Rate Law & Rate Constant — JEE Main Chemistry MCQs with Solutions
Free JEE Main 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
For $A + B \rightarrow P$: Exp 1 $[A]=0.10, [B]=0.20$, rate $=2.0\times10^{-3}$; Exp 2 $[A]=0.20, [B]=0.20$, rate $=8.0\times10^{-3}$; Exp 3 $[A]=0.20, [B]=0.40$, rate $=8.0\times10^{-3}$ (all in $mol,\,L,\,s$ units). The rate law is:
A. $Rate=k[A]^2[B]$
B. $Rate=k[A]^2$ ✓ Correct
C. $Rate=k[A]^2[B]^2$
D. $Rate=k[A][B]^2$
Solution: Exp 1$\rightarrow$2: $[A]$ doubles, rate $\times4\Rightarrow$ order $2$ in $A$. Exp 2$\rightarrow$3: $[B]$ doubles, rate unchanged $\Rightarrow$ order $0$ in $B$. Hence $Rate=k[A]^2$.
Q2 — Rate Law & Rate Constant · easy · numerical
A reaction is second order overall with $Rate=k[A]^2$. When $[A]=0.050\,M$ the rate is $4.0\times10^{-4}\,mol\,L^{-1}s^{-1}$. The rate constant $k$ is:
A. $0.16\,L\,mol^{-1}s^{-1}$ ✓ Correct
B. $8.0\times10^{-3}\,s^{-1}$
C. $1.6\times10^{-2}\,L\,mol^{-1}s^{-1}$
D. $0.16\,mol\,L^{-1}s^{-1}$
Solution: $k=\dfrac{Rate}{[A]^2}=\dfrac{4.0\times10^{-4}}{(0.050)^2}=\dfrac{4.0\times10^{-4}}{2.5\times10^{-3}}=0.16\,L\,mol^{-1}s^{-1}$.
Q3 — Rate Law & Rate Constant · easy · numerical
A reaction has the rate law $Rate=k[A]^{1/2}[B]^2$. If $[A]$ is increased 4-fold and $[B]$ is doubled, the rate becomes:
A. $6$ times the original
B. $32$ times the original
C. $16$ times the original
D. $8$ times the original ✓ Correct
Solution: Factor $=(4)^{1/2}(2)^2=2\times4=8$.
Q4 — Rate Law & Rate Constant · easy · numerical
The rate constant of a reaction of overall order $\dfrac{5}{2}$ has the units:
A. $mol^{-1/2}\,L^{1/2}\,s^{-1}$
B. $mol^{3/2}\,L^{-3/2}\,s^{-1}$
C. $mol^{-1}\,L\,s^{-1}$
D. $mol^{-3/2}\,L^{3/2}\,s^{-1}$ ✓ Correct
Solution: $k$ has units $(mol\,L^{-1})^{1-n}s^{-1}$; for $n=\dfrac{5}{2}$ this is $(mol\,L^{-1})^{-3/2}s^{-1}=mol^{-3/2}L^{3/2}s^{-1}$.
Q5 — Rate Law & Rate Constant · easy · numerical
A first-order reaction has $k=2.0\times10^{-2}\,s^{-1}$. At an instant its rate is $3.0\times10^{-4}\,mol\,L^{-1}s^{-1}$. The concentration of the reactant at that instant is:
A. $1.5\times10^{-2}\,M$ ✓ Correct
B. $66.7\,M$
C. $6.0\times10^{-6}\,M$
D. $1.5\times10^{-3}\,M$
Solution: For first order $Rate=k[A]$, so $[A]=\dfrac{Rate}{k}=\dfrac{3.0\times10^{-4}}{2.0\times10^{-2}}=1.5\times10^{-2}\,M$.
Q6 — Rate Law & Rate Constant · easy · numerical
Which one of the following statements about the rate and the rate constant of a reaction is correct?
A. Increasing the reactant concentration increases both the rate and the rate constant.
B. Adding a catalyst increases the rate but does not change the rate constant $k$.
C. The rate constant of a reaction has the same units for reactions of every order.
D. Raising the temperature increases the rate constant $k$, whereas increasing the reactant concentration increases the rate but leaves $k$ unchanged. ✓ Correct
Solution: $k$ depends on temperature (and catalyst) but not on concentration; changing concentration changes the rate, not $k$. A catalyst does change (raise) $k$, and $k$ has order-dependent units.
