Chemical Kinetics MCQs with Answers – Part 3 (NEET Chemistry)

Explanation: The measured quantity is oxygen collected rather than hydrogen peroxide remaining. The increase in gas volume is proportional to the amount of hydrogen peroxide that has reacted. For a first-order reaction, equal time intervals correspond to equal fractional reductions of the remaining reactant. Therefore, the oxygen-volume increments over successive equal intervals decrease geometrically. This observation allows the limiting oxygen volume to be inferred. The limiting gas volume can then be converted into the initial amount of hydrogen peroxide using the stoichiometric coefficients. \(\textbf{Oxygen produced in successive intervals:}\) During the first 10 min, \[ \Delta V_1=240-0 \] \[ \Delta V_1=240\ \mathrm{mL} \] During the next 10 min, \[ \Delta V_2=360-240 \] \[ \Delta V_2=120\ \mathrm{mL} \] The ratio is \[ \frac{\Delta V_2}{\Delta V_1}=\frac{120}{240} \] \[ \frac{\Delta V_2}{\Delta V_1}=\frac12 \] \(\textbf{Infer the first-order fraction remaining:}\) The oxygen formed in each equal interval is proportional to hydrogen peroxide consumed during that interval. The second increment is half the first. Therefore, the amount of hydrogen peroxide remaining is halved during each 10-minute interval. Hence, \[ t_{1/2}=10\ \mathrm{min} \] \(\textbf{Recover the limiting oxygen volume:}\) The first 10 min corresponds to one half-life. Therefore, half the initial hydrogen peroxide has decomposed. The collected \(240\ \mathrm{mL}\) of oxygen represents half the oxygen that will be produced on complete decomposition. Thus, \[ V_{\infty}=2(240) \] \[ V_{\infty}=480\ \mathrm{mL} \] \(\textbf{Convert the limiting oxygen volume into moles:}\) \[ V_{\infty}=0.480\ \mathrm L \] Using the given molar volume, \[ n_{\mathrm{O_2},\infty}=\frac{0.480}{24.0} \] \[ n_{\mathrm{O_2},\infty}=0.0200\ \mathrm{mol} \] \(\textbf{Use the decomposition stoichiometry:}\) For \[ 2\mathrm{H_2O_2}\rightarrow2\mathrm{H_2O}+\mathrm{O_2} \] two moles of hydrogen peroxide form one mole of oxygen. Therefore, \[ n_{\mathrm{H_2O_2},0}=2n_{\mathrm{O_2},\infty} \] \[ n_{\mathrm{H_2O_2},0}=2(0.0200) \] \[ n_{\mathrm{H_2O_2},0}=0.0400\ \mathrm{mol} \] \(\textbf{Calculate the initial concentration:}\) The solution volume is \[ V_{\mathrm{solution}}=0.100\ \mathrm L \] Therefore, \[ [\mathrm{H_2O_2}]_0=\frac{0.0400}{0.100} \] \[ [\mathrm{H_2O_2}]_0=0.400\ \mathrm{mol\,L^{-1}} \] \(\textbf{Check against the 20-minute observation:}\) After two half-lives, three-fourths of the initial hydrogen peroxide has decomposed. The corresponding oxygen volume should be \[ V_{20}=\frac34V_{\infty} \] \[ V_{20}=\frac34(480) \] \[ V_{20}=360\ \mathrm{mL} \] This agrees with the observation. The measured gas volume must not be treated as the remaining reactant concentration. The limiting collected volume represents complete reaction and is linked to the initial amount through the balanced equation. \(\textbf{Final answer:}\) \[ \boxed{\left([\mathrm{H_2O_2}]_0,t_{1/2}\right)=\left(0.400\ \mathrm{mol\,L^{-1}},10\ \mathrm{min}\right)} \]

Explanation: The two students focus on different features of the observations. Equal elapsed intervals alone do not identify reaction order. The decisive question is whether the reactant loses equal absolute amounts or equal fractions during those intervals. The observed concentration decreases are identical in absolute value. That pattern is characteristic of a zero-order integrated rate equation under unchanged conditions. \(\textbf{Compare the absolute concentration decreases:}\) During the first interval, \[ \Delta[\mathrm A]_1=0.80-0.60 \] \[ \Delta[\mathrm A]_1=0.20\ \mathrm{mol\,L^{-1}} \] During the second interval, \[ \Delta[\mathrm A]_2=0.60-0.40 \] \[ \Delta[\mathrm A]_2=0.20\ \mathrm{mol\,L^{-1}} \] Both changes occur over 8 min. \(\textbf{Determine the zero-order constant:}\) For zero-order disappearance, \[ k=\frac{-\Delta[\mathrm A]}{\Delta t} \] Using either interval, \[ k=\frac{0.20}{8} \] \[ k=0.025\ \mathrm{mol\,L^{-1},min^{-1}} \] \(\textbf{Predict the next interval:}\) The requested concentration decrease is \[ 0.40-0.20=0.20\ \mathrm{mol\,L^{-1}} \] At the same zero-order rate, \[ \Delta t=\frac{0.20}{0.025} \] \[ \Delta t=8\ \mathrm{min} \] \(\textbf{Test the alternative first-order interpretation:}\) For a first-order reaction, the remaining fraction should be identical during equal time intervals. During the first 8 min, \[ f_1=\frac{0.60}{0.80} \] \[ f_1=0.75 \] During the second 8 min, \[ f_2=\frac{0.40}{0.60} \] \[ f_2=\frac23 \] These fractions are not equal. Therefore, the data are inconsistent with a single first-order rate constant. \(\textbf{Physical interpretation:}\) The absolute concentration decrease remains constant because the zero-order disappearance rate is constant while its kinetic regime applies. The fractional decrease becomes progressively larger as the remaining concentration becomes smaller. The first student's reasoning confuses equal time intervals with equal fractional concentration reductions. The second student's interpretation is supported by both the observed increments and the zero-order integrated equation. \(\textbf{Final answer:}\) \[ \Delta t=8\ \mathrm{min}\quad\text{for zero-order disappearance} \]

Explanation: For first-order kinetics, the instantaneous disappearance rate is directly proportional to the remaining reactant concentration. Therefore, a specified fraction of the initial rate corresponds to the same fraction of the initial concentration. A rate reduction to one-fourth and \(75\%\) decomposition both describe one-fourth of the initial reactant remaining. The kinetic time can then be calculated using the first-order rate constant. \(\textbf{Determine the first-order constant:}\) The differential rate law is \[ r=k[\mathrm A] \] Initially, \[ r_0=k[\mathrm A]_0 \] Therefore, \[ k=\frac{r_0}{[\mathrm A]_0} \] Substituting, \[ k=\frac{0.040}{0.80} \] \[ k=0.050\ \mathrm{s^{-1}} \] \(\textbf{Concentration corresponding to the target rate:}\) At the desired rate, \[ r_t=0.010\ \mathrm{mol\,L^{-1}\,s^{-1}} \] Therefore, \[ [\mathrm A]_t=\frac{r_t}{k} \] \[ [\mathrm A]_t=\frac{0.010}{0.050} \] \[ [\mathrm A]_t=0.20\ \mathrm{mol\,L^{-1}} \] \(\textbf{Interpret the corresponding decomposition:}\) The fraction remaining is \[ f=\frac{0.20}{0.80} \] \[ f=\frac14 \] Thus, the fraction decomposed is \[ 1-f=1-\frac14 \] \[ 1-f=\frac34 \] This corresponds to \(75\%\) decomposition. \(\textbf{Determine the common elapsed time:}\) The integrated first-order equation gives \[ t=\frac1k\ln\frac{[\mathrm A]_0}{[\mathrm A]_t} \] Substituting, \[ t=\frac1{0.050}\ln\frac{0.80}{0.20} \] \[ t=\frac{\ln4}{0.050} \] Using \(\ln4=1.386\), \[ t=\frac{1.386}{0.050} \] \[ t=27.72\ \mathrm s \] \(\textbf{Rate–conversion consistency:}\) At \(75\%\) decomposition, one-fourth of A remains. The rate must therefore be one-fourth of its initial value. \[ r_t=\frac{0.040}{4} \] \[ r_t=0.010\ \mathrm{mol\,L^{-1}\,s^{-1}} \] Both observers are therefore describing the same reaction state. The agreement follows from the first-order proportionality between rate and remaining concentration. It would not follow automatically for a zero-order reaction, whose instantaneous rate is independent of the remaining concentration within its applicable regime. \(\textbf{Final answer:}\) \[ t=27.7\ \mathrm s \]

