Chemical Kinetics MCQs with Answers – Part 5 (NEET Chemistry)

Explanation: The observed instrument signal contains a concentration-dependent contribution and a constant background contribution. Because the background is not known initially, the raw signal cannot be treated directly as proportional to the concentration of A. However, the background-corrected signal is proportional to the remaining reactant concentration. For first-order kinetics, this corrected signal decreases by a constant fraction during equal time intervals. The three observations are equally spaced, allowing the unknown background to be determined before predicting the reading at a later time. \(\textbf{Step 1 — Relate the measured signal to first-order kinetics:}\) The instrument equation is \[ S_t=b+\alpha[\mathrm A]_t \] Rearranging, \[ S_t-b=\alpha[\mathrm A]_t \] For a first-order reaction, \[ [\mathrm A]_t=[\mathrm A]_0e^{-kt} \] Therefore, \[ S_t-b=\alpha[\mathrm A]_0e^{-kt} \] Since \[ S_0-b=\alpha[\mathrm A]_0 \] the measured signal satisfies \[ S_t-b=(S_0-b)e^{-kt} \] Thus, the background-corrected signal follows the same exponential decay factor as the concentration of A. \(\textbf{Step 2 — Use the equally spaced observations:}\) The readings are taken at 0, 12 and 24 min. For first-order kinetics, the fraction remaining during each 12-minute interval is the same. Therefore, \[ \frac{S_{12}-b}{S_0-b} =\frac{S_{24}-b}{S_{12}-b} \] Cross-multiplying, \[ (S_{12}-b)^2=(S_0-b)(S_{24}-b) \] Using the measured signals, \[ (9.0-b)^2=(14.0-b)(6.5-b) \] \(\textbf{Step 3 — Determine the unknown background:}\) Expanding the left-hand side, \[ (9.0-b)^2=81.0-18.0b+b^2 \] Expanding the right-hand side, \[ (14.0-b)(6.5-b)=91.0-20.5b+b^2 \] Equating both expressions, \[ 81.0-18.0b+b^2=91.0-20.5b+b^2 \] Cancelling \(b^2\) and rearranging, \[ 20.5b-18.0b=91.0-81.0 \] \[ 2.5b=10.0 \] Therefore, \[ b=4.0 \] \(\textbf{Step 4 — Determine the first-order decay factor:}\) The initial background-corrected signal is \[ S_0-b=14.0-4.0 \] \[ S_0-b=10.0 \] After 12 min, the corrected signal becomes \[ S_{12}-b=9.0-4.0 \] \[ S_{12}-b=5.0 \] Therefore, the fraction remaining after 12 min is \[ f_{12}=\frac{5.0}{10.0} \] \[ f_{12}=\frac12 \] The concentration-dependent contribution is reduced to half in 12 min. Consequently, the first-order half-life is \[ t_{1/2}=12\ \mathrm{min} \] \(\textbf{Step 5 — Predict the corrected signal after 48 min:}\) The total elapsed time is \[ t=48\ \mathrm{min} \] The number of half-lives is \[ n=\frac{48}{12} \] \[ n=4 \] For first-order decay, \[ S_t-b=(S_0-b)\left(\frac12\right)^n \] Therefore, \[ S_{48}-b=10.0\left(\frac12\right)^4 \] \[ S_{48}-b=\frac{10.0}{16} \] \[ S_{48}-b=0.625 \] \(\textbf{Step 6 — Restore the constant background:}\) The instrument records the sum of the remaining concentration-dependent signal and the constant background. Therefore, \[ S_{48}=b+0.625 \] Substituting \(b=4.0\), \[ S_{48}=4.0+0.625 \] \[ S_{48}=4.625 \] \(\textbf{Step 7 — Verify the measured signal at 24 min:}\) Twenty-four minutes corresponds to two half-lives. The expected concentration-dependent signal is \[ S_{24}-b=10.0\left(\frac12\right)^2 \] \[ S_{24}-b=2.5 \] Adding the background, \[ S_{24}=4.0+2.5 \] \[ S_{24}=6.5 \] This agrees with the recorded observation. \(\textbf{Experimental interpretation:}\) The measured signal does not approach zero as the reactant disappears because the instrument contains a constant background contribution. Instead, the signal approaches the background value \(b=4.0\). The first-order integrated equation applies directly to the background-corrected signal \(S-b\), not to the uncorrected signal S. The equality of successive fractional decreases over equal time intervals allows the background to be determined without separately knowing the calibration constant. After four half-lives, only one-sixteenth of the original concentration-dependent signal remains. The complete instrument reading is obtained by adding the constant background to this remaining contribution. \(\textbf{Final answer:}\) \[ \boxed{(b,S_{48})=(4.00,4.6250)} \]

Explanation: The two experiments have different initial concentrations but reach the same specified final concentration. Their measured times must be tested against the proposed integrated kinetic laws. For a zero-order reaction, elapsed time is proportional to the absolute concentration decrease. For a first-order reaction, elapsed time depends on the logarithm of the initial-to-final concentration ratio. Only the model that gives the same intrinsic rate constant in both experiments is consistent with the observations. \(\textbf{Test the zero-order model:}\) The zero-order integrated equation is \[ [\mathrm A]_t=[\mathrm A]_0-kt \] Therefore, \[ k=\frac{[\mathrm A]_0-[\mathrm A]_t}{t} \] For Experiment I, \[ k_{\mathrm I}=\frac{0.80-0.20}{30} \] \[ k_{\mathrm I}=0.020\ \mathrm{mol\,L^{-1}\,min^{-1}} \] For Experiment II, \[ k_{\mathrm{II}}=\frac{0.40-0.20}{10} \] \[ k_{\mathrm{II}}=0.020\ \mathrm{mol\,L^{-1}\,min^{-1}} \] Both experiments give the same zero-order constant. \(\textbf{Test the competing first-order model:}\) For first-order disappearance, \[ k=\frac1t\ln\frac{[\mathrm A]_0}{[\mathrm A]_t} \] Experiment I would give \[ k_{\mathrm I}=\frac{\ln4}{30} \] Experiment II would give \[ k_{\mathrm{II}}=\frac{\ln2}{10} \] Using \[ \ln4=2\ln2, \] the ratio of these proposed constants is \[ \frac{k_{\mathrm I}}{k_{\mathrm{II}}}=\frac{2\ln2/30}{\ln2/10} \] \[ \frac{k_{\mathrm I}}{k_{\mathrm{II}}}=\frac23 \] The first-order constants would differ, contradicting the experimental condition. Therefore, the zero-order model is supported. \(\textbf{Predict the time in Experiment III:}\) The third sample begins with \[ [\mathrm A]_0=0.60\ \mathrm{mol\,L^{-1}} \] The target concentration is \[ [\mathrm A]_t=0.10\ \mathrm{mol\,L^{-1}} \] Using \[ k=0.020\ \mathrm{mol\,L^{-1}\,min^{-1}}, \] \[ t=\frac{0.60-0.10}{0.020} \] \[ t=25.0\ \mathrm{min} \] \(\textbf{Calculate its first half-life:}\) For zero-order disappearance, \[ t_{1/2}=\frac{[\mathrm A]_0}{2k} \] Therefore, \[ t_{1/2}=\frac{0.60}{2(0.020)} \] \[ t_{1/2}=15.0\ \mathrm{min} \] \(\textbf{Check the physical sequence:}\) After 15 min, the concentration is \[ 0.60-0.020(15)=0.30\ \mathrm{mol\,L^{-1}} \] It reaches \(0.10\ \mathrm{mol\,L^{-1}}\) after a further 10 min. Thus, the target time of 25 min is consistent with the constant disappearance rate. \(\textbf{Chemical interpretation:}\) Two observations that each appear compatible with a kinetic equation can become decisive when the same rate constant must explain both. Zero-order kinetics is supported here because equal absolute concentration losses take equal times. The first half-life in the third experiment depends on its own initial concentration. This differs from the concentration-independent first-order half-life. \(\textbf{Final answer:}\) \[ \left(\text{zero order},t_{\mathrm{target}},t_{1/2}\right)=\left(\text{zero order},25.0\ \mathrm{min},15.0\ \mathrm{min}\right) \]

