301. Which observation is most consistent with first-order kinetics rather than zero-order kinetics?
ⓐ. \([R]\) decreases linearly with time and successive half-lives become shorter.
ⓑ. Rate remains independent of concentration while \([R]\) decreases uniformly.
ⓒ. A plot of \([R]\) versus \(t\) is a straight line with slope \(-k\).
ⓓ. A plot of \(\log [R]\) versus \(t\) is linear and successive half-lives are equal.
Correct Answer: A plot of \(\log [R]\) versus \(t\) is linear and successive half-lives are equal.
Explanation: First-order kinetics is identified by the linear relation between \(\log [R]\) and \(t\) and by a half-life that remains constant at fixed temperature. Zero-order kinetics instead gives a linear \([R]\) versus \(t\) plot and unequal successive half-lives.
302. A reaction has \(k\) in \(\text{s}^{-1}\) and equal half-lives at fixed temperature. Which expression should be used directly to calculate the time taken for concentration to fall from \([R]_0\) to \([R]_t\)?
ⓐ. \([R]_t = [R]_0 - kt\)
ⓑ. \(t = \frac{2.303}{k}\log\frac{[R]_0}{[R]_t}\)
ⓒ. \(\log [R]_t = \log [R]_0 - \frac{kt}{2.303}\)
ⓓ. \(\log [R]_t = \log [R]_0 + \frac{kt}{2.303}\)
Correct Answer: \(t = \frac{2.303}{k}\log\frac{[R]_0}{[R]_t}\)
Explanation: The unit \(\text{s}^{-1}\) and constant half-life indicate first-order kinetics. Rearranging the integrated first-order equation gives
\[
t = \frac{2.303}{k}\log\frac{[R]_0}{[R]_t}
\]
which directly calculates time from concentration data.
303. For ester hydrolysis with true rate law \(r = k[\text{ester}][H_2O]\), water is present in large excess. Which expression gives the pseudo-first-order rate law?
ⓐ. \(r = k'[\text{ester}]\), where \(k' = k[H_2O]\)
ⓑ. \(r = k'[H_2O]\), where \(k' = k[\text{ester}]\)
ⓒ. \(r = k[\text{ester}]^2\), because water is ignored
ⓓ. \(r = k\), because the reaction becomes zero order
Correct Answer: \(r = k'[\text{ester}]\), where \(k' = k[H_2O]\)
Explanation: When water is present in very large excess, its concentration changes negligibly during the reaction. The nearly constant \([H_2O]\) can be combined with \(k\) to form an apparent constant \(k' = k[H_2O]\). The observed rate law then becomes \(r = k'[\text{ester}]\), which has first-order form with respect to ester.
304. For the decomposition \(A(g) \rightarrow 2B(g)\), the initial pressure of pure \(A\) is \(100\,\text{mmHg}\) and the total pressure at some time is \(175\,\text{mmHg}\). What fraction of \(A\) remains undecomposed at that time?
ⓐ. \(\frac{1}{2}\)
ⓑ. \(\frac{3}{4}\)
ⓒ. \(\frac{1}{4}\)
ⓓ. \(\frac{1}{8}\)
Correct Answer: \(\frac{1}{4}\)
Explanation: \( \textbf{Given:} \)
\[
P_0 = 100\,\text{mmHg}
\]
\[
P_t = 175\,\text{mmHg}
\]
\( \textbf{Required:} \)
Fraction of \(A\) remaining
\( \textbf{Relevant principle:} \)
For
\[
A(g) \rightarrow 2B(g)
\]
the pressure of unreacted \(A\) is
\[
P_A = 2P_0 - P_t
\]
\( \textbf{Why this formula applies:} \)
Total pressure increases because one mole of \(A\) forms two moles of gaseous product.
