Learn › IISER · Chemistry › Chemical Kinetics
Chemical Kinetics — IISER Chemistry MCQs with Solutions
Free IISER 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.
▶ Practise Chemical Kinetics online (free)
Subtopics
Sample questions with solutions
Q1 — Rate of a Chemical Reaction · easy · numerical
In a reaction the concentration of a reactant decreases from $0.75\,M$ to $0.60\,M$ in $30\,s$. The average rate of disappearance of the reactant is:
A. $5.0\times10^{-2}\,mol\,L^{-1}s^{-1}$
B. $1.5\times10^{-2}\,mol\,L^{-1}s^{-1}$
C. $2.5\times10^{-3}\,mol\,L^{-1}s^{-1}$
D. $5.0\times10^{-3}\,mol\,L^{-1}s^{-1}$ ✓ Correct
Solution: Average rate $=\dfrac{0.75-0.60}{30}=\dfrac{0.15}{30}=5.0\times10^{-3}\,mol\,L^{-1}s^{-1}$.
Q2 — Rate of a Chemical Reaction · easy · numerical
If concentration is measured in $mol\,L^{-1}$ and time in minutes, the average rate of a reaction has the units:
A. $min^{-1}$
B. $mol\,L^{-1}s^{-1}$
C. $mol\,L^{-1}min^{-1}$ ✓ Correct
D. $L\,mol^{-1}min^{-1}$
Solution: Rate $=\dfrac{\Delta[\,]}{\Delta t}$; with concentration in $mol\,L^{-1}$ and time in minutes the units are $mol\,L^{-1}min^{-1}$.
Q3 — Rate of a Chemical Reaction · easy · numerical
For $2SO_2 + O_2 \rightarrow 2SO_3$, oxygen is consumed at $3\times10^{-3}\,mol\,L^{-1}s^{-1}$. The rate of formation of $SO_3$ is:
A. $1.2\times10^{-2}\,mol\,L^{-1}s^{-1}$
B. $1.5\times10^{-3}\,mol\,L^{-1}s^{-1}$
C. $3\times10^{-3}\,mol\,L^{-1}s^{-1}$
D. $6\times10^{-3}\,mol\,L^{-1}s^{-1}$ ✓ Correct
Solution: $-\dfrac{d[O_2]}{dt}=\dfrac{1}{2}\dfrac{d[SO_3]}{dt}$, so $\dfrac{d[SO_3]}{dt}=2(3\times10^{-3})=6\times10^{-3}\,mol\,L^{-1}s^{-1}$.
Q4 — Rate of a Chemical Reaction · easy · numerical
For $2SO_2 + O_2 \rightarrow 2SO_3$, $SO_2$ is consumed at $8\times10^{-3}\,mol\,L^{-1}s^{-1}$. The rate of consumption of $O_2$ is:
A. $2\times10^{-3}\,mol\,L^{-1}s^{-1}$
B. $8\times10^{-3}\,mol\,L^{-1}s^{-1}$
C. $1.6\times10^{-2}\,mol\,L^{-1}s^{-1}$
D. $4\times10^{-3}\,mol\,L^{-1}s^{-1}$ ✓ Correct
Solution: $-\dfrac{1}{2}\dfrac{d[SO_2]}{dt}=-\dfrac{d[O_2]}{dt}$, so $-\dfrac{d[O_2]}{dt}=\dfrac{1}{2}(8\times10^{-3})=4\times10^{-3}\,mol\,L^{-1}s^{-1}$.
Q5 — Rate of a Chemical Reaction · easy · numerical
For $4NO_2 + O_2 \rightarrow 2N_2O_5$, oxygen is consumed at $2.5\times10^{-3}\,mol\,L^{-1}s^{-1}$. The rate of consumption of $NO_2$ is:
A. $1.0\times10^{-2}\,mol\,L^{-1}s^{-1}$ ✓ Correct
B. $5.0\times10^{-3}\,mol\,L^{-1}s^{-1}$
C. $6.25\times10^{-4}\,mol\,L^{-1}s^{-1}$
D. $2.5\times10^{-3}\,mol\,L^{-1}s^{-1}$
Solution: $-\dfrac{1}{4}\dfrac{d[NO_2]}{dt}=-\dfrac{d[O_2]}{dt}$, so $-\dfrac{d[NO_2]}{dt}=4(2.5\times10^{-3})=1.0\times10^{-2}\,mol\,L^{-1}s^{-1}$.
