Rate Law & Rate Constant — IISER Chemistry MCQs with Solutions
Free IISER 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 reaction that is first order overall, $Rate=k[A]$. The units of the rate constant $k$ are:
A. $s^{-1}$ ✓ Correct
B. $L^2\,mol^{-2}s^{-1}$
C. $mol\,L^{-1}s^{-1}$
D. $L\,mol^{-1}s^{-1}$
Solution: $k=\dfrac{Rate}{[A]}=\dfrac{mol\,L^{-1}s^{-1}}{mol\,L^{-1}}=s^{-1}$.
Q2 — Rate Law & Rate Constant · easy · numerical
For a zero-order reaction, $Rate=k$. The units of the rate constant $k$ are:
A. $L\,mol^{-1}s^{-1}$
B. $s^{-1}$
C. $mol\,L^{-1}s^{-1}$ ✓ Correct
D. $L^2\,mol^{-2}s^{-1}$
Solution: For zero order the rate equals $k$, so $k$ has the same units as rate, $mol\,L^{-1}s^{-1}$.
Q3 — Rate Law & Rate Constant · easy · numerical
For a reaction that is second order overall, $Rate=k[A]^2$. The units of the rate constant $k$ are:
A. $mol\,L^{-1}s^{-1}$
B. $L^2\,mol^{-2}s^{-1}$
C. $L\,mol^{-1}s^{-1}$ ✓ Correct
D. $s^{-1}$
Solution: $k=\dfrac{Rate}{[A]^2}=\dfrac{mol\,L^{-1}s^{-1}}{mol^2\,L^{-2}}=L\,mol^{-1}s^{-1}$.
Q4 — Rate Law & Rate Constant · easy · numerical
A reaction has the experimentally determined rate law $Rate=k[NO]^2[H_2]$. Its overall order is:
A. $4$
B. $3$ ✓ Correct
C. $1$
D. $2$
Solution: Overall order is the sum of the powers $=2+1=3$.
Q5 — Rate Law & Rate Constant · easy · numerical
For a reaction with $Rate=k[A]^3$, if the concentration of $A$ is doubled the rate becomes:
A. $6$ times the original
B. $8$ times the original ✓ Correct
C. $3$ times the original
D. $9$ times the original
Solution: Rate $\propto[A]^3$, so doubling $[A]$ multiplies the rate by $2^3=8$.
Q6 — Rate Law & Rate Constant · easy · numerical
Which of the following quantities is independent of the concentrations of the reactants?
A. the rate of disappearance of a reactant
B. the rate of the reaction
C. the rate constant of the reaction ✓ Correct
D. the rate of formation of a product
Solution: The rate constant $k$ depends only on temperature (and catalyst), not on concentrations; the rate and the individual species rates all change as concentrations change.
Q7 — Rate Law & Rate Constant · easy · numerical
The value of the rate constant of a reaction is changed by altering the:
A. volume of the vessel at constant temperature
B. initial amount of reactant taken
C. concentration of the reactants
D. temperature ✓ Correct
Solution: $k$ depends on temperature through the Arrhenius equation; it is unaffected by the amount taken or the concentrations.
Q8 — Rate Law & Rate Constant · easy · numerical
The rate constant is called the specific reaction rate because it is numerically equal to the reaction rate when the concentration of each reactant is:
A. $0.5\,mol\,L^{-1}$
B. unity ($1\,mol\,L^{-1}$) ✓ Correct
C. the value at equilibrium
D. zero
Solution: When every reactant concentration is $1\,mol\,L^{-1}$, the concentration terms equal $1$ and $Rate=k$.
Q9 — Rate Law & Rate Constant · easy · numerical
For a reaction with $Rate=k[A]$, if the concentration of $A$ is reduced to half, the rate becomes:
A. double the original
B. one-fourth of the original
C. unchanged
D. half of the original ✓ Correct
Solution: Rate $\propto[A]$, so halving $[A]$ halves the rate.
Q10 — Rate Law & Rate Constant · easy · numerical
For a zero-order reaction, the rate of reaction is:
A. directly proportional to the reactant concentration
B. inversely proportional to the reactant concentration
C. proportional to the square of the reactant concentration
D. equal to the rate constant and independent of the reactant concentration ✓ Correct
Solution: For zero order $Rate=k[A]^0=k$, a constant independent of $[A]$.
