A. 0
B. 1
C. 2 ✓ Correct
D. 3
Solution: From Exp 1 to 2, $[A]$ doubles (with $[B]$ fixed) and the rate rises fourfold ($2^2$), so the order in $A$ is 2.
Q14 — Rate Law & Rate Constant · medium · numerical
For $2A + B \rightarrow$ products: Exp 1 $[A]=0.1, [B]=0.1$, rate $=6 \times 10^{-3}$; Exp 2 $[A]=0.2, [B]=0.1$, rate $=2.4 \times 10^{-2}$; Exp 3 $[A]=0.2, [B]=0.2$, rate $=2.4 \times 10^{-2}$ (mol, L, s). The order with respect to $B$ is:
A. 2
B. 0 ✓ Correct
C. 3
D. 1
Solution: From Exp 2 to 3, $[B]$ doubles (with $[A]$ fixed) yet the rate is unchanged, so the reaction is zero order in $B$.
Q15 — Rate Law & Rate Constant · medium · numerical
For $2A + B \rightarrow$ products the rate law is $Rate = k[A]^2$. Exp 1 gives $[A]=0.1\,M$, rate $=6 \times 10^{-3}\,mol\,L^{-1}s^{-1}$. The rate constant $k$ is:
A. $0.06\,L\,mol^{-1}s^{-1}$
B. $0.6\,s^{-1}$
C. $6\,L\,mol^{-1}s^{-1}$
D. $0.6\,L\,mol^{-1}s^{-1}$ ✓ Correct
Solution: $k = \dfrac{rate}{[A]^2} = \dfrac{6 \times 10^{-3}}{(0.1)^2} = \dfrac{6 \times 10^{-3}}{0.01} = 0.6\,L\,mol^{-1}s^{-1}$ (second-order units).
Q16 — Rate Law & Rate Constant · medium · numerical
For a reaction with rate law $Rate = k[A][B]^2$, if both $[A]$ and $[B]$ are doubled, the rate becomes:
A. 4 times the original
B. 8 times the original ✓ Correct
C. 6 times the original
D. 16 times the original
Solution: Factor $= 2^1 \times 2^2 = 2 \times 4 = 8$.
Q17 — Rate Law & Rate Constant · medium · numerical
For a reaction with rate law $Rate = k[A]^2[B]$, if $[A]$ is tripled and $[B]$ is doubled, the rate becomes:
A. 6 times the original
B. 36 times the original
C. 12 times the original
D. 18 times the original ✓ Correct
Solution: Factor $= 3^2 \times 2^1 = 9 \times 2 = 18$.
Q18 — Rate Law & Rate Constant · medium · numerical
For a reaction with rate law $Rate = k[A]^2[B]$, if $[A]$ is doubled and $[B]$ is tripled, the rate becomes:
A. 18 times the original
B. 6 times the original
C. 12 times the original ✓ Correct
D. 24 times the original
Solution: Factor $= 2^2 \times 3^1 = 4 \times 3 = 12$.
Q19 — Rate Law & Rate Constant · medium · numerical
For $A \rightarrow$ products the initial-rate data are: $[A]=0.1$, rate $=1 \times 10^{-2}$; $[A]=0.2$, rate $=4 \times 10^{-2}$; $[A]=0.4$, rate $=1.6 \times 10^{-1}$ (mol, L, s). The order of the reaction is:
A. 2 ✓ Correct
B. 1
C. 3
D. 0
Solution: Each doubling of $[A]$ quadruples the rate ($2^2$), so the reaction is second order in $A$.
Q20 — Rate Law & Rate Constant · medium · numerical
For $A \rightarrow$ products (second order in $A$), when $[A]=0.1\,M$ the rate is $1 \times 10^{-2}\,mol\,L^{-1}s^{-1}$. The rate constant $k$ is:
A. $0.1\,L\,mol^{-1}s^{-1}$
B. $1\,s^{-1}$
C. $1\,L\,mol^{-1}s^{-1}$ ✓ Correct
D. $10\,L\,mol^{-1}s^{-1}$
Solution: $k = \dfrac{rate}{[A]^2} = \dfrac{1 \times 10^{-2}}{(0.1)^2} = \dfrac{1 \times 10^{-2}}{0.01} = 1\,L\,mol^{-1}s^{-1}$.
Q21 — Rate Law & Rate Constant · medium · numerical
For $A + B \rightarrow P$ the data are: Exp 1 $[A]=1.0, [B]=1.0$, rate $=0.02$; Exp 2 $[A]=2.0, [B]=1.0$, rate $=0.08$; Exp 3 $[A]=1.0, [B]=3.0$, rate $=0.06$ (mol, L, s). The rate law is:
A. $Rate = k[A]^2[B]^2$
B. $Rate = k[A][B]^2$
C. $Rate = k[A]^2[B]$ ✓ Correct
D. $Rate = k[A][B]$
Solution: Doubling $[A]$ quadruples the rate (order 2 in $A$); tripling $[B]$ triples the rate (order 1 in $B$), giving $Rate = k[A]^2[B]$.
