Chemical Kinetics MCQs with Answers – Part 4 (NEET Chemistry)

Explanation: Compression increases gas concentration because the same number of moles occupies a smaller volume. Therefore, the instantaneous disappearance rate expressed per unit volume increases. However, the total number of moles of A does not change during compression. For a first-order reaction, the total molar disappearance rate is proportional to the amount of A remaining. Consequently, compression alone does not change that total molar rate. The subsequent chemical reaction must be calculated after establishing the new partial pressures. \(\textbf{Composition after the first half-life:}\) Initially, \[ p_{\mathrm A,0}=4.0\ \mathrm{atm} \] After 10 min, \[ p_{\mathrm A,10^-}=4.0\left(\frac12\right) \] \[ p_{\mathrm A,10^-}=2.0\ \mathrm{atm} \] Since A and B have a \(1:1\) stoichiometric relationship, \[ p_{\mathrm B,10^-}=2.0\ \mathrm{atm} \] \(\textbf{Apply the instantaneous compression:}\) The volume becomes half its previous value. At constant temperature, each partial pressure doubles because the corresponding number of moles is unchanged. Therefore, \[ p_{\mathrm A,10^+}=2(2.0)=4.0\ \mathrm{atm} \] \[ p_{\mathrm B,10^+}=2(2.0)=4.0\ \mathrm{atm} \] \(\textbf{Compare disappearance rates per unit volume:}\) The first-order concentration-based law is \[ r_V=k[\mathrm A]\] For an ideal gas, \[ [\mathrm A]=\frac{p_{\mathrm A}}{RT}\] Immediately after compression, A concentration doubles. Therefore, \[ \frac{r_{V,10^+}}{r_{V,10^-}}=2 \] \(\textbf{Compare total molar disappearance rates:}\) The total molar disappearance rate is \[ R=r_VV\] Using the first-order rate law, \[ R=k[\mathrm A]V \] \[ R=kn_{\mathrm A} \] Because no reaction occurs during compression, the amount of A remains unchanged. Therefore, \[ \frac{R_{10^+}}{R_{10^-}}=1 \] \(\textbf{Second reaction interval:}\) The reaction continues for another half-life at the new fixed volume. Thus, \[ p_{\mathrm A,20}=4.0\left(\frac12\right) \] \[ p_{\mathrm A,20}=2.0\ \mathrm{atm} \] The partial-pressure-equivalent amount of A consumed during this interval is \[ 4.0-2.0=2.0\ \mathrm{atm} \] Therefore, \[ p_{\mathrm B,20}=4.0+2.0 \] \[ p_{\mathrm B,20}=6.0\ \mathrm{atm} \] \(\textbf{Determine the requested product-to-reactant ratio:}\) \[ \frac{p_{\mathrm B,20}}{p_{\mathrm A,20}}=\frac{6.0}{2.0} \] \[ \frac{p_{\mathrm B,20}}{p_{\mathrm A,20}}=3 \] \(\textbf{Chemical interpretation:}\) The normalized volumetric reaction rate and total molar reaction rate are different physical quantities. Compression raises the local concentration but does not instantaneously change the number of reacting molecules. The first-order disappearance of the total amount of A therefore continues with the same intrinsic constant. The final product-to-reactant ratio depends on the cumulative chemical conversion, not on the compression-induced pressure increase alone. \(\textbf{Final answer:}\) \[ \boxed{\left(\frac{r_{V,10^+}}{r_{V,10^-}},\frac{R_{10^+}}{R_{10^-}},\frac{p_{\mathrm B,20}}{p_{\mathrm A,20}}\right)=(2,1,3)} \]

Explanation: The two pathways consume the same reactant A. Their independent zero-order disappearance rates can therefore be added while A remains available. However, this combined constant rate cannot continue after the finite initial amount of A has been exhausted. The average rate over 25 min must include the inactive time following complete consumption. The product concentration depends on the amount of A consumed through the P-forming pathway. \(\textbf{Determine the combined disappearance rate:}\) While both pathways operate, \[ -\frac{d[\mathrm A]}{dt}=r_P+r_Q \] Substituting, \[ -\frac{d[\mathrm A]}{dt}=0.020+0.010 \] \[ -\frac{d[\mathrm A]}{dt}=0.030\ \mathrm{mol\,L^{-1}\,min^{-1}} \] \(\textbf{Calculate the time to exhaustion:}\) For zero-order disappearance, \[ t_{\mathrm{exhaustion}}=\frac{[\mathrm A]_0}{r_P+r_Q} \] \[ t_{\mathrm{exhaustion}}=\frac{0.60}{0.030} \] \[ t_{\mathrm{exhaustion}}=20.0\ \mathrm{min} \] Thus, A is exhausted at 20 min, before the requested observation time of 25 min. \(\textbf{Determine cumulative product formation:}\) The P-forming pathway operates at its constant rate for 20 min. Therefore, \[ \Delta[\mathrm P]=r_Pt_{\mathrm{exhaustion}} \] \[ \Delta[\mathrm P]=0.020(20) \] \[ \Delta[\mathrm P]=0.400\ \mathrm{mol\,L^{-1}} \] The P concentration at 25 min is \[ [\mathrm P]_{25}=0.10+0.400 \] \[ [\mathrm P]_{25}=0.500\ \mathrm{mol\,L^{-1}} \] \(\textbf{Check formation of the competing product:}\) \[ \Delta[\mathrm Q]=0.010(20) \] \[ \Delta[\mathrm Q]=0.200\ \mathrm{mol\,L^{-1}} \] Therefore, \[ [\mathrm Q]_{25}=0.050+0.200 \] \[ [\mathrm Q]_{25}=0.250\ \mathrm{mol\,L^{-1}} \] The total product formed is \[ 0.400+0.200=0.600\ \mathrm{mol\,L^{-1}} \] This equals the initial amount of A consumed. \(\textbf{Calculate the average disappearance rate over 25 min:}\) The total concentration of A lost is \[ 0.60\ \mathrm{mol\,L^{-1}} \] The observation interval lasts 25 min. Therefore, \[ r_{\mathrm{avg}}=\frac{0.60}{25} \] \[ r_{\mathrm{avg}}=0.024\ \mathrm{mol\,L^{-1}\,min^{-1}} \] \(\textbf{Distinguish average and instantaneous rates:}\) During the first 20 min, the instantaneous total disappearance rate is \[ 0.030\ \mathrm{mol\,L^{-1}\,min^{-1}} \] After A is exhausted, the instantaneous disappearance rate is zero. Therefore, the average rate over 25 min is less than the active zero-order rate. Extending the original straight-line concentration equation to 25 min would incorrectly predict a negative A concentration. \(\textbf{Chemical interpretation:}\) Zero-order kinetics means that the rate is independent of A concentration within the specified kinetic regime. It does not mean that reactant consumption can continue after the reactant has been exhausted. Independent zero-order pathways contribute constant product-formation rates while their common reactant remains present. The final product amounts reflect the relative pathway rates and the finite initial supply of A. \(\textbf{Final answer:}\) \[ \boxed{\left(r_{\mathrm{avg}},[\mathrm P]_{25}\right)=\left(0.024\ \mathrm{mol\,L^{-1}\,min^{-1}},0.500\ \mathrm{mol\,L^{-1}}\right)} \]

Explanation: The limiting product concentration provides a stoichiometric constraint. It determines how much P is formed for each mole of A consumed. However, it does not independently determine the kinetic rate constant. The concentration of P at 12 min gives the corresponding amount of A consumed. The remaining A concentration then allows the first-order constant to be calculated. Finally, the instantaneous appearance rate of P must include the stoichiometric coefficient \(n\). \(\textbf{Recover the unknown product coefficient:}\) At complete conversion, the concentration of A consumed is \[ [\mathrm A]_0=0.40\ \mathrm{mol\,L^{-1}} \] The total increase in P concentration is \[ \Delta[\mathrm P]_{\infty}=0.90-0.10 \] \[ \Delta[\mathrm P]_{\infty}=0.80\ \mathrm{mol\,L^{-1}} \] For \[ \mathrm A\rightarrow n\mathrm P, \] \[ \Delta[\mathrm P]_{\infty}=n[\mathrm A]_0 \] Therefore, \[ n=\frac{0.80}{0.40} \] \[ n=2 \] \(\textbf{Determine A consumption at 12 min:}\) The increase in product concentration is \[ \Delta[\mathrm P]_{12}=0.50-0.10 \] \[ \Delta[\mathrm P]_{12}=0.40\ \mathrm{mol\,L^{-1}} \] Because two P units are formed per A consumed, \[ -\Delta[\mathrm A]_{12}=\frac{\Delta[\mathrm P]_{12}}2 \] \[ -\Delta[\mathrm A]_{12}=\frac{0.40}{2} \] \[ -\Delta[\mathrm A]_{12}=0.20\ \mathrm{mol\,L^{-1}} \] \(\textbf{Determine the remaining A concentration:}\) \[ [\mathrm A]_{12}=0.40-0.20 \] \[ [\mathrm A]_{12}=0.20\ \mathrm{mol\,L^{-1}} \] Thus, A has reached half its initial concentration after 12 min. \(\textbf{Calculate the first-order constant:}\) \[ t_{1/2}=12\ \mathrm{min} \] Therefore, \[ k=\frac{\ln2}{12} \] \[ k=\frac{0.693}{12} \] \[ k=0.05775\ \mathrm{min^{-1}} \] \(\textbf{Calculate the instantaneous disappearance rate of A:}\) For first-order disappearance, \[ -\frac{d[\mathrm A]}{dt}=k[\mathrm A] \] At 12 min, \[ -\left(\frac{d[\mathrm A]}{dt}\right)_{12}=0.05775(0.20) \] \[ -\left(\frac{d[\mathrm A]}{dt}\right)_{12}=0.01155\ \mathrm{mol\,L^{-1}\,min^{-1}} \] \(\textbf{Convert to the appearance rate of P:}\) The stoichiometric relationship is \[ \frac{d[\mathrm P]}{dt}=n\left(-\frac{d[\mathrm A]}{dt}\right) \] Therefore, \[ \left(\frac{d[\mathrm P]}{dt}\right)_{12}=2(0.01155) \] \[ \left(\frac{d[\mathrm P]}{dt}\right)_{12}=0.02310\ \mathrm{mol\,L^{-1}\,min^{-1}} \] \(\textbf{Check the result against the kinetic meaning:}\) The disappearance rate of A is not numerically equal to the appearance rate of P. It is half the product-formation rate because each A produces two P. The initial concentration of P changes neither the stoichiometric coefficient nor the intrinsic disappearance constant of A. \(\textbf{Chemical interpretation:}\) The limiting concentration of a stable product can reveal reaction stoichiometry when the initial reactant and product concentrations are known. The intermediate product concentration then provides reaction progress. Combining these observations with first-order kinetics allows the instantaneous product-formation rate to be calculated. The stoichiometric and kinetic calculations serve different purposes and must be applied in the correct order. \(\textbf{Final answer:}\) \[ \boxed{\left(n,\left.\frac{d[\mathrm P]}{dt}\right|_{12}\right)=\left(2,0.02310\ \mathrm{mol\,L^{-1}\,min^{-1}}\right)} \]