Q11 — Rate Law & Rate Constant · easy · numerical
For a reaction with $Rate=k[A]^2[B]$, if only $[A]$ is doubled the rate becomes:
A. $6$ times the original
B. $4$ times the original ✓ Correct
C. $2$ times the original
D. $8$ times the original
Solution: Rate $\propto[A]^2$, so doubling $[A]$ multiplies the rate by $2^2=4$.
Q12 — Rate Law & Rate Constant · easy · numerical
For a reaction with $Rate=k[A][B]^2$, if only $[B]$ is tripled the rate becomes:
A. $27$ times the original
B. $6$ times the original
C. $9$ times the original ✓ Correct
D. $3$ times the original
Solution: Rate $\propto[B]^2$, so tripling $[B]$ multiplies the rate by $3^2=9$.
Q13 — Rate Law & Rate Constant · medium · numerical
For $X + Y \rightarrow$ products the initial-rate data are: Exp 1 $[X]=0.10,\,[Y]=0.10$, rate $=1.5\times10^{-3}$; Exp 2 $[X]=0.20,\,[Y]=0.10$, rate $=3.0\times10^{-3}$; Exp 3 $[X]=0.10,\,[Y]=0.30$, rate $=1.35\times10^{-2}$ (all in $mol,\,L,\,s$ units). The rate law is:
A. $Rate=k[X][Y]$
B. $Rate=k[X][Y]^2$ ✓ Correct
C. $Rate=k[X]^2[Y]$
D. $Rate=k[X]^2[Y]^2$
Solution: Exp 1$\rightarrow$2: $[X]$ doubles, rate doubles, so order in $X=1$. Exp 1$\rightarrow$3: $[Y]$ triples, rate becomes $9\times$, so order in $Y=2$. Hence $Rate=k[X][Y]^2$.
Q14 — Rate Law & Rate Constant · medium · numerical
For $X + Y \rightarrow$ products with $Rate=k[X][Y]^2$, Exp 1 gives $[X]=0.10\,M,\,[Y]=0.10\,M$ and rate $=1.5\times10^{-3}\,mol\,L^{-1}s^{-1}$. The rate constant $k$ is:
A. $15\,L^2\,mol^{-2}s^{-1}$
B. $0.15\,L^2\,mol^{-2}s^{-1}$
C. $1.5\,L\,mol^{-1}s^{-1}$
D. $1.5\,L^2\,mol^{-2}s^{-1}$ ✓ Correct
Solution: $k=\dfrac{Rate}{[X][Y]^2}=\dfrac{1.5\times10^{-3}}{0.10\times(0.10)^2}=\dfrac{1.5\times10^{-3}}{1.0\times10^{-3}}=1.5\,L^2\,mol^{-2}s^{-1}$.
Q15 — Rate Law & Rate Constant · medium · numerical
For $2A + B \rightarrow$ products the data are: Exp 1 $[A]=0.10,\,[B]=0.20$, 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 orders with respect to $A$ and $B$ are, respectively:
A. $2$ and $0$ ✓ Correct
B. $1$ and $0$
C. $1$ and $1$
D. $2$ and $1$
Solution: Exp 1$\rightarrow$2: $[A]$ doubles, rate becomes $4\times$, so order in $A=2$. Exp 1$\rightarrow$3: $[B]$ doubles, rate unchanged, so order in $B=0$.
Q16 — Rate Law & Rate Constant · medium · numerical
For $2A + B \rightarrow$ products the rate law is $Rate=k[A]^2$. Exp 1 gives $[A]=0.10\,M$ and rate $=4.0\times10^{-3}\,mol\,L^{-1}s^{-1}$. The rate constant $k$ is:
A. $0.40\,L\,mol^{-1}s^{-1}$ ✓ Correct
B. $4.0\,L\,mol^{-1}s^{-1}$
C. $0.40\,s^{-1}$
D. $0.040\,L\,mol^{-1}s^{-1}$
Solution: $k=\dfrac{Rate}{[A]^2}=\dfrac{4.0\times10^{-3}}{(0.10)^2}=\dfrac{4.0\times10^{-3}}{1.0\times10^{-2}}=0.40\,L\,mol^{-1}s^{-1}$.