\( \textbf{Substitution:} \)
\[
P_A = 2(100) - 175 = 25\,\text{mmHg}
\]
\( \textbf{Now find the fraction remaining:} \)
\[
\frac{P_A}{P_0} = \frac{25}{100} = \frac{1}{4}
\]
\( \textbf{Final Answer:} \)
\[
\frac{1}{4}
\]
305. Reaction X is zero order with \(k = 0.020\,\text{mol L}^{-1}\text{s}^{-1}\) and \([R]_0 = 1.0\,\text{mol L}^{-1}\). Reaction Y is first order with \(k = 0.0693\,\text{s}^{-1}\) and the same initial concentration. What are the concentrations after \(20\,\text{s}\)?
ⓐ. X: \(0.20\,\text{mol L}^{-1}\), Y: \(0.60\,\text{mol L}^{-1}\)
ⓑ. X: \(0.25\,\text{mol L}^{-1}\), Y: \(0.60\,\text{mol L}^{-1}\)
ⓒ. X: \(0.60\,\text{mol L}^{-1}\), Y: \(0.50\,\text{mol L}^{-1}\)
ⓓ. X: \(0.60\,\text{mol L}^{-1}\), Y: \(0.25\,\text{mol L}^{-1}\)
Correct Answer: X: \(0.60\,\text{mol L}^{-1}\), Y: \(0.25\,\text{mol L}^{-1}\)
Explanation: \( \textbf{Given:} \)
For X:
\[
k_X = 0.020\,\text{mol L}^{-1}\text{s}^{-1}, \quad [R]_0 = 1.0\,\text{mol L}^{-1}
\]
For Y:
\[
k_Y = 0.0693\,\text{s}^{-1}, \quad [R]_0 = 1.0\,\text{mol L}^{-1}
\]
\( \textbf{Required:} \)
Concentrations after \(20\,\text{s}\)
Relevant formulas:
For zero order,
\[
[R]_t = [R]_0 - kt
\]
For first order,
\[
t_{1/2} = \frac{0.693}{k}
\]
Why these formulas apply:
Reaction X is zero order and Reaction Y is first order.
For X:
\[
[R]_t = 1.0 - (0.020)(20) = 1.0 - 0.40 = 0.60\,\text{mol L}^{-1}
\]
For Y:
\[
t_{1/2} = \frac{0.693}{0.0693} = 10\,\text{s}
\]
So \(20\,\text{s}\) corresponds to 2 half-lives:
\[
1.0 \to 0.50 \to 0.25\,\text{mol L}^{-1}
\]
\( \textbf{Final Answer:} \)
X: \(0.60\,\text{mol L}^{-1}\), Y: \(0.25\,\text{mol L}^{-1}\)
306. For a reaction, \(k_1 = 1.0 \times 10^{-3}\,\text{s}^{-1}\) at \(300\,\text{K}\) and \(E_a = 50\,\text{kJ mol}^{-1}\). What is the approximate value of \(k_2\) at \(330\,\text{K}\)?
(Use \(R = 8.314\,\text{J mol}^{-1}\text{K}^{-1}\))
ⓐ. \(3.1 \times 10^{-3}\,\text{s}^{-1}\)
ⓑ. \(6.2 \times 10^{-3}\,\text{s}^{-1}\)
ⓒ. \(1.2 \times 10^{-2}\,\text{s}^{-1}\)
ⓓ. \(2.4 \times 10^{-2}\,\text{s}^{-1}\)
Correct Answer: \(6.2 \times 10^{-3}\,\text{s}^{-1}\)
Explanation: \( \textbf{Given:} \)
\[
k_1 = 1.0 \times 10^{-3}\,\text{s}^{-1}
\]
\[
E_a = 50\,\text{kJ mol}^{-1} = 5.0 \times 10^4\,\text{J mol}^{-1}
\]
\[
T_1 = 300\,\text{K}, \quad T_2 = 330\,\text{K}
\]
\[
R = 8.314\,\text{J mol}^{-1}\text{K}^{-1}
\]
\( \textbf{Required:} \)
\[
k_2
\]
\( \textbf{Relevant formula:} \)
\[
\log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)
\]
\( \textbf{Why this formula applies:} \)
Two temperatures, one rate constant, and activation energy are given.
