301. Assertion: Successive half-lives of an ideal first-order reaction are equal at fixed temperature.
Reason: The same fraction of the reactant is consumed during each equal half-life interval.
ⓐ. Assertion is true, but Reason is false
ⓑ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓒ. Assertion is false, but Reason is true
ⓓ. Both Assertion and Reason are true, and Reason explains Assertion
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: First-order kinetics gives a constant half-life when \(k\) remains unchanged. During the first half-life, the concentration decreases from \([R]_0\) to \(\frac{[R]_0}{2}\). During the next half-life, it decreases from \(\frac{[R]_0}{2}\) to \(\frac{[R]_0}{4}\). Although the absolute amount consumed becomes smaller, the fraction consumed remains one-half of the amount present at the beginning of each interval. The Reason therefore explains why successive half-life intervals are equal.
302. On a graph of \(\ln[R]\) against time for a first-order reaction, the vertical decrease during one half-life is:
ⓐ. \(k\)
ⓑ. \(2.303\)
ⓒ. \(\ln2\)
ⓓ. \(\ln[R]_0\)
Correct Answer: \(\ln2\)
Explanation: \( \textbf{Half-life condition:} \)
At the beginning of a half-life interval, let the concentration be \([R]\). At the end of that interval, it becomes \(\frac{[R]}{2}\). The change in the logarithmic coordinate is:
\[
\ln[R]-\ln\frac{[R]}{2}
\]
\( \textbf{Time or rate-constant calculation:} \)
Use the logarithm quotient rule:
\[
\ln\frac{[R]}{[R]/2}
\]
\[
=\ln2
\]
Because the \(\ln[R]\)-against-\(t\) plot has slope \(-k\), the time corresponding to this vertical decrease is \(\frac{\ln2}{k}\). Equal logarithmic decreases therefore correspond to equal first-order half-lives.
303. Complete the first-order half-life equation.
\[
kt_{1/2}=\underline{\hspace{1.2cm}}
\]
ⓐ. \(1\)
ⓑ. \(2.303\)
ⓒ. \([R]_0\)
ⓓ. \(0.693\)
Correct Answer: \(0.693\)
Explanation: At the half-life, the concentration ratio is \(\frac{[R]_0}{[R]_t}=2\). The first-order integrated relation gives \(kt_{1/2}=\ln2\). The numerical value of \(\ln2\) is approximately \(0.693\). Thus, the product of a first-order rate constant and its half-life is constant at fixed logarithm convention. This product is dimensionless because \(k\) has inverse-time units.
304. After four half-lives of a first-order reaction, the fraction of reactant remaining is:
ⓐ. \(\frac{1}{4}\)
ⓑ. \(\frac{1}{8}\)
ⓒ. \(\frac{1}{16}\)
ⓓ. \(\frac{1}{32}\)
Correct Answer: \(\frac{1}{16}\)
Explanation: The fraction remaining after \(n\) first-order half-lives is:
\[
\left(\frac{1}{2}\right)^n
\]
For four half-lives:
\[
\left(\frac{1}{2}\right)^4
\]
\[
=\frac{1}{16}
\]
This corresponds to:
\[
\frac{1}{16}\times100=6.25\%
\]
Thus, \(6.25\%\) remains and \(93.75\%\) has decomposed after four half-lives.
305. Match the number of first-order half-lives in Column I with the percentage decomposed in Column II.
| Column I | Column II |
|---|
| P. One half-life | 1. \(87.5\%\) |
| Q. Two half-lives | 2. \(50\%\) |
| R. Three half-lives | 3. \(93.75\%\) |
| S. Four half-lives | 4. \(75\%\) |
ⓐ. P-4, Q-2, R-1, S-3
ⓑ. P-2, Q-1, R-4, S-3
ⓒ. P-2, Q-4, R-3, S-1
ⓓ. P-2, Q-4, R-1, S-3
Correct Answer: P-2, Q-4, R-1, S-3
Explanation: After one half-life, one-half remains, so \(50\%\) has decomposed. After two half-lives, \(\left(\frac{1}{2}\right)^2=\frac{1}{4}\) remains, giving \(75\%\) decomposition. After three half-lives, \(\frac{1}{8}\) remains, so \(87.5\%\) has decomposed. After four half-lives, \(\frac{1}{16}\) remains, corresponding to \(93.75\%\) decomposition. The matched sequence is therefore P-2, Q-4, R-1, S-3.
