Chemical Kinetics MCQs with Answers – Part 2 (NEET Chemistry)

Explanation: The units of a second-order rate constant contain inverse concentration and inverse time. Therefore, changing both the concentration unit and the time unit affects its numerical value. The conversions must be applied consistently to the rate constant and the subsequent rate calculation. \(\textbf{Convert the time unit:}\) The original constant is \[ k=0.60\ \mathrm{L\,mol^{-1}\,min^{-1}} \] Because one minute contains 60 seconds, \[ k=\frac{0.60}{60}\ \mathrm{L\,mol^{-1}\,s^{-1}} \] \[ k=1.0\times10^{-2}\ \mathrm{L\,mol^{-1}\,s^{-1}} \] \(\textbf{Convert the inverse concentration unit:}\) One mole contains \(1000\) millimoles. Therefore, \[ 1\ \mathrm{mol^{-1}}=\frac1{1000}\ \mathrm{mmol^{-1}} \] Substituting, \[ k=\frac{1.0\times10^{-2}}{1000}\ \mathrm{L\,mmol^{-1}\,s^{-1}} \] \[ k=1.0\times10^{-5}\ \mathrm{L\,mmol^{-1}\,s^{-1}} \] \(\textbf{Calculate the rate using the new units:}\) The reactant concentration is \[ [\mathrm A]=200\ \mathrm{mmol\,L^{-1}} \] The rate law is \[ r=k[\mathrm A]^2 \] Therefore, \[ r=(1.0\times10^{-5})(200)^2 \] \[ r=(1.0\times10^{-5})(40000) \] \[ r=0.40\ \mathrm{mmol\,L^{-1}\,s^{-1}} \] \(\textbf{Independent verification in molar units:}\) The same concentration is \[ [\mathrm A]=0.200\ \mathrm{mol\,L^{-1}} \] Using the rate constant in molar units, \[ r=(1.0\times10^{-2})(0.200)^2 \] \[ r=4.0\times10^{-4}\ \mathrm{mol\,L^{-1}\,s^{-1}} \] Converting the rate into millimoles, \[ r=0.40\ \mathrm{mmol\,L^{-1}\,s^{-1}} \] The two calculation routes agree. The physical rate has not changed; only its numerical representation in the selected units has changed. The example demonstrates why rate constants belonging to concentration-dependent rate laws must be converted together with the units used for concentration and time. \(\textbf{Final answer:}\) \[ \boxed{\left(k,r\right)=\left(1.0\times10^{-5}\ \mathrm{L\,mmol^{-1}\,s^{-1}},0.40\ \mathrm{mmol\,L^{-1}\,s^{-1}}\right)} \]

Explanation: The problem provides two measured partial concentration dependencies and the units of the rate constant. The units reveal the overall order, while the initial-rate comparisons independently reveal two of the three partial orders. The remaining partial order can therefore be recovered algebraically. The final changed-condition calculation then tests the inferred concentration dependence on C. \(\textbf{Overall order from the rate-constant units:}\) For overall order \(N\), \[ [k]=(\mathrm{mol\,L^{-1}})^{1-N}\mathrm{s^{-1}} \] The measured units are \[ [k]=\mathrm{L^2\,mol^{-2}\,s^{-1}} \] Equivalently, \[ [k]=(\mathrm{mol\,L^{-1}})^{-2}\mathrm{s^{-1}} \] Therefore, \[ 1-N=-2 \] \[ N=3 \] \(\textbf{Determine the partial order of A:}\) Doubling A causes a fourfold increase in rate. \[ 4=2^m \] Therefore, \[ m=2 \] \(\textbf{Determine the partial order of B:}\) Increasing B fourfold doubles the rate. \[ 2=4^n \] \[ 2=2^{2n} \] Therefore, \[ n=\frac12 \] \(\textbf{Recover the partial order of C:}\) The overall order is \[ N=m+n+p \] Substituting the established values, \[ 3=2+\frac12+p \] \[ p=\frac12 \] \(\textbf{Predict the response to increasing C:}\) At unchanged A, B and temperature, \[ \frac{r_2}{r_1}=\left(\frac{[\mathrm C]_2}{[\mathrm C]_1}\right)^p \] A ninefold increase in C gives \[ \frac{r_2}{r_1}=9^{1/2} \] \[ \frac{r_2}{r_1}=3 \] \(\textbf{Dimensional consistency:}\) The inferred exponents sum to \[ 2+\frac12+\frac12=3 \] This agrees with the overall order inferred independently from the dimensions of \(k\). The rate is third order overall, even though two individual partial orders are fractional. Fractional reaction order does not imply fractional molecularity because the experimentally determined overall rate law need not represent a single elementary collision. \(\textbf{Final answer:}\) \[ \boxed{\left(p,\frac{r_2}{r_1}\right)=\left(\frac12,3\right)} \]

