Order & Molecularity of a Reaction — IISER Chemistry MCQs with Solutions
Free IISER Chemistry Order & Molecularity of a Reaction 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 — 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: $2 + 1 = 3$.
Q2 — 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 $n=\dfrac{1}{2}$.
Q3 — 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).
Q4 — 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.
Q5 — 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).
Q6 — 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 $n = 3$.
Q7 — Order & Molecularity of a Reaction · easy · numerical
Which of the following could be a valid value of the molecularity of an elementary reaction?
A. $1.5$
B. $-1$
C. $0$
D. $2$ ✓ Correct
Solution: Molecularity counts the actual particles colliding in one elementary step, so it must be a positive whole number (1, 2 or 3). Only $2$ qualifies; it can never be zero, fractional or negative.
Q8 — Order & Molecularity of a Reaction · easy · numerical
For an elementary step $2A + B \rightarrow$ products, the rate law can be written directly from its stoichiometry as:
A. Rate $= k[A][B]^2$
B. Rate $= k[A][B]$
C. Rate $= k[A]^2[B]^0$
D. Rate $= k[A]^2[B]$ ✓ Correct
Solution: For an elementary reaction the exponents equal the stoichiometric coefficients, so Rate $= k[A]^2[B]$.
Q9 — Order & Molecularity of a Reaction · easy · numerical
The molecularity of the elementary reaction $NO + O_3 \rightarrow NO_2 + O_2$ is:
A. $1$ (unimolecular)
B. $4$
C. $2$ (bimolecular) ✓ Correct
D. $3$ (termolecular)
Solution: Two species ($NO$ and $O_3$) collide in the single elementary step, so the molecularity is 2 (bimolecular).
Q10 — Order & Molecularity of a Reaction · easy · numerical
For a certain reaction the rate is found to be Rate $= k$, independent of the concentration of every reactant. The overall order of the reaction is:
A. $1$
B. indeterminate
C. a fraction between $0$ and $1$
D. $0$ ✓ Correct
Solution: If the rate does not depend on any concentration, every exponent is zero, so the overall order is $0$ (a zero-order reaction).
Q11 — Order & Molecularity of a Reaction · easy · numerical
The reaction $H_2 + Br_2 \rightarrow 2HBr$ follows the rate law Rate $= k[H_2][Br_2]^{1/2}$. The order with respect to $Br_2$ is:
A. $\dfrac{1}{2}$ ✓ Correct
B. $1$
C. $\dfrac{3}{2}$
D. $2$
Solution: The exponent of $[Br_2]$ in the rate law is $\dfrac{1}{2}$, so the order with respect to $Br_2$ is $\dfrac{1}{2}$.
Q12 — Order & Molecularity of a Reaction · easy · numerical
For which kind of reaction is molecularity a meaningful quantity?
A. An elementary (single-step) reaction ✓ Correct
B. Only a first-order reaction
C. Any reaction, whether elementary or complex
D. Only a zero-order reaction
Solution: Molecularity is defined only for an elementary reaction, where it equals the number of species colliding in that single step. For a multi-step (complex) reaction only the experimental order is meaningful.
Q13 — Order & Molecularity of a Reaction · medium · numerical
For a reaction $A \rightarrow$ products, the half-life is $200\,s$ when $[A]_0 = 0.40\,M$ and $400\,s$ when $[A]_0 = 0.80\,M$. The order of the reaction is:
A. $0.5$
B. $2$
C. $0$ ✓ Correct
D. $1$
Solution: Here $t_{1/2} \propto [A]_0$ (doubling $[A]_0$ doubles $t_{1/2}$). Since $t_{1/2} \propto [A]_0^{\,1-n}$, we need $1-n = 1$, giving $n = 0$ (zero order).
Q14 — Order & Molecularity of a Reaction · medium · numerical
For a reaction $A \rightarrow$ products, reducing the initial concentration to one-third of its value makes the half-life 9 times longer. The order of the reaction is:
A. $\dfrac{5}{2}$
B. $3$ ✓ Correct
C. $1$
D. $2$
Solution: Using $t_{1/2} \propto [A]_0^{\,1-n}$: $\left(\dfrac{1}{3}\right)^{1-n} = 9 = 3^2$, so $1-n = -2$ and $n = 3$.
Q15 — Order & Molecularity of a Reaction · medium · numerical
For $A + B \rightarrow$ products, doubling $[A]$ (with $[B]$ fixed) makes the rate four times larger, while doubling $[B]$ (with $[A]$ fixed) halves the rate. The overall order of the reaction is:
A. $1$ ✓ Correct
B. $2$
C. $1.5$
D. $3$
Solution: Order in $A$: $2^a = 4 \Rightarrow a = 2$. Order in $B$: $2^b = \dfrac{1}{2} \Rightarrow b = -1$. Overall order $= 2 + (-1) = 1$.