Q17 — Rate Law & Rate Constant · medium · numerical
For $A + B \rightarrow C$ the data are: Exp 1 $[A]=0.05,\,[B]=0.05$, rate $=2.0\times10^{-3}$; Exp 2 $[A]=0.10,\,[B]=0.05$, rate $=4.0\times10^{-3}$; Exp 3 $[A]=0.10,\,[B]=0.15$, rate $=1.2\times10^{-2}$ (mol, L, s). The rate law and overall order are:
A. $Rate=k[A]^2[B]$; overall order $3$
B. $Rate=k[A][B]$; overall order $2$ ✓ Correct
C. $Rate=k[A][B]^2$; overall order $3$
D. $Rate=k[A]$; overall order $1$
Solution: Exp 1$\rightarrow$2: $[A]$ doubles, rate doubles, so first order in $A$. Exp 2$\rightarrow$3: $[B]$ triples, rate becomes $3\times$, so first order in $B$. Hence $Rate=k[A][B]$, overall order $2$.
Q18 — Rate Law & Rate Constant · medium · numerical
For $A + B \rightarrow C$ with $Rate=k[A][B]$, Exp 1 gives $[A]=0.05\,M,\,[B]=0.05\,M$ and rate $=2.0\times10^{-3}\,mol\,L^{-1}s^{-1}$. The rate constant $k$ is:
A. $8.0\,L\,mol^{-1}s^{-1}$
B. $0.08\,L\,mol^{-1}s^{-1}$
C. $0.80\,L\,mol^{-1}s^{-1}$ ✓ Correct
D. $0.80\,s^{-1}$
Solution: $k=\dfrac{Rate}{[A][B]}=\dfrac{2.0\times10^{-3}}{0.05\times0.05}=\dfrac{2.0\times10^{-3}}{2.5\times10^{-3}}=0.80\,L\,mol^{-1}s^{-1}$.
Q19 — Rate Law & Rate Constant · medium · numerical
A first-order reaction has $Rate=k[A]$ with $k=0.025\,s^{-1}$. At an instant when the rate is $5.0\times10^{-3}\,mol\,L^{-1}s^{-1}$, the concentration of $A$ is:
A. $1.25\times10^{-4}\,M$
B. $0.20\,M$ ✓ Correct
C. $0.125\,M$
D. $0.050\,M$
Solution: $[A]=\dfrac{Rate}{k}=\dfrac{5.0\times10^{-3}}{0.025}=0.20\,M$.
Q20 — Rate Law & Rate Constant · medium · numerical
A second-order reaction has $Rate=k[A]^2$ with $k=2.0\,L\,mol^{-1}s^{-1}$. When $[A]=0.30\,M$, the rate of the reaction is:
A. $0.090\,mol\,L^{-1}s^{-1}$
B. $0.18\,mol\,L^{-1}s^{-1}$ ✓ Correct
C. $0.36\,mol\,L^{-1}s^{-1}$
D. $0.60\,mol\,L^{-1}s^{-1}$
Solution: $Rate=k[A]^2=2.0\times(0.30)^2=2.0\times0.09=0.18\,mol\,L^{-1}s^{-1}$.
Q21 — Rate Law & Rate Constant · medium · numerical
For a reaction with $Rate=k[A]^2[B]$, if $[A]$ is halved and $[B]$ is quadrupled, the rate becomes:
A. half of the original
B. $4$ times the original
C. $2$ times the original
D. unchanged (equal to the original) ✓ Correct
Solution: Factor $=\left(\dfrac{1}{2}\right)^2\times4=\dfrac{1}{4}\times4=1$, so the rate is unchanged.
Q22 — Rate Law & Rate Constant · medium · numerical
A reaction has $Rate=k[A]^{1/2}[B]$. If $[A]$ is increased $4$-fold and $[B]$ is increased $3$-fold, the rate becomes:
A. $3.5$ times the original
B. $6$ times the original ✓ Correct
C. $7$ times the original
D. $12$ times the original
Solution: Factor $=4^{1/2}\times3=2\times3=6$.
Q23 — Rate Law & Rate Constant · medium · numerical
For a reaction with $Rate=k[A]^2[B]$, if $[A]$ is doubled and $[B]$ is tripled, the rate becomes:
A. $5$ times the original
B. $36$ times the original
C. $12$ times the original ✓ Correct
D. $6$ times the original
Solution: Factor $=2^2\times3=4\times3=12$.
Q24 — Rate Law & Rate Constant · medium · numerical
Which one of the following statements is correct?