Calculate reciprocal-temperature difference:
\[
\frac{1}{300} - \frac{1}{330} = \frac{30}{99000} = 3.03 \times 10^{-4}\,\text{K}^{-1}
\]
\( \textbf{Substitution:} \)
\[
\log\frac{k_2}{k_1} = \frac{5.0 \times 10^4}{2.303 \times 8.314}(3.03 \times 10^{-4})
\]
\( \textbf{Intermediate simplification:} \)
\[
2.303 \times 8.314 \approx 19.15
\]
\[
\log\frac{k_2}{k_1} \approx \frac{5.0 \times 10^4}{19.15}(3.03 \times 10^{-4}) \approx 0.791
\]
Take antilog:
\[
\frac{k_2}{k_1} = 10^{0.791} \approx 6.2
\]
Now calculate:
\[
k_2 = 6.2 \times 10^{-3}\,\text{s}^{-1}
\]
\( \textbf{Final Answer:} \)
\[
k_2 = 6.2 \times 10^{-3}\,\text{s}^{-1}
\]
307. The rate of a reaction increases by a factor of 4 for every rise of \(10\,\text{K}\). If the rate is \(r\) at \(300\,\text{K}\), what will be the rate at \(320\,\text{K}\)?
ⓐ. \(16r\)
ⓑ. \(8r\)
ⓒ. \(4r\)
ⓓ. \(12r\)
Correct Answer: \(16r\)
Explanation: \( \textbf{Given:} \)
Rate increases by a factor of 4 for each \(10\,\text{K}\) rise
\( \textbf{Required:} \)
Rate at \(320\,\text{K}\) if rate at \(300\,\text{K}\) is \(r\)
\( \textbf{Relevant principle:} \)
A rise from \(300\,\text{K}\) to \(320\,\text{K}\) consists of two equal rises of \(10\,\text{K}\).
\( \textbf{Why this principle applies:} \)
The temperature coefficient is applied once for each \(10\,\text{K}\) step.
First rise:
\[
r \to 4r
\]
Second rise:
\[
4r \to 4 \times 4r
\]
\( \textbf{Final simplification:} \)
\[
\text{new rate} = 16r
\]
\( \textbf{Final Answer:} \)
\[
16r
\]
308. Two reactions are studied at the same temperature and have the same activation energy. Reaction X has a larger frequency factor than Reaction Y. Which statement is correct?
ⓐ. Reaction X must have a smaller rate constant because more collisions are wasted.
ⓑ. Both reactions must have the same rate constant because \(E_a\) is the same.
ⓒ. Reaction X has the larger rate constant because the pre-exponential factor is larger.
ⓓ. The rate constants cannot be compared unless the balanced equations are identical.
Correct Answer: Reaction X has the larger rate constant because the pre-exponential factor is larger.
Explanation: In the Arrhenius equation,
\[
k = A e^{-E_a/RT}
\]
if \(E_a\) and \(T\) are the same, the exponential factor is the same for both reactions. Then the reaction with larger \(A\) must have the larger rate constant. So a larger frequency factor gives a larger \(k\) under these conditions.
309. Assertion: A catalyst increases the rate constant of a reaction at fixed temperature.
Reason: A catalyst provides an alternative pathway with lower activation energy.
ⓐ. Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.
ⓑ. Assertion is true, but Reason is false.
ⓒ. Assertion is false, but Reason is true.
ⓓ. Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
Correct Answer: Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
Explanation: A catalyst lowers the activation energy for an alternative pathway. From
\[
k = A e^{-E_a/RT}
\]
a lower \(E_a\) at the same temperature increases the rate constant. So both statements are true, and the reason correctly explains the assertion.