Explanation: The experiment changes two reactant concentrations simultaneously. Therefore, the observed rate response contains contributions from both partial reaction orders. The rate ratio cannot generally separate those contributions when the concentration multipliers are the same. For the proposed rate law, \[ r=k[\mathrm A]^m[\mathrm B]^n. \] When both concentrations double at constant temperature, \[ \frac{r_2}{r_1}=2^m2^n. \] The observed rate also doubles. Therefore, \[ 2=2^{m+n}. \] Hence, \[ m+n=1. \] This establishes the overall reaction order as one. However, the individual values of \(m\) and \(n\) remain undetermined. The pair \(m=1,n=0\) would satisfy the observation. The pair \(m=0,n=1\) would also satisfy it. The pair \(m=1/2,n=1/2\) is another possibility, but it is not uniquely established. To identify the partial order of A, an additional experiment should change A while keeping B concentration and the other kinetic conditions constant. Its measured rate multiplier would then depend only on \(m\). Once \(m\) is determined, \(n\) can be calculated from the known relation \(m+n=1\). Repeating the same simultaneous concentration change would provide no new independent equation for the two unknown exponents. Changing temperature alone would investigate temperature dependence rather than independently isolate either concentration exponent.

Explanation: The Assertion is true because overall reaction order is the algebraic sum of the partial orders. Different rate laws can therefore have the same overall order. For example, one reaction may follow \(r=k[\mathrm A]\). Another may follow \(r=k[\mathrm A]^2[\mathrm B]^{-1}\). The first reaction has overall order one. The second also has overall order one because \(2+(-1)=1\). However, the individual concentration dependencies are clearly different. In the first law, the rate depends linearly on A and does not involve B. In the second law, the rate depends quadratically on A and inversely on B. The Reason is true because the dimensions of the rate constant are determined by the difference between the dimensions of the rate and the combined concentration factors. For an overall order \(N\), \[ [k]=(\mathrm{concentration})^{1-N}(\mathrm{time})^{-1}. \] When \(N=1\), the concentration factor cancels. Thus, both examples have rate constants with units of inverse time when identical concentration and time units are used. The equality of dimensions does not imply that the two rate constants have equal numerical values. It also does not imply that the two reactions have identical mechanisms or identical concentration responses. The Reason directly explains the Assertion because dimensional analysis depends on the algebraic total order rather than on the separate exponents.

Explanation: Statement I is correct because the experimental rate law has a partial order of \(+1\) with respect to A and \(-1\) with respect to B. The overall order is their algebraic sum. Therefore, \(1+(-1)=0\). Statement II is correct because doubling both concentrations introduces one factor of two in the numerator and one factor of two in the denominator. These factors cancel, so the rate remains unchanged at constant temperature. Statement III is incorrect because B appears in the denominator. Doubling B alone at fixed A decreases the rate to one-half of its original value. Statement IV is correct because a zero-overall-order rate law has a concentration factor with zero net dimensions. Therefore, the rate constant must have the same dimensions as the normalized reaction rate. In conventional molar concentration units, these are \(\mathrm{mol\,L^{-1}\,s^{-1}}\). Statement V is incorrect because molecularity is the number of reacting species participating in an elementary reaction event. Molecularity cannot be zero and is not inferred from the overall kinetic order of a complex reaction. A zero overall order does not mean that no molecules participate in the chemical transformation. It also does not imply that the rate is independent of every individual reactant concentration. In this case, the rate responds positively to A and negatively to B. The zero overall order describes how the two concentration exponents combine algebraically. The rate constant remains characteristic of the reaction under the specified kinetic conditions and need not remain unchanged if temperature or the reaction pathway changes.

Explanation: The two rate expressions differ in two ways. The first uses the stoichiometrically normalized reaction rate. The second uses the unnormalized disappearance rate of A. They also use different measures of reactant abundance: concentration and partial pressure. Both differences must be included when relating their rate constants. \(\textbf{Normalize the disappearance rate:}\) For \[ 2\mathrm A(g)\rightarrow\mathrm P(g) \] the normalized reaction rate is \[ r_c=-\frac12\frac{d[\mathrm A]}{dt} \] The experimental law is \[ r_c=k_c[\mathrm A]^2 \] Therefore, \[ -\frac{d[\mathrm A]}{dt}=2k_c[\mathrm A]^2 \] \(\textbf{Convert the concentration derivative into a pressure derivative:}\) For an ideal gas at constant temperature, \[ p_{\mathrm A}=[\mathrm A]RT \] Therefore, \[ -\frac{dp_{\mathrm A}}{dt}=-RT\frac{d[\mathrm A]}{dt} \] Substituting the disappearance-rate law, \[ -\frac{dp_{\mathrm A}}{dt}=2RT\,k_c[\mathrm A]^2 \] \(\textbf{Express the rate in partial-pressure form:}\) Using \[ [\mathrm A]=\frac{p_{\mathrm A}}{RT} \] gives \[ -\frac{dp_{\mathrm A}}{dt}=2RT\,k_c\left(\frac{p_{\mathrm A}}{RT}\right)^2 \] \[ -\frac{dp_{\mathrm A}}{dt}=\frac{2k_c}{RT}p_{\mathrm A}^2 \] Comparing with the researcher's expression, \[ -\frac{dp_{\mathrm A}}{dt}=K_p p_{\mathrm A}^2 \] we obtain \[ K_p=\frac{2k_c}{RT} \] \(\textbf{Calculate the pressure-form constant:}\) \[ K_p=\frac{2(0.40)}{20.0} \] \[ K_p=0.040\ \mathrm{atm^{-1}\,s^{-1}} \] \(\textbf{Dimensional and stoichiometric check:}\) The pressure disappearance rate has units of \(\mathrm{atm\,s^{-1}}\). Since the right-hand side contains pressure squared, \(K_p\) must have units of \(\mathrm{atm^{-1}\,s^{-1}}\). The factor of two arises from the stoichiometric coefficient of A. The factor \(1/(RT)\) arises from converting the second-order concentration law into pressure units. Omitting either factor would produce a numerically incorrect constant despite using an apparently plausible pressure-based equation. \(\textbf{Final answer:}\) \[ \boxed{K_p=0.040\ \mathrm{atm^{-1}\,s^{-1}}} \]