Explanation: The total pressure increases as dinitrogen pentoxide decomposes because the products contain more gaseous molecules than the reactant. However, the pressure increase is not equal to the pressure-equivalent amount of reactant consumed. For every two moles of \(\mathrm{N_2O_5}\) consumed, five moles of gaseous products form. The reaction therefore produces a net increase of three moles of gas per two moles of reactant consumed. This stoichiometric factor must be included before the first-order conversion can be inferred from the measured total pressure. \(\textbf{Establish the initial gas pressure:}\) The initial total pressure is \[ P_0=400+100 \] \[ P_0=500\ \mathrm{torr} \] \(\textbf{Express the pressure increase through conversion:}\) Let \(x\) be the fraction of the original \(\mathrm{N_2O_5}\) decomposed. The pressure-equivalent amount of reactant consumed is \[ 400x\ \mathrm{torr} \] For the balanced reaction, \[ 2\mathrm{N_2O_5}\rightarrow4\mathrm{NO_2}+\mathrm{O_2}, \] the net gas-mole increase per mole of \(\mathrm{N_2O_5}\) consumed is \[ \frac{5-2}{2}=\frac32 \] Therefore, \[ P_t=P_0+\frac32(400x) \] \[ P_t=500+600x \] \(\textbf{Determine the conversion after 10 min:}\) Using the observed total pressure, \[ 800=500+600x_{10} \] \[ 600x_{10}=300 \] \[ x_{10}=\frac12 \] Thus, half the initial dinitrogen pentoxide has decomposed. \(\textbf{Apply the first-order half-life:}\) The reaction reaches \(50\%\) decomposition after 10 min. Therefore, \[ t_{1/2}=10\ \mathrm{min} \] After 20 min, two half-lives have elapsed. The remaining fraction is \[ f_{20}=\left(\frac12\right)^2 \] \[ f_{20}=\frac14 \] The decomposed fraction is \[ x_{20}=1-\frac14 \] \[ x_{20}=\frac34 \] \(\textbf{Determine the total pressure after 20 min:}\) \[ P_{20}=500+600\left(\frac34\right) \] \[ P_{20}=500+450 \] \[ P_{20}=950\ \mathrm{torr} \] \(\textbf{Determine the oxygen partial pressure:}\) For every two moles of \(\mathrm{N_2O_5}\) decomposed, one mole of oxygen forms. Therefore, \[ p_{\mathrm{O_2},20}=\frac12(400x_{20}) \] Substituting, \[ p_{\mathrm{O_2},20}=\frac12(400)\left(\frac34\right) \] \[ p_{\mathrm{O_2},20}=150\ \mathrm{torr} \] \(\textbf{Verify the complete gas balance:}\) The remaining reactant pressure is \[ p_{\mathrm{N_2O_5},20}=400\left(\frac14\right) \] \[ p_{\mathrm{N_2O_5},20}=100\ \mathrm{torr} \] The nitrogen dioxide pressure is \[ p_{\mathrm{NO_2},20}=2(400)\left(\frac34\right) \] \[ p_{\mathrm{NO_2},20}=600\ \mathrm{torr} \] Including the inert gas, \[ P_{20}=100+600+150+100 \] \[ P_{20}=950\ \mathrm{torr} \] \(\textbf{Chemical interpretation:}\) The pressure history of a gaseous decomposition depends on its exact stoichiometry. The first-order kinetic law determines the fraction of reactant remaining. The balanced equation then determines the amounts of all products and the resulting total pressure. Using a pressure relation derived for a different gaseous reaction would give an incorrect answer. \(\textbf{Final answer:}\) \[ \boxed{\left(P_{20},p_{\mathrm{O_2},20}\right)=\left(950\ \mathrm{torr},150\ \mathrm{torr}\right)} \]

Explanation: The movable piston allows the gas volume to change as the number of gas molecules increases. The ideal-gas equation connects this expansion with the extent of reaction. The gas volume is therefore an indirect measurement of chemical conversion. The first-order disappearance law must be applied to the fraction of A remaining, not directly to the expanding vessel volume. Once the conversion is known, the mole fraction follows from the gas-mole balance. \(\textbf{Relate volume to the fraction decomposed:}\) Let \(n_0\) be the original amount of A. If a fraction \(x\) decomposes, \[ n_{\mathrm A}=n_0(1-x) \] and \[ n_{\mathrm B}=2n_0x \] Thus, \[ n_{\mathrm{total}}=n_0(1+x) \] At constant temperature and pressure, \[ V\propto n_{\mathrm{total}} \] Therefore, \[ \frac{V_t}{V_0}=1+x \] \(\textbf{Recover the conversion at 10 min:}\) \[ \frac{1.60}{1.00}=1+x_{10} \] \[ x_{10}=0.60 \] The fraction of A remaining after 10 min is \[ f_{10}=1-0.60 \] \[ f_{10}=0.40 \] \(\textbf{Use the first-order time relationship:}\) For first-order disappearance, \[ f_t=e^{-kt} \] Therefore, \[ f_{10}=e^{-10k}=0.40 \] After a further equal interval of 10 min, \[ f_{20}=e^{-20k} \] \[ f_{20}=(e^{-10k})^2 \] \[ f_{20}=(0.40)^2 \] \[ f_{20}=0.16 \] \(\textbf{Determine the corresponding gas volume:}\) At 20 min, the fraction decomposed is \[ x_{20}=1-0.16 \] \[ x_{20}=0.84 \] Therefore, \[ V_{20}=V_0(1+x_{20}) \] \[ V_{20}=1.00(1+0.84) \] \[ V_{20}=1.84\ \mathrm L \] Thus, the requested volume is reached at 20 min. \(\textbf{Calculate the mole fraction of A:}\) At this instant, \[ n_{\mathrm A}=0.16n_0 \] The total amount of gas is \[ n_{\mathrm{total}}=1.84n_0 \] Therefore, \[ y_{\mathrm A}=\frac{n_{\mathrm A}}{n_{\mathrm{total}}} \] \[ y_{\mathrm A}=\frac{0.16}{1.84} \] \[ y_{\mathrm A}=\frac{16}{184} \] \[ y_{\mathrm A}=\frac2{23} \] \(\textbf{Check the product contribution:}\) The amount of B is \[ n_{\mathrm B}=2(0.84)n_0 \] \[ n_{\mathrm B}=1.68n_0 \] Therefore, \[ y_{\mathrm B}=\frac{1.68}{1.84} \] \[ y_{\mathrm B}=\frac{21}{23} \] The two mole fractions sum to one. \(\textbf{Chemical interpretation:}\) In a constant-pressure vessel, volume expansion can reveal reaction progress when gaseous stoichiometry is known. The fraction of original A remaining differs from its mole fraction in the expanding product-containing gas mixture. First-order kinetics governs the former; gas stoichiometry determines the latter. Confusing these fractions leads to incorrect pressure and composition predictions. \(\textbf{Final answer:}\) \[ \boxed{\left(t,y_{\mathrm A}\right)=\left(20\ \mathrm{min},\frac2{23}\right)} \]

Explanation: The decomposition changes the identities of the gas molecules but not their total number. For every two molecules of A consumed, one molecule each of B and C forms. Therefore, the total gas pressure remains unchanged at constant temperature and volume. The first-order conversion must be inferred from a species-specific measurement, such as the partial pressure of B. The total pressure alone provides no information about the extent of this reaction. \(\textbf{Use the product pressure to recover A consumption:}\) The balanced reaction is \[ 2\mathrm A\rightarrow\mathrm B+\mathrm C \] For each mole of B formed, two moles of A disappear. Therefore, after 10 min the pressure-equivalent amount of A consumed is \[ \Delta p_{\mathrm A}=2p_{\mathrm B,10} \] \[ \Delta p_{\mathrm A}=2(150) \] \[ \Delta p_{\mathrm A}=300\ \mathrm{torr} \] \(\textbf{Determine the reactant pressure after 10 min:}\) \[ p_{\mathrm A,10}=600-300 \] \[ p_{\mathrm A,10}=300\ \mathrm{torr} \] Thus, half the initial A remains. \(\textbf{Identify the first-order half-life:}\) Because the reactant concentration halves in 10 min, \[ t_{1/2}=10\ \mathrm{min} \] After 30 min, three half-lives have elapsed. \(\textbf{Calculate the remaining pressure of A:}\) \[ p_{\mathrm A,30}=600\left(\frac12\right)^3 \] \[ p_{\mathrm A,30}=600\left(\frac18\right) \] \[ p_{\mathrm A,30}=75.0\ \mathrm{torr} \] \(\textbf{Recover the amount of B formed:}\) The pressure-equivalent amount of A consumed by 30 min is \[ 600-75=525\ \mathrm{torr} \] The balanced equation gives \[ p_{\mathrm B,30}=\frac12(525) \] \[ p_{\mathrm B,30}=262.5\ \mathrm{torr} \] Likewise, \[ p_{\mathrm C,30}=262.5\ \mathrm{torr} \] \(\textbf{Calculate the total pressure:}\) \[ P_{30}=p_{\mathrm A,30}+p_{\mathrm B,30}+p_{\mathrm C,30} \] \[ P_{30}=75.0+262.5+262.5 \] \[ P_{30}=600\ \mathrm{torr} \] \(\textbf{Check the pressure invariant:}\) The initial number of gaseous molecules and the total number after reaction are equal under the reaction stoichiometry. Therefore, \[ P_t=P_0 \] at every instant under the stated physical conditions. The constant total pressure does not imply that the chemical reaction has stopped. \(\textbf{Chemical interpretation:}\) Gas-pressure kinetics requires a distinction between total pressure and reactant partial pressure. For a reaction with no net change in gaseous molecule count, total pressure remains unchanged even during substantial conversion. A product-specific pressure measurement can still reveal the remaining reactant concentration and first-order half-life. \(\textbf{Final answer:}\) \[ \boxed{\left(p_{\mathrm B,30},P_{30}\right)=\left(262.5\ \mathrm{torr},600\ \mathrm{torr}\right)} \]