Q22 — Rate Law & Rate Constant · medium · numerical
For $A + B \rightarrow P$ with $Rate = k[A]^2[B]$, Exp 1 gives $[A]=1.0\,M, [B]=1.0\,M$, rate $=0.02\,mol\,L^{-1}s^{-1}$. The rate constant $k$ is:
A. $0.02\,s^{-1}$
B. $0.02\,mol^{-2}L^2\,s^{-1}$ ✓ Correct
C. $0.02\,L\,mol^{-1}s^{-1}$
D. $2\,mol^{-2}L^2\,s^{-1}$
Solution: $k = \dfrac{rate}{[A]^2[B]} = \dfrac{0.02}{(1.0)^2(1.0)} = 0.02\,mol^{-2}L^2\,s^{-1}$ (third-order units).
Q23 — Rate Law & Rate Constant · medium · numerical
The rate constant of a third-order reaction has the units:
A. $mol^{-1}L\,s^{-1}$
B. $s^{-1}$
C. $mol\,L^{-1}s^{-1}$
D. $mol^{-2}L^2\,s^{-1}$ ✓ Correct
Solution: For $n=3$, units of $k = mol^{1-3}L^{3-1}s^{-1} = mol^{-2}L^2\,s^{-1}$.
Q24 — Rate Law & Rate Constant · medium · numerical
The rate constant of a reaction of order $3/2$ has the units:
A. $mol\,L^{-1}s^{-1}$
B. $s^{-1}$
C. $mol^{-1/2}L^{1/2}s^{-1}$ ✓ Correct
D. $mol^{1/2}L^{-1/2}s^{-1}$
Solution: Units of $k = mol^{1-n}L^{n-1}s^{-1}$; for $n=3/2$ this is $mol^{-1/2}L^{1/2}s^{-1}$.
Q25 — Rate Law & Rate Constant · medium · numerical
The rate constant of a reaction is found to have the units $mol\,L^{-1}s^{-1}$. The order of the reaction is:
A. 1
B. 2
C. 0 ✓ Correct
D. 3
Solution: Since units of $k = mol^{1-n}L^{n-1}s^{-1}$ equal $mol\,L^{-1}s^{-1}$ only when $n=0$, the reaction is zero order.
Q26 — Rate Law & Rate Constant · medium · numerical
A reaction has a rate constant with units $L\,mol^{-1}s^{-1}$. The order of the reaction is:
A. 3
B. 0
C. 1
D. 2 ✓ Correct
Solution: $L\,mol^{-1}s^{-1} = mol^{-1}L\,s^{-1}$ corresponds to $mol^{1-n}L^{n-1}s^{-1}$ with $n=2$, so the reaction is second order.
Q27 — Rate Law & Rate Constant · medium · numerical
A reaction is first order in $A$ and second order in $B$. If the concentration of $B$ alone is halved, the rate becomes:
A. unchanged
B. one-eighth of the original
C. one-half of the original
D. one-fourth of the original ✓ Correct
Solution: Rate $\propto [B]^2$, so halving $[B]$ changes the rate by $\left(\dfrac{1}{2}\right)^2 = \dfrac{1}{4}$.
Q28 — Rate Law & Rate Constant · medium · numerical
For a reaction with rate law $Rate = k[A]$, if the concentration of $A$ is reduced to one-third of its value, the rate becomes:
A. unchanged
B. three times the original
C. one-third of the original ✓ Correct
D. one-ninth of the original
Solution: The reaction is first order in $A$, so rate $\propto [A]$; reducing $[A]$ to $\dfrac{1}{3}$ makes the rate $\dfrac{1}{3}$ of the original.
Q29 — Rate Law & Rate Constant · medium · numerical
For $A + B \rightarrow P$ the data are: Exp 1 $[A]=0.05, [B]=0.05$, rate $=1.2 \times 10^{-3}$; Exp 2 $[A]=0.10, [B]=0.05$, rate $=2.4 \times 10^{-3}$; Exp 3 $[A]=0.10, [B]=0.10$, rate $=2.4 \times 10^{-3}$ (mol, L, s). The rate constant $k$ is:
A. $1.2 \times 10^{-3}\,s^{-1}$
B. $0.024\,L\,mol^{-1}s^{-1}$
C. $2.4 \times 10^{-2}\,s^{-1}$ ✓ Correct
D. $0.048\,s^{-1}$
Solution: Order in $A$ is 1 (rate doubles when $[A]$ doubles) and in $B$ is 0 (rate unchanged when $[B]$ doubles), so $Rate = k[A]$ and $k = \dfrac{1.2 \times 10^{-3}}{0.05} = 2.4 \times 10^{-2}\,s^{-1}$.
Q30 — Rate Law & Rate Constant · medium · numerical
Which one of the following changes does NOT alter the value of the rate constant of a reaction?
A. Raising the temperature of the reaction.
B. Adding a suitable catalyst.
C. Lowering the temperature of the reaction.
D. Increasing the initial concentration of a reactant. ✓ Correct
Solution: The rate constant depends only on temperature and the presence of a catalyst; changing the reactant concentration leaves $k$ unchanged (it changes the rate, not $k$).