201. A first-order reaction can sometimes be followed using a measurable quantity such as pressure, optical rotation, absorbance, or titration reading instead of directly measuring concentration. This method is valid when
ⓐ. the measured quantity is proportional to the required concentration or has a known concentration-equivalent relation.
ⓑ. the measured quantity remains constant throughout the reaction.
ⓒ. the measured quantity depends only on the catalyst used.
ⓓ. the measured quantity changes only after the reaction is complete.
Correct Answer: the measured quantity is proportional to the required concentration or has a known concentration-equivalent relation.
Explanation: The first-order integrated rate equation compares the amount or concentration of reactant remaining at different times. A measurable property may be used in place of concentration only when it tracks that concentration in a definite way. Some properties, such as optical rotation or absorbance in suitable cases, may be directly proportional to concentration. Other properties, such as total pressure in a gas reaction, may require conversion using the reaction relation before they can represent the reactant amount. Without this proportional or convertible relation, the measured value cannot be safely substituted into the first-order expression.
202. Which statement best defines pseudo-first-order kinetics?
ⓐ. A first-order reaction that becomes zero order at high temperature
ⓑ. A zero-order reaction that is treated as first order for graph plotting
ⓒ. A higher-order reaction appearing first order due to one excess reactant
ⓓ. A first-order reaction whose rate constant changes with concentration
Correct Answer: A higher-order reaction appearing first order due to one excess reactant
Explanation: In pseudo-first-order behavior, the actual rate law involves more than one reactant, but one of them is present in such large excess that its concentration changes negligibly. That nearly constant concentration is combined with the true rate constant, making the reaction appear first order in the other reactant.
203. A reaction has the true rate law \(r = k[A][B]\). If \([B]\) is kept in very large excess so that it remains nearly constant during the experiment, the apparent rate law becomes
ⓐ. \(r = k[A]^2\)
ⓑ. \(r = k[B]^2\)
ⓒ. \(r = k[A][B]\)
ⓓ. \(r = k'[A]\)
Correct Answer: \(r = k'[A]\)
Explanation: Since \([B]\) is in large excess, its concentration changes very little and can be treated as constant. Then
\[
r = k[A][B] = (k[B])[A]
\]
If \(k[B]\) is written as a new constant \(k'\), the apparent law becomes
\[
r = k'[A]
\]
So the reaction appears first order in \(A\).
204. The hydrolysis of an ester follows the rate law \(r = k[\text{ester}][H_2O]\). If water is present in large excess at \(10.0\,\text{mol L}^{-1}\) and \(k = 2.0 \times 10^{-3}\,\text{L mol}^{-1}\text{s}^{-1}\), what is the pseudo-first-order constant \(k'\)?
ⓐ. \(2.0 \times 10^{-4}\,\text{s}^{-1}\)
ⓑ. \(2.0 \times 10^{-2}\,\text{s}^{-1}\)
ⓒ. \(5.0 \times 10^{2}\,\text{s}^{-1}\)
ⓓ. \(1.2 \times 10^{-2}\,\text{s}^{-1}\)
Correct Answer: \(2.0 \times 10^{-2}\,\text{s}^{-1}\)
Explanation: \( \textbf{Given:} \)
\[
r = k[\text{ester}][H_2O]
\]
\[
[H_2O] = 10.0\,\text{mol L}^{-1}
\]
\[
k = 2.0 \times 10^{-3}\,\text{L mol}^{-1}\text{s}^{-1}
\]
\( \textbf{Required:} \)
Pseudo-first-order constant, \(k'\)
\( \textbf{Relevant formula:} \)
When water is in large excess,
\[
k' = k[H_2O]
\]
\( \textbf{Why this formula applies:} \)
The concentration of water remains nearly constant and is absorbed into the rate constant.
\( \textbf{Substitution:} \)
\[
k' = (2.0 \times 10^{-3})(10.0)
\]
\( \textbf{Intermediate simplification:} \)
\[
k' = 2.0 \times 10^{-2}
\]
\( \textbf{Unit / notation check:} \)
\[
(\text{L mol}^{-1}\text{s}^{-1})(\text{mol L}^{-1}) = \text{s}^{-1}
\]
\( \textbf{Final Answer:} \)
\[
k' = 2.0 \times 10^{-2}\,\text{s}^{-1}
\]
205. A reaction has the following half-life data at the same temperature:
At \([R]_0 = 0.20\,\text{mol L}^{-1}\), \(t_{1/2} = 30\,\text{s}\)
At \([R]_0 = 0.40\,\text{mol L}^{-1}\), \(t_{1/2} = 30\,\text{s}\)
Which order is most consistent with these data?