Explanation: The graph relates the instantaneous disappearance rate of A to the concentration of its product P. It is not an ordinary concentration–time graph. The first-order rate law relates the disappearance rate to the remaining A concentration. The balanced equation then relates that concentration to the accumulated amount of P. Combining these relationships explains why the measured graph is linear and allows its slope to determine the rate constant. \(\textbf{Connect reactant and product concentrations:}\) For \[ \mathrm A\rightarrow2\mathrm P, \] \[ [\mathrm P]_t=[\mathrm P]_0+2([\mathrm A]_0-[\mathrm A]_t) \] Rearranging, \[ [\mathrm A]_t=[\mathrm A]_0-\frac{[\mathrm P]_t-[\mathrm P]_0}{2} \] \(\textbf{Insert the stoichiometric balance into the rate law:}\) The positive disappearance rate is \[ r_A=k[\mathrm A]_t \] Therefore, \[ r_A=k\left([\mathrm A]_0-\frac{[\mathrm P]_t-[\mathrm P]_0}{2}\right) \] Expanding, \[ r_A=k\left([\mathrm A]_0+\frac{[\mathrm P]_0}{2}\right)-\frac{k}{2}[\mathrm P]_t \] Thus, the slope of the stated graph is \[ m=-\frac{k}{2} \] \(\textbf{Extract the slope from the measured points:}\) \[ m=\frac{0.020-0.040}{0.50-0.10} \] \[ m=\frac{-0.020}{0.40} \] \[ m=-0.050\ \mathrm{min^{-1}} \] Therefore, \[ -\frac{k}{2}=-0.050 \] \[ k=0.100\ \mathrm{min^{-1}} \] \(\textbf{Determine the first-order half-life:}\) \[ t_{1/2}=\frac{\ln2}{k} \] \[ t_{1/2}=\frac{0.693}{0.100} \] \[ t_{1/2}=6.93\ \mathrm{min} \] \(\textbf{Recover the initial concentration of A:}\) The first graph point represents the initial instant. Therefore, \[ r_{A,0}=k[\mathrm A]_0 \] \[ 0.040=(0.100)[\mathrm A]_0 \] \[ [\mathrm A]_0=0.400\ \mathrm{mol\,L^{-1}} \] \(\textbf{Determine the limiting product concentration:}\) Initially, \[ [\mathrm P]_0=0.10\ \mathrm{mol\,L^{-1}} \] At complete eventual conversion, each mole of A forms two moles of P. Therefore, \[ [\mathrm P]_{\infty}=[\mathrm P]_0+2[\mathrm A]_0 \] \[ [\mathrm P]_{\infty}=0.10+2(0.400) \] \[ [\mathrm P]_{\infty}=0.90\ \mathrm{mol\,L^{-1}} \] \(\textbf{Verify the second graph point:}\) When \[ [\mathrm P]=0.50\ \mathrm{mol\,L^{-1}}, \] the additional product formed is \[ 0.50-0.10=0.40\ \mathrm{mol\,L^{-1}} \] Therefore, the A concentration consumed is \[ \frac{0.40}{2}=0.20\ \mathrm{mol\,L^{-1}} \] The remaining A concentration is \[ 0.400-0.200=0.200\ \mathrm{mol\,L^{-1}} \] Its disappearance rate is \[ r_A=0.100(0.200) \] \[ r_A=0.020\ \mathrm{mol\,L^{-1}\,min^{-1}} \] This reproduces the second graph point. \(\textbf{Chemical interpretation:}\) A first-order reaction does not generally produce a straight-line reactant concentration–time graph. However, the instantaneous rate can vary linearly with product concentration when stoichiometry connects accumulated product with remaining reactant. The slope of this rate–product graph contains both the first-order constant and the product stoichiometric coefficient. Its interpretation therefore requires both kinetics and material balance. \(\textbf{Final answer:}\) \[ \boxed{\left(t_{1/2},[\mathrm P]_{\infty}\right)=\left(6.93\ \mathrm{min},0.90\ \mathrm{mol\,L^{-1}}\right)} \]

Explanation: The two reactions have identical initial concentrations and reach half those concentrations after the same time. Their average disappearance rates over that interval must therefore be equal. However, their instantaneous rates at the endpoint need not be equal. The zero-order rate remains constant during the interval. The first-order rate decreases as the reactant concentration decreases. The comparison therefore requires the differential and integrated characteristics of the two kinetic orders. \(\textbf{Determine the common average rate:}\) After the first half-life, \[ C_h=\frac{C_0}{2} \] Therefore, \[ r_{\mathrm{avg}}=\frac{C_0-C_0/2}{h} \] \[ r_{\mathrm{avg}}=\frac{C_0}{2h} \] This result applies to both reactions because their endpoint concentrations and elapsed times are identical. \(\textbf{Evaluate the zero-order endpoint rate:}\) For Reaction X, \[ r_X=k_0 \] The zero-order half-life relation is \[ h=\frac{C_0}{2k_0} \] Therefore, \[ k_0=\frac{C_0}{2h} \] Consequently, \[ r_{h,X}=\frac{C_0}{2h} \] \(\textbf{Compare the zero-order rates:}\) \[ \left(\frac{r_{\mathrm{avg}}}{r_h}\right)_X= \frac{C_0/(2h)}{C_0/(2h)} \] \[ \left(\frac{r_{\mathrm{avg}}}{r_h}\right)_X=1 \] \(\textbf{Evaluate the first-order endpoint rate:}\) For Reaction Y, \[ r_Y=k_1C \] Its first-order half-life satisfies \[ h=\frac{\ln2}{k_1} \] Therefore, \[ k_1=\frac{\ln2}{h} \] At the endpoint, \[ C_h=\frac{C_0}{2} \] Thus, \[ r_{h,Y}=k_1\frac{C_0}{2} \] \[ r_{h,Y}=\frac{C_0\ln2}{2h} \] \(\textbf{Compare the first-order rates:}\) \[ \left(\frac{r_{\mathrm{avg}}}{r_h}\right)_Y= \frac{C_0/(2h)}{C_0\ln2/(2h)} \] \[ \left(\frac{r_{\mathrm{avg}}}{r_h}\right)_Y=\frac1{\ln2} \] \(\textbf{Interpret the comparison:}\) For the zero-order reaction, the average and instantaneous endpoint rates are equal because the disappearance rate is constant. For the first-order reaction, the average rate is larger than the endpoint rate because the reaction was faster earlier in the interval. The ratio is independent of the numerical values of \(C_0\) and h. \(\textbf{Chemical interpretation:}\) Equal half-lives do not imply identical instantaneous kinetic behaviour. A half-life describes an integrated concentration change over an interval. An instantaneous rate describes the slope of the concentration–time curve at one particular instant. The comparison distinguishes the linear zero-order concentration curve from exponential first-order decay. \(\textbf{Final answer:}\) \[ \boxed{\left(\frac{r_{\mathrm{avg}}}{r_h}\right)_X=1,\qquad \left(\frac{r_{\mathrm{avg}}}{r_h}\right)_Y=\frac1{\ln2}\approx1.443} \]