306. For a first-order reaction, the time required for \(90\%\) completion is approximately:
ⓐ. \(2.303t_{1/2}\)
ⓑ. \(3.322t_{1/2}\)
ⓒ. \(4.606t_{1/2}\)
ⓓ. \(6.644t_{1/2}\)
Correct Answer: \(3.322t_{1/2}\)
Explanation: At \(90\%\) completion, the fraction remaining is:
\[
f=0.10
\]
The first-order time relation is:
\[
t=\frac{1}{k}\ln\frac{1}{f}
\]
Therefore:
\[
t_{90}=\frac{\ln10}{k}
\]
The half-life is:
\[
t_{1/2}=\frac{\ln2}{k}
\]
Divide the two:
\[
\frac{t_{90}}{t_{1/2}}
=
\frac{\ln10}{\ln2}
\]
\[
\frac{t_{90}}{t_{1/2}}
=
\frac{2.303}{0.693}
\]
\[
\frac{t_{90}}{t_{1/2}}\approx3.322
\]
Thus, \(90\%\) completion requires slightly more than three half-lives.
307. After \(500\,s\), a first-order reaction is \(96\%\) complete. Its half-life is closest to:
ⓐ. \(72.0\,s\)
ⓑ. \(144\,s\)
ⓒ. \(108\,s\)
ⓓ. \(250\,s\)
Correct Answer: \(108\,s\)
Explanation: \( \textbf{Half-life condition:} \)
If the reaction is \(96\%\) complete, the fraction remaining is:
\[
f=1-0.96=0.04
\]
The first-order rate constant is:
\[
k=\frac{1}{t}\ln\frac{1}{f}
\]
Substitute:
\[
k=\frac{1}{500}\ln\frac{1}{0.04}
\]
\[
k=\frac{\ln25}{500}
\]
\( \textbf{Time or rate-constant calculation:} \)
Using \(\ln25\approx3.219\):
\[
k=\frac{3.219}{500}
\]
\[
k\approx6.44\times10^{-3}\,s^{-1}
\]
Now calculate the half-life:
\[
t_{1/2}=\frac{0.693}{k}
\]
\[
t_{1/2}=\frac{0.693}{6.44\times10^{-3}}
\]
\[
t_{1/2}\approx108\,s
\]
The completion percentage must first be converted into the fraction remaining.
308. Starting with \(160\,mg\), a first-order reactant is allowed to decay. Which sequence gives the amounts remaining after one, two, three, and four half-lives, respectively?
ⓐ. \(80\,mg,\ 40\,mg,\ 10\,mg,\ 0\,mg\)
ⓑ. \(80\,mg,\ 40\,mg,\ 20\,mg,\ 10\,mg\)
ⓒ. \(80\,mg,\ 20\,mg,\ 10\,mg,\ 5\,mg\)
ⓓ. \(120\,mg,\ 80\,mg,\ 40\,mg,\ 20\,mg\)
Correct Answer: \(80\,mg,\ 40\,mg,\ 20\,mg,\ 10\,mg\)
Explanation: Each first-order half-life reduces the amount present at the beginning of that interval by one-half. Starting from \(160\,mg\), the amount after one half-life is \(80\,mg\). Halving again gives \(40\,mg\), followed by \(20\,mg\), and then \(10\,mg\). The absolute amount lost during each interval decreases from \(80\,mg\) to \(40\,mg\), \(20\,mg\), and \(10\,mg\). The constant feature is the fractional loss, not the absolute loss.
309. The pressure method can be used to follow a gaseous reaction at constant temperature and volume because:
ⓐ. the total pressure is always equal to the rate constant
ⓑ. pressure tracks gas moles at fixed temperature and volume
ⓒ. the activation energy is proportional to the total pressure
ⓓ. every gaseous reaction has zero-order kinetics
Correct Answer: pressure tracks gas moles at fixed temperature and volume
Explanation: For an ideal gas, \(PV=nRT\). When temperature and volume remain constant, \(R\), \(T\), and \(V\) are fixed. The pressure is then directly proportional to the number of gaseous moles. A reaction that changes the total number of gas particles can therefore be monitored through its total-pressure change. Stoichiometric reasoning is still needed to convert total pressure into the partial pressure of the reacting gas.