Q16 — Order & Molecularity of a Reaction · medium · numerical
The decomposition $2NO_2Cl \rightarrow 2NO_2 + Cl_2$ proceeds by: (slow) $NO_2Cl \rightarrow NO_2 + Cl$; (fast) $NO_2Cl + Cl \rightarrow NO_2 + Cl_2$. The rate law is:
A. Rate $= k[NO_2Cl]^2[Cl_2]^{-1}$
B. Rate $= k[NO_2Cl]$ (first order) ✓ Correct
C. Rate $= k[NO_2Cl][Cl]$
D. Rate $= k[NO_2Cl]^2$
Solution: The slow step is unimolecular in $NO_2Cl$, so Rate $= k[NO_2Cl]$; the reaction is first order overall.
Q17 — Order & Molecularity of a Reaction · medium · numerical
For $CHCl_3 + Cl_2 \rightarrow CCl_4 + HCl$ the mechanism is: fast equilibrium $Cl_2 \rightleftharpoons 2Cl$ (constant $K$); slow $Cl + CHCl_3 \rightarrow HCl + CCl_3$; fast $CCl_3 + Cl \rightarrow CCl_4$. The rate law is:
A. Rate $= k[Cl_2]^{1/2}$ (order $\dfrac{1}{2}$)
B. Rate $= k[CHCl_3]^2[Cl_2]$ (order $3$)
C. Rate $= k[CHCl_3][Cl_2]$ (order $2$)
D. Rate $= k[CHCl_3][Cl_2]^{1/2}$ (order $\dfrac{3}{2}$) ✓ Correct
Solution: From the equilibrium $[Cl] = (K[Cl_2])^{1/2}$. The slow step gives Rate $= k_2[Cl][CHCl_3] = k_2K^{1/2}[CHCl_3][Cl_2]^{1/2}$, an overall order of $\dfrac{3}{2}$.
Q18 — Order & Molecularity of a Reaction · medium · numerical
The iodide-catalysed decomposition $2H_2O_2 \rightarrow 2H_2O + O_2$ has the slow step $H_2O_2 + I^- \rightarrow H_2O + IO^-$ and a fast step $H_2O_2 + IO^- \rightarrow H_2O + O_2 + I^-$. The molecularity of the rate-determining step and the overall order of the reaction are, respectively:
A. $3$ and $2$
B. $2$ and $2$ ✓ Correct
C. $1$ and $1$
D. $2$ and $3$
Solution: The slow step involves two species ($H_2O_2$ and $I^-$), so its molecularity is 2 and Rate $= k[H_2O_2][I^-]$, giving an overall order of 2.
Q19 — Order & Molecularity of a Reaction · medium · numerical
A reaction between $X$ and $Y$ is found to be $\dfrac{3}{2}$ order overall and first order in $X$. The order with respect to $Y$ must be:
A. $\dfrac{3}{2}$
B. $2$
C. $\dfrac{1}{2}$ ✓ Correct
D. $1$
Solution: Overall order $=$ (order in $X$) $+$ (order in $Y$). Thus $\dfrac{3}{2} = 1 + (\text{order in }Y)$, giving order in $Y = \dfrac{1}{2}$.
Q20 — Order & Molecularity of a Reaction · medium · numerical
Assertion (A): A single elementary step can have a molecularity of 3 but never of 4. Reason (R): Simultaneous effective collisions of three particles are already rare, and the chance of four particles colliding at once is negligibly small. 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. A is true but R is false
D. Both A and R are true, and R is the correct explanation of A ✓ Correct
Solution: Termolecular steps are uncommon and higher molecularities are essentially forbidden precisely because the probability of a correctly oriented simultaneous collision falls off sharply with the number of particles. R correctly explains A.
Q21 — Order & Molecularity of a Reaction · medium · numerical
Assertion (A): Molecularity has no meaning for a complex (multistep) reaction. Reason (R): Molecularity is defined only for an elementary reaction that takes place in a single step.
A. A is true but R is false
B. Both A and R are true but R is not the correct explanation of A
C. A is false but R is true
D. Both A and R are true and R is the correct explanation of A ✓ Correct
Solution: Molecularity is meaningful only for an elementary (single-step) reaction; a complex reaction proceeds through several elementary steps, each with its own molecularity, so the overall reaction has none — R correctly explains A.
Q22 — Order & Molecularity of a Reaction · medium · numerical
For $2P + Q \rightarrow$ products the initial-rate data (in $mol,\,L,\,s$ units) are: Exp 1 $[P]=0.10,\,[Q]=0.20$, rate $=4.0\times10^{-3}$; Exp 2 $[P]=0.20,\,[Q]=0.20$, rate $=1.6\times10^{-2}$; Exp 3 $[P]=0.20,\,[Q]=0.40$, rate $=1.6\times10^{-2}$. The order with respect to $Q$ and the overall order are, respectively:
A. $1$ and $3$
B. $2$ and $4$
C. $0$ and $2$ ✓ Correct
D. $0$ and $3$
Solution: Exp 1$\to$2: $[P]$ doubles, rate $\times4$, so order in $P$ is 2. Exp 2$\to$3: $[Q]$ doubles, rate unchanged, so order in $Q$ is 0. Overall order $= 2 + 0 = 2$.
