201. The overall reaction has overall order \(2\), and its balanced equation is
\[
A+B\rightarrow P
\]
Which conclusion is safest?
ⓐ. The reaction must be a bimolecular elementary step because the coefficient sum is \(2\)
ⓑ. The agreement alone does not establish an elementary step
ⓒ. The molecularity must be zero because the measured overall order is not a mechanism
ⓓ. The mechanism must contain exactly two steps because the overall order equals \(2\)
Correct Answer: The agreement alone does not establish an elementary step
Explanation: The overall order is \(2\), and the sum of the reactant coefficients is also \(2\). Numerical agreement is possible for an elementary bimolecular step. However, the same agreement can occur accidentally for a multistep mechanism whose experimental rate law is second order. Mechanism evidence is required before assigning molecularity to the overall equation. Neither the number of mechanism steps nor molecularity follows from order alone.
202. A complex reaction differs from an elementary reaction because a complex reaction:
ⓐ. must involve only one reactant
ⓑ. always has fractional molecularity
ⓒ. cannot possess an experimentally measurable rate
ⓓ. proceeds through two or more elementary steps
Correct Answer: proceeds through two or more elementary steps
Explanation: An elementary reaction occurs through one molecular event and is represented by one elementary step. A complex reaction proceeds through a sequence of elementary steps called a reaction mechanism. Species may be formed in one step and consumed in a later step during this sequence. The overall balanced equation is obtained by adding the elementary steps and cancelling species that do not remain in the net reaction. Complexity refers to the number of kinetic steps, not to whether the rate can be measured.
203. A mechanism contains the steps:
\[
A+X\rightarrow AX
\]
\[
AX+B\rightarrow P+X
\]
Which description of \(X\) is most suitable?
ⓐ. \(X\) is an intermediate because it is formed before it is consumed
ⓑ. \(X\) is a catalyst because it is consumed and later regenerated
ⓒ. \(X\) is a product because it appears on the right of the second step
ⓓ. \(X\) is a reactant consumed in the overall equation
Correct Answer: \(X\) is a catalyst because it is consumed and later regenerated
Explanation: Species \(X\) is used as a reactant in the first step. It is regenerated as a product in the second step. When the two steps are added, \(X\) cancels and is absent from the overall equation. This consume-then-regenerate pattern is characteristic of a catalyst. By contrast, \(AX\) is formed in the first step and consumed in the second, making \(AX\) the intermediate.
204. Use the consecutive steps shown below.
\[
2A\rightarrow I
\]
\[
I+B\rightarrow 2P
\]
The overall reaction and the molecularity of the first step are:
ⓐ. \(2A+B\rightarrow2P\), unimolecular
ⓑ. \(A+B\rightarrow2P\), bimolecular
ⓒ. \(2A+B\rightarrow2P\), bimolecular
ⓓ. \(2A+I+B\rightarrow I+2P\), termolecular
Correct Answer: \(2A+B\rightarrow2P\), bimolecular
Explanation: \( \textbf{Add the elementary steps:} \)
\[
2A+I+B\rightarrow I+2P
\]
\( \textbf{Cancel the intermediate }I\textbf{:} \)
\[
2A+\cancel{I}+B\rightarrow\cancel{I}+2P
\]
\( \textbf{Net equation:} \)
\[
2A+B\rightarrow2P
\]
\( \textbf{First-step particle count:} \) The first elementary event involves two particles of \(A\).
\[
\text{Molecularity}=2
\]
The first step is therefore bimolecular, even though only one chemical formula appears on its reactant side.
205. A proposed mechanism is:
\[
A+B\rightleftharpoons I \qquad \text{fast}
\]
\[
I+C\rightarrow P \qquad \text{slow}
\]
Writing \(r=k[I][C]\) as the final experimentally usable rate law is incomplete because:
ⓐ. The rate law of a slow step can never contain the concentration of an intermediate
ⓑ. Intermediate \(I\) must be eliminated using a relation involving measurable species
ⓒ. \(C\) is an overall product and must therefore be removed from every kinetic expression
ⓓ. The slow step should be ignored and the fast step alone should be used as the final rate law
Correct Answer: Intermediate \(I\) must be eliminated using a relation involving measurable species
Explanation: The slow elementary step directly suggests the expression \(r=k[I][C]\). However, \(I\) is an intermediate rather than an ordinary initial reactant whose concentration is usually controlled and measured directly. A final experimental rate law should normally be expressed using stable reactants or other measurable quantities. An additional valid relationship would therefore be needed to eliminate \([I]\). Without that substitution, the expression is only the rate law of the elementary slow step, not yet a complete experimentally useful overall law.