310. Which matching between factor and collision-theory effect is correct?
ⓐ. Concentration \(\rightarrow\) lowers activation energy; Temperature \(\rightarrow\) changes molecularity; Catalyst \(\rightarrow\) raises collision frequency only
ⓑ. Concentration \(\rightarrow\) increases collision frequency; Temperature \(\rightarrow\) increases fraction above \(E_a\); Catalyst \(\rightarrow\) lowers activation barrier
ⓒ. Concentration \(\rightarrow\) changes equilibrium constant; Temperature \(\rightarrow\) lowers surface area; Catalyst \(\rightarrow\) increases reactant concentration
ⓓ. Concentration \(\rightarrow\) changes stoichiometric coefficients; Temperature \(\rightarrow\) fixes orientation; Catalyst \(\rightarrow\) removes need for collision
Correct Answer: Concentration \(\rightarrow\) increases collision frequency; Temperature \(\rightarrow\) increases fraction above \(E_a\); Catalyst \(\rightarrow\) lowers activation barrier
Explanation: Concentration mainly affects how often particles collide. Temperature strongly affects the fraction of particles with sufficient energy to react. A catalyst works by lowering the activation-energy barrier, thereby increasing the number of effective collisions.
311. Two collisions between the same pair of molecules occur with equal energy above the activation energy. One collision forms products, and the other does not. The most reasonable explanation is that
ⓐ. only one collision had suitable orientation for bond rearrangement.
ⓑ. one collision must have involved a catalyst and the other must not.
ⓒ. one collision had lower concentration than the other.
ⓓ. the molecularity changed between the two collisions.
Correct Answer: only one collision had suitable orientation for bond rearrangement.
Explanation: Even when energy is sufficient, a collision may still fail if the molecules do not approach with the correct spatial arrangement. Collision theory therefore requires both sufficient energy and proper orientation. The difference here is best explained by orientation.
312. Which set of observations is internally consistent for a first-order reaction?
ⓐ. \([R]\) versus \(t\) is linear, \(k\) has unit \(\text{mol L}^{-1}\text{s}^{-1}\), and half-life depends on \([R]_0\)
ⓑ. \(\log [R]\) versus \(t\) is linear, \(k\) has unit \(\text{s}^{-1}\), and half-life is constant
ⓒ. \(\frac{1}{[R]}\) versus \(t\) is linear, \(k\) has unit \(\text{L mol}^{-1}\text{s}^{-1}\), and half-life is constant
ⓓ. Rate is independent of concentration, and successive half-lives become shorter
Correct Answer: \(\log [R]\) versus \(t\) is linear, \(k\) has unit \(\text{s}^{-1}\), and half-life is constant
Explanation: First-order kinetics is identified by the linear relation between \(\log [R]\) and \(t\), a rate constant with unit \(\text{s}^{-1}\), and a half-life that does not depend on initial concentration. These three features are consistent with first-order behavior.
313. A plot of \(\ln k\) versus \(\frac{1}{T}\) gives intercept \(12.0\). What is the value of the frequency factor \(A\)?
ⓐ. \(1.2 \times 10^1\)
ⓑ. \(1.2 \times 10^2\)
ⓒ. \(1.6 \times 10^4\)
ⓓ. \(1.6 \times 10^5\)
Correct Answer: \(1.6 \times 10^5\)
Explanation: \( \textbf{Given:} \)
Intercept of \(\ln k\) versus \(\frac{1}{T}\) plot
\[
= 12.0
\]
\( \textbf{Required:} \)
Frequency factor, \(A\)
\( \textbf{Relevant formula:} \)
\[
\ln k = \ln A - \frac{E_a}{R}\left(\frac{1}{T}\right)
\]
\( \textbf{Why this formula applies:} \)
In this straight-line form, the intercept equals \(\ln A\).
So,
\[
\ln A = 12.0
\]
Take antilog in natural form:
\[
A = e^{12}
\]
\( \textbf{Intermediate simplification:} \)
\[
e^{12} \approx 1.6 \times 10^5
\]
\( \textbf{Final Answer:} \)
\[
A \approx 1.6 \times 10^5
\]
314. A first-order reaction has \(k = 3.465 \times 10^{-2}\,\text{min}^{-1}\) at \(300\,\text{K}\). Its activation energy is \(27.7\,\text{kJ mol}^{-1}\). What is the half-life at \(320\,\text{K}\)?
