Chemical Kinetics MCQs with Answers – Part 1 (NEET Chemistry)

Explanation: The rate of a chemical reaction describes how rapidly the concentration of a participating species changes with time. For a reactant undergoing consumption, its concentration generally decreases as the reaction proceeds. Therefore, the concentration change \(\Delta[\mathrm R]\), calculated as final concentration minus initial concentration, is negative. The reported rate of disappearance is conventionally expressed as a positive quantity. Hence, a negative sign is introduced before the concentration change. The appropriate relation is \[ r_{\mathrm{av}}=-\frac{\Delta[\mathrm R]}{\Delta t} \] Concentration has the unit \(\mathrm{mol\,L^{-1}}\), while the elapsed time is measured in seconds. Dividing concentration by time gives \(\mathrm{mol\,L^{-1}\,s^{-1}}\). The negative sign indicates the direction of concentration change; it does not make the reported rate negative. A concentration value alone cannot represent a reaction rate because the associated time interval must also be considered.

Explanation: \(\textbf{Given concentrations:}\) \[ [\mathrm R]_0=0.120\ \mathrm{mol\,L^{-1}} \] \[ [\mathrm R]_t=0.084\ \mathrm{mol\,L^{-1}} \] The elapsed time must be expressed in seconds because the required rate unit contains \(\mathrm{s^{-1}}\). \(\textbf{Time conversion:}\) \[ \Delta t=3\times60=180\ \mathrm{s} \] \(\textbf{Average-rate relation:}\) For the disappearance of a reactant with stoichiometric coefficient unity, \[ r_{\mathrm{av}}=-\frac{\Delta[\mathrm R]}{\Delta t} \] Substituting the final and initial concentrations, \[ r_{\mathrm{av}}=-\frac{0.084-0.120}{180} \] \[ r_{\mathrm{av}}=\frac{0.036}{180} \] \[ r_{\mathrm{av}}=0.00020\ \mathrm{mol\,L^{-1}\,s^{-1}} \] The concentration change is negative, but the disappearance rate is positive. The units are concentration divided by time, as required. \(\textbf{Final answer:}\) \[ \boxed{r_{\mathrm{av}}=2.0\times10^{-4}\ \mathrm{mol\,L^{-1}\,s^{-1}}} \]

Explanation: Statement I is correct because a secant joins two points corresponding to two different times on a concentration–time graph. Its slope represents the change in concentration divided by the elapsed time. For a reactant, this slope is negative when its concentration decreases. Changing its sign gives the positive average disappearance rate for the selected interval. Statement II is incorrect because a tangent describes the local slope at one particular instant. The negative of the tangent slope gives the instantaneous disappearance rate at that instant. For a curved concentration–time graph, the tangent slope generally changes from one point to another. Consequently, the instantaneous rates at the endpoints need not equal the average rate between them. For a straight-line concentration–time graph with constant slope, the instantaneous and average rates are equal. Although an instantaneous rate may equal an interval-average rate at some intermediate time, such equality cannot automatically be assigned to either endpoint. The essential distinction is that a secant describes a finite interval, whereas a tangent describes an instant.

Explanation: \(\textbf{Rate over the first interval:}\) The average rate is determined using the concentration change between \(t=0\) and \(t=20\ \mathrm{s}\). \[ r_1=-\frac{\Delta[\mathrm R]}{\Delta t} \] \[ r_1=-\frac{0.62-0.80}{20-0} \] \[ r_1=\frac{0.18}{20} \] \[ r_1=9.0\times10^{-3}\ \mathrm{mol\,L^{-1}\,s^{-1}} \] \(\textbf{Rate over the complete interval:}\) The complete interval extends from \(t=0\) to \(t=50\ \mathrm{s}\), not from 20 s to 50 s. \[ r_2=-\frac{0.47-0.80}{50-0} \] \[ r_2=\frac{0.33}{50} \] \[ r_2=6.6\times10^{-3}\ \mathrm{mol\,L^{-1}\,s^{-1}} \] \(\textbf{Comparison:}\) \[ \frac{r_1}{r_2}=\frac{9.0\times10^{-3}}{6.6\times10^{-3}} \] \[ \frac{r_1}{r_2}=\frac{90}{66}=\frac{15}{11} \] Both rates have identical units, so their ratio is dimensionless. The ratio is greater than unity, showing that disappearance was faster on average during the initial 20 s than during the complete observation period. \(\textbf{Final answer:}\) \[ \boxed{\frac{r_1}{r_2}=\frac{15}{11}} \]