Explanation: The two reactions have the same initial concentration and the same first half-life. However, these equalities do not imply equal rate constants or equal instantaneous disappearance rates. The zero-order rate remains constant throughout the applicable regime. The first-order rate decreases in proportion to the remaining concentration. At the end of one half-life, both reactions have the same remaining concentration, but their instantaneous disappearance rates are different. The half-life equations allow those rates to be compared without specifying a numerical value of \(C_0\) or \(T\). \(\textbf{Zero-order rate from its half-life:}\) For Reaction X, \[ T=\frac{C_0}{2k_X} \] Therefore, \[ k_X=\frac{C_0}{2T} \] The zero-order disappearance rate at the end of the half-life is \[ r_X=k_X \] Hence, \[ r_X=\frac{C_0}{2T} \] \(\textbf{First-order constant from its half-life:}\) For Reaction Y, \[ T=\frac{\ln2}{k_Y} \] Therefore, \[ k_Y=\frac{\ln2}{T} \] \(\textbf{First-order instantaneous rate after one half-life:}\) At \(t=T\), \[ C_Y=\frac{C_0}{2} \] Therefore, \[ r_Y=k_YC_Y \] Substituting, \[ r_Y=\frac{\ln2}{T}\frac{C_0}{2} \] \[ r_Y=\frac{C_0\ln2}{2T} \] \(\textbf{Compare the instantaneous rates:}\) \[ \frac{r_X}{r_Y}=\frac{C_0/(2T)}{C_0\ln2/(2T)} \] Cancelling common factors, \[ \frac{r_X}{r_Y}=\frac1{\ln2} \] Using \(\ln2=0.693\), \[ \frac{r_X}{r_Y}\approx1.443 \] \(\textbf{Consistency check using average rates:}\) Both reactions consume half their initial concentrations during the same elapsed time \(T\). Therefore, their average disappearance rates over the first half-life are equal. However, the zero-order instantaneous rate remains equal to its average rate. The first-order instantaneous rate decreases throughout the interval and is below its average at the endpoint. Consequently, \[ r_X\gt r_Y \] The derived ratio is greater than one, as required. \(\textbf{Chemical interpretation:}\) Equal half-lives at a particular initial concentration do not establish identical kinetic behaviour. The zero-order and first-order half-life expressions depend differently on concentration and rate constant. Even when both reactions reach half their initial concentrations simultaneously, the slopes of their concentration–time curves need not be equal. The ratio \(1/\ln2\) captures this difference independently of the common starting concentration and half-life. \(\textbf{Final answer:}\) \[ \boxed{\frac{r_X}{r_Y}=\frac1{\ln2}\approx1.443} \]

Explanation: Each compartment individually follows an ideal first-order decay law. However, the compartments have different rate constants because they are maintained at different temperatures. The reported analytical concentration is an average of two independently decaying concentrations. A sum of two exponential decays with different constants is not generally one exponential decay with a single constant. This distinction can be established directly from the predicted concentrations at successive times. \(\textbf{Concentration in Compartment I:}\) The half-life of A in Compartment I is 10 min. Therefore, \[ C_1(t)=0.80\left(\frac12\right)^{t/10} \] At 20 min, \[ C_1(20)=0.80\left(\frac12\right)^2 \] \[ C_1(20)=0.20\ \mathrm{mol\,L^{-1}} \] At 40 min, \[ C_1(40)=0.80\left(\frac12\right)^4 \] \[ C_1(40)=0.050\ \mathrm{mol\,L^{-1}} \] \(\textbf{Concentration in Compartment II:}\) The half-life in Compartment II is 20 min. Therefore, \[ C_2(t)=0.80\left(\frac12\right)^{t/20} \] At 20 min, \[ C_2(20)=0.40\ \mathrm{mol\,L^{-1}} \] At 40 min, \[ C_2(40)=0.20\ \mathrm{mol\,L^{-1}} \] \(\textbf{Analytical average at the three times:}\) Initially, \[ C_{\mathrm{avg}}(0)=0.80\ \mathrm{mol\,L^{-1}} \] At 20 min, \[ C_{\mathrm{avg}}(20)=\frac{0.20+0.40}{2} \] \[ C_{\mathrm{avg}}(20)=0.30\ \mathrm{mol\,L^{-1}} \] At 40 min, \[ C_{\mathrm{avg}}(40)=\frac{0.050+0.20}{2} \] \[ C_{\mathrm{avg}}(40)=0.125\ \mathrm{mol\,L^{-1}} \] \(\textbf{Test for a single first-order constant:}\) For one ideal first-order law, equal time intervals must produce equal fractional concentration reductions. During the first 20 min, the average concentration ratio is \[ f_1=\frac{0.30}{0.80} \] \[ f_1=0.375 \] During the second 20 min, \[ f_2=\frac{0.125}{0.30} \] \[ f_2\approx0.417 \] These fractions are unequal. Therefore, the average concentration cannot follow one ideal first-order equation with an unchanged rate constant over both intervals. \(\textbf{Interpret the observation:}\) The faster compartment contributes increasingly less to the remaining total as time passes. The slower compartment consequently becomes more important in the measured average at later times. This causes the apparent overall decay behaviour to change. The observation does not contradict the first-order kinetics within either individual compartment. It shows that averaging populations with different rate constants need not preserve a single first-order integrated law. The chemical conditions of the two populations must therefore be considered before assigning a kinetic order from pooled measurements.