Explanation: The experiment combines dimensional information, initial-rate observations and a change of temperature. The rate-constant units first establish the overall reaction order. The initial-rate observations determine the separate orders in A and B. The remaining order in C can then be inferred. Finally, the unknown concentration change in C must compensate for both the change in A and the increased rate constant. \(\textbf{Determine the overall reaction order:}\) For overall order \(N\), \[ [k]=(\mathrm{mol\,L^{-1}})^{1-N}\mathrm{s^{-1}} \] The given units are \[ [k]=\mathrm{L\,mol^{-1}\,s^{-1}} \] Therefore, \[ 1-N=-1 \] \[ N=2 \] \(\textbf{Identify the partial orders of A and B:}\) Doubling A doubles the rate while other concentrations remain constant. \[ 2=2^m \] Hence, \[ m=1 \] Doubling B leaves the rate unchanged. \[ 1=2^n \] Hence, \[ n=0 \] \(\textbf{Infer the partial order of C:}\) \[ N=m+n+p \] \[ 2=1+0+p \] \[ p=1 \] The established rate-law form is therefore \[ r=k[\mathrm A][\mathrm C] \] within the investigated kinetic conditions. \(\textbf{Rate at the initial temperature:}\) \[ r_1=k_1[\mathrm A]_1[\mathrm C]_1 \] \(\textbf{Rate after heating and concentration changes:}\) At \(T_2\), \[ k_2=3k_1 \] The new concentration of A is \[ [\mathrm A]_2=\frac12[\mathrm A]_1 \] Therefore, \[ r_2=3k_1\left(\frac{[\mathrm A]_1}{2}\right)[\mathrm C]_2 \] \(\textbf{Require the original rate to be maintained:}\) \[ r_2=r_1 \] Thus, \[ 3k_1\frac{[\mathrm A]_1}{2}[\mathrm C]_2=k_1[\mathrm A]_1[\mathrm C]_1 \] Cancelling the common non-zero factors, \[ \frac32[\mathrm C]_2=[\mathrm C]_1 \] Therefore, \[ \frac{[\mathrm C]_2}{[\mathrm C]_1}=\frac23 \] \(\textbf{Check the competing effects:}\) Heating increases the rate constant threefold. Halving A concentration reduces its concentration-dependent contribution by a factor of two. Before changing C, these effects together would multiply the rate by \(3/2\). Reducing C concentration to \(2/3\) of its original value exactly compensates for this increase. The final reaction rate therefore equals its initial value. The rate constant changes because the temperature changes, whereas the concentration exponents remain the same under the stated assumption. \(\textbf{Final answer:}\) \[ \boxed{\frac{[\mathrm C]_2}{[\mathrm C]_1}=\frac23} \]

Explanation: The problem requires identifying the dependency between successive elementary reactions. The three steps cannot be arranged arbitrarily because two of them require intermediates that are initially absent. A reaction step cannot occur until its required reacting species are available. The initial mixture contains A and B, while I and J are absent. Therefore, the first elementary reaction that can occur is Step II. In Step II, \[ \mathrm{A\rightarrow I} \] the initially available A produces intermediate I. Once I has been formed, it can react with the initially available B through Step III: \[ \mathrm{I+B\rightarrow J} \] This produces the second intermediate J. Only after J becomes available can Step I occur: \[ \mathrm{J\rightarrow P} \] Consequently, the required ordering is II → III → I. Adding the three elementary equations gives the overall reaction: \[ \mathrm{A+B\rightarrow P} \] Both I and J cancel because they are produced and subsequently consumed. The overall equation does not reveal the chronological sequence of the elementary steps. That sequence is obtained by identifying which intermediates must first be generated. The problem illustrates how reaction mechanisms contain information about molecular events that is absent from a net balanced equation.