Explanation: The measured product concentration contains contributions from two independent reactions. The amount of P formed by each pathway depends on the initial concentration of its reactant and its own first-order half-life. The two measurements therefore provide simultaneous equations for the two unknown initial concentrations. The first-order fractions remaining at 20 and 40 min must be determined before these equations are constructed. Once the initial composition is known, the product concentration at 60 min can be predicted. \(\textbf{Represent the unknown initial concentrations:}\) Let \[ [\mathrm A]_0=a \] and \[ [\mathrm B]_0=b\] Both concentrations are in \(\mathrm{mol\,L^{-1}}\). \(\textbf{First-order conversion after 20 min:}\) For A, 20 min corresponds to two half-lives. The fraction remaining is \[ f_{\mathrm A,20}=\left(\frac12\right)^2=\frac14 \] Therefore, the fraction of A consumed is \[ 1-\frac14=\frac34 \] For B, 20 min corresponds to one half-life. The fraction consumed is \[ \frac12 \] \(\textbf{Construct the first product balance:}\) The product formed by 20 min is \[ [\mathrm P]_{20}-[\mathrm P]_0=0.7000-0.050 \] \[ \Delta[\mathrm P]_{20}=0.6500\ \mathrm{mol\,L^{-1}} \] Therefore, \[ \frac34a+\frac12b=0.6500 \] Multiplying by four, \[ 3a+2b=2.6000 \] \(\textbf{First-order conversion after 40 min:}\) For A, 40 min corresponds to four half-lives. The fraction consumed is \[ 1-\frac1{16}=\frac{15}{16} \] For B, 40 min corresponds to two half-lives. The fraction consumed is \[ 1-\frac14=\frac34 \] \(\textbf{Construct the second product balance:}\) \[ \Delta[\mathrm P]_{40}=0.9125-0.050 \] \[ \Delta[\mathrm P]_{40}=0.8625\ \mathrm{mol\,L^{-1}} \] Therefore, \[ \frac{15}{16}a+\frac34b=0.8625 \] Multiplying by sixteen, \[ 15a+12b=13.800 \] \(\textbf{Solve for the initial concentrations:}\) The first equation is \[ 3a+2b=2.600 \] Multiplying by five, \[ 15a+10b=13.000 \] Subtracting from the second equation, \[ (15a+12b)-(15a+10b)=13.800-13.000 \] \[ 2b=0.800 \] \[ b=0.400\ \mathrm{mol\,L^{-1}} \] Substituting into the first equation, \[ 3a+2(0.400)=2.600 \] \[ 3a=1.800 \] \[ a=0.600\ \mathrm{mol\,L^{-1}} \] \(\textbf{Determine A remaining after 60 min:}\) Sixty minutes corresponds to six half-lives of A. \[ [\mathrm A]_{60}=0.600\left(\frac12\right)^6 \] \[ [\mathrm A]_{60}=0.009375\ \mathrm{mol\,L^{-1}} \] The amount of A consumed is \[ 0.600-0.009375=0.590625\ \mathrm{mol\,L^{-1}} \] \(\textbf{Determine B remaining after 60 min:}\) Sixty minutes corresponds to three half-lives of B. \[ [\mathrm B]_{60}=0.400\left(\frac12\right)^3 \] \[ [\mathrm B]_{60}=0.0500\ \mathrm{mol\,L^{-1}} \] The amount of B consumed is \[ 0.400-0.0500=0.3500\ \mathrm{mol\,L^{-1}} \] \(\textbf{Predict the product concentration:}\) \[ [\mathrm P]_{60}=[\mathrm P]_0+\Delta[\mathrm P]_{\mathrm A}+\Delta[\mathrm P]_{\mathrm B} \] \[ [\mathrm P]_{60}=0.050+0.590625+0.3500 \] \[ [\mathrm P]_{60}=0.990625\ \mathrm{mol\,L^{-1}} \] \(\textbf{Limiting-concentration check:}\) At complete eventual conversion of both reactants, \[ [\mathrm P]_{\infty}=0.050+0.600+0.400 \] \[ [\mathrm P]_{\infty}=1.050\ \mathrm{mol\,L^{-1}} \] The calculated product concentration at 60 min is below this limit. \(\textbf{Chemical interpretation:}\) A single product-concentration measurement cannot independently determine both initial reactant concentrations. The second measurement supplies the additional independent information because A and B have different first-order constants. The resulting product history is a sum of two different first-order contributions. It should not automatically be treated as one first-order process with a single half-life. \(\textbf{Final answer:}\) \[ \boxed{\left([\mathrm A]_0,[\mathrm P]_{60}\right)=\left(0.600,\ 0.990625\right)\ \mathrm{mol\,L^{-1}}} \]

Explanation: Both reactions begin at the same concentration and reach half that concentration after 10 min. However, their concentration–time curves have different shapes. The zero-order concentration decreases linearly, whereas the first-order concentration decreases exponentially. Therefore, equal starting concentrations, endpoint concentrations and half-lives do not imply equal areas under the two curves. The area can be calculated using the respective integrated kinetic equations. Dividing by the interval duration gives the time-averaged concentration. \(\textbf{Zero-order concentration history:}\) For Reaction X, \[ C_X(t)=C_0-k_Xt \] Its half-life is \[ t_{1/2}=\frac{C_0}{2k_X} \] Therefore, \[ k_X=\frac{0.80}{2(10)} \] \[ k_X=0.040\ \mathrm{mol\,L^{-1}\,min^{-1}} \] The concentration decreases linearly from \[ 0.80\ \mathrm{mol\,L^{-1}} \] to \[ 0.40\ \mathrm{mol\,L^{-1}} \] during the first 10 min. \(\textbf{Calculate the zero-order graph area:}\) The area is the area of a trapezium: \[ I_X=\frac{C_0+C_{10}}2(10) \] Substituting, \[ I_X=\frac{0.80+0.40}{2}(10) \] \[ I_X=0.60(10) \] \[ I_X=6.00\ \mathrm{mol\,min\,L^{-1}} \] \(\textbf{First-order concentration history:}\) For Reaction Y, \[ C_Y(t)=C_0e^{-k_Yt} \] The first-order constant is \[ k_Y=\frac{\ln2}{10} \] \[ k_Y=0.0693\ \mathrm{min^{-1}} \] \(\textbf{Calculate the first-order graph area:}\) \[ I_Y=\int_0^{10}C_0e^{-k_Yt}\,dt \] Integrating, \[ I_Y=\frac{C_0}{k_Y}(1-e^{-10k_Y}) \] After one half-life, \[ e^{-10k_Y}=\frac12 \] Therefore, \[ I_Y=\frac{0.80}{k_Y}\left(1-\frac12\right) \] \[ I_Y=\frac{0.40}{0.0693} \] \[ I_Y\approx5.772\ \mathrm{mol\,min\,L^{-1}} \] \(\textbf{Compare the areas:}\) \[ \frac{I_X}{I_Y}=\frac{6.00}{5.772} \] \[ \frac{I_X}{I_Y}\approx1.040 \] Equivalently, \[ \frac{I_X}{I_Y}=\frac{3\ln2}{2} \] \(\textbf{Determine the time-averaged first-order concentration:}\) \[ \overline C_Y=\frac{I_Y}{10} \] \[ \overline C_Y=\frac{5.772}{10} \] \[ \overline C_Y\approx0.577\ \mathrm{mol\,L^{-1}} \] \(\textbf{Interpret the graphical result:}\) The zero-order curve is a straight line connecting the initial and half-life concentrations. The first-order curve is not that straight line. Its area over the same interval is slightly smaller under the stated equal-endpoint conditions. The arithmetic mean of the initial and final concentrations is therefore the time average for the zero-order line, but not for the exponential first-order curve. \(\textbf{Chemical interpretation:}\) A half-life fixes one feature of the concentration history, not the complete shape of the curve. The area under a concentration–time graph represents concentration accumulated over time. It is not the amount of reactant consumed. The consumed amount is determined by the difference between endpoint concentrations. These quantities must not be confused when interpreting kinetic graphs. \(\textbf{Final answer:}\) \[ \boxed{\left(\frac{I_X}{I_Y},\overline C_Y\right)=\left(1.040,0.577\ \mathrm{mol\,L^{-1}}\right)} \]