Explanation: Both intervals involve the same absolute decrease in A concentration. However, first-order disappearance does not proceed at a constant absolute rate. The instantaneous rate becomes smaller as A concentration decreases. Therefore, the second equal concentration decrease requires more time than the first. The integrated first-order equation must be applied separately to the two concentration intervals. \(\textbf{Determine the first-order rate constant:}\) For first-order kinetics, \[ t_{1/2}=\frac{\ln2}{k} \] Therefore, \[ k=\frac{\ln2}{10} \] \[ k=\frac{0.693}{10} \] \[ k=0.0693\ \mathrm{min^{-1}} \] \(\textbf{Time for the first concentration decrease:}\) The concentration decreases from \(0.80\) to \(0.60\ \mathrm{mol\,L^{-1}}\). For a first-order reaction, \[ t_1=\frac1k\ln\frac{0.80}{0.60} \] Simplifying, \[ t_1=\frac1k\ln\frac43 \] Substitution gives \[ t_1=\frac{0.2877}{0.0693} \] \[ t_1=4.15\ \mathrm{min} \] \(\textbf{Time for the second concentration decrease:}\) The concentration now decreases from \(0.60\) to \(0.40\ \mathrm{mol\,L^{-1}}\). Therefore, \[ t_2=\frac1k\ln\frac{0.60}{0.40} \] \[ t_2=\frac1k\ln\frac32 \] Substituting, \[ t_2=\frac{0.4055}{0.0693} \] \[ t_2=5.85\ \mathrm{min} \] \(\textbf{Check the combined elapsed time:}\) \[ t_1+t_2=4.15+5.85 \] \[ t_1+t_2=10.00\ \mathrm{min} \] The concentration has decreased from \(0.80\) to \(0.40\ \mathrm{mol\,L^{-1}}\). This is exactly one half-life, consistent with the given 10-minute half-life. \(\textbf{Compare the fractional decreases:}\) The first interval removes the fraction \[ \frac{0.80-0.60}{0.80}=\frac14 \] of the concentration present at its beginning. The second interval removes \[ \frac{0.60-0.40}{0.60}=\frac13 \] of the concentration present at its beginning. Thus, the second interval involves a larger fractional reduction even though both absolute decreases equal \(0.20\ \mathrm{mol\,L^{-1}}\). \(\textbf{Chemical interpretation:}\) Equal absolute concentration losses require equal times in an ideal zero-order reaction while its constant-rate regime applies. For a first-order reaction, equal fractional concentration losses require equal times. Here, the second interval takes longer because the reaction becomes slower as A concentration decreases. This comparison distinguishes absolute concentration change from fractional decomposition. \(\textbf{Final answer:}\) \[ \boxed{\left(t_1,t_2\right)=\left(4.15\ \mathrm{min},5.85\ \mathrm{min}\right)} \]

Explanation: Graph I uses the remaining concentration of reactant A. For a first-order reaction, the natural logarithm of this concentration decreases linearly with time. Graph II uses a different quantity: the amount by which the product concentration remains below its eventual limiting value. This product deficit is directly related to the amount of reactant still available to form additional product. The initially present product must be included when establishing the limiting concentration. However, it does not alter the exponential disappearance law of the remaining reactant. \(\textbf{Integrated law for the reactant:}\) For first-order disappearance, \[ [\mathrm A]_t=[\mathrm A]_0e^{-kt} \] Taking natural logarithms with a fixed concentration reference, \[ \ln\!\left(\frac{[\mathrm A]_t}{c^\circ}\right)=\ln\!\left(\frac{[\mathrm A]_0}{c^\circ}\right)-kt \] Therefore, Graph I has slope \[ m_{\mathrm I}=-k \] \(\textbf{Relate product formation to reactant consumption:}\) For \[ \mathrm A\rightarrow\mathrm P \] the concentration balance is \[ [\mathrm P]_t=[\mathrm P]_0+[\mathrm A]_0-[\mathrm A]_t \] At complete eventual conversion, \[ [\mathrm P]_{\infty}=[\mathrm P]_0+[\mathrm A]_0 \] Subtracting the product concentration at time \(t\), \[ [\mathrm P]_{\infty}-[\mathrm P]_t=[\mathrm A]_t \] \(\textbf{Construct the product-deficit graph:}\) Substituting the first-order reactant concentration, \[ [\mathrm P]_{\infty}-[\mathrm P]_t=[\mathrm A]_0e^{-kt} \] Taking natural logarithms, \[ \ln\!\left(\frac{[\mathrm P]_{\infty}-[\mathrm P]_t}{c^\circ}\right)=\ln\!\left(\frac{[\mathrm A]_0}{c^\circ}\right)-kt \] Therefore, Graph II has slope \[ m_{\mathrm{II}}=-k \] Thus, \[ m_{\mathrm I}=m_{\mathrm{II}} \] \(\textbf{Interpret the apparently different plotted quantities:}\) The concentration of P increases during the reaction. However, its deficit relative to the limiting product concentration decreases. Therefore, the logarithm of that deficit has a negative slope rather than a positive one. The presence of product initially changes the limiting product concentration but not the equality between the remaining reactant and remaining product-forming capacity. The slope of Graph II is therefore independent of whether the initial product concentration is zero. The result illustrates how a suitable transformation of product-concentration measurements can recover the same kinetic information as direct measurements of a disappearing reactant. The quantity being logarithmically plotted must be identified correctly before interpreting a graphical slope.

Explanation: Statement I is correct because the first-order half-life is determined by the rate constant: \[ t_{1/2}=\frac{\ln2}{k}. \] Both experiments have the same rate constant at the same temperature. Therefore, their half-lives are identical despite their different initial concentrations. Statement II is correct because 12 min corresponds to one half-life in both experiments. In Experiment I, the concentration decreases from \(0.80\) to \(0.40\ \mathrm{mol\,L^{-1}}\). The amount consumed per litre is therefore \(0.40\ \mathrm{mol\,L^{-1}}\). In Experiment II, the concentration decreases from \(0.40\) to \(0.20\ \mathrm{mol\,L^{-1}}\). The amount consumed per litre is \(0.20\ \mathrm{mol\,L^{-1}}\). Thus, Experiment I consumes twice as much A per litre during the same interval. Statement III is incorrect because the instantaneous rates retain a constant ratio at every common finite time. The reactant concentrations follow \[ [\mathrm A]_{\mathrm I}(t)=0.80e^{-kt} \] and \[ [\mathrm A]_{\mathrm{II}}(t)=0.40e^{-kt}. \] Consequently, their instantaneous disappearance rates satisfy \[ \frac{r_{\mathrm I}(t)}{r_{\mathrm{II}}(t)}=\frac{k(0.80)e^{-kt}}{k(0.40)e^{-kt}}=2. \] They both approach zero, but they do not become equal at a finite time under the given ideal model. Statement IV is therefore correct. The initial concentration ratio remains preserved in the concentration and rate ratios because both samples share the same exponential factor. Statement V is incorrect because ideal first-order exponential decay approaches zero concentration asymptotically. It does not predict exact complete disappearance at a finite elapsed time. After 24 min, two half-lives have passed. Each experiment retains one-fourth of its own initial reactant concentration rather than reaching zero. The statements distinguish equal fractional conversion from equal absolute consumption and equal half-lives from equal instantaneous rates. They also emphasize that first-order decay preserves the ratio between two independently reacting samples having the same rate constant.