310. A gaseous decomposition follows
\[
A(g)\rightarrow B(g)+C(g).
\]
If the initial pressure of pure \(A\) is \(P_0\) and the total pressure at time \(t\) is \(P_t\), the partial pressure of unreacted \(A\) at time \(t\) is:
ⓐ. \(2P_0-P_t\)
ⓑ. \(P_t-P_0\)
ⓒ. \(P_t+P_0\)
ⓓ. \(2P_t-P_0\)
Correct Answer: \(2P_0-P_t\)
Explanation: \( \textbf{Pressure-to-reactant relation:} \)
Let the pressure-equivalent amount of \(A\) decomposed by time \(t\) be \(x\). The pressure of unreacted \(A\) is then \(P_0-x\). The products \(B\) and \(C\) each contribute pressure \(x\). Therefore:
\[
P_t=(P_0-x)+x+x
\]
\[
P_t=P_0+x
\]
Hence:
\[
x=P_t-P_0
\]
\( \textbf{First-order evaluation:} \)
The partial pressure of \(A\) remaining is:
\[
P_A=P_0-x
\]
\[
P_A=P_0-(P_t-P_0)
\]
\[
P_A=2P_0-P_t
\]
311. For first-order decomposition
\[
A(g)\rightarrow B(g)+C(g),
\]
starting with pure \(A\) at pressure \(P_0\), the appropriate pressure-based rate equation is:
ⓐ. \(k=\frac{2.303}{t}\log\frac{P_t}{P_0}\)
ⓑ. \(k=\frac{2.303}{t}\log\frac{P_t-P_0}{P_0}\)
ⓒ. \(k=\frac{2.303}{t}\log\frac{P_0}{P_t-P_0}\)
ⓓ. \(k=\frac{2.303}{t}\log\frac{P_0}{2P_0-P_t}\)
Correct Answer: \(k=\frac{2.303}{t}\log\frac{P_0}{2P_0-P_t}\)
Explanation: The first-order equation requires the initial and remaining partial pressures of the reactant. Initially, the partial pressure of \(A\) is \(P_0\). At time \(t\), the remaining partial pressure is \(2P_0-P_t\). Substitution into the first-order integrated law gives:
\[
k=\frac{2.303}{t}\log\frac{P_{A,0}}{P_{A,t}}
\]
\[
k=\frac{2.303}{t}\log\frac{P_0}{2P_0-P_t}
\]
The total pressure \(P_t\) cannot be used directly as though it were the remaining pressure of \(A\).
312. Assertion: During the decomposition \(A(g)\rightarrow B(g)+C(g)\) at constant temperature and volume, the total pressure increases even though the partial pressure of \(A\) decreases.
Reason: Each mole of \(A\) consumed produces two moles of gaseous products.
ⓐ. Assertion is true, but Reason is false
ⓑ. Both Assertion and Reason are true, and Reason explains Assertion
ⓒ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: The reaction converts one gaseous particle unit of \(A\) into two gaseous product particle units. As \(A\) reacts, its partial pressure decreases. However, the total number of gaseous moles increases because two product moles replace each mole of \(A\). At constant temperature and volume, this increase in gaseous moles raises the total pressure. The Reason therefore explains the simultaneous decrease in reactant pressure and increase in total pressure.
313. The following pressure data are obtained for \(A(g)\rightarrow B(g)+C(g)\), starting with pure \(A\).
| Time | Total pressure |
|---|
| \(0\,s\) | \(400\,mmHg\) |
| \(200\,s\) | \(500\,mmHg\) |
| \(400\,s\) | \(575\,mmHg\) |
The first-order rate constant calculated from the data is closest to:
ⓐ. \(1.44\times10^{-3}\,s^{-1}\)
ⓑ. \(7.19\times10^{-4}\,s^{-1}\)
ⓒ. \(2.88\times10^{-3}\,s^{-1}\)
ⓓ. \(5.75\times10^{-3}\,s^{-1}\)
Correct Answer: \(1.44\times10^{-3}\,s^{-1}\)
Explanation: \( \textbf{Stoichiometric pressure step:} \)
At \(200\,s\), the remaining pressure of \(A\) is:
\[
P_{A,200}=2P_0-P_{200}
\]
\[
P_{A,200}=800-500=300\,mmHg
\]
Use the first-order equation:
\[
k=\frac{1}{t}\ln\frac{P_{A,0}}{P_{A,t}}
\]
\[
k=\frac{1}{200}\ln\frac{400}{300}
\]
\( \textbf{Rate-constant calculation:} \)
\[
k=\frac{\ln(4/3)}{200}
\]
Using \(\ln(4/3)\approx0.2877\):
\[
k\approx\frac{0.2877}{200}
\]
\[
k\approx1.44\times10^{-3}\,s^{-1}
\]
At \(400\,s\):
\[
P_{A,400}=800-575=225\,mmHg
\]
\[
k=\frac{1}{400}\ln\frac{400}{225}
\]
This gives the same value, supporting first-order behaviour.