Explanation: The experiment provides two instantaneous rates and the total concentration decrease between their measurement times. It does not provide either the initial concentration or the elapsed time. For first-order kinetics, the instantaneous disappearance rate is proportional to the remaining concentration. Therefore, the difference between the measured rates can be related to the known concentration decrease. This allows the intrinsic constant to be determined without first knowing the initial concentration. The rate ratio then determines the elapsed time. \(\textbf{Relate each instantaneous rate to concentration:}\) The first-order differential law is \[ r=k[\mathrm A] \] At the beginning of the observation interval, \[ r_0=k[\mathrm A]_0 \] At the end, \[ r_t=k[\mathrm A]_t \] \(\textbf{Subtract the rate expressions:}\) \[ r_0-r_t=k([\mathrm A]_0-[\mathrm A]_t) \] The concentration decrease is known: \[ [\mathrm A]_0-[\mathrm A]_t=0.900\ \mathrm{mol\,L^{-1}} \] Therefore, \[ 0.0800-0.0200=k(0.900) \] \[ 0.0600=0.900k \] \(\textbf{Calculate the first-order constant:}\) \[ k=\frac{0.0600}{0.900} \] \[ k=0.06667\ \mathrm{min^{-1}} \] \(\textbf{Recover the initial concentration:}\) Using \[ r_0=k[\mathrm A]_0, \] \[ [\mathrm A]_0=\frac{0.0800}{0.06667} \] \[ [\mathrm A]_0=1.20\ \mathrm{mol\,L^{-1}} \] \(\textbf{Recover the final concentration:}\) \[ [\mathrm A]_t=[\mathrm A]_0-0.900 \] \[ [\mathrm A]_t=1.20-0.900 \] \[ [\mathrm A]_t=0.300\ \mathrm{mol\,L^{-1}} \] \(\textbf{Determine the remaining fraction:}\) \[ f=\frac{[\mathrm A]_t}{[\mathrm A]_0} \] \[ f=\frac{0.300}{1.20} \] \[ f=\frac14 \] \(\textbf{Apply the first-order integrated law:}\) \[ kt=\ln\frac{[\mathrm A]_0}{[\mathrm A]_t} \] \[ kt=\ln4 \] Therefore, \[ t=\frac{1.386}{0.06667} \] \[ t\approx20.79\ \mathrm{min} \] \(\textbf{Verify the final instantaneous rate:}\) \[ r_t=k[\mathrm A]_t \] \[ r_t=0.06667(0.300) \] \[ r_t\approx0.0200\ \mathrm{mol\,L^{-1}\,min^{-1}} \] The reconstructed values satisfy the measured final rate. \(\textbf{Chemical interpretation:}\) The ratio of first-order instantaneous rates equals the ratio of the corresponding reactant concentrations. Their difference also contains useful concentration information when the amount consumed is known. Combining a differential measurement with a material balance can therefore recover both an unknown rate constant and an unknown time interval. \(\textbf{Final answer:}\) \[ \boxed{(k,t)=\left(0.0667\ \mathrm{min^{-1}},20.79\ \mathrm{min}\right)} \]

Explanation: The reaction follows zero-order disappearance in both stages, but the constant changes after catalyst addition. The concentration–time history is therefore composed of two straight-line segments having different slopes. The initial disappearance constant must be used only for the first six minutes. The second constant must then be applied to the concentration remaining at the moment of catalyst addition. Product formation depends on the cumulative concentration of A consumed across both stages. \(\textbf{Calculate A consumed before catalyst addition:}\) For zero-order disappearance, \[ [\mathrm A]_t=[\mathrm A]_0-kt \] During the first six minutes, \[ -\Delta[\mathrm A]_1=k_1t_1 \] \[ -\Delta[\mathrm A]_1=0.020(6) \] \[ -\Delta[\mathrm A]_1=0.120\ \mathrm{mol\,L^{-1}} \] \(\textbf{Determine A remaining at the transition:}\) \[ [\mathrm A]_{6^-}=0.600-0.120 \] \[ [\mathrm A]_{6^-}=0.480\ \mathrm{mol\,L^{-1}} \] No reaction occurs during catalyst addition. Therefore, \[ [\mathrm A]_{6^+}=0.480\ \mathrm{mol\,L^{-1}} \] \(\textbf{Determine the remaining reaction duration:}\) After catalyst addition, \[ k_2=0.060\ \mathrm{mol\,L^{-1}\,min^{-1}} \] The remaining time to consume all available A is \[ t_2=\frac{[\mathrm A]_{6^+}}{k_2} \] \[ t_2=\frac{0.480}{0.060} \] \[ t_2=8.0\ \mathrm{min} \] \(\textbf{Calculate the complete exhaustion time:}\) \[ t_{\mathrm{exhaustion}}=6.0+8.0 \] \[ t_{\mathrm{exhaustion}}=14.0\ \mathrm{min} \] \(\textbf{Calculate consumption between 6 and 10 min:}\) The second observation interval lasts \[ 10-6=4\ \mathrm{min} \] During this interval, \[ -\Delta[\mathrm A]_2=0.060(4) \] \[ -\Delta[\mathrm A]_2=0.240\ \mathrm{mol\,L^{-1}} \] \(\textbf{Determine cumulative A consumption:}\) \[ -\Delta[\mathrm A]_{\mathrm{total}}=0.120+0.240 \] \[ -\Delta[\mathrm A]_{\mathrm{total}}=0.360\ \mathrm{mol\,L^{-1}} \] The remaining A concentration at 10 min is \[ [\mathrm A]_{10}=0.600-0.360 \] \[ [\mathrm A]_{10}=0.240\ \mathrm{mol\,L^{-1}} \] \(\textbf{Convert consumption into product formation:}\) For \[ \mathrm A\rightarrow2\mathrm P, \] \[ \Delta[\mathrm P]=2(-\Delta[\mathrm A]) \] Therefore, \[ \Delta[\mathrm P]=2(0.360) \] \[ \Delta[\mathrm P]=0.720\ \mathrm{mol\,L^{-1}} \] Including the initially present product, \[ [\mathrm P]_{10}=0.050+0.720 \] \[ [\mathrm P]_{10}=0.770\ \mathrm{mol\,L^{-1}} \] \(\textbf{Check the remaining duration at 10 min:}\) At 10 min, the reactant concentration is \(0.240\ \mathrm{mol\,L^{-1}}\). Under the second-stage zero-order law, \[ t_{\mathrm{remaining}}=\frac{0.240}{0.060} \] \[ t_{\mathrm{remaining}}=4.0\ \mathrm{min} \] Thus, exhaustion occurs at 14 min, agreeing with the earlier calculation. \(\textbf{Chemical interpretation:}\) A change in active catalyst availability can alter the zero-order disappearance constant. The rate can change even though it remains independent of reactant concentration within each stated regime. The concentration itself remains continuous across an instantaneous catalyst addition that does not consume or supply reactant. The total conversion is obtained by combining the separately calculated reaction intervals. \(\textbf{Final answer:}\) \[ \boxed{\left(t_{\mathrm{exhaustion}},[\mathrm P]_{10}\right)=\left(14.0\ \mathrm{min},0.770\ \mathrm{mol\,L^{-1}}\right)} \]

Explanation: Statement I is correct because zero-order disappearance follows \[ [\mathrm X]_t=[\mathrm X]_0-k_0t. \] If the constant rate remains valid until the reactant is exhausted, the predicted exhaustion time is \[ t_{\mathrm{exhaustion}}=\frac{[\mathrm X]_0}{k_0}. \] This time is finite for positive initial concentration and positive \(k_0\). Statement II is incorrect because the first-order integrated equation is \[ [\mathrm Y]_t=[\mathrm Y]_0e^{-k_1t}. \] For finite positive k and finite elapsed time, the exponential factor remains positive. Therefore, the ideal first-order equation approaches zero concentration asymptotically without reaching exactly zero at a finite time. Statement III is correct because the concentration consumed by a first-order reaction in a fixed time t is \[ \Delta C=C_0(1-e^{-k_1t}). \] At unchanged k and t, the fractional conversion is unchanged. Doubling \(C_0\) therefore doubles the absolute concentration consumed. Statement IV is correct because the zero-order time to achieve a conversion fraction x is \[ t=\frac{xC_0}{k_0}. \] For a fixed conversion fraction and unchanged \(k_0\), this time is directly proportional to initial concentration. Statement V is incorrect because the zero-order disappearance rate is constant. The first half-life consumes \[ \frac{C_0}{2} \] of reactant. The subsequent decrease from \(C_0/2\) to \(C_0/4\) consumes only \[ \frac{C_0}{4}. \] At the same constant disappearance rate, this second interval takes half as long as the first. Thus, successive zero-order concentration halvings require progressively shorter times. The statements distinguish complete exhaustion, absolute consumption, fractional conversion and successive half-life intervals. The correct statements are I, III and IV only.