Q7 — Rate Law & Rate Constant · easy · numerical
For $2A + B \rightarrow P$: Exp 1 $[A]=0.10, [B]=0.10$, rate $=1.0\times10^{-3}$; Exp 2 $[A]=0.20, [B]=0.20$, rate $=8.0\times10^{-3}$; Exp 3 $[A]=0.10, [B]=0.20$, rate $=4.0\times10^{-3}$ (mol, L, s). The overall order of the reaction is:
A. $2$
B. $3$ ✓ Correct
C. $1$
D. $4$
Solution: Exp 1$\rightarrow$3: $[B]$ doubles, rate $\times4\Rightarrow$ order $2$ in $B$. Exp 1$\rightarrow$2: both doubled, rate $\times8=2^a\times4\Rightarrow a=1$. Overall order $=1+2=3$.
Q8 — Rate Law & Rate Constant · easy · numerical
For a reaction $Rate=k[A]^2[B]^{-1}$, if $[A]$ is doubled and $[B]$ is quadrupled, the rate becomes:
A. $\dfrac{1}{4}$ times the original
B. $16$ times the original
C. unchanged ($1$ times the original) ✓ Correct
D. $4$ times the original
Solution: Factor $=(2)^2(4)^{-1}=4\times\dfrac{1}{4}=1$, so the rate is unchanged.
Q9 — Rate Law & Rate Constant · easy · numerical
A reaction shows a rate that is independent of reactant concentration: at $[A]=0.20\,M$ the rate is $5.0\times10^{-3}\,mol\,L^{-1}s^{-1}$, and at $[A]=0.40\,M$ the rate is still $5.0\times10^{-3}\,mol\,L^{-1}s^{-1}$. The rate constant is:
A. $5.0\times10^{-3}\,s^{-1}$
B. $2.5\times10^{-2}\,s^{-1}$
C. $1.25\times10^{-2}\,s^{-1}$
D. $5.0\times10^{-3}\,mol\,L^{-1}s^{-1}$ ✓ Correct
Solution: The reaction is zero order, so $Rate=k$ and $k=5.0\times10^{-3}\,mol\,L^{-1}s^{-1}$ (the units are those of rate).
Q10 — Rate Law & Rate Constant · easy · numerical
A reaction $A_2 + B \rightarrow$ products proceeds by: (slow) $A_2 \rightarrow 2A$; (fast) $2A + B \rightarrow$ products. The expected rate law and the molecularity of the rate-determining step are:
A. $Rate=k[A_2]$; molecularity of the slow step $=1$ ✓ Correct
B. $Rate=k[A][B]$; molecularity of the slow step $=2$
C. $Rate=k[A_2][B]$; molecularity of the slow step $=2$
D. $Rate=k[A_2]^2[B]$; molecularity of the slow step $=3$
Solution: The rate is set by the slow step $A_2\rightarrow2A$, which is unimolecular, so $Rate=k[A_2]$ and the molecularity of the slow step is $1$.
Q11 — Rate Law & Rate Constant · easy · numerical
A reaction is first order in $X$ and second order in $Y$. If $[X]$ is halved while $[Y]$ is tripled, the rate becomes:
A. $1.5$ times the original
B. $6$ times the original
C. $4.5$ times the original ✓ Correct
D. $9$ times the original
Solution: Factor $=\left(\dfrac{1}{2}\right)^1(3)^2=\dfrac{1}{2}\times9=4.5$.
Q12 — Rate Law & Rate Constant · easy · numerical
For a reaction $Rate=k[A][B]^{1/2}$, the units of the rate constant $k$ (using $mol,\,L,\,s$) are:
A. $mol^{1/2}\,L^{-1/2}\,s^{-1}$
B. $mol^{-1/2}\,L^{1/2}\,s^{-1}$ ✓ Correct
C. $L\,mol^{-1}s^{-1}$
D. $mol^{-3/2}\,L^{3/2}\,s^{-1}$
Solution: Overall order $=1+\dfrac{1}{2}=\dfrac{3}{2}$, so $k$ has units $(mol\,L^{-1})^{-1/2}s^{-1}=mol^{-1/2}L^{1/2}s^{-1}$.