Explanation: \(\textbf{Graph interpretation:}\) The slope of the tangent to a concentration–time curve gives the instantaneous concentration derivative. Because A is a reactant, its concentration derivative is negative. \[ \frac{d[\mathrm A]}{dt}=-6.0\times10^{-3}\ \mathrm{mol\,L^{-1}\,s^{-1}} \] Therefore, the positive instantaneous disappearance rate of A is \[ -\frac{d[\mathrm A]}{dt}=6.0\times10^{-3}\ \mathrm{mol\,L^{-1}\,s^{-1}} \] \(\textbf{Stoichiometric normalization:}\) For \(2\mathrm A\rightarrow3\mathrm P\), the common reaction rate is \[ r=-\frac12\frac{d[\mathrm A]}{dt}=\frac13\frac{d[\mathrm P]}{dt} \] Substituting the reactant derivative, \[ r=\frac12(6.0\times10^{-3}) \] \[ r=3.0\times10^{-3}\ \mathrm{mol\,L^{-1}\,s^{-1}} \] \(\textbf{Product-formation rate:}\) The stoichiometric coefficient of P is three, so its appearance rate is three times the normalized reaction rate. \[ \frac{d[\mathrm P]}{dt}=3r \] \[ \frac{d[\mathrm P]}{dt}=3(3.0\times10^{-3}) \] \[ \frac{d[\mathrm P]}{dt}=9.0\times10^{-3}\ \mathrm{mol\,L^{-1}\,s^{-1}} \] The product-formation derivative is positive, consistent with product accumulation. The two species need not have equal concentration-change rates because their stoichiometric coefficients differ. \(\textbf{Final answer:}\) \[ \boxed{\left(\frac{d[\mathrm P]}{dt},r\right)=\left(9.0\times10^{-3},3.0\times10^{-3}\right)\ \mathrm{mol\,L^{-1}\,s^{-1}}} \]

Explanation: The Assertion is true because the species-concentration rates in a chemical reaction are related through their stoichiometric coefficients. For the reaction \(2\mathrm A\rightarrow3\mathrm B\), two stoichiometric units of A are consumed for every three units of B formed. Consequently, the concentration of B increases \(3/2\) times as rapidly as the concentration of A decreases, provided the rates refer to the same instant. The Reason is also true because the normalized rate is obtained by dividing each species rate by its stoichiometric coefficient. For this reaction, \[ -\frac12\frac{d[\mathrm A]}{dt}=\frac13\frac{d[\mathrm B]}{dt}=r. \] Rearranging the equality gives \[ \frac{d[\mathrm B]}{dt}=\frac32\left(-\frac{d[\mathrm A]}{dt}\right). \] The quantity within parentheses is the positive disappearance rate of A. Therefore, the appearance rate of B is \(1.5\) times the disappearance rate of A. The Reason directly establishes the numerical relationship stated in the Assertion. This stoichiometric relationship should not be confused with a rate law: the coefficients determine relative species-concentration changes but do not necessarily determine the concentration exponents in the kinetic rate equation.