Explanation: The overall balanced equation contains one molecule of nitrogen dioxide and one molecule of carbon monoxide. However, the proposed mechanism contains two elementary steps rather than one. Therefore, the rate law must be determined from the specified mechanism instead of being inferred from the overall stoichiometric coefficients. The first elementary step is identified as slow. It involves two molecules of nitrogen dioxide. Under the stated rate-determining-step assumption, its ordinary mass-action rate expression is \[ r=k[\mathrm{NO_2}]^2 \] The concentration of carbon monoxide does not appear in this predicted rate expression. This is because CO participates in the subsequent fast step rather than in the controlling elementary step. Doubling the concentration of nitrogen dioxide gives \[ \frac{r_2}{r_1}=\left(\frac{2[\mathrm{NO_2}]}{[\mathrm{NO_2}]}\right)^2 \] \[ \frac{r_2}{r_1}=4 \] Thus, the predicted rate becomes four times its original value. Doubling CO alone would not change the rate within the validity of the proposed mechanism. Doubling both reactants would produce a fourfold increase, not an eightfold increase. Halving nitrogen dioxide would reduce the rate to one-fourth, not one-half. The intermediate \(\mathrm{NO_3}\) is produced in the first step and consumed in the second. Adding the elementary equations and cancelling the intermediate also reproduces the stated overall reaction. This illustrates how a complex reaction may exhibit a rate law whose concentration exponents differ from its overall stoichiometric coefficients.

Explanation: The slow elementary step determines the rate of product formation under the stated approximation. However, its rate expression contains the intermediate I. The concentration of I is not independently specified. It must first be related to the concentration of A using the preceding fast equilibrium. The resulting expression can then be written entirely in terms of the stable reactants. \(\textbf{Rate expression for the slow step:}\) For \[ \mathrm{I+B\rightarrow P} \] the elementary-step rate expression is \[ r=k_2[\mathrm I][\mathrm B] \] \(\textbf{Concentration of the intermediate:}\) The first step establishes \[ 2\mathrm A\rightleftharpoons\mathrm I \] The equilibrium relation is \[ K_1=\frac{[\mathrm I]}{[\mathrm A]^2} \] Therefore, \[ [\mathrm I]=K_1[\mathrm A]^2 \] Substituting the stated values, \[ [\mathrm I]=10.0(0.10)^2 \] \[ [\mathrm I]=0.10\ \mathrm{mol\,L^{-1}} \] \(\textbf{Construct the effective overall rate law:}\) Substitute the equilibrium expression into the slow-step rate law. \[ r=k_2K_1[\mathrm A]^2[\mathrm B] \] Defining \[ k_{\mathrm{eff}}=k_2K_1 \] gives \[ r=k_{\mathrm{eff}}[\mathrm A]^2[\mathrm B] \] \(\textbf{Calculate the effective rate constant:}\) \[ k_{\mathrm{eff}}=(0.20)(10.0) \] \[ k_{\mathrm{eff}}=2.0\ \mathrm{L^2\,mol^{-2}\,s^{-1}} \] The units follow from multiplying the units of \(k_2\) and \(K_1\). \(\textbf{Calculate the instantaneous reaction rate:}\) \[ r=2.0(0.10)^2(0.20) \] \[ r=2.0(0.010)(0.20) \] \[ r=4.0\times10^{-3}\ \mathrm{mol\,L^{-1}\,s^{-1}} \] \(\textbf{Check the reaction order and dimensions:}\) The effective rate law is second order in A and first order in B. Its overall order is three. Therefore, the effective rate constant must have dimensions of concentration raised to the power \(-2\), multiplied by inverse time. The calculated units agree with this requirement. The slow elementary step itself is bimolecular, although the resulting overall rate law is third order. This distinction arises because the intermediate concentration depends quadratically on A. \(\textbf{Final answer:}\) \[ \boxed{\left(k_{\mathrm{eff}},r\right)=\left(2.0\ \mathrm{L^2\,mol^{-2}\,s^{-1}},4.0\times10^{-3}\ \mathrm{mol\,L^{-1}\,s^{-1}}\right)} \]

Explanation: The reaction rate depends on the concentrations of A and B through experimentally determined powers. A logarithmic representation provides a way to identify these concentration exponents from measured rate data. Taking the logarithm of the rate law gives \[ \log r=\log k+m\log[\mathrm A]+n\log[\mathrm B] \] where the logarithmic quantities are interpreted using consistent reference units. When B concentration is fixed, the terms involving \(k\) and B are constant. The slope of a plot of \(\log r\) against \(\log[\mathrm A]\) therefore gives \(m\). From Plot I, \[ m=\frac32 \] Similarly, when A concentration is fixed, the slope of \(\log r\) against \(\log[\mathrm B]\) gives \(n\). From Plot II, \[ n=-\frac12 \] The negative slope in Plot II indicates that the rate decreases as B concentration increases. The overall order is the algebraic sum of the partial orders. \[ N=m+n \] \[ N=\frac32-\frac12 \] \[ N=1 \] For a first-order overall rate law, the rate constant has units of inverse time. Therefore, \[ [k]=\mathrm{s^{-1}} \] The reaction nevertheless has different individual concentration dependencies. Its rate increases with A concentration but decreases with B concentration. The overall first-order classification does not imply that the rate depends on only one reacting species. It also does not imply that the reaction is necessarily a unimolecular elementary process.