Explanation: The reaction increases the total number of gaseous molecules because one A forms two B. At constant temperature and volume, the slope of total pressure against time is proportional to the instantaneous disappearance rate of A. The ratio of measured pressure slopes therefore reveals the fraction of A remaining after 10 min. This determines the first-order constant. The magnitude of the initial slope can then be used to recover the initially unknown partial pressure of A. The inert-gas pressure follows from the initial total pressure. \(\textbf{Relate total-pressure change to A disappearance:}\) For \[ \mathrm A\rightarrow2\mathrm B, \] one mole of A decomposed produces a net increase of one mole of gas. At constant temperature and volume, \[ P_{\mathrm{total}}=P_0+(p_{\mathrm A,0}-p_{\mathrm A}) \] Therefore, \[ \frac{dP_{\mathrm{total}}}{dt}=-\frac{dp_{\mathrm A}}{dt} \] \(\textbf{Apply the first-order pressure law:}\) For first-order disappearance, \[ -\frac{dp_{\mathrm A}}{dt}=kp_{\mathrm A} \] Consequently, \[ \frac{dP_{\mathrm{total}}}{dt}=kp_{\mathrm A} \] \(\textbf{Infer the fraction of A remaining after 10 min:}\) Because k remains unchanged, \[ \frac{(dP_{\mathrm{total}}/dt)_{10}}{(dP_{\mathrm{total}}/dt)_0}=\frac{p_{\mathrm A,10}}{p_{\mathrm A,0}} \] Substituting, \[ \frac{0.0693}{0.1386}=\frac{p_{\mathrm A,10}}{p_{\mathrm A,0}} \] \[ \frac{p_{\mathrm A,10}}{p_{\mathrm A,0}}=\frac12 \] Thus, 10 min corresponds to one half-life. \(\textbf{Determine the first-order constant:}\) \[ k=\frac{\ln2}{10} \] \[ k=\frac{0.693}{10} \] \[ k=0.0693\ \mathrm{min^{-1}} \] \(\textbf{Recover the initial partial pressure of A:}\) At time zero, \[ \left(\frac{dP_{\mathrm{total}}}{dt}\right)_0=kp_{\mathrm A,0} \] Therefore, \[ p_{\mathrm A,0}=\frac{0.1386}{0.0693} \] \[ p_{\mathrm A,0}=2.0\ \mathrm{atm} \] \(\textbf{Determine the initial inert-gas pressure:}\) Initially, \[ P_0=p_{\mathrm A,0}+p_{\mathrm{inert},0} \] Therefore, \[ p_{\mathrm{inert},0}=6.0-2.0 \] \[ p_{\mathrm{inert},0}=4.0\ \mathrm{atm} \] \(\textbf{Predict A remaining after 20 min:}\) Twenty minutes corresponds to two half-lives. Therefore, \[ p_{\mathrm A,20}=2.0\left(\frac12\right)^2 \] \[ p_{\mathrm A,20}=0.50\ \mathrm{atm} \] \(\textbf{Determine the total-pressure increase:}\) The pressure-equivalent amount of A decomposed is \[ x_{20}=2.0-0.50 \] \[ x_{20}=1.50\ \mathrm{atm} \] Therefore, \[ P_{20}=6.0+1.50 \] \[ P_{20}=7.50\ \mathrm{atm} \] \(\textbf{Limiting-pressure check:}\) At complete eventual decomposition of A, \[ P_{\infty}=6.0+2.0 \] \[ P_{\infty}=8.0\ \mathrm{atm} \] The calculated pressure at 20 min is below this limit. \(\textbf{Experimental interpretation:}\) The total-pressure slope provides instantaneous kinetic information, whereas total pressure itself records cumulative gas formation. The slope ratio determines the first-order remaining fraction without initially knowing the separate gas pressures. The initial slope magnitude then determines the initial amount of reacting gas. This distinction between a measured signal and its time derivative is essential when interpreting pressure-based kinetic experiments. \(\textbf{Final answer:}\) \[ \boxed{\left(p_{\mathrm{inert},0},P_{20}\right)=\left(4.0\ \mathrm{atm},7.50\ \mathrm{atm}\right)} \]

Explanation: The measurements provide product-to-reactant ratios rather than absolute concentrations. The reaction balance connects each product concentration with the amount of A consumed. Because the initial product concentration is unknown, one ratio measurement is insufficient to determine the first-order decay factor uniquely. The two measurements at successive equal time intervals provide the additional relationship needed. Once the fractional decay is determined, the initial product concentration and the later product-to-reactant ratio can be reconstructed. \(\textbf{Represent the first-order decay over 10 min:}\) Let \(f\) denote the fraction of A remaining after 10 min. Therefore, \[ [\mathrm A]_{10}=0.60f \] After 20 min, \[ [\mathrm A]_{20}=0.60f^2 \] Let the unknown initial product concentration be \(p_0\). \(\textbf{Apply the product balance at 10 min:}\) For \[ \mathrm A\rightarrow\mathrm P, \] \[ [\mathrm P]_{10}=p_0+0.60(1-f) \] The measured ratio is \[ \frac{[\mathrm P]_{10}}{[\mathrm A]_{10}}=2 \] Therefore, \[ \frac{p_0+0.60(1-f)}{0.60f}=2 \] Multiplying, \[ p_0+0.60-0.60f=1.20f \] Thus, \[ p_0=1.80f-0.60 \] \(\textbf{Apply the product balance at 20 min:}\) At 20 min, \[ [\mathrm P]_{20}=p_0+0.60(1-f^2) \] The measured ratio is five. Therefore, \[ \frac{p_0+0.60(1-f^2)}{0.60f^2}=5 \] Multiplying, \[ p_0+0.60-0.60f^2=3.00f^2 \] Therefore, \[ p_0=3.60f^2-0.60 \] \(\textbf{Determine the decay factor:}\) Equating the two expressions for \(p_0\), \[ 1.80f-0.60=3.60f^2-0.60 \] \[ 1.80f=3.60f^2 \] Because \(f\) is positive, \[ f=\frac{1.80}{3.60} \] \[ f=\frac12 \] Thus, the first-order half-life is 10 min. \(\textbf{Recover the initial product concentration:}\) Using \[ p_0=1.80f-0.60, \] \[ p_0=1.80\left(\frac12\right)-0.60 \] \[ p_0=0.90-0.60 \] \[ p_0=0.30\ \mathrm{mol\,L^{-1}} \] \(\textbf{Determine A after 30 min:}\) Thirty minutes corresponds to three half-lives. Therefore, \[ [\mathrm A]_{30}=0.60\left(\frac12\right)^3 \] \[ [\mathrm A]_{30}=0.075\ \mathrm{mol\,L^{-1}} \] \(\textbf{Determine P after 30 min:}\) The amount of A consumed is \[ 0.60-0.075=0.525\ \mathrm{mol\,L^{-1}} \] Therefore, \[ [\mathrm P]_{30}=0.30+0.525 \] \[ [\mathrm P]_{30}=0.825\ \mathrm{mol\,L^{-1}} \] \(\textbf{Predict the requested ratio:}\) \[ \frac{[\mathrm P]_{30}}{[\mathrm A]_{30}}=\frac{0.825}{0.075} \] \[ \frac{[\mathrm P]_{30}}{[\mathrm A]_{30}}=11 \] \(\textbf{Verify the observed ratios:}\) At 10 min, \[ [\mathrm A]_{10}=0.30,\qquad [\mathrm P]_{10}=0.60\ \mathrm{mol\,L^{-1}} \] Their ratio is two. At 20 min, \[ [\mathrm A]_{20}=0.15,\qquad [\mathrm P]_{20}=0.75\ \mathrm{mol\,L^{-1}} \] Their ratio is five. Both match the observations. \(\textbf{Chemical interpretation:}\) The product-to-reactant ratio does not itself decay exponentially. It changes because product accumulates while reactant disappears. Two time-separated ratio measurements can nevertheless reveal the underlying first-order fraction remaining. The initially present product must be included in the material balance. Ignoring it would give an incorrect kinetic reconstruction. \(\textbf{Final answer:}\) \[ \boxed{\left([\mathrm P]_0,\frac{[\mathrm P]_{30}}{[\mathrm A]_{30}}\right)=\left(0.30\ \mathrm{mol\,L^{-1}},11\right)} \]

Explanation: Statement I is correct because the ammonia pressure doubles from \(5.0\) to \(10.0\ \mathrm{atm}\) while the measured rate remains unchanged. The active platinum surface area and temperature are identical. This is consistent with zero-order dependence on ammonia pressure over that tested range. A plausible surface interpretation is that the catalytically active sites are effectively saturated under those conditions. Statement II is incorrect because a rate law established over one pressure interval need not apply at every possible pressure. Experiment IV itself gives a lower rate at a lower ammonia pressure using the same relative platinum surface area. The observations therefore do not support an unrestricted zero-order conclusion. Statement III is correct because the rate at \(1.0\ \mathrm{atm}\) is smaller than the rate at \(5.0\ \mathrm{atm}\). This demonstrates that the higher-pressure rate independence need not persist at lower pressure. However, a single lower-pressure observation does not determine the precise reaction order throughout the lower-pressure region. Statement IV is incorrect because molecularity is assigned to an elementary reaction event. It cannot be zero. An experimentally observed zero-order dependence in a complex surface reaction does not imply an elementary event involving zero molecules. Statement V is correct because doubling the available active platinum surface area at the same ammonia pressure doubles the measured rate in Experiments II and III. This is consistent with a greater number of active catalytic sites supporting more reaction events per unit time. The observation distinguishes the influence of catalyst surface availability from the influence of ammonia pressure. Zero-order behaviour in a reactant does not mean that all experimental conditions are irrelevant to the rate. Catalytic surface area and temperature can remain important even when increasing reactant pressure has no further effect within a saturated regime. The correct statements are I, III and V only.