Explanation: The graph does not show concentration against time. Instead, it shows how the first half-life changes with the initial reactant concentration. For a zero-order disappearance law, the half-life is directly proportional to the initial concentration. Therefore, the slope of this graph determines the inverse of twice the zero-order disappearance constant. The constant can then be used to calculate the amount of A consumed in the second experiment. The concentration of P must subsequently be obtained using the stoichiometric relationship. \(\textbf{Identify the half-life relationship:}\) For zero-order disappearance, \[ -\frac{d[\mathrm A]}{dt}=k_{\mathrm{dis}} \] The integrated equation is \[ [\mathrm A]_t=[\mathrm A]_0-k_{\mathrm{dis}}t \] At the first half-life, \[ [\mathrm A]_{t_{1/2}}=\frac{[\mathrm A]_0}{2} \] Therefore, \[ t_{1/2}=\frac{[\mathrm A]_0}{2k_{\mathrm{dis}}} \] \(\textbf{Interpret the graph slope:}\) The slope of the half-life graph is \[ m=\frac{t_{1/2}}{[\mathrm A]_0} \] Using the marked point, \[ m=\frac{10.0}{0.40} \] \[ m=25.0\ \mathrm{L\,min\,mol^{-1}} \] Since \[ m=\frac1{2k_{\mathrm{dis}}} \] we obtain \[ k_{\mathrm{dis}}=\frac1{2m} \] \[ k_{\mathrm{dis}}=\frac1{2(25.0)} \] \[ k_{\mathrm{dis}}=0.020\ \mathrm{mol\,L^{-1}\,min^{-1}} \] \(\textbf{Calculate A consumption during the new experiment:}\) The disappearance constant is unchanged. During 10 min, \[ -\Delta[\mathrm A]=k_{\mathrm{dis}}t \] \[ -\Delta[\mathrm A]=0.020(10) \] \[ -\Delta[\mathrm A]=0.200\ \mathrm{mol\,L^{-1}} \] The remaining concentration of A is \[ [\mathrm A]_{10}=0.60-0.200 \] \[ [\mathrm A]_{10}=0.400\ \mathrm{mol\,L^{-1}} \] \(\textbf{Convert reactant consumption into product formation:}\) For \[ 2\mathrm A\rightarrow3\mathrm P \] the stoichiometric relationship is \[ \Delta[\mathrm P]=\frac32(-\Delta[\mathrm A]) \] Therefore, \[ \Delta[\mathrm P]=\frac32(0.200) \] \[ \Delta[\mathrm P]=0.300\ \mathrm{mol\,L^{-1}} \] \(\textbf{Include the initially present product:}\) \[ [\mathrm P]_{10}=[\mathrm P]_0+\Delta[\mathrm P] \] \[ [\mathrm P]_{10}=0.050+0.300 \] \[ [\mathrm P]_{10}=0.350\ \mathrm{mol\,L^{-1}} \] \(\textbf{Check the kinetic normalization:}\) The disappearance constant describes \[ -\frac{d[\mathrm A]}{dt}=0.020\ \mathrm{mol\,L^{-1}\,min^{-1}} \] The stoichiometrically normalized reaction rate is half the disappearance rate because the coefficient of A is two. Thus, the normalized zero-order constant would have half the numerical value. The question explicitly asks for the disappearance constant, so no additional factor is applied to the graph-derived result. \(\textbf{Final answer:}\) \[ \boxed{\left(k_{\mathrm{dis}},[\mathrm P]_{10}\right)=\left(0.020\ \mathrm{mol\,L^{-1}\,min^{-1}},0.350\ \mathrm{mol\,L^{-1}}\right)} \]

Explanation: Both experiments expose A to the same two temperatures for equal total durations. However, the order of the temperature intervals is reversed. For first-order kinetics, the concentration remaining after consecutive intervals is determined by the product of their exponential-decay factors. These factors can be multiplied in either order without changing the final concentration. The intermediate concentrations are nevertheless different. Consequently, the instantaneous reaction rates immediately after switching temperatures need not be equal. \(\textbf{First interval in Experiment X:}\) Experiment X initially follows the rate constant \[ k_1=\frac{\ln2}{10}\ \mathrm{min^{-1}} \] After 10 min, \[ [\mathrm A]_{X,10}=0.80e^{-k_1(10)} \] \[ [\mathrm A]_{X,10}=0.80e^{-\ln2} \] \[ [\mathrm A]_{X,10}=0.80\left(\frac12\right) \] \[ [\mathrm A]_{X,10}=0.40\ \mathrm{mol\,L^{-1}} \] \(\textbf{First interval in Experiment Y:}\) Experiment Y initially follows \[ k_2=\frac{3\ln2}{10}\ \mathrm{min^{-1}} \] After 10 min, \[ [\mathrm A]_{Y,10}=0.80e^{-k_2(10)} \] \[ [\mathrm A]_{Y,10}=0.80e^{-3\ln2} \] \[ [\mathrm A]_{Y,10}=0.80\left(\frac18\right) \] \[ [\mathrm A]_{Y,10}=0.10\ \mathrm{mol\,L^{-1}} \] \(\textbf{Instantaneous rates immediately after switching:}\) Experiment X switches to \(T_2\), so its new rate constant is \(k_2\). Therefore, \[ r_X=k_2[\mathrm A]_{X,10} \] \[ r_X=\frac{3\ln2}{10}(0.40) \] Experiment Y switches to \(T_1\), so its new rate constant is \(k_1\). \[ r_Y=k_1[\mathrm A]_{Y,10} \] \[ r_Y=\frac{\ln2}{10}(0.10) \] \(\textbf{Compare the rates:}\) \[ \frac{r_X}{r_Y}=\frac{(3\ln2/10)(0.40)}{(\ln2/10)(0.10)} \] Cancelling common factors, \[ \frac{r_X}{r_Y}=\frac{3(0.40)}{0.10} \] \[ \frac{r_X}{r_Y}=12 \] \(\textbf{Second interval in Experiment X:}\) At \(T_2\), the concentration decreases by a factor of eight in 10 min. Therefore, \[ [\mathrm A]_{X,20}=0.40\left(\frac18\right) \] \[ [\mathrm A]_{X,20}=0.050\ \mathrm{mol\,L^{-1}} \] \(\textbf{Second interval in Experiment Y:}\) At \(T_1\), the concentration decreases by a factor of two in 10 min. Therefore, \[ [\mathrm A]_{Y,20}=0.10\left(\frac12\right) \] \[ [\mathrm A]_{Y,20}=0.050\ \mathrm{mol\,L^{-1}} \] \(\textbf{Explain the invariant final concentration:}\) For either experiment, the overall concentration multiplier is \[ e^{-10k_1}e^{-10k_2} \] Therefore, \[ [\mathrm A]_{20}=0.80e^{-10(k_1+k_2)} \] The order of the two exponential factors does not affect their product. This explains the equal final concentrations. \(\textbf{Chemical interpretation:}\) The two experiments have identical final conversion despite following different concentration histories. Immediately after the temperature changes, Experiment X has both the higher reactant concentration and the higher rate constant. Its instantaneous rate is consequently twelve times the corresponding rate in Experiment Y. The equality of final conversions must not be confused with equality of instantaneous rates throughout the experiment. \(\textbf{Final answer:}\) \[ \boxed{\left([\mathrm A]_{20},\frac{r_X}{r_Y}\right)=\left(0.050\ \mathrm{mol\,L^{-1}},12\right)} \]