Explanation: The two proposed mechanisms have the same overall stoichiometric equation. They can also produce the same experimentally observed concentration dependence. For Mechanism I, the elementary-step mass-action expression is \[ r=k_{\mathrm I}[\mathrm A][\mathrm B]. \] For Mechanism II, the slow step gives \[ r=k_2[\mathrm I][\mathrm B]. \] The preceding fast equilibrium relates I to A: \[ K_1=\frac{[\mathrm I]}{[\mathrm A]}. \] Therefore, \[ [\mathrm I]=K_1[\mathrm A] \] and \[ r=k_2K_1[\mathrm A][\mathrm B]. \] Defining \(k_{\mathrm{II}}=k_2K_1\) produces the same concentration dependence as Mechanism I. Consequently, doubling A alone or doubling B alone gives the same predicted rate multiplier for both mechanisms. Their overall orders are also identical, so their effective rate constants have identical dimensions. These kinetic observations cannot uniquely distinguish the two proposed pathways. Mechanism II explicitly involves the formation of intermediate I. Mechanism I does not contain I. Therefore, directly detecting a transient population of I under conditions excluding alternative sources would provide evidence favouring Mechanism II. The question illustrates why agreement with a measured rate law does not uniquely prove a proposed molecular mechanism. Additional evidence about intermediates or reaction pathways can be necessary to discriminate between kinetically equivalent proposals.

Explanation: The experimental table provides total pressure and mole fractions rather than individual reactant concentrations. The relevant gaseous concentrations are determined by the partial pressures of A and B. At constant temperature, the ideal-gas relationship makes concentration proportional to partial pressure. The partial pressures must therefore be determined before the individual reaction orders are inferred. \(\textbf{Convert total pressure into reactant partial pressures:}\) For each gas, \[ p_i=x_iP_{\mathrm{total}} \] Experiment I: \[ p_{\mathrm A,I}=0.25(4.0)=1.0\ \mathrm{atm} \] \[ p_{\mathrm B,I}=0.50(4.0)=2.0\ \mathrm{atm} \] Experiment II: \[ p_{\mathrm A,II}=0.25(8.0)=2.0\ \mathrm{atm} \] \[ p_{\mathrm B,II}=0.25(8.0)=2.0\ \mathrm{atm} \] Experiment III: \[ p_{\mathrm A,III}=0.125(8.0)=1.0\ \mathrm{atm} \] \[ p_{\mathrm B,III}=0.50(8.0)=4.0\ \mathrm{atm} \] \(\textbf{Determine the order with respect to A:}\) Experiments I and II have the same partial pressure of B. The partial pressure of A doubles, and the observed rate doubles. Therefore, \[ \frac{r_{\mathrm{II}}}{r_{\mathrm I}}=2^m \] \[ \frac{0.024}{0.012}=2^m \] \[ m=1 \] \(\textbf{Determine the order with respect to B:}\) Experiments I and III have the same partial pressure of A. The partial pressure of B doubles, and the rate becomes four times its original value. \[ \frac{r_{\mathrm{III}}}{r_{\mathrm I}}=2^n \] \[ \frac{0.048}{0.012}=2^n \] \[ 4=2^n \] \[ n=2 \] \(\textbf{Rate law under the stated conditions:}\) \[ r=k[\mathrm A][\mathrm B]^2 \] \(\textbf{Convert the new mixture into partial pressures:}\) The total pressure is \(6.0\ \mathrm{atm}\). Therefore, \[ p_{\mathrm A,new}=\frac13(6.0)=2.0\ \mathrm{atm} \] \[ p_{\mathrm B,new}=\frac13(6.0)=2.0\ \mathrm{atm} \] These are the same reactant partial pressures as in Experiment II. Because the temperature is unchanged, the corresponding concentrations of A and B are identical. The reaction rates must therefore be equal. \[ r_{\mathrm{new}}=r_{\mathrm{II}} \] \[ r_{\mathrm{new}}=0.024\ \mathrm{mol\,L^{-1}\,s^{-1}} \] \(\textbf{Experimental interpretation:}\) The new mixture has a different total pressure from Experiment II. However, its kinetically relevant reactant partial pressures are identical. The difference in total pressure is associated with the inert-gas contribution. Because the rate law contains only the concentrations of A and B, the inert-gas difference does not change the rate under the stated conditions. The comparison illustrates why mole fractions must be combined with total pressure before gaseous rate data are interpreted. \(\textbf{Final answer:}\) \[ \boxed{r_{\mathrm{new}}=0.024\ \mathrm{mol\,L^{-1}\,s^{-1}}} \]

Explanation: The experimentally determined rate law contains a fractional exponent for each reactant. The partial order with respect to A is one-half. The partial order with respect to B is three-halves. Their algebraic sum is two, so the overall reaction order is indeed two. However, reaction order and molecularity describe different chemical properties. Reaction order characterizes the experimentally observed dependence of rate on concentration. Molecularity describes the number of reacting species participating in one elementary molecular event. For an elementary reaction under the ordinary mass-action treatment, the concentration exponents correspond to the numbers of participating reactant species. These elementary-step coefficients must be non-negative integers. The fractional exponents in the supplied experimental rate law therefore do not represent the ordinary mass-action expression of a single bimolecular elementary step involving A and B. The overall order being two does not eliminate this distinction. Complex reactions may display fractional reaction orders even when their individual elementary steps have integer molecularities. Consequently, the observed overall order alone cannot determine the molecularity of a complete reaction. The experimental observations must be considered together with mechanistic evidence before assigning an elementary pathway. The student's conclusion confuses equality of a numerical overall order with proof of a particular molecular event.