Explanation: The directly measured rates belong to different species in two different reactions. They cannot be compared as normalized reaction rates without accounting for their stoichiometric coefficients. \(\textbf{Reaction I:}\) For \(2\mathrm A\rightarrow\mathrm P\), \[ r_{\mathrm I}=-\frac12\frac{d[\mathrm A]}{dt} \] The supplied disappearance rate of A is positive. \[ r_{\mathrm I}=\frac12(1.4\times10^{-2}) \] \[ r_{\mathrm I}=7.0\times10^{-3}\ \mathrm{mol\,L^{-1}\,s^{-1}} \] \(\textbf{Reaction II:}\) For \(\mathrm Q\rightarrow3\mathrm R\), \[ r_{\mathrm{II}}=\frac13\frac{d[\mathrm R]}{dt} \] \[ r_{\mathrm{II}}=\frac13(1.8\times10^{-2}) \] \[ r_{\mathrm{II}}=6.0\times10^{-3}\ \mathrm{mol\,L^{-1}\,s^{-1}} \] \(\textbf{Rate comparison:}\) \[ r_{\mathrm I}:r_{\mathrm{II}}=7.0\times10^{-3}:6.0\times10^{-3} \] \[ r_{\mathrm I}:r_{\mathrm{II}}=7:6 \] Reaction I has the greater normalized rate, even though the observed appearance rate of R in Reaction II is numerically larger. The comparison demonstrates why raw species rates can be misleading when different stoichiometric coefficients are involved. \(\textbf{Final answer:}\) \[ \boxed{r_{\mathrm I}:r_{\mathrm{II}}=7:6} \]

Explanation: Statement I is correct because the stoichiometric coefficient of A is two. For the balanced reaction, the normalized reaction rate is \[ r=-\frac12\frac{d[\mathrm A]}{dt}=-\frac{d[\mathrm B]}{dt}=\frac13\frac{d[\mathrm C]}{dt}. \] Hence, the positive disappearance rate of A equals \(2r\). Statement II is correct because C has a stoichiometric coefficient of three. Its appearance rate is therefore \(3r\), not simply \(r\). Statement III is correct because two stoichiometric units of A disappear for each unit of B consumed. A decrease of \(0.10\ \mathrm{mol\,L^{-1}}\) in B corresponds to a decrease of \(0.20\ \mathrm{mol\,L^{-1}}\) in A. Statement IV is incorrect because the formation of C follows the coefficient ratio \(3:2\) relative to the disappearance of A. Consumption of \(0.20\ \mathrm{mol\,L^{-1}}\) of A therefore produces an increase of \(0.30\ \mathrm{mol\,L^{-1}}\) in C. The ratio applies to concentration changes measured over the same interval at constant volume. It also applies to instantaneous derivatives at a common instant. The important distinction is that stoichiometric normalization makes the reaction rate common to all species, while their individual appearance and disappearance rates can have different magnitudes.

Explanation: \(\textbf{Information from the concentration changes:}\) The reaction is the only process consuming X and forming Y and Z. Therefore, their concentration changes must be proportional to their stoichiometric coefficients. The magnitude of the concentration decrease of X is \(0.18\ \mathrm{mol\,L^{-1}}\). The increases in Y and Z are \(0.12\ \mathrm{mol\,L^{-1}}\) and \(0.30\ \mathrm{mol\,L^{-1}}\). \(\textbf{Stoichiometric inference:}\) \[ a:b:c=0.18:0.12:0.30 \] Dividing each quantity by \(0.06\), \[ a:b:c=3:2:5 \] Thus, the reaction may be represented using the simplest stoichiometric coefficients as \[ 3\mathrm X\rightarrow2\mathrm Y+5\mathrm Z \] \(\textbf{Instantaneous-rate relationship:}\) Although the coefficients were inferred from an earlier finite interval, the same stoichiometric relationship applies to instantaneous concentration derivatives. \[ r=-\frac13\frac{d[\mathrm X]}{dt}=\frac12\frac{d[\mathrm Y]}{dt}=\frac15\frac{d[\mathrm Z]}{dt} \] The required rate is the instantaneous appearance rate of Z. \[ \frac{d[\mathrm Z]}{dt}=\frac52\frac{d[\mathrm Y]}{dt} \] \(\textbf{Substitution:}\) \[ \frac{d[\mathrm Z]}{dt}=\frac52(8.0\times10^{-3}) \] \[ \frac{d[\mathrm Z]}{dt}=20.0\times10^{-3} \] \[ \frac{d[\mathrm Z]}{dt}=2.0\times10^{-2}\ \mathrm{mol\,L^{-1}\,s^{-1}} \] The positive sign is consistent with the formation of Z. The appearance rate of Z is greater than that of Y because their stoichiometric coefficients are in the ratio \(5:2\). \(\textbf{Final answer:}\) \[ \boxed{\frac{d[\mathrm Z]}{dt}=2.0\times10^{-2}\ \mathrm{mol\,L^{-1}\,s^{-1}}} \]