Explanation: The collected gas volume measures the cumulative amount of B formed rather than the remaining concentration of A. The linear gas-volume graph indicates a constant gas-formation rate while the reaction remains active. The molar volume converts this measured gas-volume rate into a molar formation rate. The balanced equation then converts B formation into A disappearance. The exhaustion time supplies the total initial amount of A. The gas volume at 25 min can be calculated because the reaction remains in the linear zero-order regime until 30 min. \(\textbf{Determine the gas-volume formation rate:}\) Between 5 and 15 min, \[ \Delta V_{\mathrm B}=0.60-0.20 \] \[ \Delta V_{\mathrm B}=0.40\ \mathrm L \] The elapsed time is \[ \Delta t=15-5 \] \[ \Delta t=10\ \mathrm{min} \] Therefore, \[ \frac{dV_{\mathrm B}}{dt}=\frac{0.40}{10} \] \[ \frac{dV_{\mathrm B}}{dt}=0.040\ \mathrm{L\,min^{-1}} \] \(\textbf{Convert gas-volume rate into molar formation rate:}\) At the stated temperature and pressure, \[ \frac{dn_{\mathrm B}}{dt}=\frac1{V_m}\frac{dV_{\mathrm B}}{dt} \] Substituting, \[ \frac{dn_{\mathrm B}}{dt}=\frac{0.040}{20.0} \] \[ \frac{dn_{\mathrm B}}{dt}=0.0020\ \mathrm{mol\,min^{-1}} \] \(\textbf{Use the reaction stoichiometry:}\) The balanced equation is \[ 2\mathrm A\rightarrow\mathrm B \] Therefore, two moles of A disappear for every mole of B formed. The molar disappearance rate of A is \[ -\frac{dn_{\mathrm A}}{dt}=2\frac{dn_{\mathrm B}}{dt} \] \[ -\frac{dn_{\mathrm A}}{dt}=2(0.0020) \] \[ -\frac{dn_{\mathrm A}}{dt}=0.0040\ \mathrm{mol\,min^{-1}} \] \(\textbf{Convert to concentration-based disappearance:}\) The solution volume is \[ V_{\mathrm{solution}}=1.00\ \mathrm L \] Therefore, \[ k_{\mathrm{dis}}=\frac{0.0040}{1.00} \] \[ k_{\mathrm{dis}}=0.0040\ \mathrm{mol\,L^{-1}\,min^{-1}} \] \(\textbf{Recover the initial concentration from exhaustion:}\) The reaction stops because A is exhausted at 30 min. For zero-order disappearance, \[ [\mathrm A]_0=k_{\mathrm{dis}}t_{\mathrm{exhaustion}} \] Substituting, \[ [\mathrm A]_0=0.0040(30) \] \[ [\mathrm A]_0=0.120\ \mathrm{mol\,L^{-1}} \] \(\textbf{Calculate the cumulative gas volume after 25 min:}\) The constant gas-volume formation rate is \[ 0.040\ \mathrm{L\,min^{-1}} \] Therefore, \[ V_{\mathrm B,25}=0.040(25) \] \[ V_{\mathrm B,25}=1.00\ \mathrm L \] \(\textbf{Stoichiometric verification:}\) The initial amount of A is \[ n_{\mathrm A,0}=0.120(1.00) \] \[ n_{\mathrm A,0}=0.120\ \mathrm{mol} \] At complete conversion, the amount of B produced is \[ n_{\mathrm B,\infty}=\frac{0.120}{2} \] \[ n_{\mathrm B,\infty}=0.0600\ \mathrm{mol} \] Its limiting gas volume is \[ V_{\mathrm B,\infty}=0.0600(20.0) \] \[ V_{\mathrm B,\infty}=1.20\ \mathrm L \] Since gas evolution continues until 30 min, a volume of \(1.00\ \mathrm L\) at 25 min is physically consistent with this limiting value. \(\textbf{Chemical interpretation:}\) Linear cumulative gas production can provide evidence of zero-order behaviour when gas evolution is stoichiometrically related to reactant disappearance. The measured gas-volume slope must not be confused directly with the normalized chemical reaction rate. Gas molar volume, reaction stoichiometry and solution volume are needed to convert the measurement into a reactant-disappearance constant. The linear relationship must terminate once A is exhausted. \(\textbf{Final answer:}\) \[ \boxed{\left([\mathrm A]_0,V_{\mathrm B,25}\right)=\left(0.120\ \mathrm{mol\,L^{-1}},1.00\ \mathrm L\right)} \]

Explanation: The normalized reaction rate is zero order, but the disappearance rates of A and B are different because their stoichiometric coefficients differ. The first half-life of A must therefore be calculated using the disappearance rate of A, not the normalized reaction rate directly. The reaction also contains a limiting reactant. Its stopping time must be established before the mathematical zero-order equation is extrapolated to complete disappearance of A. The final product concentration follows from the amount of reaction progress achieved before the limiting reactant is exhausted. \(\textbf{Write the stoichiometric rate relationship:}\) For \[ 2\mathrm A+\mathrm B\rightarrow3\mathrm P, \] \[ r=-\frac12\frac{d[\mathrm A]}{dt} \] \[ r=-\frac{d[\mathrm B]}{dt} \] and \[ r=\frac13\frac{d[\mathrm P]}{dt} \] \(\textbf{Determine A disappearance rate:}\) Because \[ r=k_0=0.010\ \mathrm{mol\,L^{-1}\,min^{-1}}, \] \[ -\frac{d[\mathrm A]}{dt}=2k_0 \] \[ -\frac{d[\mathrm A]}{dt}=0.020\ \mathrm{mol\,L^{-1}\,min^{-1}} \] \(\textbf{Determine the first half-life of A:}\) Half the initial A concentration is \[ \frac{0.50}{2}=0.25\ \mathrm{mol\,L^{-1}} \] The concentration of A consumed to reach this state is \[ 0.50-0.25=0.25\ \mathrm{mol\,L^{-1}} \] Therefore, \[ t_A=\frac{0.25}{0.020} \] \[ t_A=12.5\ \mathrm{min} \] \(\textbf{Check whether B remains available at that time:}\) The normalized reaction progress is \[ x=k_0t_A \] \[ x=0.010(12.5) \] \[ x=0.125\ \mathrm{mol\,L^{-1}} \] The concentration of B remaining is \[ [\mathrm B]_{t_A}=0.20-0.125 \] \[ [\mathrm B]_{t_A}=0.075\ \mathrm{mol\,L^{-1}} \] Therefore, B is still present when A reaches half its initial concentration. The calculated half-life is physically attainable. \(\textbf{Determine the limiting reactant:}\) Complete consumption of A alone would require normalized reaction progress \[ x_A=\frac{0.50}{2} \] \[ x_A=0.25\ \mathrm{mol\,L^{-1}} \] Complete consumption of B requires \[ x_B=0.20\ \mathrm{mol\,L^{-1}} \] Since \[ x_B\lt x_A, \] B is the limiting reactant. \(\textbf{Determine the stopping time:}\) B disappears at the rate \[ -\frac{d[\mathrm B]}{dt}=k_0 \] Therefore, \[ t_{\mathrm{stop}}=\frac{0.20}{0.010} \] \[ t_{\mathrm{stop}}=20.0\ \mathrm{min} \] \(\textbf{Determine final product formation:}\) At stopping, \[ x_{\mathrm{stop}}=0.20\ \mathrm{mol\,L^{-1}} \] The product coefficient is three. Therefore, \[ \Delta[\mathrm P]=3x_{\mathrm{stop}} \] \[ \Delta[\mathrm P]=3(0.20) \] \[ \Delta[\mathrm P]=0.60\ \mathrm{mol\,L^{-1}} \] Including initially present product, \[ [\mathrm P]_{\mathrm{stop}}=0.050+0.60 \] \[ [\mathrm P]_{\mathrm{stop}}=0.650\ \mathrm{mol\,L^{-1}} \] \(\textbf{Verify the final reactant amounts:}\) The amount of A consumed is \[ 2x_{\mathrm{stop}}=0.40\ \mathrm{mol\,L^{-1}} \] Therefore, \[ [\mathrm A]_{\mathrm{stop}}=0.50-0.40 \] \[ [\mathrm A]_{\mathrm{stop}}=0.10\ \mathrm{mol\,L^{-1}} \] Some A remains because B becomes exhausted first. \(\textbf{Chemical interpretation:}\) The zero-order normalized rate does not numerically equal the disappearance rate of a reactant with stoichiometric coefficient two. The finite supply of the limiting reactant determines when the overall chemical reaction must stop. A mathematical continuation of the zero-order law beyond this point would predict chemically impossible consumption of B. The first half-life of A and the final product concentration must therefore be calculated using both kinetics and stoichiometry. \(\textbf{Final answer:}\) \[ \boxed{\left(t_A,[\mathrm P]_{\mathrm{stop}}\right)=\left(12.5\ \mathrm{min},0.650\ \mathrm{mol\,L^{-1}}\right)} \]