Explanation: The actual rate law depends on the concentrations of both A and B. Pseudo-first-order treatment becomes appropriate when one reactant is sufficiently abundant that its concentration changes negligibly over the observation interval. Here, B is initially present in substantial excess. However, two moles of B are consumed for every mole of A consumed. Therefore, the allowable conversion of A must be calculated using the stoichiometric relationship rather than by comparing initial concentrations alone. \(\textbf{Maximum permitted decrease in B:}\) Initially, \[ [\mathrm B]_0=3.00\ \mathrm{mol\,L^{-1}} \] The maximum permitted fractional decrease is \[ f_{\mathrm B}=0.010 \] Therefore, \[ -\Delta[\mathrm B]_{\max}=0.010(3.00) \] \[ -\Delta[\mathrm B]_{\max}=0.0300\ \mathrm{mol\,L^{-1}} \] \(\textbf{Convert B consumption into A consumption:}\) The balanced equation is \[ \mathrm{A+2B\rightarrow P} \] The stoichiometric concentration changes satisfy \[ -\Delta[\mathrm A]=\frac{-\Delta[\mathrm B]}{2} \] Therefore, \[ -\Delta[\mathrm A]_{\max}=\frac{0.0300}{2} \] \[ -\Delta[\mathrm A]_{\max}=0.0150\ \mathrm{mol\,L^{-1}} \] \(\textbf{Calculate the permitted conversion of A:}\) The initial concentration of A is \[ [\mathrm A]_0=0.150\ \mathrm{mol\,L^{-1}} \] The maximum fraction of A consumed is \[ f_{\mathrm A}=\frac{0.0150}{0.150} \] \[ f_{\mathrm A}=0.100 \] Thus, \[ 100f_{\mathrm A}=10.0\% \] \(\textbf{Verify the excess-reactant criterion:}\) At this conversion, the concentration of B becomes \[ [\mathrm B]=3.00-0.0300 \] \[ [\mathrm B]=2.970\ \mathrm{mol\,L^{-1}} \] The decrease relative to its initial concentration is \[ \frac{3.00-2.970}{3.00}\times100=1.0\% \] Therefore, the stated limit is satisfied. \(\textbf{Kinetic interpretation:}\) The observed pseudo-first-order constant is approximately \[ k_{\mathrm{obs}}\approx k[\mathrm B]_0 \] A decrease of \(1.0\%\) in B produces a corresponding \(1.0\%\) decrease in this concentration-dependent factor. The pseudo-first-order approximation is therefore controlled by the permitted fractional depletion of B. The result does not imply that B is chemically inactive or completely unconsumed. It shows how stoichiometry and relative initial concentrations determine the useful range of a kinetic approximation. \(\textbf{Final answer:}\) \[ \boxed{10.0\%} \]

Explanation: The observed pseudo-first-order rate constant incorporates the nearly constant concentration of the excess reactant B. Consequently, changing the excess concentration of B can change the observed constant even when the underlying rate constant remains unchanged. The two experiments therefore provide information about the partial reaction order with respect to B. Once this exponent is known, the intrinsic rate constant can be determined. \(\textbf{Pseudo-first-order representation:}\) The underlying rate law is \[ r=k[\mathrm A][\mathrm B]^n \] Since B is effectively constant during each experiment, \[ r=k_{\mathrm{obs}}[\mathrm A] \] where \[ k_{\mathrm{obs}}=k[\mathrm B]^n \] \(\textbf{Compare the observed constants:}\) The ratio of pseudo-first-order constants is \[ \frac{k_{\mathrm{obs},2}}{k_{\mathrm{obs},1}}=\left(\frac{[\mathrm B]_2}{[\mathrm B]_1}\right)^n \] Substitution gives \[ \frac{0.320}{0.080}=\left(\frac{0.40}{0.20}\right)^n \] \[ 4=2^n \] Therefore, \[ n=2 \] \(\textbf{Establish the underlying rate law:}\) \[ r=k[\mathrm A][\mathrm B]^2 \] The overall order is three. \(\textbf{Calculate the underlying rate constant:}\) Using the first experiment, \[ k=\frac{k_{\mathrm{obs},1}}{[\mathrm B]_1^2} \] \[ k=\frac{0.080}{(0.20)^2} \] \[ k=\frac{0.080}{0.040} \] \[ k=2.00\ \mathrm{L^2\,mol^{-2}\,s^{-1}} \] \(\textbf{Verify against the second experiment:}\) \[ k_{\mathrm{obs},2}=k[\mathrm B]_2^2 \] \[ k_{\mathrm{obs},2}=2.00(0.40)^2 \] \[ k_{\mathrm{obs},2}=2.00(0.160) \] \[ k_{\mathrm{obs},2}=0.320\ \mathrm{s^{-1}} \] The predicted value agrees with the observation. \(\textbf{Dimensional verification:}\) Because the underlying reaction is third order overall, \[ [k]=(\mathrm{mol\,L^{-1}})^{-2}\mathrm{s^{-1}} \] Hence, \[ [k]=\mathrm{L^2\,mol^{-2}\,s^{-1}} \] Multiplying these units by the square of concentration gives \(\mathrm{s^{-1}}\), as required for \(k_{\mathrm{obs}}\). Thus, the observed pseudo-first-order constant and the underlying rate constant have different dimensions. The difference arises because the observed constant includes the excess-reactant concentration. \(\textbf{Final answer:}\) \[ \boxed{\left(n,k\right)=\left(2,\ 2.00\ \mathrm{L^2\,mol^{-2}\,s^{-1}}\right)} \]