Explanation: The total-pressure measurements contain enough information to determine the kinetic fractional decay and the initially unknown amount of A. The decomposition increases total gaseous moles because one molecule of A produces two molecules of B. At constant temperature and volume, each pressure increase is proportional to the amount of A consumed during the corresponding interval. For first-order kinetics, the fraction of A remaining after equal time intervals is constant. Therefore, successive equal-time pressure increments form a geometric sequence. This provides a way to identify the fraction remaining without initially knowing the separate gas partial pressures. \(\textbf{Identify successive total-pressure increases:}\) During the first 10 min, \[ \Delta P_1=7.0-6.0 \] \[ \Delta P_1=1.0\ \mathrm{atm} \] During the next 10 min, \[ \Delta P_2=7.5-7.0 \] \[ \Delta P_2=0.50\ \mathrm{atm} \] Their ratio is \[ \frac{\Delta P_2}{\Delta P_1}=\frac{0.50}{1.0} \] \[ \frac{\Delta P_2}{\Delta P_1}=\frac12 \] \(\textbf{Infer the first-order fraction remaining:}\) Let \(f\) denote the fraction of A remaining after each 10-minute interval. For first-order kinetics, the amount consumed during successive equal intervals decreases by the factor \(f\). Therefore, \[ f=\frac{\Delta P_2}{\Delta P_1} \] \[ f=\frac12 \] Thus, the first-order half-life is 10 min. \(\textbf{Recover the initial partial pressure of A:}\) Let the initial partial pressure of A be \(p_{\mathrm A,0}\). During the first half-life, half the initial A decomposes. For \[ \mathrm A\rightarrow2\mathrm B \] the total-pressure increase equals the pressure-equivalent amount of A consumed. Therefore, \[ \Delta P_1=\frac{p_{\mathrm A,0}}2 \] Substituting, \[ 1.0=\frac{p_{\mathrm A,0}}2 \] \[ p_{\mathrm A,0}=2.0\ \mathrm{atm} \] \(\textbf{Infer the initial inert-gas contribution:}\) Initially, the total pressure is \(6.0\ \mathrm{atm}\). Therefore, \[ p_{\mathrm{inert},0}=6.0-2.0 \] \[ p_{\mathrm{inert},0}=4.0\ \mathrm{atm} \] \(\textbf{Predict the remaining A after 30 min:}\) Thirty minutes corresponds to three half-lives. Therefore, \[ p_{\mathrm A,30}=2.0\left(\frac12\right)^3 \] \[ p_{\mathrm A,30}=0.25\ \mathrm{atm} \] \(\textbf{Calculate the total decomposition:}\) The pressure-equivalent amount of A consumed is \[ x_{30}=2.0-0.25 \] \[ x_{30}=1.75\ \mathrm{atm} \] The total pressure increases by this amount because the reaction creates one net additional gaseous molecule per A consumed. Therefore, \[ P_{30}=P_0+x_{30} \] \[ P_{30}=6.0+1.75 \] \[ P_{30}=7.75\ \mathrm{atm} \] \(\textbf{Limiting-pressure verification:}\) Complete decomposition of the initial A would increase the total pressure by \(2.0\ \mathrm{atm}\). Therefore, \[ P_{\infty}=6.0+2.0 \] \[ P_{\infty}=8.0\ \mathrm{atm} \] The predicted pressure at 30 min lies below this limit. The total-pressure increments become progressively smaller as the amount of A available for decomposition decreases. The calculation shows how multiple total-pressure observations can reveal both an initially unknown reacting-gas amount and its first-order kinetic behaviour. \(\textbf{Final answer:}\) \[ \boxed{\left(p_{\mathrm A,0},P_{30}\right)=\left(2.0\ \mathrm{atm},7.75\ \mathrm{atm}\right)} \]

Explanation: The statements describe four different experimental representations of the same reaction. They involve the initial differential rate law, two integrated graphical forms, a concentration-dependent rate change and product formation. Their consistency must be checked using one common first-order rate constant. The stoichiometric coefficient of P must also be considered when interpreting its appearance rate. \(\textbf{Determine the first-order disappearance constant:}\) For first-order disappearance, \[ r_0=k[\mathrm A]_0 \] Therefore, \[ k=\frac{r_0}{[\mathrm A]_0} \] \[ k=\frac{0.040}{0.80} \] \[ k=0.050\ \mathrm{s^{-1}} \] Statement I is correct because the integrated first-order equation is \[ \ln[\mathrm A]_t=\ln[\mathrm A]_0-kt \] Therefore, its logarithmic slope is \[ m_{\ln[\mathrm A]}=-k \] \[ m_{\ln[\mathrm A]}=-0.050\ \mathrm{s^{-1}} \] Statement II is also correct because the common-logarithm form is \[ \log_{10}[\mathrm A]_t=\log_{10}[\mathrm A]_0-\frac{k}{2.303}t \] Thus, \[ m_{\log_{10}[\mathrm A]}=-\frac{0.050}{2.303} \] \[ m_{\log_{10}[\mathrm A]}\approx-0.0217\ \mathrm{s^{-1}} \] Statement III is correct because the instantaneous disappearance rate is directly proportional to A concentration. When the rate becomes half its initial value, \[ [\mathrm A]=\frac{0.80}{2} \] \[ [\mathrm A]=0.40\ \mathrm{mol\,L^{-1}} \] The amount of A consumed is \[ -\Delta[\mathrm A]=0.80-0.40 \] \[ -\Delta[\mathrm A]=0.40\ \mathrm{mol\,L^{-1}} \] For \[ \mathrm A\rightarrow2\mathrm P \] the product formed is \[ \Delta[\mathrm P]=2(0.40) \] \[ \Delta[\mathrm P]=0.80\ \mathrm{mol\,L^{-1}} \] Including the initial product, \[ [\mathrm P]=0.10+0.80 \] \[ [\mathrm P]=0.90\ \mathrm{mol\,L^{-1}} \] Statement IV is incorrect because the appearance rate of P is twice the disappearance rate of A. At half the initial A concentration, \[ -\frac{d[\mathrm A]}{dt}=0.050(0.40) \] \[ -\frac{d[\mathrm A]}{dt}=0.020\ \mathrm{mol\,L^{-1}\,s^{-1}} \] However, \[ \frac{d[\mathrm P]}{dt}=2\left(-\frac{d[\mathrm A]}{dt}\right) \] Therefore, \[ \frac{d[\mathrm P]}{dt}=0.040\ \mathrm{mol\,L^{-1}\,s^{-1}} \] The value \(0.020\ \mathrm{mol\,L^{-1}\,s^{-1}}\) is the disappearance rate of A, not the appearance rate of P. The example demonstrates how graphical slopes, instantaneous rates and stoichiometric product formation provide mutually connected checks on a proposed kinetic description. The correct statements are I, II and III only.

Explanation: The underlying reaction is first order in both A and B. However, B is initially present at a concentration twenty times that of A. Its fractional change can therefore remain small while A undergoes a substantial fractional change. The allowable observation interval is determined by the permitted depletion of B. Stoichiometry first converts this limit into the corresponding consumption of A. The pseudo-first-order integrated equation can then estimate the time required for that consumption. Because B does change slightly, the resulting time is an approximation rather than the exact time predicted by the full two-reactant rate law. \(\textbf{Maximum permitted consumption of B:}\) Initially, \[ [\mathrm B]_0=2.00\ \mathrm{mol\,L^{-1}} \] The maximum permitted fractional decrease is \[ f_{\mathrm B}=0.020 \] Therefore, \[ -\Delta[\mathrm B]_{\max}=0.020(2.00) \] \[ -\Delta[\mathrm B]_{\max}=0.0400\ \mathrm{mol\,L^{-1}} \] \(\textbf{Convert B consumption into A consumption:}\) The balanced equation is \[ \mathrm{A+B\rightarrow P} \] A and B are consumed in a \(1:1\) ratio. Therefore, \[ -\Delta[\mathrm A]_{\max}=-\Delta[\mathrm B]_{\max} \] \[ -\Delta[\mathrm A]_{\max}=0.0400\ \mathrm{mol\,L^{-1}} \] \(\textbf{Determine A at the approximation limit:}\) \[ [\mathrm A]_{\mathrm{limit}}=0.100-0.0400 \] \[ [\mathrm A]_{\mathrm{limit}}=0.0600\ \mathrm{mol\,L^{-1}} \] \(\textbf{Calculate the observed first-order constant:}\) Within the constant-B approximation, \[ k_{\mathrm{obs}}=k[\mathrm B]_0 \] Substituting, \[ k_{\mathrm{obs}}=0.200(2.00) \] \[ k_{\mathrm{obs}}=0.400\ \mathrm{min^{-1}} \] \(\textbf{Apply the approximate integrated equation:}\) For pseudo-first-order disappearance, \[ t\approx\frac1{k_{\mathrm{obs}}}\ln\frac{[\mathrm A]_0}{[\mathrm A]_t} \] At the permitted depletion limit, \[ t_{\max}\approx\frac1{0.400}\ln\frac{0.100}{0.0600} \] \[ t_{\max}\approx\frac1{0.400}\ln\frac53 \] Using the supplied logarithm, \[ t_{\max}\approx\frac{0.5108}{0.400} \] \[ t_{\max}\approx1.277\ \mathrm{min} \] \(\textbf{Check the excess-reactant condition:}\) At this estimated endpoint, \[ [\mathrm B]=2.00-0.0400 \] \[ [\mathrm B]=1.960\ \mathrm{mol\,L^{-1}} \] The relative depletion is \[ \frac{2.00-1.960}{2.00}\times100 \] \[ =2.0\% \] Therefore, the chosen endpoint corresponds to the specified depletion limit. \(\textbf{Interpret the approximation:}\) The intrinsic rate constant remains \(0.200\ \mathrm{L\,mol^{-1}\,min^{-1}}\). The observed pseudo-first-order constant incorporates the approximately constant concentration of B. As B is consumed, the actual instantaneous value of \(k[\mathrm B]\) decreases slightly. Consequently, the exact reaction requires slightly more time to reach the same endpoint than the constant-B estimate. The question explicitly asks for the pseudo-first-order estimate, so this small change is intentionally neglected. The calculation demonstrates how reaction stoichiometry and an allowed experimental tolerance can define the useful time range of a kinetic approximation. \(\textbf{Final answer:}\) \[ \boxed{t_{\max}\approx1.28\ \mathrm{min}} \]