Q13 — Rate Law & Rate Constant · easy · numerical
For a second-order reaction $Rate=k[A]^2$ with $k=5.0\,L\,mol^{-1}s^{-1}$, the concentration of $A$ at which the rate equals $0.80\,mol\,L^{-1}s^{-1}$ is:
A. $0.20\,M$
B. $0.16\,M$
C. $0.40\,M$ ✓ Correct
D. $0.63\,M$
Solution: $[A]=\sqrt{\dfrac{Rate}{k}}=\sqrt{\dfrac{0.80}{5.0}}=\sqrt{0.16}=0.40\,M$.
Q14 — Rate Law & Rate Constant · easy · numerical
The rate constant of a reaction is also called its "specific reaction rate" because it equals the rate of the reaction when:
A. the reaction is exactly half complete
B. the concentrations of all reactants are zero
C. the temperature is raised to $298\,K$
D. the molar concentration of every reactant in the rate law is unity ($1\,mol\,L^{-1}$) ✓ Correct
Solution: When each concentration in the rate law equals $1\,mol\,L^{-1}$, all concentration terms become $1$ and $Rate=k$; hence $k$ is the specific reaction rate.
Q15 — Rate Law & Rate Constant · easy · numerical
For a reaction, doubling $[A]$ (other concentrations fixed) increases the rate 8-fold, while doubling $[B]$ (other concentrations fixed) halves the rate. The rate law is:
A. $Rate=k[A]^{-1}[B]^3$
B. $Rate=k[A]^8[B]^{-1}$
C. $Rate=k[A]^3[B]$
D. $Rate=k[A]^3[B]^{-1}$ ✓ Correct
Solution: Doubling $[A]$: $2^a=8\Rightarrow a=3$. Doubling $[B]$: $2^b=\dfrac{1}{2}\Rightarrow b=-1$. Hence $Rate=k[A]^3[B]^{-1}$.
Q16 — Rate Law & Rate Constant · medium · numerical
For $A + B \rightarrow P$: Exp 1 $[A]=0.10, [B]=0.10$, rate $=2.0\times10^{-4}$; Exp 2 $[A]=0.30, [B]=0.20$, rate $=3.6\times10^{-3}$; Exp 3 $[A]=0.30, [B]=0.40$, rate $=7.2\times10^{-3}$ (mol, L, s). The rate law and rate constant are:
A. $Rate=k[A]^2[B]$, $k=0.20\,L^2mol^{-2}s^{-1}$ ✓ Correct
B. $Rate=k[A][B]^2$, $k=0.20\,L^2mol^{-2}s^{-1}$
C. $Rate=k[A]^2[B]^2$, $k=20\,L^3mol^{-3}s^{-1}$
D. $Rate=k[A]^2[B]$, $k=2.0\,L^2mol^{-2}s^{-1}$
Solution: Exp 2$\rightarrow$3: $[B]\times2$, rate $\times2\Rightarrow$ order $1$ in $B$. Exp 1$\rightarrow$2: $[A]\times3,[B]\times2$, rate $\times18=3^a\times2\Rightarrow a=2$. $k=\dfrac{2.0\times10^{-4}}{(0.10)^2(0.10)}=0.20\,L^2mol^{-2}s^{-1}$.
Q17 — Rate Law & Rate Constant · medium · numerical
For $A + B \rightarrow P$: Exp 1 $[A]=0.20, [B]=0.25$, rate $=1.0\times10^{-2}$; Exp 2 $[A]=0.20, [B]=1.00$, rate $=2.0\times10^{-2}$; Exp 3 $[A]=0.80, [B]=1.00$, rate $=8.0\times10^{-2}$ (mol, L, s). The order with respect to $B$ and the overall order are, respectively:
A. order in $B=\dfrac{1}{2}$; overall order $=\dfrac{3}{2}$ ✓ Correct
B. order in $B=1$; overall order $=2$
C. order in $B=\dfrac{1}{2}$; overall order $=1$
D. order in $B=2$; overall order $=3$
Solution: Exp 1$\rightarrow$2: $[B]\times4$, rate $\times2\Rightarrow 4^b=2\Rightarrow b=\dfrac{1}{2}$. Exp 2$\rightarrow$3: $[A]\times4$, rate $\times4\Rightarrow$ order $1$ in $A$. Overall $=1+\dfrac{1}{2}=\dfrac{3}{2}$.