Explanation: The apparatus measures the instantaneous rate of gas-volume evolution. It does not directly measure the amount of A remaining or the cumulative volume already collected. However, gas formation is stoichiometrically related to A disappearance. At constant gas-collection temperature and pressure, gas-volume evolution rate is proportional to the molar formation rate of B. Because A follows first-order disappearance, this gas-volume evolution rate also decreases exponentially with time. The ratio of measured gas-volume rates therefore determines the first-order constant. The cumulative gas volume is then obtained from the area under the gas-volume-rate curve. \(\textbf{Relate gas evolution to first-order disappearance:}\) For \[ 2\mathrm A\rightarrow\mathrm B, \] the amount of B formed is proportional to the amount of A consumed. The disappearance of A follows \[ -\frac{dn_{\mathrm A}}{dt}=kn_{\mathrm A} \] under the stated constant-volume conditions. Therefore, \[ n_{\mathrm A}(t)=n_{\mathrm A,0}e^{-kt} \] The gas-evolution rate has the corresponding form \[ q(t)=q_0e^{-kt} \] where \(q_0\) is the initial gas-volume evolution rate. \(\textbf{Use the two measured rates:}\) The initial gas-volume rate is \[ q_0=13.860\ \mathrm{mL\,min^{-1}} \] At 20 min, \[ q_{20}=3.465\ \mathrm{mL\,min^{-1}} \] Therefore, \[ \frac{q_{20}}{q_0}=\frac{3.465}{13.860} \] \[ \frac{q_{20}}{q_0}=\frac14 \] \(\textbf{Determine the first-order rate constant:}\) From the exponential relation, \[ \frac{q_{20}}{q_0}=e^{-20k} \] Thus, \[ \frac14=e^{-20k} \] Taking natural logarithms, \[ 20k=\ln4 \] \[ k=\frac{\ln4}{20} \] Since \[ \ln4=2\ln2, \] \[ k=\frac{2(0.693)}{20} \] \[ k=0.0693\ \mathrm{min^{-1}} \] \(\textbf{Determine the corresponding half-life:}\) \[ t_{1/2}=\frac{\ln2}{k} \] \[ t_{1/2}=\frac{0.693}{0.0693} \] \[ t_{1/2}=10.0\ \mathrm{min} \] Thus, 20 min corresponds to two half-lives. \(\textbf{Relate instantaneous gas evolution to collected volume:}\) The cumulative gas volume is \[ V_{\mathrm B}(t)=\int_0^tq(t')\,dt' \] Substituting the first-order rate expression, \[ V_{\mathrm B}(t)=\int_0^tq_0e^{-kt'}\,dt' \] Integrating, \[ V_{\mathrm B}(t)=\frac{q_0}{k}(1-e^{-kt}) \] \(\textbf{Determine the limiting collected volume:}\) At complete eventual conversion, \[ e^{-kt}\rightarrow0 \] Therefore, \[ V_{\mathrm B,\infty}=\frac{q_0}{k} \] Substituting, \[ V_{\mathrm B,\infty}=\frac{13.860}{0.0693} \] \[ V_{\mathrm B,\infty}=200\ \mathrm{mL} \] \(\textbf{Calculate gas collected during the first 20 min:}\) At 20 min, \[ e^{-20k}=\frac14 \] Therefore, \[ V_{\mathrm B,20}=200\left(1-\frac14\right) \] \[ V_{\mathrm B,20}=200\left(\frac34\right) \] \[ V_{\mathrm B,20}=150\ \mathrm{mL} \] \(\textbf{Check the distinction between instantaneous rate and accumulated volume:}\) At 20 min, the gas-volume evolution rate is one-fourth its initial value. However, three-fourths of the limiting gas volume has already been collected. These fractions describe different physical quantities. The first concerns the rate of gas formation at one instant. The second concerns the total gas formed from the beginning of the experiment. \(\textbf{Chemical interpretation:}\) A first-order reaction can be monitored through a stoichiometrically related gas-volume-rate measurement. The logarithmic decrease of the gas-volume evolution rate gives the first-order constant. The accumulated gas volume corresponds to the area under that rate–time curve. The gas stoichiometry affects the proportionality between A disappearance and B formation but does not change the exponential decay constant. \(\textbf{Final answer:}\) \[ \boxed{\left(k,V_{\mathrm B,20}\right)=\left(0.0693\ \mathrm{min^{-1}},150\ \mathrm{mL}\right)} \]

Explanation: The measurements provide product formation over two unequal time intervals. Therefore, the product increments cannot be compared as though both intervals represented the same duration. For an irreversible \(1:1\) reaction, the increase in P concentration equals the decrease in A concentration. The first-order law determines the fraction of A remaining after a specified time. The second interval lasts twice as long as the first, so its remaining fraction is the square of the first interval's remaining fraction. This relationship allows the unknown decay factor to be recovered from the observed product increments. \(\textbf{Represent the first-order decay factor:}\) Let the initial concentration of A be \(C_0\). Let \(f\) denote the fraction of A remaining after 10 min. Therefore, \[ [\mathrm A]_{10}=C_0f \] After 30 min, \[ [\mathrm A]_{30}=C_0f^3 \] because 30 min contains three equal 10-minute intervals. \(\textbf{Use product formation during the first interval:}\) The concentration of A consumed from 0 to 10 min is \[ C_0-C_0f=C_0(1-f) \] Therefore, \[ C_0(1-f)=0.360 \] \(\textbf{Use product formation during the next 20 min:}\) The concentration of A consumed between 10 and 30 min is \[ C_0f-C_0f^3 \] Factorizing, \[ C_0f(1-f^2)=0.270 \] \(\textbf{Eliminate the unknown initial concentration:}\) Divide the second equation by the first: \[ \frac{0.270}{0.360}=\frac{C_0f(1-f^2)}{C_0(1-f)} \] Since \[ 1-f^2=(1-f)(1+f), \] \[ 0.75=f(1+f) \] Therefore, \[ f^2+f-0.75=0 \] Factorizing, \[ (f-0.5)(f+1.5)=0 \] The physically admissible fraction is \[ f=0.50 \] The negative root is impossible for a remaining concentration fraction. \(\textbf{Recover the initial concentration:}\) Using the first product increment, \[ C_0(1-0.50)=0.360 \] \[ 0.50C_0=0.360 \] \[ C_0=0.720\ \mathrm{mol\,L^{-1}} \] \(\textbf{Determine the half-life:}\) Because half the original A remains after 10 min, \[ t_{1/2}=10\ \mathrm{min} \] \(\textbf{Predict the concentration after 40 min:}\) Forty minutes corresponds to four half-lives. Therefore, \[ [\mathrm A]_{40}=0.720\left(\frac12\right)^4 \] \[ [\mathrm A]_{40}=\frac{0.720}{16} \] \[ [\mathrm A]_{40}=0.0450\ \mathrm{mol\,L^{-1}} \] \(\textbf{Verify the unequal-interval observations:}\) At 10 min, \[ [\mathrm A]_{10}=0.360\ \mathrm{mol\,L^{-1}} \] At 30 min, \[ [\mathrm A]_{30}=0.0900\ \mathrm{mol\,L^{-1}} \] The consumption between these instants is \[ 0.360-0.0900=0.270\ \mathrm{mol\,L^{-1}} \] This matches the second measured product increment. \(\textbf{Chemical interpretation:}\) Product increments over unequal time intervals must be connected using the integrated rate equation. The later interval is longer but produces less P because the remaining concentration of A has decreased. The reaction consumes equal fractions during equal intervals, not equal absolute concentrations. The two product increments are sufficient to determine the first-order decay factor and the initially unknown reactant concentration. \(\textbf{Final answer:}\) \[ \boxed{\left([\mathrm A]_0,[\mathrm A]_{40}\right)=\left(0.720,\ 0.0450\right)\ \mathrm{mol\,L^{-1}}} \]

Explanation: The same reaction follows two different kinetic regimes at different reactant concentrations. Therefore, one integrated rate equation cannot be applied over the entire concentration interval. The zero-order equation must be used until A reaches the specified transition concentration. The first-order equation then determines the additional time required to reach the final concentration. The instantaneous rates calculated from both laws must agree at the transition because the problem specifies continuity. \(\textbf{Identify the initial kinetic regime:}\) Initially, \[ [\mathrm A]_0=0.80\ \mathrm{mol\,L^{-1}} \] Since this exceeds the transition concentration, the reaction initially follows the zero-order equation: \[ [\mathrm A]_t=[\mathrm A]_0-k_0t \] where \[ k_0=0.040\ \mathrm{mol\,L^{-1}\,min^{-1}} \] \(\textbf{Calculate the time to reach the transition:}\) At the transition, \[ [\mathrm A]=0.40\ \mathrm{mol\,L^{-1}} \] Therefore, \[ 0.40=0.80-0.040t_1 \] Rearranging, \[ 0.040t_1=0.40 \] \[ t_1=10.0\ \mathrm{min} \] \(\textbf{Apply first-order kinetics below the transition:}\) After the transition, \[ -\frac{d[\mathrm A]}{dt}=k_1[\mathrm A] \] with \[ k_1=0.100\ \mathrm{min^{-1}} \] The starting concentration for this second interval is \[ C_{\mathrm{start}}=0.40\ \mathrm{mol\,L^{-1}} \] The required final concentration is \[ C_{\mathrm{final}}=0.10\ \mathrm{mol\,L^{-1}} \] \(\textbf{Calculate the time in the first-order regime:}\) The integrated equation is \[ k_1t_2=\ln\frac{C_{\mathrm{start}}}{C_{\mathrm{final}}} \] Substituting, \[ 0.100t_2=\ln\frac{0.40}{0.10} \] \[ 0.100t_2=\ln4 \] Using the supplied value, \[ t_2=\frac{1.386}{0.100} \] \[ t_2=13.86\ \mathrm{min} \] \(\textbf{Determine the total elapsed time:}\) \[ t_{\mathrm{total}}=t_1+t_2 \] \[ t_{\mathrm{total}}=10.0+13.86 \] \[ t_{\mathrm{total}}=23.86\ \mathrm{min} \] \(\textbf{Check rate continuity at the transition:}\) From the zero-order side, \[ r_{\mathrm{transition}}=k_0 \] \[ r_{\mathrm{transition}}=0.040\ \mathrm{mol\,L^{-1}\,min^{-1}} \] From the first-order side, \[ r_{\mathrm{transition}}=k_1[\mathrm A]_{\mathrm{transition}} \] \[ r_{\mathrm{transition}}=0.100(0.40) \] \[ r_{\mathrm{transition}}=0.040\ \mathrm{mol\,L^{-1}\,min^{-1}} \] The two values agree. \(\textbf{Chemical interpretation:}\) A surface-catalysed reaction can display different concentration dependencies under different surface-coverage conditions. Zero-order behaviour at higher concentration does not establish zero-order behaviour at every concentration. The transition concentration determines where the integrated equation must change. The continuity check confirms that the supplied model does not predict an artificial jump in the instantaneous rate at the boundary. \(\textbf{Final answer:}\) \[ \boxed{\left(t_{\mathrm{total}},r_{\mathrm{transition}}\right)=\left(23.86\ \mathrm{min},0.040\ \mathrm{mol\,L^{-1}\,min^{-1}}\right)} \]