ⓐ. Zero order
ⓑ. Second order
ⓒ. First order
ⓓ. Order cannot be inferred
Correct Answer: First order
Explanation: For a first-order reaction,
\[
t_{1/2} = \frac{0.693}{k}
\]
so the half-life does not depend on the initial concentration. Since the half-life remains the same even when the initial concentration doubles, the behavior is most consistent with first-order kinetics.
206. The unit of the rate constant for a reaction is \(\text{L mol}^{-1}\text{s}^{-1}\). What is the overall order of the reaction?
ⓐ. Second order
ⓑ. First order
ⓒ. Zeroth order
ⓓ. Third order
Correct Answer: Second order
Explanation: For a reaction of overall order \(n\), the unit of \(k\) depends on \(n\). The standard unit pattern is:
Zero order:
\[
\text{mol L}^{-1}\text{s}^{-1}
\]
First order:
\[
\text{s}^{-1}
\]
Second order:
\[
\text{L mol}^{-1}\text{s}^{-1}
\]
Since the given unit is \(\text{L mol}^{-1}\text{s}^{-1}\), the reaction is second order.
207. A reaction has a rate constant with unit \(\text{s}^{-1}\), and its half-life is independent of the initial concentration. Which statement is correct?
ⓐ. The reaction is zero order.
ⓑ. The reaction is second order.
ⓒ. The reaction is pseudo-zero-order.
ⓓ. The reaction is first order.
Correct Answer: The reaction is first order.
Explanation: Two key facts are given. First, the unit \(\text{s}^{-1}\) is characteristic of a first-order rate constant. Second, a first-order reaction has half-life
\[
t_{1/2} = \frac{0.693}{k}
\]
which is independent of initial concentration. Both observations point to first-order kinetics.
208. Two reactions are studied at the same temperature.
Reaction X:
At \(0.20\,\text{mol L}^{-1}\), \(t_{1/2} = 10\,\text{s}\)
At \(0.40\,\text{mol L}^{-1}\), \(t_{1/2} = 20\,\text{s}\)
Reaction Y:
At \(0.20\,\text{mol L}^{-1}\), \(t_{1/2} = 15\,\text{s}\)
At \(0.40\,\text{mol L}^{-1}\), \(t_{1/2} = 15\,\text{s}\)
Which conclusion is correct?
ⓐ. Both X and Y are zero order.
ⓑ. X is zero order and Y is first order.
ⓒ. X is first order and Y is zero order.
ⓓ. Both X and Y are first order.
Correct Answer: X is zero order and Y is first order.
Explanation: \( \textbf{Given:} \)
Reaction X has half-life that doubles when initial concentration doubles.
Reaction Y has constant half-life even when initial concentration changes.
\( \textbf{Required:} \)
Identify the likely orders of X and Y
Relevant principles:
For zero-order kinetics,
\[
t_{1/2} = \frac{[R]_0}{2k}
\]
so half-life depends on initial concentration.
For first-order kinetics,
\[
t_{1/2} = \frac{0.693}{k}
\]
so half-life is independent of initial concentration.
Apply to Reaction X:
Half-life changes from \(10\,\text{s}\) to \(20\,\text{s}\) when concentration doubles.
This matches zero-order behavior.
Apply to Reaction Y:
Half-life remains \(15\,\text{s}\) at both concentrations.
This matches first-order behavior.
\( \textbf{Final Answer:} \)
Reaction X is zero order and Reaction Y is first order.
209. A reaction has the true rate law \(r = k[A][B]\), but during the experiment \([B]\) is kept very large compared with \([A]\). Why can the reaction behave as pseudo-first-order?
ⓐ. Because \([A]\) becomes exactly zero throughout the reaction
ⓑ. Because the true rate law changes permanently to zero order
ⓒ. Because \([B]\) is nearly constant and included in a new constant
ⓓ. Because the balanced equation becomes that of a one-reactant reaction
Correct Answer: Because \([B]\) is nearly constant and included in a new constant
Explanation: When one reactant is present in large excess, its concentration changes very little during the reaction. In that case, for
\[
r = k[A][B]
\]
the term \([B]\) may be treated as constant. Then
\[
r = k[B][A] = k'[A]
\]
where \(k' = k[B]\). The reaction then appears first order in \(A\) under those conditions.
210. For the reaction \(A + B \rightarrow \text{products}\) with rate law \(r = k[A][B]\), reactant \(B\) is taken in very large excess. Which apparent rate law is correct during most of the reaction?
ⓐ. \(r = k'[A]\)
ⓑ. \(r = k'[B]\)
ⓒ. \(r = k'[A]^2\)
ⓓ. \(r = k'\)
Correct Answer: \(r = k'[A]\)
Explanation: Since \([B]\) is very large, it changes negligibly and can be treated as nearly constant. Therefore,
\[
r = k[A][B]
\]
becomes
\[
r = (k[B])[A] = k'[A]
\]
So the reaction appears first order with respect to \(A\).