206. Match each term in Column I with its description in Column II.
| Column I | Column II |
|---|
| P. Elementary step | 1. Species formed and later consumed |
| Q. Intermediate | 2. Sequence of elementary steps |
| R. Reaction mechanism | 3. Single molecular event |
| S. Rate-determining step | 4. Step that limits the overall rate in a simple mechanism |
ⓐ. P-2, Q-1, R-3, S-4
ⓑ. P-3, Q-1, R-2, S-4
ⓒ. P-3, Q-4, R-2, S-1
ⓓ. P-1, Q-3, R-4, S-2
Correct Answer: P-3, Q-1, R-2, S-4
Explanation: An elementary step represents one molecular event, so P matches 3. An intermediate is produced and then consumed, giving Q-1. A reaction mechanism is the ordered sequence of elementary steps, so R-2 is appropriate. In a simple bottleneck description, the rate-determining step limits the observed overall rate, giving S-4. The matching separates the individual event, the sequence, the temporary species, and the controlling step.
207. The following mechanism is proposed:
\[
A\rightarrow I+B \qquad \text{slow}
\]
\[
I+C\rightarrow D \qquad \text{fast}
\]
The overall equation and predicted simple rate law are:
ⓐ. \(A+C\rightarrow B+D\) and \(r=k[A]\)
ⓑ. \(A+I+C\rightarrow I+B+D\) and \(r=k[I][C]\)
ⓒ. \(A+C\rightarrow B+D\) and \(r=k[A][C]\)
ⓓ. \(A\rightarrow I+B\) and \(r=k[C]\)
Correct Answer: \(A+C\rightarrow B+D\) and \(r=k[A]\)
Explanation: \( \textbf{Add the two steps:} \)
\[
A+I+C\rightarrow I+B+D
\]
\( \textbf{Cancel the intermediate:} \)
\[
A+\cancel{I}+C\rightarrow\cancel{I}+B+D
\]
\( \textbf{Overall equation:} \)
\[
A+C\rightarrow B+D
\]
\( \textbf{Identify the controlling step:} \) The first step is labelled slow.
\( \textbf{Elementary slow-step law:} \)
\[
r=k[A]
\]
The later reactant \(C\) appears in the overall equation but not in the simple predicted law because it enters only after the slow unimolecular event.
208. Complete the missing species so that it behaves as an intermediate.
\[
A+B\rightarrow \underline{\hspace{1cm}}
\]
\[
\underline{\hspace{1cm}}+C\rightarrow P
\]
ⓐ. \(A\)
ⓑ. \(P\)
ⓒ. \(I\)
ⓓ. \(C\)
Correct Answer: \(I\)
Explanation: The same species must be formed in the first step and consumed in the second step. Writing \(I\) in both blanks produces that pattern. When the equations are added, \(I\) appears on both sides and cancels. The resulting overall equation is \(A+B+C\rightarrow P\). A stable initial reactant or final product would not satisfy the defining formed-then-consumed behaviour.
209. The slowest step in a simple consecutive mechanism is often compared with a narrow section of a pipeline because:
ⓐ. the narrow section limits the overall flow
ⓑ. every molecule must remain permanently in the slow step
ⓒ. the faster steps stop occurring completely
ⓓ. the slow step changes the equilibrium constant
Correct Answer: the narrow section limits the overall flow
Explanation: Consecutive elementary steps must collectively process material from reactants to products. If one step is much slower than the others, material cannot pass through the entire sequence faster than that step permits. The faster steps may adjust rapidly before or after the bottleneck. They do not necessarily stop, and the equilibrium constant is not changed merely because one step is slow. The analogy captures how one kinetically limiting event can control the observed overall rate.
210. A reaction-coordinate profile for a proposed mechanism contains two energy maxima separated by one minimum. The profile most naturally represents:
ⓐ. one elementary step with no intermediate
ⓑ. three elementary steps with two intermediates
ⓒ. two overall reactions occurring at equilibrium
ⓓ. two elementary steps with one intermediate
Correct Answer: two elementary steps with one intermediate
Explanation: Each energy maximum corresponds qualitatively to a transition-state region for an elementary step. Two maxima therefore indicate two elementary steps. The minimum between them represents a species or configuration with finite relative stability between the two barriers. This minimum is associated with an intermediate. The profile does not by itself identify the chemical formula of that intermediate, but it shows the multistep character of the pathway.
211. A two-step mechanism has a first activation barrier much higher than the second barrier. Under otherwise comparable conditions, the most reasonable qualitative inference is:
ⓐ. the second step must determine the equilibrium constant
ⓑ. the first step is likely slower and rate-controlling
ⓒ. both steps must have identical rates
ⓓ. the reaction must be zero order
Correct Answer: the first step is likely slower and rate-controlling
Explanation: A higher activation barrier generally corresponds to a smaller rate constant when the other kinetic factors are comparable. The first elementary event would therefore be expected to proceed more slowly than the second. It may act as the rate-determining step and limit the overall sequence. This inference is qualitative because pre-exponential factors and detailed conditions can also influence rate. The barrier comparison does not determine the overall reaction order or equilibrium constant by itself.