Explanation: The catalyst surface is the same in both experiments, and its measured activity is reported per unit surface area. Therefore, the total molar disappearance rate of A is the same in both vessels. However, the concentration-based disappearance rate is obtained by dividing the total molar rate by the solution volume. Consequently, the two vessels have different concentration-based zero-order constants despite having the same active catalyst surface. The first half-life also depends on the initial concentration and the applicable concentration-based constant. \(\textbf{Determine the total molar disappearance rate:}\) The surface activity is \[ \sigma=0.0400\ \mathrm{mol\,m^{-2}\,min^{-1}} \] The active surface area is \[ S=0.500\ \mathrm{m^2} \] Therefore, \[ -\frac{dn_{\mathrm A}}{dt}=\sigma S \] Substituting, \[ -\frac{dn_{\mathrm A}}{dt}=0.0400(0.500) \] \[ -\frac{dn_{\mathrm A}}{dt}=0.0200\ \mathrm{mol\,min^{-1}} \] This total molar disappearance rate applies to both vessels. \(\textbf{Determine the zero-order constant in Vessel I:}\) For a constant solution volume, \[ -\frac{d[\mathrm A]}{dt}=-\frac1V\frac{dn_{\mathrm A}}{dt} \] Therefore, \[ k_{\mathrm I}=\frac{0.0200}{1.00} \] \[ k_{\mathrm I}=0.0200\ \mathrm{mol\,L^{-1}\,min^{-1}} \] \(\textbf{Calculate the first half-life in Vessel I:}\) For zero-order disappearance, \[ t_{1/2}=\frac{[\mathrm A]_0}{2k} \] Therefore, \[ t_{1/2,\mathrm I}=\frac{0.300}{2(0.0200)} \] \[ t_{1/2,\mathrm I}=\frac{0.300}{0.0400} \] \[ t_{1/2,\mathrm I}=7.50\ \mathrm{min} \] \(\textbf{Determine the zero-order constant in Vessel II:}\) Vessel II has twice the solution volume but the same total active catalytic surface. Therefore, \[ k_{\mathrm{II}}=\frac{0.0200}{2.00} \] \[ k_{\mathrm{II}}=0.0100\ \mathrm{mol\,L^{-1}\,min^{-1}} \] \(\textbf{Check the corresponding half-life in Vessel II:}\) \[ t_{1/2,\mathrm{II}}=\frac{0.300}{2(0.0100)} \] \[ t_{1/2,\mathrm{II}}=15.0\ \mathrm{min} \] This is twice the half-life in Vessel I. \(\textbf{Verify using the initial amounts:}\) Initially, Vessel I contains \[ n_{\mathrm A,I}=0.300(1.00) \] \[ n_{\mathrm A,I}=0.300\ \mathrm{mol} \] Vessel II contains \[ n_{\mathrm A,II}=0.300(2.00) \] \[ n_{\mathrm A,II}=0.600\ \mathrm{mol} \] Vessel II therefore contains twice as much A, while the two vessels consume equal molar amounts per minute. Its first half-life must consequently be twice as long. \(\textbf{Chemical interpretation:}\) Zero-order behaviour refers to the independence of the disappearance rate from reactant concentration within a specified kinetic regime. For a surface-controlled reaction, the measured concentration-based constant can also depend on the ratio of active catalyst area to solution volume. Equal catalyst surfaces do not necessarily give equal disappearance rates per litre when the solution volumes differ. The distinction between total molar rate and concentration-normalized rate is essential for comparing heterogeneous catalytic experiments. \(\textbf{Final answer:}\) \[ \boxed{\left(t_{1/2,\mathrm I},k_{\mathrm{II}}\right)=\left(7.50\ \mathrm{min},0.0100\ \mathrm{mol\,L^{-1}\,min^{-1}}\right)} \]

Explanation: The instrument measures instantaneous rates rather than concentrations. For a first-order reaction, these rates are proportional to the remaining concentration of A. Therefore, the rate itself decreases exponentially when the intrinsic constant remains unchanged. The confirmed measurements show a repeated halving of rate over equal time intervals. This determines the first-order half-life and permits the incorrect fourth measurement to be corrected. The initial concentration can then be recovered from the differential rate law. \(\textbf{Identify the rate-decay pattern:}\) From 0 to 10 min, \[ \frac{r_{10}}{r_0}=\frac{0.01386}{0.02772} \] \[ \frac{r_{10}}{r_0}=\frac12 \] From 10 to 20 min, \[ \frac{r_{20}}{r_{10}}=\frac{0.00693}{0.01386} \] \[ \frac{r_{20}}{r_{10}}=\frac12 \] Thus, the rate halves during each 10-minute interval. \(\textbf{Determine the first-order constant:}\) The rate and concentration have the same exponential dependence. Therefore, \[ t_{1/2}=10.0\ \mathrm{min} \] Using \[ k=\frac{\ln2}{t_{1/2}}, \] \[ k=\frac{0.693}{10.0} \] \[ k=0.0693\ \mathrm{min^{-1}} \] \(\textbf{Correct the fourth instantaneous rate:}\) Another 10-minute interval elapses between 20 and 30 min. Therefore, \[ r_{30}=r_{20}\left(\frac12\right) \] \[ r_{30}=0.00693\left(\frac12\right) \] \[ r_{30}=0.003465\ \mathrm{mol\,L^{-1}\,min^{-1}} \] The reported value \(0.00693\) is inconsistent with the established first-order decay pattern. \(\textbf{Recover the initial concentration:}\) The first-order differential rate law gives \[ r_0=k[\mathrm A]_0 \] Therefore, \[ [\mathrm A]_0=\frac{r_0}{k} \] \[ [\mathrm A]_0=\frac{0.02772}{0.0693} \] \[ [\mathrm A]_0=0.400\ \mathrm{mol\,L^{-1}} \] \(\textbf{Determine A remaining after 30 min:}\) Thirty minutes corresponds to three half-lives. Therefore, \[ [\mathrm A]_{30}=0.400\left(\frac12\right)^3 \] \[ [\mathrm A]_{30}=0.400\left(\frac18\right) \] \[ [\mathrm A]_{30}=0.0500\ \mathrm{mol\,L^{-1}} \] \(\textbf{Calculate cumulative disappearance:}\) \[ \Delta[\mathrm A]_{\mathrm{consumed}}=[\mathrm A]_0-[\mathrm A]_{30} \] \[ \Delta[\mathrm A]_{\mathrm{consumed}}=0.400-0.0500 \] \[ \Delta[\mathrm A]_{\mathrm{consumed}}=0.350\ \mathrm{mol\,L^{-1}} \] \(\textbf{Verify the corrected rate directly:}\) \[ r_{30}=k[\mathrm A]_{30} \] \[ r_{30}=0.0693(0.0500) \] \[ r_{30}=0.003465\ \mathrm{mol\,L^{-1}\,min^{-1}} \] This agrees with the rate predicted from successive halvings. \(\textbf{Experimental interpretation:}\) For first-order kinetics, an instantaneous disappearance-rate measurement contains information about the remaining reactant concentration. A series of such measurements can test the consistency of the kinetic law even when concentrations are not reported directly. The rate at 30 min and the total concentration consumed over 30 min are distinct quantities. The first comes from the differential rate law; the second comes from the integrated concentration history. \(\textbf{Final answer:}\) \[ \boxed{\left(r_{30},\Delta[\mathrm A]_{\mathrm{consumed}}\right)=\left(0.003465\ \mathrm{mol\,L^{-1}\,min^{-1}},0.350\ \mathrm{mol\,L^{-1}}\right)} \]

Explanation: The concentrations of the four species must change in the proportions fixed by the balanced chemical equation. The absolute concentration changes are not necessarily equal because their stoichiometric coefficients differ. A normalized reaction rate is obtained by dividing each concentration-change rate by the appropriate coefficient. The three verified measurements allow the reaction extent to be determined independently. This extent can then be used to correct the inconsistent measurement of D. \(\textbf{Calculate the verified reactant consumption:}\) For A, \[ -\Delta[\mathrm A]=0.900-0.700 \] \[ -\Delta[\mathrm A]=0.200\ \mathrm{mol\,L^{-1}} \] For B, \[ -\Delta[\mathrm B]=1.200-0.900 \] \[ -\Delta[\mathrm B]=0.300\ \mathrm{mol\,L^{-1}} \] \(\textbf{Normalize the concentration changes:}\) The balanced equation requires \[ -\frac{\Delta[\mathrm A]}2=-\frac{\Delta[\mathrm B]}3=\frac{\Delta[\mathrm C]}4=\Delta[\mathrm D] \] The concentration-based reaction extent is therefore \[ \xi=\frac{0.200}{2} \] \[ \xi=0.100\ \mathrm{mol\,L^{-1}} \] Using B gives \[ \xi=\frac{0.300}{3} \] \[ \xi=0.100\ \mathrm{mol\,L^{-1}} \] \(\textbf{Verify the product C measurement:}\) The increase in C is \[ \Delta[\mathrm C]=0.500-0.100 \] \[ \Delta[\mathrm C]=0.400\ \mathrm{mol\,L^{-1}} \] Its normalized extent is \[ \frac{\Delta[\mathrm C]}4=\frac{0.400}{4} \] \[ \frac{\Delta[\mathrm C]}4=0.100\ \mathrm{mol\,L^{-1}} \] Thus, A, B and C are mutually consistent. \(\textbf{Correct the final concentration of D:}\) The coefficient of D is one. Consequently, \[ \Delta[\mathrm D]=\xi \] \[ \Delta[\mathrm D]=0.100\ \mathrm{mol\,L^{-1}} \] Since \[ [\mathrm D]_0=0.050\ \mathrm{mol\,L^{-1}}, \] \[ [\mathrm D]_{5}=0.050+0.100 \] \[ [\mathrm D]_{5}=0.150\ \mathrm{mol\,L^{-1}} \] \(\textbf{Determine the average normalized reaction rate:}\) The elapsed time is \[ \Delta t=5.0\ \mathrm{min} \] Therefore, \[ r_{\mathrm{avg}}=\frac{\xi}{\Delta t} \] \[ r_{\mathrm{avg}}=\frac{0.100}{5.0} \] \[ r_{\mathrm{avg}}=0.0200\ \mathrm{mol\,L^{-1}\,min^{-1}} \] \(\textbf{Experimental interpretation:}\) The incorrect D reading implies an increase of \(0.140\ \mathrm{mol\,L^{-1}}\), which violates the stoichiometric balance established by the other three species. The discrepancy cannot be explained by using a different disappearance rate for one species. All species share the same normalized reaction rate for the specified balanced reaction. The measured concentration changes differ only through their stoichiometric coefficients. \(\textbf{Final answer:}\) \[ \boxed{\left([\mathrm D]_{5},r_{\mathrm{avg}}\right)=\left(0.150\ \mathrm{mol\,L^{-1}},0.0200\ \mathrm{mol\,L^{-1}\,min^{-1}}\right)} \]