A. Both the rate and the rate constant decrease as the reaction proceeds.
B. As a reaction proceeds its rate falls because reactant concentrations fall, but the rate constant stays the same at constant temperature. ✓ Correct
C. The rate constant depends on the reactant concentrations, whereas the rate does not.
D. The rate and the rate constant of a reaction are always numerically equal.
Solution: The rate depends on concentrations, which fall with time; the rate constant depends only on temperature, so it is unchanged at constant $T$.
Q25 — Rate Law & Rate Constant · medium · numerical
Which of the following statements about the order of a reaction is INCORRECT?
A. The order with respect to a species may be zero, fractional, or a whole number.
B. The order of a reaction can always be obtained from the coefficients of its balanced chemical equation. ✓ Correct
C. The order of a reaction is determined experimentally from the rate law.
D. The overall order is the sum of the powers of the concentration terms in the rate law.
Solution: Order is fixed by the experimental rate law, not by the balanced equation (they coincide only for an elementary step); so that statement is incorrect.
Q26 — Rate Law & Rate Constant · medium · numerical
Assertion (A): The rate constant of a reaction has a definite value only when the temperature is specified. Reason (R): The rate constant depends on temperature through the Arrhenius equation but is independent of the reactant concentrations. Which is correct?
A. Both A and R are true, but R is not the correct explanation of A.
B. A is false but R is true.
C. Both A and R are true, and R is the correct explanation of A. ✓ Correct
D. A is true but R is false.
Solution: $k=Ae^{-E_a/RT}$ depends only on $T$ (with $E_a$ and $A$), not on concentrations; hence $k$ is fixed only once $T$ is fixed, and R correctly explains A.
Q27 — Rate Law & Rate Constant · medium · numerical
For a reaction with $Rate=k[A]^2$, the temperature is raised so that $k$ doubles while at the same time $[A]$ is halved. The new rate compared with the original is:
A. equal to the original
B. one-fourth of the original
C. double the original
D. half of the original ✓ Correct
Solution: New rate $=(2k)\left(\dfrac{[A]}{2}\right)^2=2k\times\dfrac{[A]^2}{4}=\dfrac{1}{2}k[A]^2$, i.e. half the original.
Q28 — Rate Law & Rate Constant · medium · numerical
For $A + B \rightarrow$ products the data are: Exp 1 $[A]=0.10,\,[B]=0.10$, rate $=2.0\times10^{-2}$; Exp 2 $[A]=0.30,\,[B]=0.10$, rate $=1.8\times10^{-1}$; Exp 3 $[A]=0.10,\,[B]=0.20$, rate $=2.0\times10^{-2}$ (mol, L, s). The overall order of the reaction is:
A. $3$
B. $1$
C. $0$
D. $2$ ✓ Correct
Solution: Exp 1$\rightarrow$2: $[A]$ tripled, rate becomes $9\times=3^2$, so order in $A=2$. Exp 1$\rightarrow$3: $[B]$ doubled, rate unchanged, so order in $B=0$. Overall order $=2$.
Q29 — Rate Law & Rate Constant · medium · numerical
For the reaction $A + B \rightarrow$ products above, with $Rate=k[A]^2$, Exp 1 gives $[A]=0.10\,M$ and rate $=2.0\times10^{-2}\,mol\,L^{-1}s^{-1}$. The rate constant $k$ is:
A. $20\,L\,mol^{-1}s^{-1}$
B. $0.20\,L\,mol^{-1}s^{-1}$
C. $2.0\,L\,mol^{-1}s^{-1}$ ✓ Correct
D. $2.0\,s^{-1}$
Solution: $k=\dfrac{Rate}{[A]^2}=\dfrac{2.0\times10^{-2}}{(0.10)^2}=\dfrac{2.0\times10^{-2}}{1.0\times10^{-2}}=2.0\,L\,mol^{-1}s^{-1}$.
Q30 — Rate Law & Rate Constant · medium · numerical
For the same reaction ($Rate=k[A]^2$, $k=2.0\,L\,mol^{-1}s^{-1}$), the initial rate when $[A]=0.20\,M$ is:
A. $8.0\times10^{-2}\,mol\,L^{-1}s^{-1}$ ✓ Correct
B. $4.0\times10^{-2}\,mol\,L^{-1}s^{-1}$
C. $1.6\times10^{-1}\,mol\,L^{-1}s^{-1}$
D. $2.0\times10^{-2}\,mol\,L^{-1}s^{-1}$
Solution: $Rate=k[A]^2=2.0\times(0.20)^2=2.0\times0.04=8.0\times10^{-2}\,mol\,L^{-1}s^{-1}$.