314. A student uses
\[
k=\frac{2.303}{t}\log\frac{P_0}{P_t}
\]
for the reaction \(A(g)\rightarrow B(g)+C(g)\), where \(P_t\) is the total pressure. The procedure is incorrect because:
ⓐ. pressure cannot be used in any integrated rate equation
ⓑ. \(P_t\) is the total pressure, not the partial pressure of unreacted \(A\)
ⓒ. first-order rate constants cannot be calculated from one time reading
ⓓ. logarithms apply only to concentrations measured in \(mol\,L^{-1}\)
Correct Answer: \(P_t\) is the total pressure, not the partial pressure of unreacted \(A\)
Explanation: The first-order equation must compare the initial and remaining quantities of the reacting species. During \(A\rightarrow B+C\), the total pressure includes unreacted \(A\) and both gaseous products. It therefore increases as the reaction proceeds. The partial pressure of \(A\), by contrast, decreases and is given by \(2P_0-P_t\) for a pure initial sample. Pressure is valid in the logarithmic ratio only after the required stoichiometric conversion has been made.
315. For a first-order gas reaction \(A(g)\rightarrow B(g)+C(g)\), the pressure of \(A\) remaining at time \(t\) is represented by the blank:
\[
P_{A,t}=\underline{\hspace{1.5cm}}
\]
when \(P_\infty\) is the final total pressure.
ⓐ. \(P_t-P_0\)
ⓑ. \(P_\infty-P_0\)
ⓒ. \(P_t+P_\infty\)
ⓓ. \(P_\infty-P_t\)
Correct Answer: \(P_\infty-P_t\)
Explanation: The final pressure represents the total pressure after all of \(A\) has reacted. At an intermediate time, the pressure has not yet reached this value because some \(A\) remains. The remaining amount of \(A\) corresponds to the additional pressure increase still possible. That difference is \(P_\infty-P_t\). This quantity can replace concentration in the first-order integrated equation.
316. In \(A(g)\rightarrow B(g)+C(g)\) at constant temperature and volume, match each quantity in Column I with the expression in Column II.
| Column I | Column II |
|---|
| P. Initial partial pressure of \(A\) | 1. \(P_\infty-P_t\) |
| Q. Pressure-equivalent amount of \(A\) decomposed by time \(t\) | 2. \(\frac{P_t-P_0}{P_\infty-P_0}\) |
| R. Partial pressure of \(A\) remaining at time \(t\) | 3. \(P_\infty-P_0\) |
| S. Fraction of \(A\) decomposed | 4. \(P_t-P_0\) |
ⓐ. P-1, Q-4, R-3, S-2
ⓑ. P-3, Q-4, R-1, S-2
ⓒ. P-3, Q-1, R-4, S-2
ⓓ. P-4, Q-3, R-1, S-2
Correct Answer: P-3, Q-4, R-1, S-2
Explanation: The initial partial pressure of \(A\) is the total pressure increase possible from the initial state to completion, so it is \(P_\infty-P_0\). The pressure-equivalent amount decomposed by time \(t\) is the observed increase \(P_t-P_0\). The reactant pressure remaining is the unachieved pressure increase \(P_\infty-P_t\). Dividing the reacted pressure-equivalent amount by the initial partial pressure gives the fraction decomposed. These relations remain useful even when an inert gas contributes to the measured total pressure.
317. Pressure-method statements for first-order kinetics are listed below.
Statement I: Temperature and volume should remain constant during the measurements.
Statement II: The measured total pressure can always be used directly as the remaining reactant pressure.
Statement III: The final pressure helps convert total-pressure data into the partial pressure of unreacted gas.