Q6 — Rate of a Chemical Reaction · easy · numerical
For $4NO_2 + O_2 \rightarrow 2N_2O_5$, $N_2O_5$ is formed at $1.5\times10^{-3}\,mol\,L^{-1}s^{-1}$. The rate of consumption of $O_2$ is:
A. $7.5\times10^{-3}\,mol\,L^{-1}s^{-1}$
B. $3.0\times10^{-3}\,mol\,L^{-1}s^{-1}$
C. $1.5\times10^{-3}\,mol\,L^{-1}s^{-1}$
D. $7.5\times10^{-4}\,mol\,L^{-1}s^{-1}$ ✓ Correct
Solution: $\dfrac{1}{2}\dfrac{d[N_2O_5]}{dt}=-\dfrac{d[O_2]}{dt}$, so $-\dfrac{d[O_2]}{dt}=\dfrac{1}{2}(1.5\times10^{-3})=7.5\times10^{-4}\,mol\,L^{-1}s^{-1}$.
Q7 — Rate of a Chemical Reaction · easy · numerical
For the decomposition $2N_2O \rightarrow 2N_2 + O_2$, $N_2O$ disappears at $8\times10^{-3}\,mol\,L^{-1}s^{-1}$. The rate of formation of $O_2$ is:
A. $4\times10^{-3}\,mol\,L^{-1}s^{-1}$ ✓ Correct
B. $2\times10^{-3}\,mol\,L^{-1}s^{-1}$
C. $8\times10^{-3}\,mol\,L^{-1}s^{-1}$
D. $1.6\times10^{-2}\,mol\,L^{-1}s^{-1}$
Solution: $-\dfrac{1}{2}\dfrac{d[N_2O]}{dt}=\dfrac{d[O_2]}{dt}$, so $\dfrac{d[O_2]}{dt}=\dfrac{1}{2}(8\times10^{-3})=4\times10^{-3}\,mol\,L^{-1}s^{-1}$.
Q8 — Rate of a Chemical Reaction · easy · numerical
For the decomposition $2NOBr \rightarrow 2NO + Br_2$, $Br_2$ is formed at $3\times10^{-3}\,mol\,L^{-1}s^{-1}$. The rate of disappearance of $NOBr$ is:
A. $6\times10^{-3}\,mol\,L^{-1}s^{-1}$ ✓ Correct
B. $3\times10^{-3}\,mol\,L^{-1}s^{-1}$
C. $1.2\times10^{-2}\,mol\,L^{-1}s^{-1}$
D. $1.5\times10^{-3}\,mol\,L^{-1}s^{-1}$
Solution: $-\dfrac{1}{2}\dfrac{d[NOBr]}{dt}=\dfrac{d[Br_2]}{dt}$, so $-\dfrac{d[NOBr]}{dt}=2(3\times10^{-3})=6\times10^{-3}\,mol\,L^{-1}s^{-1}$.
Q9 — Rate of a Chemical Reaction · easy · numerical
For $N_2 + 3H_2 \rightarrow 2NH_3$, ammonia is formed at $4\times10^{-3}\,mol\,L^{-1}s^{-1}$. The rate of disappearance of $N_2$ is:
A. $2\times10^{-3}\,mol\,L^{-1}s^{-1}$ ✓ Correct
B. $8\times10^{-3}\,mol\,L^{-1}s^{-1}$
C. $1\times10^{-3}\,mol\,L^{-1}s^{-1}$
D. $4\times10^{-3}\,mol\,L^{-1}s^{-1}$
Solution: $-\dfrac{d[N_2]}{dt}=\dfrac{1}{2}\dfrac{d[NH_3]}{dt}=\dfrac{1}{2}(4\times10^{-3})=2\times10^{-3}\,mol\,L^{-1}s^{-1}$.
Q10 — Rate of a Chemical Reaction · easy · numerical
For $2SO_2 + O_2 \rightarrow 2SO_3$, the ratio of the rate of consumption of $SO_2$ to the rate of consumption of $O_2$ at any instant is:
A. $3 : 1$
B. $1 : 2$
C. $1 : 1$
D. $2 : 1$ ✓ Correct
Solution: From the stoichiometry $SO_2$ is consumed twice as fast as $O_2$, so the ratio is $2 : 1$.