Explanation: Both samples follow the same first-order law but begin at different reactant concentrations. The time required to reach a common final concentration is therefore different. The difference between their elapsed times can be related directly to the ratio of their initial concentrations. Because both experiments have the same rate constant, the common target concentration cancels when their integrated rate equations are subtracted. This allows the half-life to be determined without knowing the individual target times in advance. \(\textbf{Write the first-order target time for Experiment I:}\) The integrated equation gives \[ t_{\mathrm I}=\frac1k\ln\frac{[\mathrm A]_{0,\mathrm I}}{[\mathrm A]_{\mathrm{target}}} \] Therefore, \[ t_{\mathrm I}=\frac1k\ln\frac{0.80}{0.050} \] \[ t_{\mathrm I}=\frac{\ln16}{k} \] \(\textbf{Write the corresponding time for Experiment II:}\) \[ t_{\mathrm{II}}=\frac1k\ln\frac{0.20}{0.050} \] \[ t_{\mathrm{II}}=\frac{\ln4}{k} \] \(\textbf{Use the measured difference in arrival times:}\) The problem states that \[ t_{\mathrm I}-t_{\mathrm{II}}=20\ \mathrm{min} \] Therefore, \[ \frac{\ln16-\ln4}{k}=20 \] Using logarithm rules, \[ \ln16-\ln4=\ln\frac{16}{4} \] \[ \ln16-\ln4=\ln4 \] Thus, \[ \frac{\ln4}{k}=20 \] \(\textbf{Determine the first-order constant:}\) \[ k=\frac{\ln4}{20} \] Since \[ \ln4=2\ln2 \] we obtain \[ k=\frac{\ln2}{10} \] \(\textbf{Calculate the common half-life:}\) \[ t_{1/2}=\frac{\ln2}{k} \] Substitution gives \[ t_{1/2}=\frac{\ln2}{\ln2/10} \] \[ t_{1/2}=10.0\ \mathrm{min} \] \(\textbf{Determine the time for Experiment I:}\) The concentration must decrease from \(0.80\) to \(0.050\ \mathrm{mol\,L^{-1}}\). The remaining fraction is \[ f_{\mathrm I}=\frac{0.050}{0.80} \] \[ f_{\mathrm I}=\frac1{16} \] Therefore, four half-lives are required. \[ t_{\mathrm I}=4t_{1/2} \] \[ t_{\mathrm I}=4(10.0) \] \[ t_{\mathrm I}=40.0\ \mathrm{min} \] \(\textbf{Verify Experiment II:}\) Experiment II decreases from \(0.20\) to \(0.050\ \mathrm{mol\,L^{-1}}\). This is a reduction to one-fourth. Therefore, two half-lives are required. \[ t_{\mathrm{II}}=2(10.0) \] \[ t_{\mathrm{II}}=20.0\ \mathrm{min} \] The difference is \[ 40.0-20.0=20.0\ \mathrm{min} \] as observed. \(\textbf{Chemical interpretation:}\) A sample beginning at a larger concentration takes longer to reach the same absolute target concentration. However, the additional time depends logarithmically on the ratio of the initial concentrations. For samples having the same first-order constant, this additional time does not depend on which common positive target concentration is chosen, provided both samples can reach it under the same ideal law. This is a consequence of the exponential nature of first-order disappearance. \(\textbf{Final answer:}\) \[ \boxed{\left(t_{1/2},t_{\mathrm I}\right)=\left(10.0\ \mathrm{min},40.0\ \mathrm{min}\right)} \]

Explanation: The reaction contains two first-order intervals separated by an instantaneous addition of reactant. The added A increases the concentration immediately but does not represent chemical regeneration of A from P. The first-order half-life remains unchanged because the reaction conditions and rate constant remain the same. Therefore, each ten-minute interval reduces the A present at the beginning of that interval to half. The final product concentration must be calculated from the total amount of A chemically consumed, including both the originally present and subsequently introduced reactant. \(\textbf{First reaction interval:}\) Initially, \[ [\mathrm A]_0=0.80\ \mathrm{mol\,L^{-1}} \] After one half-life, \[ [\mathrm A]_{10^-}=0.80\left(\frac12\right) \] \[ [\mathrm A]_{10^-}=0.40\ \mathrm{mol\,L^{-1}} \] \(\textbf{Product formed during the first interval:}\) The amount of A consumed per litre is \[ -\Delta[\mathrm A]_1=0.80-0.40 \] \[ -\Delta[\mathrm A]_1=0.40\ \mathrm{mol\,L^{-1}} \] Since the reaction is \[ \mathrm A\rightarrow\mathrm P \] the product concentration increases by the same amount. Therefore, \[ [\mathrm P]_{10^-}=0.10+0.40 \] \[ [\mathrm P]_{10^-}=0.50\ \mathrm{mol\,L^{-1}} \] \(\textbf{Instantaneous addition of A:}\) The concentration increases by \[ 0.20\ \mathrm{mol\,L^{-1}} \] Therefore, \[ [\mathrm A]_{10^+}=0.40+0.20 \] \[ [\mathrm A]_{10^+}=0.60\ \mathrm{mol\,L^{-1}} \] The product concentration remains unchanged during the addition. \[ [\mathrm P]_{10^+}=0.50\ \mathrm{mol\,L^{-1}} \] \(\textbf{Second reaction interval:}\) The second interval lasts another half-life. Therefore, \[ [\mathrm A]_{20}=0.60\left(\frac12\right) \] \[ [\mathrm A]_{20}=0.30\ \mathrm{mol\,L^{-1}} \] \(\textbf{Product formed during the second interval:}\) The A consumed during this interval is \[ -\Delta[\mathrm A]_2=0.60-0.30 \] \[ -\Delta[\mathrm A]_2=0.30\ \mathrm{mol\,L^{-1}} \] Therefore, \[ [\mathrm P]_{20}=0.50+0.30 \] \[ [\mathrm P]_{20}=0.80\ \mathrm{mol\,L^{-1}} \] \(\textbf{Check the total material balance:}\) The total A supplied per litre is \[ 0.80+0.20=1.00\ \mathrm{mol\,L^{-1}} \] The A remaining at the end is \[ 0.30\ \mathrm{mol\,L^{-1}} \] Therefore, the total A consumed is \[ 1.00-0.30=0.70\ \mathrm{mol\,L^{-1}} \] Adding the initial product gives \[ [\mathrm P]_{20}=0.10+0.70 \] \[ [\mathrm P]_{20}=0.80\ \mathrm{mol\,L^{-1}} \] This independently confirms the product calculation. \(\textbf{Chemical interpretation:}\) The concentration does not follow one uninterrupted exponential decay from the original initial condition because an additional amount of reactant is introduced midway. Instead, the new concentration after addition becomes the starting concentration for the second first-order interval. The rate constant remains unchanged even though the instantaneous reaction rate increases when reactant is added. The final product concentration depends on total reactant consumption rather than only on the difference between the original and final A concentrations. \(\textbf{Final answer:}\) \[ \boxed{\left([\mathrm A]_{20},[\mathrm P]_{20}\right)=\left(0.30,\ 0.80\right)\ \mathrm{mol\,L^{-1}}} \]