212. A simple rate-determining-step treatment is assessed through the statements below.
Statement I: The slow step may control the observed overall rate.
Statement II: A rate law taken from a slow elementary step may require removal of an intermediate concentration.
Statement III: The overall balanced equation alone always identifies the slow step.
The valid statements are:
ⓐ. I and III only
ⓑ. II and III only
ⓒ. I and II only
ⓓ. I, II and III
Correct Answer: I and II only
Explanation: A much slower elementary step can serve as the kinetic bottleneck of a mechanism. Its elementary rate expression may provide the starting point for constructing the overall rate law. If that expression contains an intermediate, an additional allowed relation is needed to rewrite it in terms of measurable species. The net balanced equation does not reveal the relative speeds of hidden elementary steps. Experimental kinetic evidence is therefore required before identifying a rate-determining step.
213. A mechanism is proposed as:
\[
A+B\rightarrow I \qquad \text{fast}
\]
\[
I+B\rightarrow P \qquad \text{slow}
\]
A claim states that the overall rate law must be \(r=k[A][B]^2\) simply because the net equation is \(A+2B\rightarrow P\). The claim is unreliable because:
ⓐ. the intermediate \(I\) must appear in the overall balanced equation
ⓑ. every complex reaction is necessarily zero order
ⓒ. the slow step cannot influence the observed rate
ⓓ. net stoichiometry alone cannot determine an overall rate law
Correct Answer: net stoichiometry alone cannot determine an overall rate law
Explanation: Adding the steps does give the overall equation \(A+2B\rightarrow P\). However, this net equation does not describe one elementary three-particle event. The slow step directly suggests dependence on \([I][B]\), not automatically on \([A][B]^2\). A further valid relationship would be required to replace the intermediate concentration \([I]\). The proposed final rate law might or might not result after proper mechanism treatment, but it cannot be justified merely by copying the overall coefficients.
214. In the initial-rate method, the reaction rate is measured as close as practical to \(t=0\) mainly because:
ⓐ. the reaction has already reached equilibrium at that time
ⓑ. reactant concentrations are near their initial values
ⓒ. the rate constant is zero before appreciable product forms
ⓓ. all reactant concentrations are necessarily equal at \(t=0\)
Correct Answer: reactant concentrations are near their initial values
Explanation: The initial-rate method relates the measured starting rate to known initial reactant concentrations. Near \(t=0\), only a small fraction of the reactants has usually been consumed. Their concentrations are therefore close to the accurately prepared values. Product concentrations are also low, reducing complications from reverse reaction or product-related effects where relevant. Measuring only after substantial progress would make the concentration conditions less certain and less comparable between experiments.
215. The experiments below are planned for a reaction involving \(A\) and \(B\).
| Experiment | \([A]_0\) | \([B]_0\) |
|---|
| P | \(0.10\,mol\,L^{-1}\) | \(0.20\,mol\,L^{-1}\) |
| Q | \(0.20\,mol\,L^{-1}\) | \(0.20\,mol\,L^{-1}\) |
| R | \(0.20\,mol\,L^{-1}\) | \(0.40\,mol\,L^{-1}\) |
| S | \(0.40\,mol\,L^{-1}\) | \(0.60\,mol\,L^{-1}\) |
The pair most directly suited to determining the order with respect to \(A\) is:
ⓐ. Q and R
ⓑ. P and Q
ⓒ. P and R
ⓓ. R and S
Correct Answer: P and Q
Explanation: To isolate the effect of \(A\), the initial concentration of \(B\) should remain constant. Experiments P and Q both have \([B]_0=0.20\,mol\,L^{-1}\). Between these experiments, \([A]_0\) alone doubles from \(0.10\,mol\,L^{-1}\) to \(0.20\,mol\,L^{-1}\). The corresponding initial-rate ratio can therefore be attributed to the change in \(A\). Q and R would instead isolate the effect of \(B\).
216. Assertion: Initial-rate experiments should be performed at the same temperature when concentration effects are being compared.
Reason: A temperature change can alter the rate constant and obscure the concentration dependence.
ⓐ. Both Assertion and Reason are true, and Reason explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: Initial-rate comparisons are intended to reveal how concentration changes affect rate. Temperature influences the rate constant \(k\), often quite strongly. If temperature differs between runs, the rate ratio may reflect both concentration and rate-constant changes. Maintaining the same temperature keeps \(k\) comparable across the selected experiments. The Reason therefore explains why temperature control is essential for valid concentration-order determination.