Q11 — Rate of a Chemical Reaction · easy · numerical
For $4NO_2 + O_2 \rightarrow 2N_2O_5$, the ratio of the rate of consumption of $NO_2$ to the rate of formation of $N_2O_5$ is:
A. $4 : 1$
B. $2 : 1$ ✓ Correct
C. $1 : 4$
D. $1 : 2$
Solution: The coefficients are $4$ for $NO_2$ and $2$ for $N_2O_5$, so the ratio $=4:2=2:1$.
Q12 — Rate of a Chemical Reaction · easy · numerical
The instantaneous rate of a reaction is best defined as:
A. the rate constant multiplied by the elapsed time
B. the rate at a particular instant, equal to the slope of the tangent to the concentration–time curve at that instant ✓ Correct
C. the change in concentration divided by the total time of the reaction
D. the average rate taken over the whole course of the reaction
Solution: The instantaneous rate is $-\dfrac{d[R]}{dt}$ at a given moment, i.e. the slope of the tangent to the concentration–time curve at that point.
Q13 — 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: $k=\dfrac{Rate}{[A]}=\dfrac{mol\,L^{-1}s^{-1}}{mol\,L^{-1}}=s^{-1}$.
Q14 — 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, $mol\,L^{-1}s^{-1}$.
Q15 — 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: $k=\dfrac{Rate}{[A]^2}=\dfrac{mol\,L^{-1}s^{-1}}{mol^2\,L^{-2}}=L\,mol^{-1}s^{-1}$.
Q16 — 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 $=2+1=3$.
Q17 — 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 $2^3=8$.
Q18 — 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.
Q19 — 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.
Q20 — 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 $Rate=k$.
Q21 — 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.
Q22 — 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 $[A]$.
Q23 — 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 $2^2=4$.
Q24 — 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 $3^2=9$.
Q25 — Order & Molecularity of a Reaction · easy · numerical
For the reaction $2NO + 2H_2 \rightarrow N_2 + 2H_2O$ the experimentally established rate law is Rate $= k[NO]^2[H_2]$. The overall order of the reaction is:
A. $2$
B. $3$ ✓ Correct
C. $4$
D. $5$
Solution: Overall order is the sum of the exponents in the experimental rate law: $2 + 1 = 3$.
Q26 — Order & Molecularity of a Reaction · easy · numerical
The rate constant of a reaction is found to have the units $mol^{1/2}\,L^{-1/2}\,s^{-1}$. The order of the reaction is:
A. $\dfrac{1}{2}$ ✓ Correct
B. $\dfrac{3}{2}$
C. $2$
D. $1$
Solution: For order $n$ the units of $k$ are $mol^{1-n}L^{\,n-1}s^{-1}$. Matching $mol^{1/2}$ gives $1-n=\dfrac{1}{2}$, so $n=\dfrac{1}{2}$.
Q27 — Order & Molecularity of a Reaction · easy · numerical
The molecularity of the elementary reaction $O_3 \rightarrow O_2 + O$ is:
A. $3$ (termolecular)
B. $1$ (unimolecular) ✓ Correct
C. $2$ (bimolecular)
D. $0$
Solution: Only one $O_3$ molecule takes part in this single elementary step, so the molecularity is 1 (unimolecular).
Q28 — Order & Molecularity of a Reaction · easy · numerical
For a reaction the rate law is Rate $= k\dfrac{[A]^2}{[B]}$. The order with respect to $B$ is:
A. $1$
B. $0$
C. $-1$ ✓ Correct
D. $2$
Solution: The exponent of $[B]$ is $-1$ (it appears in the denominator), so the order with respect to $B$ is $-1$; increasing $[B]$ actually slows the reaction.
Q29 — Order & Molecularity of a Reaction · easy · numerical
The thermal decomposition of acetaldehyde, $CH_3CHO \rightarrow CH_4 + CO$, obeys the rate law Rate $= k[CH_3CHO]^{3/2}$. The overall order of the reaction is:
A. $2$
B. $\dfrac{3}{2}$ ✓ Correct
C. $3$
D. $1$
Solution: The single exponent in the rate law is $\dfrac{3}{2}$, so the reaction is of order $\dfrac{3}{2}$ (a fractional order signals a complex, multi-step mechanism).
Q30 — Order & Molecularity of a Reaction · easy · numerical
When the concentration of $A$ is tripled (all other concentrations held constant), the rate of a reaction increases 27-fold. The order with respect to $A$ is:
A. $27$
B. $3$ ✓ Correct
C. $2$
D. $1$
Solution: If the order is $n$, then $3^n = 27 = 3^3$, so $n = 3$.