(Use \(R = 8.314\,\text{J mol}^{-1}\text{K}^{-1}\))
ⓐ. \(20\,\text{min}\)
ⓑ. \(10\,\text{min}\)
ⓒ. \(5\,\text{min}\)
ⓓ. \(40\,\text{min}\)
Correct Answer: \(10\,\text{min}\)
Explanation: \( \textbf{Given:} \)
\[
k_1 = 3.465 \times 10^{-2}\,\text{min}^{-1}
\]
\[
E_a = 27.7\,\text{kJ mol}^{-1}
\]
\[
T_1 = 300\,\text{K}, \quad T_2 = 320\,\text{K}
\]
\[
R = 8.314\,\text{J mol}^{-1}\text{K}^{-1}
\]
\( \textbf{Required:} \)
Half-life at \(320\,\text{K}\)
Relevant formulas:
\[
\log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)
\]
and
\[
t_{1/2} = \frac{0.693}{k}
\]
Why these formulas apply:
The first formula finds the new rate constant after heating, and the second gives half-life for first-order kinetics.
First find the rate-constant ratio:
This activation energy and temperature rise correspond to
\[
\frac{k_2}{k_1} = 2
\]
So,
\[
k_2 = 2k_1 = 2(3.465 \times 10^{-2}) = 6.93 \times 10^{-2}\,\text{min}^{-1}
\]
Now find half-life at \(320\,\text{K}\):
\[
t_{1/2} = \frac{0.693}{6.93 \times 10^{-2}}
\]
\( \textbf{Final simplification:} \)
\[
t_{1/2} = 10\,\text{min}
\]
\( \textbf{Final Answer:} \)
\[
t_{1/2} = 10\,\text{min}
\]
315. A reaction has the true rate law \(r = k[A][B]\). When \([B] = 4.0\,\text{mol L}^{-1}\) is kept in large excess, the pseudo-first-order constant is \(k' = 0.060\,\text{s}^{-1}\). If a new experiment is carried out at the same temperature with \([B] = 8.0\,\text{mol L}^{-1}\) still in large excess, what will be the new pseudo-first-order constant?
ⓐ. \(0.120\,\text{s}^{-1}\)
ⓑ. \(0.060\,\text{s}^{-1}\)
ⓒ. \(0.030\,\text{s}^{-1}\)
ⓓ. \(0.240\,\text{s}^{-1}\)
Correct Answer: \(0.120\,\text{s}^{-1}\)
Explanation: \( \textbf{Given:} \)
\[
r = k[A][B]
\]
\[
[B]_1 = 4.0\,\text{mol L}^{-1}
\]
\[
k'_1 = 0.060\,\text{s}^{-1}
\]
\[
[B]_2 = 8.0\,\text{mol L}^{-1}
\]
\( \textbf{Required:} \)
New pseudo-first-order constant, \(k'_2\)
\( \textbf{Relevant formula / law / principle:} \)
When \([B]\) is in large excess,
\[
k' = k[B]
\]
\( \textbf{Why this formula applies:} \)
The concentration of \(B\) remains nearly constant during each experiment and is absorbed into the apparent rate constant.
Compare the two experiments:
\[
\frac{k'_2}{k'_1} = \frac{[B]_2}{[B]_1} = \frac{8.0}{4.0} = 2
\]
So,
\[
k'_2 = 2 \times 0.060
\]
\( \textbf{Final simplification:} \)
\[
k'_2 = 0.120\,\text{s}^{-1}
\]
\( \textbf{Final Answer:} \)
\[
k'_2 = 0.120\,\text{s}^{-1}
\]
316. The decomposition \(A(g) \rightarrow B(g) + C(g)\) is first order. The initial pressure of pure \(A\) is \(P_0\). What will be the total pressure when the reaction has reached one half-life?