Explanation: The two graph slopes describe different mathematical quantities. The slope of the logarithmic concentration–time graph gives the negative first-order rate constant. The tangent slope of the actual concentration–time curve gives the instantaneous concentration-change rate. These slopes cannot be substituted for one another. Combining them reveals the concentration of A at the tangent point. The integrated first-order equation then reconstructs the initial concentration. Finally, the balanced equation converts the disappearance of A into the formation of P. \(\textbf{Extract the rate constant from Graph I:}\) For first-order disappearance, \[ \ln[\mathrm A]_t=\ln[\mathrm A]_0-kt \] Therefore, \[ \text{slope of Graph I}=-k \] The measured slope gives \[ k=0.0693\ \mathrm{min^{-1}} \] \(\textbf{Interpret the tangent to Graph II:}\) The differential first-order equation is \[ -\frac{d[\mathrm A]}{dt}=k[\mathrm A] \] At 10 min, the tangent slope is \[ \left(\frac{d[\mathrm A]}{dt}\right)_{10}=-0.01386\ \mathrm{mol\,L^{-1}\,min^{-1}} \] Therefore, \[ 0.01386=k[\mathrm A]_{10} \] Substituting the rate constant, \[ 0.01386=0.0693[\mathrm A]_{10} \] \[ [\mathrm A]_{10}=\frac{0.01386}{0.0693} \] \[ [\mathrm A]_{10}=0.200\ \mathrm{mol\,L^{-1}} \] \(\textbf{Reconstruct the initial concentration:}\) The integrated first-order relationship is \[ [\mathrm A]_{10}=[\mathrm A]_0e^{-10k} \] Since \[ 10k=0.693, \] \[ e^{-10k}=\frac12 \] Therefore, \[ 0.200=\frac{[\mathrm A]_0}{2} \] \[ [\mathrm A]_0=0.400\ \mathrm{mol\,L^{-1}} \] \(\textbf{Determine the chemical conversion:}\) The concentration of A consumed during the first 10 min is \[ -\Delta[\mathrm A]=0.400-0.200 \] \[ -\Delta[\mathrm A]=0.200\ \mathrm{mol\,L^{-1}} \] \(\textbf{Apply the stoichiometric product coefficient:}\) For \[ 2\mathrm A\rightarrow3\mathrm P, \] \[ \Delta[\mathrm P]=\frac32(-\Delta[\mathrm A]) \] Substituting, \[ \Delta[\mathrm P]=\frac32(0.200) \] \[ \Delta[\mathrm P]=0.300\ \mathrm{mol\,L^{-1}} \] \(\textbf{Include initially present product:}\) \[ [\mathrm P]_{10}=0.050+0.300 \] \[ [\mathrm P]_{10}=0.350\ \mathrm{mol\,L^{-1}} \] \(\textbf{Graphical verification:}\) The first-order half-life is \[ t_{1/2}=\frac{0.693}{0.0693} \] \[ t_{1/2}=10.0\ \mathrm{min} \] Thus, A should decrease from \(0.400\) to \(0.200\ \mathrm{mol\,L^{-1}}\) during the measured interval. The differential rate at the endpoint is then \[ 0.0693(0.200)=0.01386\ \mathrm{mol\,L^{-1}\,min^{-1}} \] This reproduces the measured tangent magnitude. \(\textbf{Chemical interpretation:}\) A logarithmic concentration–time graph and an ordinary concentration–time graph contain complementary kinetic information. The logarithmic graph provides the rate constant directly through its slope. The ordinary graph provides an instantaneous rate through its tangent. Using both measurements allows an initially unknown concentration to be reconstructed without directly measuring the graph intercept. \(\textbf{Final answer:}\) \[ \boxed{\left([\mathrm A]_0,[\mathrm P]_{10}\right)=\left(0.400,\ 0.350\right)\ \mathrm{mol\,L^{-1}}} \]

Explanation: The reaction is first order in the amount of A remaining. However, concentration is the amount of A divided by the changing solution volume. Therefore, concentration need not decrease monotonically when the solvent volume decreases. The integrated first-order equation must first be applied to the number of moles of A. The measured volume at each time is then used to calculate concentration. Using the ordinary constant-volume concentration equation without accounting for evaporation would be incorrect. \(\textbf{Determine the original concentration:}\) \[ [\mathrm A]_0=\frac{n_{\mathrm A,0}}{V_0} \] \[ [\mathrm A]_0=\frac{0.800}{2.00} \] \[ [\mathrm A]_0=0.400\ \mathrm{mol\,L^{-1}} \] \(\textbf{Apply the first-order amount law:}\) The integrated equation is \[ n_{\mathrm A}(t)=n_{\mathrm A,0}e^{-kt} \] Using the half-life, \[ n_{\mathrm A}(t)=n_{\mathrm A,0}\left(\frac12\right)^{t/t_{1/2}} \] \(\textbf{Determine the amount remaining after 10 min:}\) Ten minutes corresponds to one half-life. Therefore, \[ n_{\mathrm A,10}=0.800\left(\frac12\right) \] \[ n_{\mathrm A,10}=0.400\ \mathrm{mol} \] \(\textbf{Convert the amount into concentration:}\) The solution volume at 10 min is \[ V_{10}=0.80\ \mathrm L \] Therefore, \[ [\mathrm A]_{10}=\frac{0.400}{0.80} \] \[ [\mathrm A]_{10}=0.500\ \mathrm{mol\,L^{-1}} \] \(\textbf{Determine the amount remaining after 20 min:}\) Twenty minutes corresponds to two half-lives. Therefore, \[ n_{\mathrm A,20}=0.800\left(\frac12\right)^2 \] \[ n_{\mathrm A,20}=0.200\ \mathrm{mol} \] \(\textbf{Convert using the later measured volume:}\) At 20 min, \[ V_{20}=0.50\ \mathrm L \] Therefore, \[ [\mathrm A]_{20}=\frac{0.200}{0.50} \] \[ [\mathrm A]_{20}=0.400\ \mathrm{mol\,L^{-1}} \] \(\textbf{Compare amount and concentration trends:}\) The amount of A decreases continuously: \[ 0.800\rightarrow0.400\rightarrow0.200\ \mathrm{mol} \] However, the corresponding concentrations are \[ 0.400\rightarrow0.500\rightarrow0.400\ \mathrm{mol\,L^{-1}} \] The concentration initially increases because the solution volume decreases more rapidly than the amount of A during that interval. \(\textbf{Chemical interpretation:}\) The disappearance of reactant molecules and the measured concentration are not identical quantities when the reacting volume changes. The ordinary first-order concentration equation assumes a suitable constant-volume description. Under changing-volume conditions, the amount-based relation supplied in the problem must be used. An apparent concentration increase does not imply that A is being chemically produced. It can arise from solvent loss even while A continues to decompose. \(\textbf{Final answer:}\) \[ \boxed{\left([\mathrm A]_{10},[\mathrm A]_{20}\right)=\left(0.500,\ 0.400\right)\ \mathrm{mol\,L^{-1}}} \]