217. A reaction obeys \(r=k[A][B]^2\), where \(k=1.5\,L^2\,mol^{-2}\,s^{-1}\). Its initial rate when \([A]_0=0.30\,mol\,L^{-1}\) and \([B]_0=0.20\,mol\,L^{-1}\) is:
ⓐ. \(9.0\times10^{-3}\,mol\,L^{-1}\,s^{-1}\)
ⓑ. \(1.8\times10^{-2}\,mol\,L^{-1}\,s^{-1}\)
ⓒ. \(4.5\times10^{-2}\,mol\,L^{-1}\,s^{-1}\)
ⓓ. \(1.2\times10^{-1}\,mol\,L^{-1}\,s^{-1}\)
Correct Answer: \(1.8\times10^{-2}\,mol\,L^{-1}\,s^{-1}\)
Explanation: \( \textbf{Rate law:} \)
\[
r=k[A][B]^2
\]
\( \textbf{Given values:} \)
\[
k=1.5\,L^2\,mol^{-2}\,s^{-1}
\]
\[
[A]=0.30\,mol\,L^{-1},\qquad[B]=0.20\,mol\,L^{-1}
\]
\( \textbf{Square the concentration of }B\textbf{:} \)
\[
[B]^2=(0.20)^2=0.040\,(mol\,L^{-1})^2
\]
\( \textbf{Substitution:} \)
\[
r=(1.5)(0.30)(0.040)
\]
\( \textbf{Intermediate product:} \)
\[
(0.30)(0.040)=0.012
\]
\( \textbf{Rate calculation:} \)
\[
r=1.5(0.012)=0.018
\]
\( \textbf{Scientific notation:} \)
\[
r=1.8\times10^{-2}\,mol\,L^{-1}\,s^{-1}
\]
The concentration and rate-constant units combine to give the required concentration-per-time rate unit.
218. Increasing \([A]_0\) by a factor of \(4\) increases the initial rate by a factor of \(8\), while all other variables remain unchanged. The partial order with respect to \(A\) is:
ⓐ. \(\frac{1}{2}\)
ⓑ. \(1\)
ⓒ. \(\frac{3}{2}\)
ⓓ. \(2\)
Correct Answer: \(\frac{3}{2}\)
Explanation: \( \textbf{Rate-ratio equation:} \)
\[
\frac{r_2}{r_1}
=
\left(\frac{[A]_2}{[A]_1}\right)^m
\]
\( \textbf{Insert the observed factors:} \)
\[
8=4^m
\]
\( \textbf{Express both numbers with base }2\textbf{:} \)
\[
8=2^3
\]
\[
4^m=(2^2)^m=2^{2m}
\]
\( \textbf{Equate exponents:} \)
\[
2m=3
\]
\[
m=\frac{3}{2}
\]
\( \textbf{Verification:} \)
\[
4^{3/2}=(\sqrt{4})^3=2^3=8
\]
The observed concentration response therefore represents half-integral order in \(A\).
219. When \([B]\) is held constant for the rate law \(r=k[A]^m[B]^n\), the missing exponent in the rate-ratio relation is ______.
\[
\frac{r_2}{r_1}
=
\left(\frac{[A]_2}{[A]_1}\right)^{\underline{\hspace{1cm}}}
\]
ⓐ. \(m\)
ⓑ. \(n\)
ⓒ. \(m+n\)
ⓓ. \(\frac{1}{m}\)
Correct Answer: \(m\)
Explanation: The two rate equations contain the same value of \(k\) because the temperature and catalyst condition are unchanged. They also contain the same value of \([B]^n\) because \([B]\) is held constant. These common factors cancel when one rate is divided by the other. Only the changing concentration factor involving \(A\) remains. Its exponent is the partial order \(m\), giving the displayed rate-ratio relation.
220. Two initial-rate experiments change \([A]\) and \([B]\) simultaneously. Their data alone generally cannot determine the separate partial orders because:
ⓐ. initial rates cannot be measured when two reactants are present
ⓑ. the overall order must always equal the coefficient sum
ⓒ. the rate constant changes whenever concentration changes
ⓓ. the rate factor combines both concentration changes
Correct Answer: the rate factor combines both concentration changes
Explanation: For \(r=k[A]^m[B]^n\), changing both concentrations produces the factor \(\left(\frac{[A]_2}{[A]_1}\right)^m\left(\frac{[B]_2}{[B]_1}\right)^n\). A single measured rate ratio then contains both unknown exponents. Without another independent comparison, the individual contributions cannot usually be separated. Controlled experimental design therefore changes one reactant concentration while holding the others fixed. The rate constant remains unchanged when temperature and catalyst conditions are maintained.