ⓐ. \(1.0P_0\)
ⓑ. \(1.5P_0\)
ⓒ. \(2.0P_0\)
ⓓ. \(0.5P_0\)
Correct Answer: \(1.5P_0\)
Explanation: \( \textbf{Given:} \)
\[
A(g) \rightarrow B(g) + C(g)
\]
\( \textbf{Required:} \)
Total pressure at one half-life
\( \textbf{Relevant principle:} \)
At one half-life, half of \(A\) has decomposed and half remains.
\( \textbf{Why this principle applies:} \)
Half-life means the amount of reactant falls to half its initial value.
Initial pressure of \(A\):
\[
P_0
\]
After one half-life:
Pressure of unreacted \(A\):
\[
\frac{P_0}{2}
\]
Decomposed pressure-equivalent of \(A\):
\[
P_0 - \frac{P_0}{2} = \frac{P_0}{2}
\]
Since each mole of \(A\) gives two moles of gaseous products in total,
pressure of \(B\) formed:
\[
\frac{P_0}{2}
\]
Pressure of \(C\) formed:
\[
\frac{P_0}{2}
\]
Total pressure:
\[
P_{\text{total}} = \frac{P_0}{2} + \frac{P_0}{2} + \frac{P_0}{2} = \frac{3P_0}{2}
\]
\( \textbf{Final Answer:} \)
\[
\frac{3P_0}{2}
\]
317. Assertion: For a complex reaction, the exponents in the rate law need not match the stoichiometric coefficients in the overall balanced equation.
Reason: Only for an elementary step can the stoichiometric participation directly reflect the molecular event controlling the rate.
ⓐ. Both Assertion and Reason are false.
ⓑ. Assertion is true, but Reason is false.
ⓒ. Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
ⓓ. Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.
Correct Answer: Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
Explanation: The overall balanced equation for a complex reaction summarizes only the net change and does not reveal the actual stepwise mechanism. Therefore its coefficients cannot generally be used directly as exponents in the rate law. For an elementary step, however, the molecular event is represented directly, so stoichiometric participation can reflect the kinetic dependence.
318. For a reaction, the following data are obtained:
Experiment 1: \([A] = 0.10\,\text{mol L}^{-1}\), \([B] = 0.20\,\text{mol L}^{-1}\), rate \(= 4.0 \times 10^{-4}\,\text{mol L}^{-1}\text{s}^{-1}\)
Experiment 2: \([A] = 0.20\,\text{mol L}^{-1}\), \([B] = 0.20\,\text{mol L}^{-1}\), rate \(= 1.6 \times 10^{-3}\,\text{mol L}^{-1}\text{s}^{-1}\)
Experiment 3: \([A] = 0.20\,\text{mol L}^{-1}\), \([B] = 0.40\,\text{mol L}^{-1}\), rate \(= 3.2 \times 10^{-3}\,\text{mol L}^{-1}\text{s}^{-1}\)
What will be the rate when \([A] = 0.30\,\text{mol L}^{-1}\) and \([B] = 0.10\,\text{mol L}^{-1}\)?
ⓐ. \(1.2 \times 10^{-3}\,\text{mol L}^{-1}\text{s}^{-1}\)
ⓑ. \(1.8 \times 10^{-3}\,\text{mol L}^{-1}\text{s}^{-1}\)
ⓒ. \(2.4 \times 10^{-3}\,\text{mol L}^{-1}\text{s}^{-1}\)
ⓓ. \(9.0 \times 10^{-4}\,\text{mol L}^{-1}\text{s}^{-1}\)
Correct Answer: \(1.8 \times 10^{-3}\,\text{mol L}^{-1}\text{s}^{-1}\)
Explanation: \( \textbf{Given:} \)
Rate data for three experiments
\( \textbf{Required:} \)
Rate at
\[
[A] = 0.30\,\text{mol L}^{-1}, \quad [B] = 0.10\,\text{mol L}^{-1}
\]
\( \textbf{Relevant principle:} \)
Determine the rate law first, then calculate the required rate.
\( \textbf{Why this principle applies:} \)
The rate depends on concentration according to experimentally determined exponents.
Compare Experiments 1 and 2:
\([B]\) constant, \([A]\) doubles, rate becomes four times.