Explanation: The balanced equation determines the chemical rates of formation and consumption of the participating species. However, the measured increase in gaseous water concentration is not necessarily equal to the chemical rate of water formation. Some water is being transferred from the gaseous phase into the liquid phase. The difference between chemical formation and net vapour accumulation determines the condensation rate. The oxygen disappearance rate follows independently from the stoichiometric relationship with NO formation. \(\textbf{Normalize the measured NO formation rate:}\) For the balanced reaction, \[ r=\frac14\frac{d[\mathrm{NO}]}{dt} \] Substituting, \[ r=\frac{0.0800}{4} \] \[ r=0.0200\ \mathrm{mol\,L^{-1}\,min^{-1}} \] \(\textbf{Determine the chemical formation rate of water:}\) Water has stoichiometric coefficient six. Therefore, \[ r_{\mathrm{H_2O,formed}}=6r \] \[ r_{\mathrm{H_2O,formed}}=6(0.0200) \] \[ r_{\mathrm{H_2O,formed}}=0.1200\ \mathrm{mol\,L^{-1}\,min^{-1}} \] \(\textbf{Distinguish formation from vapour accumulation:}\) The measured net accumulation of water vapour is \[ r_{\mathrm{vapour,net}}=0.0150\ \mathrm{mol\,L^{-1}\,min^{-1}} \] The gaseous-phase material balance is \[ r_{\mathrm{vapour,net}}=r_{\mathrm{H_2O,formed}}-r_{\mathrm{condensation}} \] Rearranging, \[ r_{\mathrm{condensation}}=r_{\mathrm{H_2O,formed}}-r_{\mathrm{vapour,net}} \] Substituting, \[ r_{\mathrm{condensation}}=0.1200-0.0150 \] \[ r_{\mathrm{condensation}}=0.1050\ \mathrm{mol\,L^{-1}\,min^{-1}} \] \(\textbf{Determine the oxygen disappearance rate:}\) The normalized chemical rate satisfies \[ r=-\frac15\frac{d[\mathrm{O_2}]}{dt} \] Therefore, \[ -\frac{d[\mathrm{O_2}]}{dt}=5r \] \[ -\frac{d[\mathrm{O_2}]}{dt}=5(0.0200) \] \[ -\frac{d[\mathrm{O_2}]}{dt}=0.1000\ \mathrm{mol\,L^{-1}\,min^{-1}} \] \(\textbf{Check the reaction-rate consistency:}\) The positive disappearance rate of ammonia is \[ -\frac{d[\mathrm{NH_3}]}{dt}=4r \] \[ -\frac{d[\mathrm{NH_3}]}{dt}=0.0800\ \mathrm{mol\,L^{-1}\,min^{-1}} \] This matches the stoichiometric \(1:1\) relationship between ammonia consumed and NO formed. \(\textbf{Chemical interpretation:}\) A measured species concentration may reflect both chemical reaction and physical transfer between phases. The stoichiometric rate relation refers to the chemical formation of water, including water that subsequently condenses. It does not apply directly to the net accumulation of water vapour when condensation occurs. A complete material balance is required to separate these processes. \(\textbf{Final answer:}\) \[ \boxed{\left(r_{\mathrm{condensation}},-\frac{d[\mathrm{O_2}]}{dt}\right)=\left(0.1050,\ 0.1000\right)\ \mathrm{mol\,L^{-1}\,min^{-1}}} \]

Explanation: The measured total-pressure slope is not the disappearance rate of dinitrogen pentoxide. A reaction event consumes two gaseous reactant molecules and produces five gaseous product molecules. Consequently, each reaction event increases the total gas amount by three molecules. The total-pressure change must first be normalized by this net gaseous stoichiometric change. The species-specific partial-pressure rates then follow from their coefficients. \(\textbf{Determine the net gaseous change:}\) The balanced reaction is \[ 2\mathrm{N_2O_5}\rightarrow4\mathrm{NO_2}+\mathrm{O_2} \] The total number of gaseous product molecules per reaction event is \[ 4+1=5 \] The number of reactant molecules consumed is two. Therefore, \[ \Delta\nu_{\mathrm{gas}}=5-2 \] \[ \Delta\nu_{\mathrm{gas}}=3 \] \(\textbf{Connect total pressure with reaction progress:}\) Let \(\dot{\xi}_p\) represent the normalized reaction-progress rate in pressure units. At constant temperature and volume, \[ \frac{dP_{\mathrm{total}}}{dt}=\Delta\nu_{\mathrm{gas}}\dot{\xi}_p \] Therefore, \[ 12.0=3\dot{\xi}_p \] \[ \dot{\xi}_p=4.0\ \mathrm{torr\,min^{-1}} \] \(\textbf{Determine the nitrogen dioxide appearance rate:}\) The coefficient of \(\mathrm{NO_2}\) is four. Therefore, \[ \frac{dp_{\mathrm{NO_2}}}{dt}=4\dot{\xi}_p \] \[ \frac{dp_{\mathrm{NO_2}}}{dt}=4(4.0) \] \[ \frac{dp_{\mathrm{NO_2}}}{dt}=16.0\ \mathrm{torr\,min^{-1}} \] \(\textbf{Determine the reactant disappearance rate:}\) The coefficient of \(\mathrm{N_2O_5}\) is two. Therefore, \[ -\frac{dp_{\mathrm{N_2O_5}}}{dt}=2\dot{\xi}_p \] \[ -\frac{dp_{\mathrm{N_2O_5}}}{dt}=2(4.0) \] \[ -\frac{dp_{\mathrm{N_2O_5}}}{dt}=8.0\ \mathrm{torr\,min^{-1}} \] \(\textbf{Verify the total-pressure slope:}\) The oxygen appearance rate is \[ \frac{dp_{\mathrm{O_2}}}{dt}=4.0\ \mathrm{torr\,min^{-1}} \] Adding the partial-pressure derivatives, \[ \frac{dP_{\mathrm{total}}}{dt}=-8.0+16.0+4.0 \] \[ \frac{dP_{\mathrm{total}}}{dt}=+12.0\ \mathrm{torr\,min^{-1}} \] This reproduces the measured total-pressure slope. \(\textbf{Chemical interpretation:}\) The total-pressure derivative represents the net rate of change of gaseous amount. Individual reactant and product derivatives need not equal that total-pressure derivative. Their relationship depends on the net gaseous stoichiometric coefficient. An inert gas already present in a rigid vessel contributes to total pressure but not to its chemical rate of change. \(\textbf{Final answer:}\) \[ \boxed{\left(\frac{dp_{\mathrm{NO_2}}}{dt},-\frac{dp_{\mathrm{N_2O_5}}}{dt}\right)=\left(16.0,\ 8.0\right)\ \mathrm{torr\,min^{-1}}} \]