Explanation: Both pathways produce the same product P. Therefore, the total rate of appearance of P is the sum of its formation rates through the two independent pathways. However, the pathways have different concentration dependencies. The first is first order in A and independent of B. The second is first order in each of A and B. Consequently, doubling both reactant concentrations affects the two pathways by different factors. \(\textbf{Initial rate through Pathway I:}\) \[ r_{1,0}=k_1[\mathrm A]_0 \] \[ r_{1,0}=0.040(0.10) \] \[ r_{1,0}=4.0\times10^{-3}\ \mathrm{mol\,L^{-1}\,s^{-1}} \] \(\textbf{Initial rate through Pathway II:}\) \[ r_{2,0}=k_2[\mathrm A]_0[\mathrm B]_0 \] \[ r_{2,0}=0.100(0.10)(0.20) \] \[ r_{2,0}=2.0\times10^{-3}\ \mathrm{mol\,L^{-1}\,s^{-1}} \] \(\textbf{Total product-formation rate in the first mixture:}\) \[ r_{\mathrm{P},0}=r_{1,0}+r_{2,0} \] \[ r_{\mathrm{P},0}=(4.0+2.0)\times10^{-3} \] \[ r_{\mathrm{P},0}=6.0\times10^{-3}\ \mathrm{mol\,L^{-1}\,s^{-1}} \] \(\textbf{Changed concentrations in the second mixture:}\) \[ [\mathrm A]'=0.20\ \mathrm{mol\,L^{-1}} \] \[ [\mathrm B]'=0.40\ \mathrm{mol\,L^{-1}} \] \(\textbf{New rate through Pathway I:}\) \[ r_1'=0.040(0.20) \] \[ r_1'=8.0\times10^{-3}\ \mathrm{mol\,L^{-1}\,s^{-1}} \] \(\textbf{New rate through Pathway II:}\) \[ r_2'=0.100(0.20)(0.40) \] \[ r_2'=8.0\times10^{-3}\ \mathrm{mol\,L^{-1}\,s^{-1}} \] \(\textbf{Total product-formation rate in the second mixture:}\) \[ r_{\mathrm P}'=8.0\times10^{-3}+8.0\times10^{-3} \] \[ r_{\mathrm P}'=1.60\times10^{-2}\ \mathrm{mol\,L^{-1}\,s^{-1}} \] \(\textbf{Compare the two total rates:}\) \[ \frac{r_{\mathrm P}'}{r_{\mathrm{P},0}}=\frac{1.60\times10^{-2}}{6.0\times10^{-3}} \] \[ \frac{r_{\mathrm P}'}{r_{\mathrm{P},0}}=\frac83 \] \(\textbf{Chemical interpretation:}\) Pathway I doubles its rate, whereas Pathway II increases its rate fourfold. The total product-formation rate therefore increases by a factor between two and four. It is not correct to assign a single concentration exponent to the combined process merely by adding the separate pathway orders. The relative contributions of the two pathways change with reactant concentrations. This is why an observed overall concentration dependence can be more complicated when several independent reaction pathways contribute to the same product. \(\textbf{Final answer:}\) \[ \boxed{\frac{r_{\mathrm P}'}{r_{\mathrm{P},0}}=\frac83} \]

Explanation: The slow elementary step involves X and A. However, the supplied concentration information refers to \(\mathrm{X_2}\), not directly to X. The concentration of X must therefore be determined using the preceding fast equilibrium. Because the equilibrium involves dissociation of one molecule into two, the equilibrium relationship contains the square of X concentration. Taking the square root produces a fractional concentration dependence in the overall rate law. \(\textbf{Rate expression of the slow elementary step:}\) \[ r=k_2[\mathrm X][\mathrm A] \] \(\textbf{Use the fast dissociation equilibrium:}\) For \[ \mathrm{X_2\rightleftharpoons2X} \] the equilibrium expression is \[ K_1=\frac{[\mathrm X]^2}{[\mathrm{X_2}]} \] Rearranging, \[ [\mathrm X]^2=K_1[\mathrm{X_2}] \] Therefore, \[ [\mathrm X]=\sqrt{K_1[\mathrm{X_2}]} \] \(\textbf{Calculate the intermediate concentration:}\) \[ [\mathrm X]=\sqrt{(0.040)(0.25)} \] \[ [\mathrm X]=\sqrt{0.0100} \] \[ [\mathrm X]=0.100\ \mathrm{mol\,L^{-1}} \] \(\textbf{Construct the effective rate law:}\) Substituting the equilibrium relationship into the slow-step rate law, \[ r=k_2[\mathrm A]\sqrt{K_1[\mathrm{X_2}]} \] \[ r=k_2K_1^{1/2}[\mathrm A][\mathrm{X_2}]^{1/2} \] Therefore, the effective partial order with respect to \(\mathrm{X_2}\) is \[ n_{\mathrm{X_2}}=\frac12 \] \(\textbf{Calculate the instantaneous rate:}\) Using the intermediate concentration, \[ r=0.30(0.100)(0.20) \] \[ r=6.0\times10^{-3}\ \mathrm{mol\,L^{-1}\,s^{-1}} \] \(\textbf{Mechanistic interpretation:}\) The slow elementary step is bimolecular because it involves X and A. The effective overall rate law is first order in A and half order in \(\mathrm{X_2}\). Its overall kinetic order is therefore \(3/2\). There is no fractional molecularity in the elementary steps. The fractional experimental order arises because the intermediate concentration is controlled by a preceding dissociation equilibrium. This demonstrates how a complex mechanism can generate a non-integer overall reaction order even though its elementary steps involve integer numbers of molecules. \(\textbf{Final answer:}\) \[ \boxed{\left(n_{\mathrm{X_2}},r\right)=\left(\frac12,\ 6.0\times10^{-3}\ \mathrm{mol\,L^{-1}\,s^{-1}}\right)} \]