Explanation: The concentration of A decreases linearly while the zero-order regime is active. However, the additional supply of A produces an instantaneous upward change in its concentration. The subsequent reaction starts from this new concentration without changing the disappearance constant. The concentration history must therefore be divided into intervals before and after addition. Product formation depends on cumulative chemical consumption of A, not on the net change between the original and final A concentrations alone. \(\textbf{First interval — Determine A consumed:}\) The zero-order disappearance constant is \[ k_0=0.020\ \mathrm{mol\,L^{-1}\,min^{-1}} \] During the first 10 min, \[ -\Delta[\mathrm A]_1=k_0t_1 \] \[ -\Delta[\mathrm A]_1=0.020(10) \] \[ -\Delta[\mathrm A]_1=0.200\ \mathrm{mol\,L^{-1}} \] Therefore, \[ [\mathrm A]_{10^-}=0.300-0.200 \] \[ [\mathrm A]_{10^-}=0.100\ \mathrm{mol\,L^{-1}} \] \(\textbf{Apply the instantaneous addition:}\) The concentration increase is \[ 0.200\ \mathrm{mol\,L^{-1}} \] Therefore, \[ [\mathrm A]_{10^+}=0.100+0.200 \] \[ [\mathrm A]_{10^+}=0.300\ \mathrm{mol\,L^{-1}} \] \(\textbf{Calculate the additional time to exhaustion:}\) After addition, the disappearance constant remains unchanged. Therefore, \[ t_{\mathrm{remaining}}=\frac{[\mathrm A]_{10^+}}{k_0} \] \[ t_{\mathrm{remaining}}=\frac{0.300}{0.020} \] \[ t_{\mathrm{remaining}}=15.0\ \mathrm{min} \] \(\textbf{Determine the total exhaustion time:}\) \[ t_{\mathrm{exhaustion}}=10.0+15.0 \] \[ t_{\mathrm{exhaustion}}=25.0\ \mathrm{min} \] \(\textbf{Determine A consumed between 10 and 20 min:}\) This interval lasts another 10 min. Therefore, \[ -\Delta[\mathrm A]_2=0.020(10) \] \[ -\Delta[\mathrm A]_2=0.200\ \mathrm{mol\,L^{-1}} \] At 20 min, \[ [\mathrm A]_{20}=0.300-0.200 \] \[ [\mathrm A]_{20}=0.100\ \mathrm{mol\,L^{-1}} \] \(\textbf{Calculate cumulative chemical consumption:}\) The reaction has consumed \[ 0.200\ \mathrm{mol\,L^{-1}} \] during the first interval and \[ 0.200\ \mathrm{mol\,L^{-1}} \] during the second. Therefore, \[ [\mathrm A]_{\mathrm{consumed,total}}=0.200+0.200 \] \[ [\mathrm A]_{\mathrm{consumed,total}}=0.400\ \mathrm{mol\,L^{-1}} \] \(\textbf{Convert consumption into product formation:}\) For \[ \mathrm A\rightarrow2\mathrm P, \] \[ \Delta[\mathrm P]=2[\mathrm A]_{\mathrm{consumed,total}} \] \[ \Delta[\mathrm P]=2(0.400) \] \[ \Delta[\mathrm P]=0.800\ \mathrm{mol\,L^{-1}} \] Including the initial product, \[ [\mathrm P]_{20}=0.050+0.800 \] \[ [\mathrm P]_{20}=0.850\ \mathrm{mol\,L^{-1}} \] \(\textbf{Material-balance verification:}\) The total amount of A supplied, expressed as concentration in the unchanged volume, is \[ 0.300+0.200=0.500\ \mathrm{mol\,L^{-1}} \] By 20 min, \(0.400\ \mathrm{mol\,L^{-1}}\) has reacted. Therefore, \[ [\mathrm A]_{20}=0.500-0.400 \] \[ [\mathrm A]_{20}=0.100\ \mathrm{mol\,L^{-1}} \] This agrees with the interval calculation. \(\textbf{Chemical interpretation:}\) Adding reactant changes the amount available for further reaction but does not necessarily change a zero-order disappearance rate within the same kinetic regime. The exhaustion time is extended because more A is present. The concentration of product records the cumulative amount chemically consumed, including consumption that occurred before reactant replenishment. \(\textbf{Final answer:}\) \[ \boxed{\left(t_{\mathrm{exhaustion}},[\mathrm P]_{20}\right)=\left(25.0\ \mathrm{min},0.850\ \mathrm{mol\,L^{-1}}\right)} \]

Explanation: B is formed by the first reaction and consumed by the second. Its concentration therefore depends on the difference between its formation and disappearance rates. Initially, the first-order formation rate of B is relatively high because A concentration is high. This formation rate decreases as A is consumed. The zero-order disappearance rate of B remains constant within the stipulated regime. Consequently, B reaches a maximum when its instantaneous formation and disappearance rates become equal. The concentration at that time follows from the cumulative amount of A converted and the cumulative amount of B consumed. \(\textbf{Determine the first-order constant for A:}\) \[ k_1=\frac{\ln2}{t_{1/2,\mathrm A}} \] \[ k_1=\frac{0.693}{10} \] \[ k_1=0.0693\ \mathrm{min^{-1}} \] \(\textbf{Construct the instantaneous formation rate of B:}\) Because \[ \mathrm{A\rightarrow B}, \] \[ r_{\mathrm{B,formation}}=k_1[\mathrm A] \] \(\textbf{Express the net rate of change of B:}\) The zero-order consumption rate is \[ r_{\mathrm{B,consumption}}=0.01386\ \mathrm{mol\,L^{-1}\,min^{-1}} \] Therefore, \[ \frac{d[\mathrm B]}{dt}=k_1[\mathrm A]-0.01386 \] \(\textbf{Apply the maximum-concentration condition:}\) At the maximum, \[ \frac{d[\mathrm B]}{dt}=0 \] Therefore, \[ k_1[\mathrm A]_{\max}=0.01386 \] Substituting, \[ 0.0693[\mathrm A]_{\max}=0.01386 \] \[ [\mathrm A]_{\max}=\frac{0.01386}{0.0693} \] \[ [\mathrm A]_{\max}=0.200\ \mathrm{mol\,L^{-1}} \] \(\textbf{Determine when A reaches this concentration:}\) Initially, \[ [\mathrm A]_0=0.80\ \mathrm{mol\,L^{-1}} \] The remaining fraction is \[ \frac{0.200}{0.80}=\frac14 \] Thus, two half-lives have elapsed. Therefore, \[ t_{\max}=2(10) \] \[ t_{\max}=20.0\ \mathrm{min} \] \(\textbf{Calculate B formed by 20 min:}\) The concentration of A consumed is \[ -\Delta[\mathrm A]=0.80-0.20 \] \[ -\Delta[\mathrm A]=0.60\ \mathrm{mol\,L^{-1}} \] Because each A produces one B, \[ [\mathrm B]_{\mathrm{formed}}=0.60\ \mathrm{mol\,L^{-1}} \] \(\textbf{Calculate B consumed through the second step:}\) The second pathway consumes B at the constant rate \[ 0.01386\ \mathrm{mol\,L^{-1}\,min^{-1}} \] Therefore, during 20 min, \[ [\mathrm B]_{\mathrm{consumed}}=0.01386(20) \] \[ [\mathrm B]_{\mathrm{consumed}}=0.2772\ \mathrm{mol\,L^{-1}} \] \(\textbf{Determine the maximum B concentration:}\) \[ [\mathrm B]_{\max}=[\mathrm B]_0+[\mathrm B]_{\mathrm{formed}}-[\mathrm B]_{\mathrm{consumed}} \] Substituting, \[ [\mathrm B]_{\max}=0.10+0.60-0.2772 \] \[ [\mathrm B]_{\max}=0.4228\ \mathrm{mol\,L^{-1}} \] \(\textbf{Check that this is a maximum:}\) Initially, the B-formation rate is \[ 0.0693(0.80)=0.05544\ \mathrm{mol\,L^{-1}\,min^{-1}} \] This exceeds the zero-order disappearance rate. Therefore, B initially accumulates. After 20 min, A concentration falls below \(0.200\ \mathrm{mol\,L^{-1}}\). Its corresponding B-formation rate then becomes smaller than the constant B-disappearance rate. Consequently, B begins to decrease. The equality at 20 min therefore identifies a maximum rather than a minimum. \(\textbf{Chemical interpretation:}\) An intermediate can attain a maximum concentration even when the reaction forming it has not stopped. The maximum occurs when its instantaneous rate of formation equals its instantaneous rate of consumption. These rates need not have the same reaction order. The integrated first-order law determines the remaining precursor concentration, while the zero-order law determines the cumulative intermediate consumption. \(\textbf{Final answer:}\) \[ \boxed{\left(t_{\max},[\mathrm B]_{\max}\right)=\left(20.0\ \mathrm{min},0.4228\ \mathrm{mol\,L^{-1}}\right)} \]