Explanation: Both A and B are consumed, but B disappears twice as rapidly as A because of its stoichiometric coefficient. Consequently, the difference \([\mathrm A]-[\mathrm B]\) can increase even while both individual concentrations decrease. The measured increase in this difference is not evidence that A is being formed. The difference signal can instead be related directly to the normalized reaction extent. Once that extent is recovered, the rate and product concentration follow. \(\textbf{Represent the concentration changes:}\) Let \(\xi\) be the concentration-based extent of reaction over the observation interval. For \[ \mathrm{A+2B\rightarrow3P}, \] \[ \Delta[\mathrm A]=-\xi \] \[ \Delta[\mathrm B]=-2\xi \] \[ \Delta[\mathrm P]=+3\xi \] \(\textbf{Relate the difference signal to reaction extent:}\) The signal is \[ S=[\mathrm A]-[\mathrm B] \] Therefore, \[ \Delta S=\Delta[\mathrm A]-\Delta[\mathrm B] \] Substituting the stoichiometric changes, \[ \Delta S=-\xi-(-2\xi) \] \[ \Delta S=\xi \] \(\textbf{Use the observed signal change:}\) \[ \Delta S=S_f-S_i \] \[ \Delta S=-0.040-(-0.100) \] \[ \Delta S=+0.060\ \mathrm{mol\,L^{-1}} \] Thus, \[ \xi=0.060\ \mathrm{mol\,L^{-1}} \] \(\textbf{Calculate the average normalized rate:}\) \[ r_{\mathrm{avg}}=\frac{\xi}{\Delta t} \] Substituting, \[ r_{\mathrm{avg}}=\frac{0.060}{6.0} \] \[ r_{\mathrm{avg}}=0.0100\ \mathrm{mol\,L^{-1}\,min^{-1}} \] \(\textbf{Determine product formation:}\) The coefficient of P is three. Therefore, \[ \Delta[\mathrm P]=3(0.060) \] \[ \Delta[\mathrm P]=0.180\ \mathrm{mol\,L^{-1}} \] Including initially present P, \[ [\mathrm P]_f=0.100+0.180 \] \[ [\mathrm P]_f=0.280\ \mathrm{mol\,L^{-1}} \] \(\textbf{Verify the final reactant concentrations:}\) For A, \[ [\mathrm A]_f=0.500-0.060 \] \[ [\mathrm A]_f=0.440\ \mathrm{mol\,L^{-1}} \] For B, \[ [\mathrm B]_f=0.600-2(0.060) \] \[ [\mathrm B]_f=0.480\ \mathrm{mol\,L^{-1}} \] The final difference is \[ [\mathrm A]_f-[\mathrm B]_f=0.440-0.480 \] \[ [\mathrm A]_f-[\mathrm B]_f=-0.040\ \mathrm{mol\,L^{-1}} \] This matches the final signal. \(\textbf{Chemical interpretation:}\) A derived experimental signal may increase even when every reactant in the expression is being consumed. Its direction of change depends on the stoichiometric coefficients and the signs used in constructing the signal. The normalized reaction extent provides a consistent way to interpret such measurements. \(\textbf{Final answer:}\) \[ \boxed{\left(r_{\mathrm{avg}},[\mathrm P]_f\right)=\left(0.0100\ \mathrm{mol\,L^{-1}\,min^{-1}},\ 0.280\ \mathrm{mol\,L^{-1}}\right)} \]

Explanation: The first vessel has already undergone one half-life before the solutions are combined. The second vessel contributes additional unreacted A and initially present P. Mixing conserves the moles of each species but changes their concentrations because the total volume increases. The intrinsic first-order rate constant remains unchanged. Therefore, the instantaneous rate changes at mixing according to the change in A concentration, and the subsequent reaction follows the same half-life. \(\textbf{Determine the contents of Vessel I before mixing:}\) After one half-life, \[ [\mathrm A]_{\mathrm{I},10}=0.80\left(\frac12\right) \] \[ [\mathrm A]_{\mathrm{I},10}=0.40\ \mathrm{mol\,L^{-1}} \] The concentration of P produced in Vessel I is \[ [\mathrm P]_{\mathrm{I},10}=0.80-0.40 \] \[ [\mathrm P]_{\mathrm{I},10}=0.40\ \mathrm{mol\,L^{-1}} \] Because its volume is \(1.00\ \mathrm L\), \[ n_{\mathrm{A,I}}=0.40\ \mathrm{mol} \] \[ n_{\mathrm{P,I}}=0.40\ \mathrm{mol} \] \(\textbf{Determine the amounts supplied by Vessel II:}\) For A, \[ n_{\mathrm{A,II}}=0.20(2.00) \] \[ n_{\mathrm{A,II}}=0.40\ \mathrm{mol} \] For P, \[ n_{\mathrm{P,II}}=0.10(2.00) \] \[ n_{\mathrm{P,II}}=0.20\ \mathrm{mol} \] \(\textbf{Determine the concentrations immediately after mixing:}\) The combined volume is \[ V=1.00+2.00 \] \[ V=3.00\ \mathrm L \] The total amount of A is \[ n_{\mathrm A}=0.40+0.40 \] \[ n_{\mathrm A}=0.80\ \mathrm{mol} \] Thus, \[ [\mathrm A]_{10^+}=\frac{0.80}{3.00} \] \[ [\mathrm A]_{10^+}=\frac4{15}\ \mathrm{mol\,L^{-1}} \] The total amount of P immediately after mixing is \[ n_{\mathrm P}=0.40+0.20 \] \[ n_{\mathrm P}=0.60\ \mathrm{mol} \] \(\textbf{Compare the instantaneous rates across mixing:}\) The first-order disappearance rate is \[ r=k[\mathrm A] \] Immediately before mixing, \[ [\mathrm A]_{10^-}=0.40\ \mathrm{mol\,L^{-1}} \] Immediately after mixing, \[ [\mathrm A]_{10^+}=\frac4{15}\ \mathrm{mol\,L^{-1}} \] Therefore, \[ \frac{r_{10^+}}{r_{10^-}}=\frac{(4/15)}{0.40} \] \[ \frac{r_{10^+}}{r_{10^-}}=\frac23 \] \(\textbf{Apply another half-life after mixing:}\) The mixed solution reacts for 10 min. Therefore, half its A is consumed. The amount of A consumed is \[ \Delta n_{\mathrm A,\mathrm{consumed}}=\frac{0.80}{2} \] \[ \Delta n_{\mathrm A,\mathrm{consumed}}=0.40\ \mathrm{mol} \] For \[ \mathrm A\rightarrow\mathrm P, \] the amount of additional P formed is \(0.40\ \mathrm{mol}\). \(\textbf{Determine the final product concentration:}\) The final amount of P is \[ n_{\mathrm P,f}=0.60+0.40 \] \[ n_{\mathrm P,f}=1.00\ \mathrm{mol} \] Therefore, \[ [\mathrm P]_f=\frac{1.00}{3.00} \] \[ [\mathrm P]_f=0.333\ \mathrm{mol\,L^{-1}} \] \(\textbf{Chemical interpretation:}\) Mixing requires a mole balance before any concentration-based rate law is applied. A first-order half-life remains unchanged at the same temperature, but an instantaneous concentration change alters the instantaneous rate. Initially present product must also be included when determining the final accumulated product concentration. \(\textbf{Final answer:}\) \[ \boxed{\left([\mathrm P]_f,\frac{r_{10^+}}{r_{10^-}}\right)=\left(0.333\ \mathrm{mol\,L^{-1}},\frac23\right)} \]

Explanation: Neither optical signal measures the concentration of A alone. Each combines contributions from reactant and product. Therefore, the individual concentrations must first be recovered from the simultaneous signal equations. The first-order half-life can then be determined from the concentration measurements at the two times. Finally, the remaining reactant and accumulated product concentrations predict the later signal. \(\textbf{Recover the initial concentrations:}\) At the initial instant, \[ 2[\mathrm A]_0+[\mathrm P]_0=1.50 \] and \[ [\mathrm A]_0+2[\mathrm P]_0=1.20 \] Multiplying the second equation by two, \[ 2[\mathrm A]_0+4[\mathrm P]_0=2.40 \] Subtracting the first equation, \[ 3[\mathrm P]_0=0.90 \] \[ [\mathrm P]_0=0.300\ \mathrm{mol\,L^{-1}} \] Substituting, \[ 2[\mathrm A]_0+0.300=1.50 \] \[ [\mathrm A]_0=0.600\ \mathrm{mol\,L^{-1}} \] \(\textbf{Recover the concentrations after 8 min:}\) The signal equations become \[ 2[\mathrm A]_8+[\mathrm P]_8=1.20 \] and \[ [\mathrm A]_8+2[\mathrm P]_8=1.50 \] Solving, \[ [\mathrm A]_8=0.300\ \mathrm{mol\,L^{-1}} \] \[ [\mathrm P]_8=0.600\ \mathrm{mol\,L^{-1}} \] \(\textbf{Determine the first-order half-life:}\) The remaining reactant fraction after 8 min is \[ f_8=\frac{0.300}{0.600} \] \[ f_8=\frac12 \] Therefore, \[ t_{1/2}=8.0\ \mathrm{min} \] \(\textbf{Calculate A remaining after 24 min:}\) Twenty-four minutes corresponds to three half-lives. Therefore, \[ [\mathrm A]_{24}=0.600\left(\frac12\right)^3 \] \[ [\mathrm A]_{24}=0.0750\ \mathrm{mol\,L^{-1}} \] \(\textbf{Apply the reaction material balance:}\) The initially present total concentration of A and P is \[ [\mathrm A]_0+[\mathrm P]_0=0.600+0.300 \] \[ [\mathrm A]_0+[\mathrm P]_0=0.900\ \mathrm{mol\,L^{-1}} \] Because \[ \mathrm A\rightarrow\mathrm P \] and P is stable, \[ [\mathrm A]_t+[\mathrm P]_t=0.900\ \mathrm{mol\,L^{-1}} \] Therefore, \[ [\mathrm P]_{24}=0.900-0.0750 \] \[ [\mathrm P]_{24}=0.825\ \mathrm{mol\,L^{-1}} \] \(\textbf{Predict the later optical signal:}\) \[ S_{1,24}=2[\mathrm A]_{24}+[\mathrm P]_{24} \] \[ S_{1,24}=2(0.0750)+0.825 \] \[ S_{1,24}=0.975 \] \(\textbf{Experimental interpretation:}\) Two calibrated signals containing different contributions from A and P can identify both concentrations even when neither signal measures a single species directly. The reconstructed concentration history establishes the first-order half-life. The chemical material balance then provides a consistency check and supports prediction at a later time. \(\textbf{Final answer:}\) \[ \boxed{\left([\mathrm A]_0,S_{1,24}\right)=\left(0.600\ \mathrm{mol\,L^{-1}},0.975\right)} \]