Explanation: \(\textbf{Stoichiometric consistency:}\) The reaction consumes three stoichiometric units of A for every two units of B formed. Since the vessel volume is constant, concentration changes follow the corresponding stoichiometric ratio. \[ -\frac{\Delta[\mathrm A]}{3}=\frac{\Delta[\mathrm B]}{2} \] Therefore, \[ \Delta[\mathrm B]=-\frac23\Delta[\mathrm A] \] The initial concentration of B is \(0.10\ \mathrm{mol\,L^{-1}}\), so this initial amount must be retained in calculating later concentrations. \(\textbf{Verification of recorded observations:}\) At 20 s, the amount of A consumed is \[ 0.80-0.74=0.06\ \mathrm{mol\,L^{-1}} \] The expected concentration of B is \[ [\mathrm B]_{20}=0.10+\frac23(0.06)=0.14\ \mathrm{mol\,L^{-1}} \] This agrees with the recorded value. At 40 s, \[ [\mathrm B]_{40}=0.10+\frac23(0.80-0.62)=0.22\ \mathrm{mol\,L^{-1}} \] This also agrees with the table. At 80 s, \[ [\mathrm B]_{80}=0.10+\frac23(0.80-0.38)=0.38\ \mathrm{mol\,L^{-1}} \] Therefore, the observations at 20 s, 40 s and 80 s are consistent with the reaction stoichiometry. \(\textbf{Correction of the inconsistent reading:}\) At 60 s, the concentration decrease of A is \[ \Delta[\mathrm A]=0.50-0.80=-0.30\ \mathrm{mol\,L^{-1}} \] The corresponding increase in B must be \[ \Delta[\mathrm B]=-\frac23(-0.30)=0.20\ \mathrm{mol\,L^{-1}} \] Adding this increase to the initial concentration, \[ [\mathrm B]_{60}=0.10+0.20=0.30\ \mathrm{mol\,L^{-1}} \] The recorded value \(0.32\ \mathrm{mol\,L^{-1}}\) would imply formation of more B than permitted by the measured consumption of A. The correction follows from stoichiometric consistency and does not require assuming that the reaction proceeds at a constant rate. \(\textbf{Final answer:}\) \[ \boxed{[\mathrm B]_{60}=0.30\ \mathrm{mol\,L^{-1}}} \]