Q18 — Rate Law & Rate Constant · medium · numerical
For a reaction with $Rate=k[A][B]^{1/2}$ (established from data), Exp 1 gives $[A]=0.20\,M, [B]=0.25\,M$ and rate $=1.0\times10^{-2}\,mol\,L^{-1}s^{-1}$. The rate constant $k$ is:
A. $0.10\,mol^{-1/2}\,L^{1/2}\,s^{-1}$ ✓ Correct
B. $1.0\,mol^{-1/2}\,L^{1/2}\,s^{-1}$
C. $0.20\,mol^{-1/2}\,L^{1/2}\,s^{-1}$
D. $0.10\,L\,mol^{-1}s^{-1}$
Solution: $k=\dfrac{Rate}{[A][B]^{1/2}}=\dfrac{1.0\times10^{-2}}{(0.20)(0.25)^{1/2}}=\dfrac{1.0\times10^{-2}}{(0.20)(0.5)}=0.10\,mol^{-1/2}L^{1/2}s^{-1}$.
Q19 — Rate Law & Rate Constant · medium · numerical
For $A + B \rightarrow P$: Exp 1 $[A]=0.10, [B]=0.10$, rate $=4.0\times10^{-3}$; Exp 2 $[A]=0.20, [B]=0.20$, rate $=1.6\times10^{-2}$; Exp 3 $[A]=0.10, [B]=0.40$, rate $=4.0\times10^{-3}$ (mol, L, s). The predicted initial rate when $[A]=0.30\,M$ and $[B]=0.50\,M$ is:
A. $3.6\times10^{-2}\,mol\,L^{-1}s^{-1}$ ✓ Correct
B. $1.2\times10^{-2}\,mol\,L^{-1}s^{-1}$
C. $1.8\times10^{-1}\,mol\,L^{-1}s^{-1}$
D. $3.6\times10^{-3}\,mol\,L^{-1}s^{-1}$
Solution: Exp 1$\rightarrow$3: $[B]\times4$, rate unchanged $\Rightarrow$ order $0$ in $B$. Exp 1$\rightarrow$2: both $\times2$, rate $\times4\Rightarrow$ order $2$ in $A$. $k=\dfrac{4.0\times10^{-3}}{(0.10)^2}=0.40$; at $[A]=0.30$, rate $=0.40(0.30)^2=3.6\times10^{-2}\,mol\,L^{-1}s^{-1}$.
Q20 — Rate Law & Rate Constant · medium · numerical
The reaction $2NO + O_2 \rightarrow 2NO_2$ is proposed to occur by: (fast equilibrium) $2NO \rightleftharpoons N_2O_2$ (constant $K$); (slow) $N_2O_2 + O_2 \rightarrow 2NO_2$ (constant $k_2$). The predicted rate law is:
A. $Rate=k[NO]^2[O_2]$ (third order overall) ✓ Correct
B. $Rate=k[NO]^2$ (second order overall)
C. $Rate=k[N_2O_2][O_2]$ (order $2$, independent of $[NO]$)
D. $Rate=k[NO][O_2]$ (second order overall)
Solution: The pre-equilibrium gives $[N_2O_2]=K[NO]^2$; the slow step rate is $k_2[N_2O_2][O_2]=k_2K[NO]^2[O_2]$, i.e. $Rate=k[NO]^2[O_2]$.
Q21 — Rate Law & Rate Constant · medium · numerical
A reaction $A_2 + B_2 \rightarrow$ products is proposed to occur by: (fast equilibrium) $A_2 \rightleftharpoons 2A$ (constant $K$); (slow) $A + B_2 \rightarrow AB + B$. The order of the reaction with respect to $A_2$ is:
A. $\dfrac{3}{2}$
B. $\dfrac{1}{2}$ ✓ Correct
C. $1$
D. $2$
Solution: From $A_2\rightleftharpoons2A$, $[A]=(K[A_2])^{1/2}\propto[A_2]^{1/2}$; the slow step rate $\propto[A][B_2]$, so the order in $A_2$ is $\dfrac{1}{2}$.