Q23 — Order & Molecularity of a Reaction · medium · numerical
A reaction has the rate law Rate $= k[A]^2[B]$. When $[A]=0.20\,M$ and $[B]=0.10\,M$ the rate is $4.0\times10^{-3}\,mol\,L^{-1}s^{-1}$. The value of the rate constant $k$ is:
A. $0.5\,L^2mol^{-2}s^{-1}$
B. $4.0\,L^2mol^{-2}s^{-1}$
C. $1.0\,L^2mol^{-2}s^{-1}$ ✓ Correct
D. $1.0\,L\,mol^{-1}s^{-1}$
Solution: $k = \dfrac{\text{rate}}{[A]^2[B]} = \dfrac{4.0\times10^{-3}}{(0.20)^2(0.10)} = \dfrac{4.0\times10^{-3}}{4.0\times10^{-3}} = 1.0$. A third-order reaction has $k$ in $L^2mol^{-2}s^{-1}$.
Q24 — Order & Molecularity of a Reaction · medium · numerical
For a reaction $A \rightarrow$ products, increasing $[A]$ nine-fold increases the rate 27-fold. The order of the reaction with respect to $A$ is:
A. $1$
B. $\dfrac{3}{2}$ ✓ Correct
C. $2$
D. $3$
Solution: If the order is $n$, then $9^n = 27$, i.e. $3^{2n} = 3^3$, so $2n = 3$ and $n = \dfrac{3}{2}$.
Q25 — Order & Molecularity of a Reaction · medium · numerical
A reaction obeys the rate law Rate $= k[A][B]^{-1}$. If $[B]$ is doubled while $[A]$ is held constant, the rate becomes:
A. reduced to one-fourth
B. unchanged
C. doubled
D. halved ✓ Correct
Solution: Because $B$ has order $-1$, the rate is inversely proportional to $[B]$; doubling $[B]$ therefore halves the rate. Such species act as inhibitors.
Q26 — Order & Molecularity of a Reaction · medium · numerical
For an elementary reaction $A + 2B \rightarrow$ products, the predicted rate law and molecularity are:
A. Rate $= k[A][B]^2$; molecularity $= 2$
B. Rate $= k[A][B]$; molecularity $= 2$
C. Rate $= k[A][B]^2$; molecularity $= 3$ ✓ Correct
D. the rate law cannot be predicted from the stoichiometry
Solution: For an elementary step the exponents equal the coefficients, so Rate $= k[A][B]^2$ and three molecules react in the step, giving a molecularity of 3 (order $=$ molecularity $=3$).
Q27 — Order & Molecularity of a Reaction · medium · numerical
For the complex chain reaction $H_2 + Cl_2 \rightarrow 2HCl$, the order cannot be predicted from the balanced equation because:
A. it is not an elementary reaction; its rate law and order must be found experimentally from the mechanism ✓ Correct
B. it is elementary, so the order is fixed at 2
C. the order of any reaction always equals the sum of the stoichiometric coefficients
D. gas-phase reactions are always first order
Solution: Order equals the sum of coefficients only for elementary reactions. $H_2 + Cl_2$ is a multi-step chain reaction, so its order is an experimental quantity determined by the rate-determining step, not by the overall stoichiometry.
Q28 — Order & Molecularity of a Reaction · medium · numerical
The rate constant of a reaction $A \rightarrow$ products has the units $L\,mol^{-1}\,s^{-1}$. If the concentration of $A$ is doubled, by what factor does the rate change?
A. $4$ ✓ Correct
B. $1$
C. $2$
D. $8$
Solution: Units $L\,mol^{-1}s^{-1} = mol^{1-n}L^{\,n-1}s^{-1}$ give $1-n=-1$, i.e. $n=2$ (second order). Doubling $[A]$ changes the rate by $2^2 = 4$.
Q29 — Order & Molecularity of a Reaction · medium · numerical
For a reaction $A \rightarrow$ products, a plot of the rate against $[A]^2$ is a straight line passing through the origin. The order of the reaction with respect to $A$ is:
A. $\dfrac{1}{2}$
B. $1$
C. $0$
D. $2$ ✓ Correct
Solution: A straight line through the origin means rate $\propto [A]^2$, i.e. Rate $= k[A]^2$, so the reaction is second order in $A$.
Q30 — Order & Molecularity of a Reaction · medium · numerical
For a zero-order reaction $A \rightarrow$ products, a plot of the rate of reaction against the concentration of $A$ is:
A. a straight line through the origin with positive slope
B. a straight line with negative slope
C. an exponentially rising curve
D. a horizontal line (the rate is independent of concentration) ✓ Correct
Solution: For a zero-order reaction Rate $= k[A]^0 = k$, a constant. A plot of rate versus $[A]$ is therefore a horizontal line, independent of concentration.