Explanation: The reaction changes kinetic behaviour as A concentration decreases. Initially, the catalytic surface is saturated and A disappears at a constant rate. At lower concentration, the rate becomes directly proportional to A concentration. Unlike a problem in which the transition concentration is given, the transition must first be inferred from the equality of the two rate expressions. The zero-order and first-order integrated equations are then applied over their appropriate concentration intervals. \(\textbf{Determine the transition concentration:}\) At the transition, the two instantaneous rates are equal. Therefore, \[ k_0=k_1[\mathrm A]_{\mathrm{transition}} \] The supplied constants are \[ k_0=0.030\ \mathrm{mol\,L^{-1}\,min^{-1}} \] and \[ k_1=0.060\ \mathrm{min^{-1}} \] Hence, \[ [\mathrm A]_{\mathrm{transition}}=\frac{k_0}{k_1} \] \[ [\mathrm A]_{\mathrm{transition}}=\frac{0.030}{0.060} \] \[ [\mathrm A]_{\mathrm{transition}}=0.500\ \mathrm{mol\,L^{-1}} \] \(\textbf{Time spent in the zero-order regime:}\) Initially, \[ [\mathrm A]_0=0.80\ \mathrm{mol\,L^{-1}} \] The concentration decreases to \[ 0.500\ \mathrm{mol\,L^{-1}} \] at the transition. For zero-order disappearance, \[ t_1=\frac{[\mathrm A]_0-[\mathrm A]_{\mathrm{transition}}}{k_0} \] Substituting, \[ t_1=\frac{0.80-0.500}{0.030} \] \[ t_1=\frac{0.300}{0.030} \] \[ t_1=10.0\ \mathrm{min} \] \(\textbf{Time spent in the first-order regime:}\) The concentration then decreases from \(0.500\) to \(0.125\ \mathrm{mol\,L^{-1}}\). The first-order integrated equation gives \[ t_2=\frac1{k_1}\ln\frac{[\mathrm A]_{\mathrm{transition}}}{[\mathrm A]_{\mathrm{target}}} \] Substituting, \[ t_2=\frac1{0.060}\ln\frac{0.500}{0.125} \] \[ t_2=\frac{\ln4}{0.060} \] Using the supplied logarithm, \[ t_2=\frac{1.386}{0.060} \] \[ t_2=23.1\ \mathrm{min} \] \(\textbf{Total elapsed time:}\) \[ t_{\mathrm{total}}=t_1+t_2 \] \[ t_{\mathrm{total}}=10.0+23.1 \] \[ t_{\mathrm{total}}=33.1\ \mathrm{min} \] \(\textbf{Check continuity at the transition:}\) In the zero-order regime, the rate is \[ r_0=0.030\ \mathrm{mol\,L^{-1}\,min^{-1}} \] In the first-order regime at the transition concentration, \[ r_1=0.060(0.500) \] \[ r_1=0.030\ \mathrm{mol\,L^{-1}\,min^{-1}} \] The two predicted rates agree. \(\textbf{Physical interpretation:}\) The kinetic law changes because the controlling surface condition changes as the reactant concentration decreases. The transition concentration is not arbitrary in the model supplied. It is fixed by the rate-continuity condition. The reaction initially consumes A linearly with time and subsequently follows exponential disappearance. The combined elapsed time must therefore be determined by adding the two physically distinct stages. \(\textbf{Final answer:}\) \[ \boxed{t_{\mathrm{total}}=33.1\ \mathrm{min}} \]

Explanation: The vessel expands as the reaction produces more gaseous molecules. Therefore, the concentration of A depends on both the amount of unreacted A and the changing vessel volume. The differential law must first be interpreted carefully. Because the rate per unit volume is proportional to concentration, the amount-disappearance rate is proportional to the amount of A remaining. However, this does not mean that its concentration necessarily follows the same simple exponential function when the volume changes. Statement I is correct. The supplied local rate law is \[ -\frac1V\frac{dn_{\mathrm A}}{dt}=k[\mathrm A]. \] Since \[ [\mathrm A]=\frac{n_{\mathrm A}}V, \] substitution gives \[ -\frac1V\frac{dn_{\mathrm A}}{dt}=k\frac{n_{\mathrm A}}V. \] Multiplying by the instantaneous volume, \[ -\frac{dn_{\mathrm A}}{dt}=kn_{\mathrm A}. \] Therefore, the amount remaining follows \[ n_{\mathrm A}(t)=n_0e^{-kt}. \] Statement II is correct because the amount of A becomes \(n_0/2\) after one amount-based half-life. At that instant, half the initial A has decomposed. For \[ \mathrm A\rightarrow2\mathrm B, \] the amount of B formed is \[ n_{\mathrm B}=2\left(\frac{n_0}{2}\right). \] Thus, \[ n_{\mathrm B}=n_0. \] The total gaseous amount is \[ n_{\mathrm{total}}=\frac{n_0}{2}+n_0 \] \[ n_{\mathrm{total}}=\frac{3n_0}{2}. \] At constant temperature and total pressure, the volume is proportional to the total gaseous amount. Therefore, \[ V=\frac32V_0. \] The A concentration at this instant is \[ [\mathrm A]=\frac{n_0/2}{3V_0/2} \] \[ [\mathrm A]=\frac{n_0}{3V_0}. \] Initially, \[ [\mathrm A]_0=\frac{n_0}{V_0}. \] Hence, \[ \frac{[\mathrm A]}{[\mathrm A]_0}=\frac13. \] Statement III is incorrect because the volume increases as the reaction proceeds. The amount remaining follows an exponential law, but concentration is the amount divided by the changing volume. For the stated reaction and conditions, \[ [\mathrm A](t)=\frac{n_0e^{-kt}}{V_0(2-e^{-kt})}. \] Therefore, \[ \frac{[\mathrm A](t)}{[\mathrm A]_0}=\frac{e^{-kt}}{2-e^{-kt}}. \] The denominator also changes with time. Consequently, \(\ln[\mathrm A]\) is not generally a straight-line function of time with slope \(-k\). Statement IV is correct because the total number of gaseous moles becomes \(3/2\) times its initial value after half the initial A decomposes. The constant-pressure piston therefore expands the vessel volume by the same factor. The problem illustrates the distinction between the exponential disappearance of an amount of reactant and the concentration measured in a vessel whose volume changes. The correct statements are I, II and IV.