Explanation: The total appearance rate of P is the sum of its formation rates through the two independent pathways. Initially, both reactions contribute to product formation. When A becomes exhausted, only the B pathway continues. Therefore, the change in product-appearance rate provides information about the individual zero-order constants. The product concentration must be calculated using the appropriate rate in each time interval. \(\textbf{Identify the remaining pathway rate:}\) After A is exhausted, only B forms P. Therefore, \[ k_B=0.010\ \mathrm{mol\,L^{-1}\,min^{-1}} \] \(\textbf{Determine the initial contribution from A:}\) While both reactants remain available, \[ \frac{d[\mathrm P]}{dt}=k_A+k_B \] Substituting, \[ 0.040=k_A+0.010 \] Therefore, \[ k_A=0.030\ \mathrm{mol\,L^{-1}\,min^{-1}} \] \(\textbf{Verify the exhaustion time of A:}\) For zero-order disappearance, \[ t_{\mathrm{exhaustion,A}}=\frac{[\mathrm A]_0}{k_A} \] \[ t_{\mathrm{exhaustion,A}}=\frac{0.180}{0.030} \] \[ t_{\mathrm{exhaustion,A}}=6.0\ \mathrm{min} \] This agrees with the observed rate change. \(\textbf{Determine the product concentration after 6 min:}\) During the first six minutes, \[ \Delta[\mathrm P]_1=0.040(6) \] \[ \Delta[\mathrm P]_1=0.240\ \mathrm{mol\,L^{-1}} \] Therefore, \[ [\mathrm P]_6=0.050+0.240 \] \[ [\mathrm P]_6=0.290\ \mathrm{mol\,L^{-1}} \] \(\textbf{Determine the remaining time to reach the target:}\) The target concentration is \[ 0.400\ \mathrm{mol\,L^{-1}} \] The additional product required after 6 min is \[ 0.400-0.290=0.110\ \mathrm{mol\,L^{-1}} \] Only B continues reacting, at the constant rate \[ 0.010\ \mathrm{mol\,L^{-1}\,min^{-1}} \] Therefore, \[ \Delta t=\frac{0.110}{0.010} \] \[ \Delta t=11.0\ \mathrm{min} \] \(\textbf{Determine the target time:}\) \[ t_{\mathrm{target}}=6.0+11.0 \] \[ t_{\mathrm{target}}=17.0\ \mathrm{min} \] \(\textbf{Check that B remains available:}\) The exhaustion time of B is \[ t_{\mathrm{exhaustion,B}}=\frac{0.240}{0.010} \] \[ t_{\mathrm{exhaustion,B}}=24.0\ \mathrm{min} \] Thus, B remains available at 17 min, validating the second-stage calculation. \(\textbf{Determine the limiting product concentration:}\) Complete conversion produces one mole of P per mole of A and one mole of P per mole of B. Therefore, \[ [\mathrm P]_{\infty}=[\mathrm P]_0+[\mathrm A]_0+[\mathrm B]_0 \] \[ [\mathrm P]_{\infty}=0.050+0.180+0.240 \] \[ [\mathrm P]_{\infty}=0.470\ \mathrm{mol\,L^{-1}} \] \(\textbf{Chemical interpretation:}\) Independent zero-order product-forming pathways can produce a piecewise-constant total appearance rate. A downward change in this rate can occur when one reactant becomes exhausted. The remaining pathway may continue forming product even after the first pathway stops. The final product concentration depends on the total initial reactant amounts, whereas the time required to reach a specified concentration depends on their separate kinetic constants. \(\textbf{Final answer:}\) \[ \boxed{\left(t_{\mathrm{target}},[\mathrm P]_{\infty}\right)=\left(17.0\ \mathrm{min},0.470\ \mathrm{mol\,L^{-1}}\right)} \]

Explanation: Removing a well-mixed portion reduces the total amounts of both A and P in the reacting vessel. However, it does not change their concentrations in the solution that remains. The first-order half-life also remains unchanged because temperature and kinetic conditions are unchanged. To calculate cumulative product formation, the product carried away in the withdrawn sample must be included. Initially present P must then be subtracted because it was not produced during the experiment. \(\textbf{Determine the amounts initially present:}\) The initial volume is \[ V_0=2.00\ \mathrm L \] Therefore, \[ n_{\mathrm A,0}=0.800(2.00) \] \[ n_{\mathrm A,0}=1.600\ \mathrm{mol} \] Initially present product is \[ n_{\mathrm P,0}=0.100(2.00) \] \[ n_{\mathrm P,0}=0.200\ \mathrm{mol} \] \(\textbf{Determine the composition after the first half-life:}\) At 10 min, \[ [\mathrm A]_{10}=0.800\left(\frac12\right) \] \[ [\mathrm A]_{10}=0.400\ \mathrm{mol\,L^{-1}} \] The amount of A consumed during the first interval is \[ 1.600-0.800=0.800\ \mathrm{mol} \] Therefore, the amount of P present immediately before withdrawal is \[ n_{\mathrm P,10^-}=0.200+0.800 \] \[ n_{\mathrm P,10^-}=1.000\ \mathrm{mol} \] Its concentration is \[ [\mathrm P]_{10^-}=\frac{1.000}{2.00} \] \[ [\mathrm P]_{10^-}=0.500\ \mathrm{mol\,L^{-1}} \] \(\textbf{Account for the withdrawal:}\) Half the solution is withdrawn. Therefore, the withdrawn sample contains \[ n_{\mathrm A,\mathrm{withdrawn}}=0.400(1.00) \] \[ n_{\mathrm A,\mathrm{withdrawn}}=0.400\ \mathrm{mol} \] and \[ n_{\mathrm P,\mathrm{withdrawn}}=0.500(1.00) \] \[ n_{\mathrm P,\mathrm{withdrawn}}=0.500\ \mathrm{mol} \] The retained solution has the same concentrations as immediately before withdrawal. \(\textbf{Apply the second half-life to the retained solution:}\) The retained A concentration initially equals \[ 0.400\ \mathrm{mol\,L^{-1}} \] After another 10 min, \[ [\mathrm A]_{20}=0.400\left(\frac12\right) \] \[ [\mathrm A]_{20}=0.200\ \mathrm{mol\,L^{-1}} \] The amount of A consumed in the retained \(1.00\ \mathrm L\) during the second interval is \[ \Delta n_{\mathrm A,2}=0.400-0.200 \] \[ \Delta n_{\mathrm A,2}=0.200\ \mathrm{mol} \] \(\textbf{Determine the final product amount in the vessel:}\) Immediately after withdrawal, the retained solution contains \[ n_{\mathrm P,10^+}=0.500\ \mathrm{mol} \] The second interval forms another \(0.200\ \mathrm{mol}\) of P. Therefore, \[ n_{\mathrm P,20}=0.500+0.200 \] \[ n_{\mathrm P,20}=0.700\ \mathrm{mol} \] \(\textbf{Calculate cumulative product formation:}\) The total amount of P present in the withdrawn and retained portions at 20 min is \[ 0.500+0.700=1.200\ \mathrm{mol} \] Subtract the amount of P initially present: \[ n_{\mathrm P,\mathrm{formed}}=1.200-0.200 \] \[ n_{\mathrm P,\mathrm{formed}}=1.000\ \mathrm{mol} \] \(\textbf{Verify using reactant consumption:}\) A consumed during the first interval is \(0.800\ \mathrm{mol}\). A consumed during the second interval is \(0.200\ \mathrm{mol}\). Therefore, \[ n_{\mathrm A,\mathrm{consumed}}=0.800+0.200 \] \[ n_{\mathrm A,\mathrm{consumed}}=1.000\ \mathrm{mol} \] This agrees with the cumulative formation of P. \(\textbf{Chemical interpretation:}\) Removing part of a well-mixed solution changes the total amount of material but not the concentrations in the remaining liquid. A first-order concentration half-life therefore remains unchanged. However, an overall product-yield calculation must account for the withdrawn material and any product present initially. The product concentration in the final vessel alone does not represent all product generated by the experiment. \(\textbf{Final answer:}\) \[ \boxed{\left([\mathrm A]_{20},n_{\mathrm P,\mathrm{formed}}\right)=\left(0.200\ \mathrm{mol\,L^{-1}},1.000\ \mathrm{mol}\right)} \]