211. Which statement correctly compares successive half-lives of zero-order and first-order reactions at the same temperature?
ⓐ. Both have equal successive half-lives.
ⓑ. Both have increasing successive half-lives.
ⓒ. Zero-order has equal successive half-lives, but first-order does not.
ⓓ. First-order has equal successive half-lives, but zero-order does not.
Correct Answer: First-order has equal successive half-lives, but zero-order does not.
Explanation: For a first-order reaction,
\[
t_{1/2} = \frac{0.693}{k}
\]
so each half-life is constant at fixed temperature. For a zero-order reaction,
\[
t_{1/2} = \frac{[R]_0}{2k}
\]
so the half-life depends on the starting concentration and changes from one stage to the next.
212. The unit of the rate constant for a reaction is \(\text{mol L}^{-1}\text{s}^{-1}\). What is the overall order of the reaction?
ⓐ. First order
ⓑ. Zero order
ⓒ. Second order
ⓓ. Third order
Correct Answer: Zero order
Explanation: In a zero-order reaction, the rate law is
\[
r = k
\]
So the rate constant has the same unit as the rate itself. Since reaction rate has unit concentration per unit time, the unit of \(k\) becomes
\[
\text{mol L}^{-1}\text{s}^{-1}
\]
which identifies zero-order kinetics.
213. A zero-order reaction has \([R]_0 = 0.50\,\text{mol L}^{-1}\) and \(k = 0.025\,\text{mol L}^{-1}\text{s}^{-1}\). What is its half-life?
ⓐ. \(10\,\text{s}\)
ⓑ. \(20\,\text{s}\)
ⓒ. \(0.10\,\text{s}\)
ⓓ. \(40\,\text{s}\)
Correct Answer: \(10\,\text{s}\)
Explanation: \( \textbf{Given:} \)
\[
[R]_0 = 0.50\,\text{mol L}^{-1}
\]
\[
k = 0.025\,\text{mol L}^{-1}\text{s}^{-1}
\]
\( \textbf{Required:} \)
Half-life, \(t_{1/2}\)
\( \textbf{Relevant formula:} \)
\[
t_{1/2} = \frac{[R]_0}{2k}
\]
\( \textbf{Why this formula applies:} \)
The reaction is zero order.
\( \textbf{Substitution:} \)
\[
t_{1/2} = \frac{0.50}{2 \times 0.025}
\]
\( \textbf{Intermediate simplification:} \)
\[
t_{1/2} = \frac{0.50}{0.050}
\]
\( \textbf{Final simplification:} \)
\[
t_{1/2} = 10\,\text{s}
\]
\( \textbf{Unit / notation check:} \)
The concentration units cancel, leaving time in seconds.
\( \textbf{Final Answer:} \)
\[
t_{1/2} = 10\,\text{s}
\]
214. Which rate law has a rate constant with unit \(\text{s}^{-1}\) when rate is in \(\text{mol L}^{-1}\text{s}^{-1}\) and concentration is in \(\text{mol L}^{-1}\)?
ⓐ. \(r = k\)
ⓑ. \(r = k[A]^2\)
ⓒ. \(r = k[A][B]\)
ⓓ. \(r = k[A]\)
Correct Answer: \(r = k[A]\)
Explanation: For a first-order rate law,
\[
r = k[A]
\]
so
\[
k = \frac{r}{[A]}
\]
Using the standard units of rate and concentration,
\[
k = \frac{\text{mol L}^{-1}\text{s}^{-1}}{\text{mol L}^{-1}} = \text{s}^{-1}
\]
Thus a rate law of the form \(r = k[A]\) gives \(k\) the unit \(\text{s}^{-1}\).
215. A reaction has the true rate law \(r = k[A][B]\). If \(k = 4.0 \times 10^{-3}\,\text{L mol}^{-1}\text{s}^{-1}\) and \([B] = 25\,\text{mol L}^{-1}\) remains effectively constant, what is the pseudo-first-order constant \(k'\)?
ⓐ. \(1.0 \times 10^{-4}\,\text{s}^{-1}\)
ⓑ. \(1.0 \times 10^{-1}\,\text{s}^{-1}\)
ⓒ. \(2.9 \times 10^{-2}\,\text{s}^{-1}\)
ⓓ. \(6.3 \times 10^{-3}\,\text{s}^{-1}\)
Correct Answer: \(1.0 \times 10^{-1}\,\text{s}^{-1}\)
Explanation: \( \textbf{Given:} \)
\[
r = k[A][B]
\]
\[
k = 4.0 \times 10^{-3}\,\text{L mol}^{-1}\text{s}^{-1}
\]
\[
[B] = 25\,\text{mol L}^{-1}
\]
\( \textbf{Required:} \)
Pseudo-first-order constant, \(k'\)
\( \textbf{Relevant formula:} \)
\[
k' = k[B]
\]
\( \textbf{Why this formula applies:} \)
Since \([B]\) is in large excess, it remains nearly constant and is absorbed into the rate constant.