Explanation: The total pressure increases as A decomposes because each mole of A forms two moles of B. However, the experiment also includes a separate pressure increase caused by the addition of inert gas. This externally introduced pressure contribution must not be mistaken for additional chemical decomposition. The total pressure before inert-gas addition determines the fraction of A consumed during the first 10 min. The first-order equation then predicts the remaining amount of A after 30 min. Finally, both chemical gas formation and the added inert gas must be included in the total-pressure balance. \(\textbf{Determine the original total pressure:}\) Initially, \[ P_0=6.0+2.0 \] \[ P_0=8.0\ \mathrm{atm} \] \(\textbf{Relate pressure increase to decomposition:}\) For \[ \mathrm A\rightarrow2\mathrm B, \] each mole of A decomposed produces a net increase of one mole of gas. At constant temperature and volume, the chemical increase in total pressure equals the pressure-equivalent amount of A consumed. Therefore, \[ P_t=P_0+(p_{\mathrm A,0}-p_{\mathrm A,t}) \] before any external gas addition. \(\textbf{Determine A remaining after 10 min:}\) At 10 min, before the inert-gas addition, \[ P_{10^-}=11.0\ \mathrm{atm} \] Therefore, \[ 11.0=8.0+(6.0-p_{\mathrm A,10}) \] Rearranging, \[ p_{\mathrm A,10}=3.0\ \mathrm{atm} \] \(\textbf{Infer the first-order half-life:}\) Initially, \[ p_{\mathrm A,0}=6.0\ \mathrm{atm} \] After 10 min, \[ p_{\mathrm A,10}=3.0\ \mathrm{atm} \] Thus, half the original A remains. Therefore, \[ t_{1/2}=10\ \mathrm{min} \] \(\textbf{Calculate the first-order constant:}\) \[ k=\frac{\ln2}{t_{1/2}} \] \[ k=\frac{0.693}{10} \] \[ k=0.0693\ \mathrm{min^{-1}} \] \(\textbf{Account for the inert-gas addition:}\) The pressure contribution from the added inert gas is \[ 2.0\ \mathrm{atm} \] Immediately after addition, \[ P_{10^+}=11.0+2.0 \] \[ P_{10^+}=13.0\ \mathrm{atm} \] The partial pressures of A and B do not change during the instantaneous addition because the vessel volume and temperature remain fixed. \(\textbf{Determine A remaining after 30 min:}\) Thirty minutes corresponds to three half-lives. Therefore, \[ p_{\mathrm A,30}=6.0\left(\frac12\right)^3 \] \[ p_{\mathrm A,30}=0.750\ \mathrm{atm} \] \(\textbf{Determine the total chemical pressure increase:}\) The pressure-equivalent amount of A decomposed is \[ x_{30}=6.0-0.750 \] \[ x_{30}=5.250\ \mathrm{atm} \] This produces an equal net increase in total pressure through the chemical reaction. \(\textbf{Calculate the final total pressure:}\) \[ P_{30}=P_0+\Delta P_{\mathrm{inert}}+x_{30} \] Substituting, \[ P_{30}=8.0+2.0+5.250 \] \[ P_{30}=15.250\ \mathrm{atm} \] \(\textbf{Verify the physical pressure limit:}\) At complete eventual decomposition, the chemical pressure increase would be \[ 6.0\ \mathrm{atm} \] Therefore, the limiting pressure after inert-gas addition is \[ P_{\infty}=8.0+2.0+6.0 \] \[ P_{\infty}=16.0\ \mathrm{atm} \] The predicted pressure at 30 min is below this limit. \(\textbf{Chemical interpretation:}\) Adding inert gas at constant temperature and volume changes total pressure without changing the concentrations of the reacting gases. Consequently, it does not change their instantaneous first-order disappearance rate. A pressure-based kinetic calculation must distinguish pressure changes caused by chemical reaction from those caused by external gas addition. The correct pressure balance includes both contributions, but only chemical conversion is used to determine the first-order disappearance of A. \(\textbf{Final answer:}\) \[ \boxed{\left(k,P_{30}\right)=\left(0.0693\ \mathrm{min^{-1}},15.250\ \mathrm{atm}\right)} \]

Explanation: The two reactions have the same initial concentration and the same remaining fraction after 10 min. However, this single matching observation does not mean their subsequent concentration histories are identical. The zero-order reaction loses a constant amount of reactant per unit time. The first-order reaction loses a constant fraction over equal intervals. Their concentrations after another 10 min must therefore be calculated using different integrated equations. Their instantaneous disappearance rates must also be obtained from their respective differential rate laws. \(\textbf{Determine the concentration after 10 min:}\) Both reactions retain \(75\%\) of the initial concentration. Therefore, \[ C_{10}=0.75(0.80) \] \[ C_{10}=0.60\ \mathrm{mol\,L^{-1}} \] \(\textbf{Determine the zero-order constant:}\) For Reaction X, \[ C_X(t)=C_0-k_Xt \] At 10 min, \[ 0.60=0.80-10k_X \] Therefore, \[ k_X=\frac{0.20}{10} \] \[ k_X=0.020\ \mathrm{mol\,L^{-1}\,min^{-1}} \] \(\textbf{Calculate the zero-order concentration after 20 min:}\) \[ C_X(20)=0.80-0.020(20) \] \[ C_X(20)=0.400\ \mathrm{mol\,L^{-1}} \] \(\textbf{Determine the first-order constant:}\) For Reaction Y, \[ k_Y=\frac1t\ln\frac{C_0}{C_t} \] Substituting, \[ k_Y=\frac1{10}\ln\frac{0.80}{0.60} \] \[ k_Y=\frac1{10}\ln\frac43 \] Using the supplied value, \[ k_Y=0.02877\ \mathrm{min^{-1}} \] \(\textbf{Calculate the first-order concentration after 20 min:}\) The remaining fraction after each 10-minute interval is \(3/4\). Therefore, \[ C_Y(20)=0.80\left(\frac34\right)^2 \] \[ C_Y(20)=0.80\left(\frac9{16}\right) \] \[ C_Y(20)=0.450\ \mathrm{mol\,L^{-1}} \] \(\textbf{Determine the concentration difference:}\) \[ C_Y(20)-C_X(20)=0.450-0.400 \] \[ C_Y(20)-C_X(20)=0.050\ \mathrm{mol\,L^{-1}} \] \(\textbf{Calculate the zero-order instantaneous rate:}\) While its kinetic regime remains valid, \[ r_X(20)=k_X \] \[ r_X(20)=0.020\ \mathrm{mol\,L^{-1}\,min^{-1}} \] \(\textbf{Calculate the first-order instantaneous rate:}\) \[ r_Y(20)=k_YC_Y(20) \] Substituting, \[ r_Y(20)=0.02877(0.450) \] \[ r_Y(20)=0.01295\ \mathrm{mol\,L^{-1}\,min^{-1}} \] \(\textbf{Compare the instantaneous rates:}\) \[ \frac{r_Y(20)}{r_X(20)}=\frac{0.01295}{0.020} \] \[ \frac{r_Y(20)}{r_X(20)}\approx0.647 \] \(\textbf{Chemical interpretation:}\) The zero-order reaction loses \(0.20\ \mathrm{mol\,L^{-1}}\) during each 10-minute interval. The first-order reaction loses \(25\%\) of the amount remaining at the beginning of each such interval. Consequently, the first-order concentration after 20 min exceeds the zero-order concentration. Nevertheless, its instantaneous disappearance rate is smaller at that time. Equal fractional conversion at one instant does not establish identical reaction order or identical subsequent rates. \(\textbf{Final answer:}\) \[ \boxed{\left(C_Y(20)-C_X(20),\frac{r_Y(20)}{r_X(20)}\right)=\left(0.050\ \mathrm{mol\,L^{-1}},0.647\right)} \]

Explanation: The gas density changes because the total amount of gas increases as A decomposes. The total mass remains unchanged, while the piston allows the vessel volume to increase at constant temperature and pressure. Therefore, the density ratio can be related to the change in total gaseous amount. The balanced equation connects that change with the fraction of A remaining. The first-order half-life then determines the elapsed time. Finally, the amounts of A and B provide the mole fraction of product. \(\textbf{Represent the composition using the fraction remaining:}\) Let the initial amount of A be \(n_0\). Let \(f\) be the fraction of A remaining. Therefore, \[ n_{\mathrm A}=fn_0 \] The amount of A decomposed is \[ n_0(1-f) \] For \[ \mathrm A\rightarrow2\mathrm B, \] the amount of B formed is \[ n_{\mathrm B}=2n_0(1-f) \] \(\textbf{Determine the total gaseous amount:}\) \[ n_{\mathrm{total}}=fn_0+2n_0(1-f) \] Factorizing, \[ n_{\mathrm{total}}=n_0(2-f) \] \(\textbf{Relate density to total gaseous amount:}\) Gas density is \[ \rho=\frac{m}{V} \] The total gas mass \(m\) is conserved. At constant temperature and pressure, the ideal-gas equation gives \[ V=\frac{n_{\mathrm{total}}RT}{P} \] Therefore, \[ \rho\propto\frac1{n_{\mathrm{total}}} \] Hence, \[ \frac{\rho}{\rho_0}=\frac{n_0}{n_{\mathrm{total}}} \] Substituting the total amount, \[ \frac{\rho}{\rho_0}=\frac1{2-f} \] \(\textbf{Apply the measured density ratio:}\) \[ \frac47=\frac1{2-f} \] Cross-multiplying, \[ 4(2-f)=7 \] \[ 8-4f=7 \] \[ 4f=1 \] \[ f=\frac14 \] Thus, one-fourth of the original A remains. \(\textbf{Determine the elapsed time:}\) For first-order disappearance, \[ f=\left(\frac12\right)^{t/t_{1/2}} \] Because \[ \frac14=\left(\frac12\right)^2, \] two half-lives have elapsed. Therefore, \[ t=2(8.0) \] \[ t=16.0\ \mathrm{min} \] \(\textbf{Calculate the mole fraction of B:}\) At this instant, \[ n_{\mathrm A}=\frac14n_0 \] The amount of B is \[ n_{\mathrm B}=2n_0\left(1-\frac14\right) \] \[ n_{\mathrm B}=\frac32n_0 \] The total gaseous amount is \[ n_{\mathrm{total}}=\frac14n_0+\frac32n_0 \] \[ n_{\mathrm{total}}=\frac74n_0 \] Therefore, \[ x_{\mathrm B}=\frac{n_{\mathrm B}}{n_{\mathrm{total}}} \] \[ x_{\mathrm B}=\frac{(3/2)n_0}{(7/4)n_0} \] \[ x_{\mathrm B}=\frac67 \] \(\textbf{Check the density relationship:}\) The total gaseous amount increases to \(7/4\) of its original value. At constant temperature and pressure, volume increases by the same factor. Therefore, density becomes \(4/7\) of its original value, agreeing with the measurement. \(\textbf{Chemical interpretation:}\) Gas density can provide kinetic information when the gaseous stoichiometry, physical constraints and mass conservation are known. The fraction of A remaining is not identical to its mole fraction in the product-containing mixture. The measured density first determines total gas expansion. Stoichiometry then gives chemical conversion, and first-order kinetics converts that conversion into elapsed time. \(\textbf{Final answer:}\) \[ \boxed{\left(t,x_{\mathrm B}\right)=\left(16.0\ \mathrm{min},\frac67\right)} \]