Q22 — Rate Law & Rate Constant · medium · numerical
A reaction has $Rate=k\dfrac{[A]^2[B]}{[C]^{1/2}}$. If $[A]$, $[B]$ and $[C]$ are each doubled simultaneously, the rate becomes (take $\sqrt2=1.414$):
A. $\approx 5.66$ times the original ✓ Correct
B. $16$ times the original
C. $8$ times the original
D. $\approx 2.83$ times the original
Solution: Factor $=2^2\cdot2^1\cdot2^{-1/2}=\dfrac{8}{\sqrt2}=\dfrac{8}{1.414}\approx5.66$.
Q23 — Rate Law & Rate Constant · medium · numerical
For $A + B \rightarrow P$, $Rate=k[A][B]^2$ with $k=50\,L^2mol^{-2}s^{-1}$. When $[B]=0.10\,M$ the observed rate is $2.0\times10^{-2}\,mol\,L^{-1}s^{-1}$. The concentration of $A$ is:
A. $0.040\,M$ ✓ Correct
B. $0.020\,M$
C. $4.0\times10^{-3}\,M$
D. $0.40\,M$
Solution: $[A]=\dfrac{Rate}{k[B]^2}=\dfrac{2.0\times10^{-2}}{50\times(0.10)^2}=\dfrac{2.0\times10^{-2}}{0.50}=0.040\,M$.
Q24 — Rate Law & Rate Constant · medium · numerical
At $300\,K$ a catalyst lowers the activation energy of a reaction so that its rate constant increases by a factor of $e^{2}$ ($e^2=7.39$), while the reactant concentrations are held fixed. The factor by which the reaction rate changes, and whether $k$ itself changes, are:
A. both the rate and $k$ increase only $2$-fold
B. the rate increases $7.39$-fold and $k$ increases $7.39$-fold ✓ Correct
C. the rate is unchanged but $k$ increases $7.39$-fold
D. the rate increases $7.39$-fold but $k$ is unchanged
Solution: A catalyst raises $k$; with the concentrations fixed $Rate=k\times(\text{constant})$, so both the rate and $k$ rise by the factor $e^2=7.39$.
Q25 — Rate Law & Rate Constant · medium · numerical
For a reaction $Rate=k[A]^2$, the temperature is raised from $300\,K$ to $310\,K$ (which doubles $k$) and at the same time $[A]$ is halved. The new rate compared with the original rate is:
A. one-quarter of the original
B. unchanged
C. doubled
D. halved ($\dfrac{1}{2}$ of the original) ✓ Correct
Solution: New rate $=(2k)\left(\dfrac{[A]}{2}\right)^2=2k\dfrac{[A]^2}{4}=\dfrac{1}{2}k[A]^2$, i.e. half the original.
Q26 — Rate Law & Rate Constant · medium · numerical
For $A + B \rightarrow P$: Exp 1 $[A]=0.10, [B]=0.20$, rate $=3.0\times10^{-3}$; Exp 2 $[A]=0.20, [B]=0.10$, rate $=1.2\times10^{-2}$; Exp 3 $[A]=0.20, [B]=0.20$, rate $=1.2\times10^{-2}$ (mol, L, s). The rate law and the rate constant are:
A. $Rate=k[A][B]$, $k=0.15\,L\,mol^{-1}s^{-1}$
B. $Rate=k[A]^2$, $k=0.30\,L\,mol^{-1}s^{-1}$ ✓ Correct
C. $Rate=k[A]^2[B]$, $k=1.5\,L^2mol^{-2}s^{-1}$
D. $Rate=k[A]^2$, $k=3.0\,L\,mol^{-1}s^{-1}$
Solution: Exp 2$\rightarrow$3: $[B]\times2$, rate unchanged $\Rightarrow$ order $0$ in $B$. Exp 1$\rightarrow$3: $[A]\times2$, rate $\times4\Rightarrow$ order $2$ in $A$. $k=\dfrac{3.0\times10^{-3}}{(0.10)^2}=0.30\,L\,mol^{-1}s^{-1}$.