\( \textbf{Substitution:} \)
\[
k' = (4.0 \times 10^{-3})(25)
\]
\( \textbf{Intermediate simplification:} \)
\[
k' = 100 \times 10^{-3}
\]
\( \textbf{Final simplification:} \)
\[
k' = 1.0 \times 10^{-1}\,\text{s}^{-1}
\]
\( \textbf{Unit / notation check:} \)
\[
(\text{L mol}^{-1}\text{s}^{-1})(\text{mol L}^{-1}) = \text{s}^{-1}
\]
\( \textbf{Final Answer:} \)
\[
k' = 1.0 \times 10^{-1}\,\text{s}^{-1}
\]
216. A reaction shows the following behavior at the same temperature:
Sample 1: initial concentration \(0.20\,\text{mol L}^{-1}\), half-life \(8\,\text{s}\)
Sample 2: initial concentration \(0.40\,\text{mol L}^{-1}\), half-life \(16\,\text{s}\)
Which order is most consistent with these data?
ⓐ. First order
ⓑ. Second order
ⓒ. Fractional order
ⓓ. Zero order
Correct Answer: Zero order
Explanation: In zero-order kinetics,
\[
t_{1/2} = \frac{[R]_0}{2k}
\]
so the half-life is directly proportional to the initial concentration. Here, when the initial concentration doubles from \(0.20\) to \(0.40\,\text{mol L}^{-1}\), the half-life also doubles from \(8\) to \(16\,\text{s}\). This is the expected zero-order pattern.
217. The unit of the rate constant for a reaction is \(\text{L}^2\text{mol}^{-2}\text{s}^{-1}\). What is the overall order of the reaction?
ⓐ. First order
ⓑ. Second order
ⓒ. Third order
ⓓ. Zero order
Correct Answer: Third order
Explanation: The unit of \(k\) changes with the overall order. The standard pattern is: zero order \(\rightarrow \text{mol L}^{-1}\text{s}^{-1}\), first order \(\rightarrow \text{s}^{-1}\), second order \(\rightarrow \text{L mol}^{-1}\text{s}^{-1}\), and third order \(\rightarrow \text{L}^2\text{mol}^{-2}\text{s}^{-1}\). Therefore the given unit corresponds to a third-order reaction.
218. A reaction has a rate constant with unit \(\text{s}^{-1}\), and its half-life remains the same when the initial concentration is changed. Which order is most consistent with these facts?
ⓐ. First order
ⓑ. Zero order
ⓒ. Second order
ⓓ. Third order
Correct Answer: First order
Explanation: A first-order reaction has rate constant unit \(\text{s}^{-1}\) and half-life \(t_{1/2} = \frac{0.693}{k}\), which is independent of initial concentration. Both clues point to first-order kinetics.
219. For a zero-order reaction at fixed temperature, the initial concentration is tripled. What happens to the half-life?
ⓐ. It becomes one-third.
ⓑ. It becomes one-half.
ⓒ. It remains the same.
ⓓ. It triples.
Correct Answer: It triples.
Explanation: For a zero-order reaction, \(t_{1/2} = \frac{[R]_0}{2k}\). Since \(k\) remains constant at fixed temperature, the half-life is directly proportional to the initial concentration. Therefore tripling \([R]_0\) triples the half-life.
220. Which combination is most characteristic of a zero-order reaction?
ⓐ. \(\log [R]\) versus \(t\) is linear and \(k\) has unit \(\text{s}^{-1}\)
ⓑ. \([R]\) versus \(t\) is linear and \(k\) has unit \(\text{mol L}^{-1}\text{s}^{-1}\)
ⓒ. Half-life is independent of initial concentration and \(k\) has unit \(\text{s}^{-1}\)
ⓓ. \(\frac{1}{[R]}\) versus \(t\) is linear and \(k\) has unit \(\text{L mol}^{-1}\text{s}^{-1}\)
Correct Answer: \([R]\) versus \(t\) is linear and \(k\) has unit \(\text{mol L}^{-1}\text{s}^{-1}\)
Explanation: A zero-order reaction follows \([R]_t = [R]_0 - kt\), so a plot of \([R]\) versus \(t\) is a straight line. Also, since the rate law is \(r = k\), the unit of \(k\) is the same as the unit of rate, namely concentration per unit time.