So,
\[
2^m = 4
\]
\[
m = 2
\]
Compare Experiments 2 and 3:
\([A]\) constant, \([B]\) doubles, rate becomes two times.
So,
\[
2^n = 2
\]
\[
n = 1
\]
Hence,
\[
r = k[A]^2[B]
\]
Now find \(k\) using Experiment 1:
\[
4.0 \times 10^{-4} = k(0.10)^2(0.20)
\]
\[
4.0 \times 10^{-4} = k(0.002)
\]
\[
k = \frac{4.0 \times 10^{-4}}{0.002} = 0.20\,\text{L}^2\text{mol}^{-2}\text{s}^{-1}
\]
\( \textbf{Now calculate the required rate:} \)
\[
r = 0.20(0.30)^2(0.10)
\]
\[
r = 0.20(0.09)(0.10)
\]
\[
r = 0.20(0.009)
\]
\[
r = 0.0018\,\text{mol L}^{-1}\text{s}^{-1}
\]
\( \textbf{Final Answer:} \)
\[
1.8 \times 10^{-3}\,\text{mol L}^{-1}\text{s}^{-1}
\]
319. A first-order reaction has \(k_1 = 3.465 \times 10^{-2}\,\text{min}^{-1}\) at \(300\,\text{K}\) and activation energy \(27.7\,\text{kJ mol}^{-1}\). When the temperature is raised to \(320\,\text{K}\), which statement about its half-life is correct?
(Use \(R = 8.314\,\text{J mol}^{-1}\text{K}^{-1}\))
ⓐ. It doubles from \(20\,\text{min}\) to \(40\,\text{min}\).
ⓑ. It remains \(20\,\text{min}\).
ⓒ. It becomes \(5\,\text{min}\).
ⓓ. It halves from \(20\,\text{min}\) to \(10\,\text{min}\).
Correct Answer: It halves from \(20\,\text{min}\) to \(10\,\text{min}\).
Explanation: \( \textbf{Given:} \)
\[
k_1 = 3.465 \times 10^{-2}\,\text{min}^{-1}
\]
\[
E_a = 27.7\,\text{kJ mol}^{-1}
\]
\[
T_1 = 300\,\text{K}, \quad T_2 = 320\,\text{K}
\]
\( \textbf{Required:} \)
Change in half-life at \(320\,\text{K}\)
\( \textbf{Relevant formulas:} \)
\[
\log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)
\]
and
\[
t_{1/2} = \frac{0.693}{k}
\]
\( \textbf{Why these formulas apply:} \)
The Arrhenius relation gives the new rate constant, and the first-order half-life is inversely proportional to \(k\).
For \(E_a = 27.7\,\text{kJ mol}^{-1}\) and \(300\,\text{K} \to 320\,\text{K}\), the rate constant doubles:
\[
\frac{k_2}{k_1} = 2
\]
At \(300\,\text{K}\):
\[
t_{1/2,1}=\frac{0.693}{3.465\times 10^{-2}}=20\,\text{min}
\]
When \(k\) doubles, the first-order half-life becomes half:
\[
t_{1/2,2}=10\,\text{min}
\]
\( \textbf{Final Answer:} \)
It halves from \(20\,\text{min}\) to \(10\,\text{min}\).
320. Two reactions are studied at the same temperature. They have the same activation energy but different frequency factors. Which statement is correct?
ⓐ. The reaction with the larger frequency factor has the larger rate constant.
ⓑ. The reaction with the smaller frequency factor has the larger rate constant.
ⓒ. Both reactions must have the same rate constant.
ⓓ. Their rate constants cannot be compared without knowing concentration.
Correct Answer: The reaction with the larger frequency factor has the larger rate constant.
Explanation: From the Arrhenius equation,
\[
k = A e^{-E_a/RT}
\]
if \(E_a\) and \(T\) are the same, the exponential factor is the same for both reactions. Therefore the value of \(k\) depends directly on \(A\). The reaction with larger \(A\) has the larger rate constant.