Q27 — Rate Law & Rate Constant · medium · numerical
For a reaction $Rate=k[A]^{3/2}[B]^{-1/2}$, if $[A]$ is increased to $4$ times and $[B]$ is increased to $9$ times, the rate becomes:
A. $6$ times the original
B. $\dfrac{8}{3}$ ($\approx2.67$) times the original ✓ Correct
C. $24$ times the original
D. $\dfrac{8}{9}$ times the original
Solution: Factor $=(4)^{3/2}(9)^{-1/2}=8\times\dfrac{1}{3}=\dfrac{8}{3}\approx2.67$.
Q28 — Rate Law & Rate Constant · medium · numerical
For $2NO + Cl_2 \rightarrow 2NOCl$: Exp 1 $[NO]=0.10, [Cl_2]=0.10$, rate $=1.0\times10^{-3}$; Exp 2 $[NO]=0.20, [Cl_2]=0.10$, rate $=4.0\times10^{-3}$; Exp 3 $[NO]=0.20, [Cl_2]=0.30$, rate $=1.2\times10^{-2}$ (mol, L, s). The overall order and the rate constant $k$ are:
A. overall order $3$; $k=10\,L^2mol^{-2}s^{-1}$
B. overall order $3$; $k=1.0\times10^{-3}\,L^2mol^{-2}s^{-1}$
C. overall order $2$; $k=0.10\,L\,mol^{-1}s^{-1}$
D. overall order $3$; $k=1.0\,L^2mol^{-2}s^{-1}$ ✓ Correct
Solution: Exp 1$\rightarrow$2: $[NO]\times2$, rate $\times4\Rightarrow$ order $2$ in $NO$. Exp 2$\rightarrow$3: $[Cl_2]\times3$, rate $\times3\Rightarrow$ order $1$ in $Cl_2$. Overall $=3$; $k=\dfrac{1.0\times10^{-3}}{(0.10)^2(0.10)}=1.0\,L^2mol^{-2}s^{-1}$.
Q29 — Rate Law & Rate Constant · medium · numerical
For a first-order reaction $A \rightarrow P$ with $k=0.020\,s^{-1}$, the rate of reaction at $[A]=0.50\,M$ and at $[A]=0.25\,M$, together with the value of $k$ at these two instants, are:
A. rates equal at both instants; $k=0.020\,s^{-1}$
B. rates $1.0\times10^{-2}$ and $5.0\times10^{-3}\,mol\,L^{-1}s^{-1}$; $k=0.020\,s^{-1}$ at both ✓ Correct
C. rate $=k=0.020$ at both instants
D. rates $1.0\times10^{-2}$ and $5.0\times10^{-3}\,mol\,L^{-1}s^{-1}$; $k=0.020$ and $0.010\,s^{-1}$
Solution: $Rate=k[A]$ gives $0.020\times0.50=1.0\times10^{-2}$ and $0.020\times0.25=5.0\times10^{-3}$; the rate falls with $[A]$, but $k=0.020\,s^{-1}$ is constant (independent of concentration).
Q30 — Rate Law & Rate Constant · medium · numerical
For $A + B \rightarrow P$: Exp 1 $[A]=0.10, [B]=0.10$, rate $=1.0\times10^{-3}$; Exp 2 $[A]=0.20, [B]=0.20$, rate $=4.0\times10^{-3}$; Exp 3 $[A]=0.40, [B]=0.20$, rate $=1.6\times10^{-2}$ (mol, L, s). The order in $A$, the order in $B$ and the overall order are, respectively:
A. order in $A=2$, order in $B=2$, overall $=4$
B. order in $A=2$, order in $B=1$, overall $=3$
C. order in $A=2$, order in $B=0$, overall $=2$ ✓ Correct
D. order in $A=1$, order in $B=1$, overall $=2$
Solution: Exp 2$\rightarrow$3: $[A]\times2$ ($[B]$ fixed), rate $\times4\Rightarrow$ order $2$ in $A$. Exp 1$\rightarrow$2: both $\times2$, rate $\times4=2^2\Rightarrow$ order in $B=0$. Overall $=2$.