Explanation: The experiments begin at different concentrations but follow the same first-order kinetic law. For any common target concentration below the smaller initial concentration, both experiments can reach that target. Their individual elapsed times depend logarithmically on the ratios of initial to target concentration. However, subtracting the two elapsed-time expressions eliminates the target concentration. The remaining time difference depends only on the ratio of their initial concentrations and the common first-order constant. For Experiment I, \[ t_{\mathrm I}=\frac1k\ln\frac{0.80}{C_{\mathrm{target}}}. \] For Experiment II, \[ t_{\mathrm{II}}=\frac1k\ln\frac{0.40}{C_{\mathrm{target}}}. \] Subtracting, \[ t_{\mathrm I}-t_{\mathrm{II}}=\frac1k\left(\ln\frac{0.80}{C_{\mathrm{target}}}-\ln\frac{0.40}{C_{\mathrm{target}}}\right). \] Using the logarithm quotient rule, \[ t_{\mathrm I}-t_{\mathrm{II}}=\frac1k\ln\frac{0.80}{0.40}. \] Therefore, \[ t_{\mathrm I}-t_{\mathrm{II}}=\frac{\ln2}{k}. \] For first-order kinetics, \[ t_{1/2}=\frac{\ln2}{k}. \] Thus, the difference between the two target times is exactly one half-life. This difference does not depend on the chosen positive target concentration. Option A is incorrect because the ratio of elapsed times is not generally fixed at two. Options B and D are incorrect because the difference does not grow or shrink as the target is lowered. The result can also be interpreted through the concentration histories. Experiment I begins with twice as much A per litre as Experiment II. After one half-life, Experiment I has the same concentration that Experiment II had initially. From that point onward, Experiment I follows the same remaining concentration sequence as Experiment II, delayed by one half-life. Therefore, it reaches every common positive target one half-life later. This time-shift property is a consequence of ideal first-order kinetics and does not depend on the numerical value of the target concentration.

Explanation: The measurements provide instantaneous rates rather than reactant concentrations. For a first-order reaction, the instantaneous disappearance rate is proportional to the remaining concentration. Therefore, the rate itself decreases exponentially with time. The ratio of instantaneous rates at two observation times can be used to determine the rate constant without knowing either corresponding reactant concentration. The initial rate can then be recovered by extrapolating the same exponential relationship backward from a measured instant. \(\textbf{Express the first-order rate history:}\) For first-order disappearance, \[ [\mathrm A]_t=[\mathrm A]_0e^{-kt} \] The instantaneous rate is \[ r_t=k[\mathrm A]_t \] Substituting, \[ r_t=k[\mathrm A]_0e^{-kt} \] Since \[ r_0=k[\mathrm A]_0, \] we obtain \[ r_t=r_0e^{-kt}\] \(\textbf{Compare the two measured rates:}\) At 10 min, \[ r_{10}=r_0e^{-10k} \] At 30 min, \[ r_{30}=r_0e^{-30k} \] Dividing, \[ \frac{r_{10}}{r_{30}}=e^{20k} \] Substituting the observations, \[ \frac{0.020}{0.0050}=e^{20k} \] \[ 4=e^{20k} \] Taking natural logarithms, \[ \ln4=20k \] \(\textbf{Determine the first-order constant:}\) \[ k=\frac{\ln4}{20} \] Using \[ \ln4=2\ln2, \] we obtain \[ k=\frac{\ln2}{10} \] \(\textbf{Calculate the half-life:}\) \[ t_{1/2}=\frac{\ln2}{k} \] Therefore, \[ t_{1/2}=\frac{\ln2}{\ln2/10} \] \[ t_{1/2}=10.0\ \mathrm{min} \] \(\textbf{Recover the initial instantaneous rate:}\) At \(t=10\ \mathrm{min}\), one half-life has elapsed. Therefore, \[ r_{10}=\frac{r_0}{2} \] Hence, \[ r_0=2r_{10} \] \[ r_0=2(0.020) \] \[ r_0=0.040\ \mathrm{mol\,L^{-1}\,min^{-1}} \] \(\textbf{Check the later measurement:}\) Thirty minutes corresponds to three half-lives. Therefore, \[ r_{30}=r_0\left(\frac12\right)^3 \] \[ r_{30}=0.040\left(\frac18\right) \] \[ r_{30}=0.0050\ \mathrm{mol\,L^{-1}\,min^{-1}} \] This agrees with the observation. \(\textbf{Chemical interpretation:}\) The instantaneous rate of a first-order reaction follows the same fractional decay pattern as the remaining reactant concentration. Two rate measurements at known times can therefore determine the first-order constant without direct concentration measurements. The time between the two observations is 20 min, corresponding to two half-lives because the rate falls to one-fourth. The initial rate is obtained by reversing one halving from the measured rate at 10 min. \(\textbf{Final answer:}\) \[ \boxed{\left(t_{1/2},r_0\right)=\left(10.0\ \mathrm{min},0.040\ \mathrm{mol\,L^{-1}\,min^{-1}}\right)} \]

Explanation: The slow elementary step involves monomeric A rather than dimeric \(\mathrm{A_2}\). Therefore, the formation rate of AB cannot be calculated by directly substituting the concentration of \(\mathrm{A_2}\) for that of A. The fast equilibrium must first be used to determine the monomer concentration. This introduces a square-root relationship between the monomer and dimer concentrations. Consequently, the effective rate law can have a fractional partial order even though the individual elementary steps have integral molecularities. \(\textbf{Rate law of the controlling elementary step:}\) The slow step is \[ \mathrm{A+B\rightarrow AB} \] Its elementary rate law is \[ r_{\mathrm{AB}}=k_2[\mathrm A][\mathrm B] \] \(\textbf{Use the preceding equilibrium:}\) For \[ \mathrm{A_2\rightleftharpoons2A} \] the equilibrium relationship is \[ K_1=\frac{[\mathrm A]^2}{[\mathrm{A_2}]} \] Rearranging, \[ [\mathrm A]^2=K_1[\mathrm{A_2}] \] Therefore, \[ [\mathrm A]=\sqrt{K_1[\mathrm{A_2}]} \] \(\textbf{Calculate the monomer concentration:}\) Substituting the given values, \[ [\mathrm A]=\sqrt{(0.040)(0.25)} \] \[ [\mathrm A]=\sqrt{0.0100} \] \[ [\mathrm A]=0.100\ \mathrm{mol\,L^{-1}} \] \(\textbf{Determine the product-formation rate:}\) \[ r_{\mathrm{AB}}=k_2[\mathrm A][\mathrm B] \] \[ r_{\mathrm{AB}}=0.50(0.100)(0.20) \] \[ r_{\mathrm{AB}}=0.0100\ \mathrm{mol\,L^{-1}\,s^{-1}} \] \(\textbf{Derive the effective rate law:}\) Substituting the equilibrium expression for A, \[ r_{\mathrm{AB}}=k_2\sqrt{K_1[\mathrm{A_2}]}[\mathrm B] \] Rearranging, \[ r_{\mathrm{AB}}=k_2K_1^{1/2}[\mathrm{A_2}]^{1/2}[\mathrm B] \] Therefore, \[ r_{\mathrm{AB}}\propto[\mathrm{A_2}]^{1/2}[\mathrm B] \] The effective partial orders are \(1/2\) in \(\mathrm{A_2}\) and one in B. Their sum is \[ N=\frac12+1 \] \[ N=\frac32 \] \(\textbf{Molecularity interpretation:}\) The forward dissociation of \(\mathrm{A_2}\) is unimolecular. Its reverse association is bimolecular. The slow reaction between A and B is also bimolecular. None of these elementary steps has fractional molecularity. The fractional overall kinetic order arises from the equilibrium relationship that connects the intermediate monomer concentration with the observable dimer concentration. The complex overall reaction itself is not assigned one molecularity. \(\textbf{Final answer:}\) \[ \boxed{r_{\mathrm{AB}}=0.0100\ \mathrm{mol\,L^{-1}\,s^{-1}},\qquad r_{\mathrm{AB}}\propto[\mathrm{A_2}]^{1/2}[\mathrm B]} \]