Explanation: The graph provides instantaneous concentration derivatives at two different reactant concentrations. Because A is consumed, both tangent slopes are negative. The positive disappearance rates are obtained by reversing their signs. These instantaneous rates can be compared to determine the reaction order. Unlike the usual initial-rate method, this comparison uses two points from the same concentration–time curve. The method is valid because the problem states that the same rate law and rate constant apply throughout the interval. \(\textbf{Extract the positive disappearance rates:}\) At \[ [\mathrm A]_1=0.80\ \mathrm{mol\,L^{-1}} \] the disappearance rate is \[ r_1=3.20\times10^{-2}\ \mathrm{mol\,L^{-1}\,s^{-1}} \] At \[ [\mathrm A]_2=0.40\ \mathrm{mol\,L^{-1}} \] the disappearance rate is \[ r_2=8.00\times10^{-3}\ \mathrm{mol\,L^{-1}\,s^{-1}} \] \(\textbf{Infer the kinetic order:}\) The rate law is \[ r=k[\mathrm A]^n \] Therefore, \[ \frac{r_1}{r_2}=\left(\frac{[\mathrm A]_1}{[\mathrm A]_2}\right)^n \] Substitution gives \[ \frac{3.20\times10^{-2}}{8.00\times10^{-3}}=\left(\frac{0.80}{0.40}\right)^n \] \[ 4=2^n \] Thus, \[ n=2 \] \(\textbf{Calculate the rate constant:}\) \[ k=\frac{r_1}{[\mathrm A]_1^2} \] \[ k=\frac{3.20\times10^{-2}}{(0.80)^2} \] \[ k=\frac{3.20\times10^{-2}}{0.64} \] \[ k=5.00\times10^{-2}\ \mathrm{L\,mol^{-1}\,s^{-1}} \] \(\textbf{Predict the rate at the new concentration:}\) At \[ [\mathrm A]=0.20\ \mathrm{mol\,L^{-1}} \] the rate is \[ r=k[\mathrm A]^2 \] \[ r=(5.00\times10^{-2})(0.20)^2 \] \[ r=(5.00\times10^{-2})(0.040) \] \[ r=2.00\times10^{-3}\ \mathrm{mol\,L^{-1}\,s^{-1}} \] \(\textbf{Graphical consistency:}\) When A concentration decreases from \(0.40\) to \(0.20\ \mathrm{mol\,L^{-1}}\), it becomes half its earlier value. For second-order kinetics, the disappearance rate should consequently become one-fourth. One-fourth of \(8.00\times10^{-3}\) is \(2.00\times10^{-3}\ \mathrm{mol\,L^{-1}\,s^{-1}}\), agreeing with the calculated result. The predicted tangent slope of the concentration–time curve would be negative, but the disappearance rate is reported as a positive quantity. \(\textbf{Final answer:}\) \[ \boxed{-\frac{d[\mathrm A]}{dt}=2.00\times10^{-3}\ \mathrm{mol\,L^{-1}\,s^{-1}}} \]

Explanation: Situation P describes a rate that is independent of B concentration within the investigated kinetic range. In a power-law expression, this corresponds to an exponent of zero for B. Therefore, P matches conclusion 3. Situation Q gives the units of the rate constant rather than a direct concentration-response observation. For a normalized rate expressed in concentration per unit time, a rate constant with units of inverse time corresponds to overall order one. Therefore, Q matches conclusion 1. Situation R describes the response to changing A concentration while other factors are fixed. If the partial order in A is \(m\), an eightfold rate increase on doubling A gives \[ 8=2^m \] Therefore, \[ m=3 \] and R matches conclusion 2. Situation S explicitly states that the rate is proportional to the square root of A concentration. Therefore, the partial order in A is one-half, and S matches conclusion 4. The observations provide different types of kinetic information. A concentration-independent response determines one partial order. Rate-constant dimensions determine the overall order. A measured rate multiplier can establish an individual exponent. A directly supplied fractional concentration dependence identifies the corresponding fractional order. These deductions do not require the partial orders to equal the coefficients in an overall balanced chemical equation. They also do not establish elementary molecularity without additional mechanistic information.