The valid statements are:
ⓐ. I and II only
ⓑ. II and III only
ⓒ. I, II and III
ⓓ. I and III only
Correct Answer: I and III only
Explanation: Constant temperature and volume are required so that pressure remains proportional to the number of gaseous moles. The measured total pressure includes all gaseous reactants and products, so it is not generally equal to the pressure of the unreacted species. The final pressure provides the stoichiometric reference needed to obtain \(P_\infty-P_t\). Statement II ignores the composition of the gas mixture and is therefore invalid.
318. A vessel initially contains gaseous \(A\) and an inert gas at a total pressure of \(700\,mmHg\). The reaction \(A(g)\rightarrow B(g)+C(g)\) is first order. The final total pressure is \(1000\,mmHg\), and after \(400\,s\) the pressure is \(850\,mmHg\). The rate constant is:
ⓐ. \(1.73\times10^{-3}\,s^{-1}\)
ⓑ. \(8.66\times10^{-4}\,s^{-1}\)
ⓒ. \(3.47\times10^{-3}\,s^{-1}\)
ⓓ. \(5.78\times10^{-3}\,s^{-1}\)
Correct Answer: \(1.73\times10^{-3}\,s^{-1}\)
Explanation: \( \textbf{Stoichiometric pressure step:} \)
The inert gas contributes equally to the initial, intermediate, and final total pressures. The initial partial pressure of \(A\) is:
\[
P_{A,0}=P_\infty-P_0
\]
\[
P_{A,0}=1000-700
\]
\[
P_{A,0}=300\,mmHg
\]
The partial pressure of \(A\) after \(400\,s\) is:
\[
P_{A,t}=P_\infty-P_t
\]
\[
P_{A,t}=1000-850
\]
\( \textbf{Rate-constant calculation:} \)
\[
P_{A,t}=150\,mmHg
\]
Apply the first-order equation:
\[
k=\frac{1}{t}\ln\frac{P_{A,0}}{P_{A,t}}
\]
\[
k=\frac{1}{400}\ln\frac{300}{150}
\]
\[
k=\frac{\ln2}{400}
\]
\[
k=\frac{0.693}{400}
\]
\[
k\approx1.73\times10^{-3}\,s^{-1}
\]
319. Assertion: The expression
\[
k=\frac{2.303}{t}\log\frac{P_\infty-P_0}{P_\infty-P_t}
\]
can remain valid when an inert gas is initially present.
Reason: The constant inert-gas pressure cancels when the pressure differences are formed.
ⓐ. Assertion is true, but Reason is false
ⓑ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓒ. Assertion is false, but Reason is true
ⓓ. Both Assertion and Reason are true, and Reason explains Assertion
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: An inert gas contributes to every measured total pressure but does not participate in the reaction. The difference \(P_\infty-P_0\) isolates the initial partial pressure of the reacting gas. Similarly, \(P_\infty-P_t\) gives its partial pressure remaining at time \(t\). The unchanged inert-gas contribution disappears from both differences. The Reason therefore explains why the pressure-difference form remains applicable.
320. In the first-order decomposition \(A(g)\rightarrow B(g)+C(g)\), which graph should be linear?
ⓐ. \(\log(P_\infty-P_t)\) against \(t\), with slope \(-\frac{k}{2.303}\)
ⓑ. \(\log P_t\) against \(t\), with slope \(+\frac{k}{2.303}\)
ⓒ. \(\log(P_t-P_0)\) against \(t\), with slope \(-\frac{k}{2.303}\)
ⓓ. \(\log(P_\infty+P_t)\) against \(t\), with slope \(+\frac{k}{2.303}\)
Correct Answer: \(\log(P_\infty-P_t)\) against \(t\), with slope \(-\frac{k}{2.303}\)
Explanation: The remaining partial pressure of \(A\) is \(P_\infty-P_t\). For first-order kinetics:
\[
\log P_{A,t}
=
\log P_{A,0}
-
\frac{k}{2.303}t
\]
Substituting the pressure differences gives:
\[
\log(P_\infty-P_t)
=
\log(P_\infty-P_0)
-
\frac{k}{2.303}t
\]
This has the form of a straight line. Its slope is \(-\frac{k}{2.303}\), and its intercept is \(\log(P_\infty-P_0)\).