Explanation: Two different types of reaction-rate information are supplied. The concentrations at the endpoints determine the average disappearance rate of R over the complete interval. The tangent slope determines the instantaneous disappearance rate of R at \(40\ \mathrm{s}\). The required comparison additionally involves the stoichiometric conversion from R to P. \(\textbf{Average disappearance rate of R:}\) \[ r_{\mathrm{av,R}}=-\frac{[\mathrm R]_{40}-[\mathrm R]_0}{40-0} \] \[ r_{\mathrm{av,R}}=-\frac{0.46-0.70}{40} \] \[ r_{\mathrm{av,R}}=\frac{0.24}{40} \] \[ r_{\mathrm{av,R}}=6.0\times10^{-3}\ \mathrm{mol\,L^{-1}\,s^{-1}} \] \(\textbf{Instantaneous rate from the graph:}\) At \(t=40\ \mathrm{s}\), \[ \left.\frac{d[\mathrm R]}{dt}\right|_{40}=-3.0\times10^{-3}\ \mathrm{mol\,L^{-1}\,s^{-1}} \] For the reaction \(2\mathrm R\rightarrow\mathrm P\), \[ -\frac12\frac{d[\mathrm R]}{dt}=\frac{d[\mathrm P]}{dt} \] Therefore, the instantaneous product-formation rate is \[ \left.\frac{d[\mathrm P]}{dt}\right|_{40}=\frac12(3.0\times10^{-3}) \] \[ \left.\frac{d[\mathrm P]}{dt}\right|_{40}=1.5\times10^{-3}\ \mathrm{mol\,L^{-1}\,s^{-1}} \] \(\textbf{Required comparison:}\) \[ \frac{r_{\mathrm{av,R}}}{\left(d[\mathrm P]/dt\right)_{40}}=\frac{6.0\times10^{-3}}{1.5\times10^{-3}} \] \[ \frac{r_{\mathrm{av,R}}}{\left(d[\mathrm P]/dt\right)_{40}}=4 \] The average disappearance rate of R is greater than its instantaneous disappearance rate at 40 s, consistent with the progressively flattening reactant curve. The appearance rate of P is additionally smaller than the disappearance rate of R at that instant because of the \(2:1\) stoichiometric ratio. \(\textbf{Final answer:}\) \[ \boxed{\frac{r_{\mathrm{av,R}}}{\left(d[\mathrm P]/dt\right)_{40}}=4} \]

Explanation: \(\textbf{Interpretation of tangent slopes:}\) The instantaneous derivative of A is negative because A is consumed. The instantaneous derivative of B is positive because B is formed. Both slopes refer to the same instant, so their ratio is governed by the stoichiometric coefficients. \(\textbf{Determine the unknown coefficient:}\) For \(2\mathrm A\rightarrow b\mathrm B+\text{other products}\), \[ -\frac12\frac{d[\mathrm A]}{dt}=\frac1b\frac{d[\mathrm B]}{dt} \] Substituting the measured derivatives, \[ -\frac12(-4.0\times10^{-3})=\frac{1.0\times10^{-2}}{b} \] \[ 2.0\times10^{-3}=\frac{1.0\times10^{-2}}{b} \] Rearranging, \[ b=\frac{1.0\times10^{-2}}{2.0\times10^{-3}} \] \[ b=5 \] \(\textbf{Apply the inferred stoichiometry to the interval:}\) The reaction therefore consumes two units of A for every five units of B formed. The decrease in A concentration during the separate interval is \[ 0.84-0.72=0.12\ \mathrm{mol\,L^{-1}} \] The corresponding increase in B is \[ \Delta[\mathrm B]=\frac52(0.12) \] \[ \Delta[\mathrm B]=0.30\ \mathrm{mol\,L^{-1}} \] The tangent measurements were used to determine the stoichiometric ratio, while the finite concentration change was used to calculate the amount of product formed. These two operations remain valid even if the instantaneous reaction rate changes between observations. \(\textbf{Final answer:}\) \[ \boxed{\left(b,\Delta[\mathrm B]\right)=\left(5,\ 0.30\ \mathrm{mol\,L^{-1}}\right)} \]

Explanation: The two average rates apply to intervals of unequal duration. Therefore, their simple arithmetic mean cannot represent the overall average disappearance rate. The total concentration decrease must first be obtained by multiplying each interval-average rate by its corresponding duration. \(\textbf{First interval:}\) For \(0-30\ \mathrm{s}\), \[ r_1=6.0\times10^{-3}\ \mathrm{mol\,L^{-1}\,s^{-1}} \] The amount of R consumed, expressed as concentration change, is \[ -\Delta[\mathrm R]_1=r_1\Delta t_1 \] \[ -\Delta[\mathrm R]_1=(6.0\times10^{-3})(30) \] \[ -\Delta[\mathrm R]_1=0.18\ \mathrm{mol\,L^{-1}} \] \(\textbf{Second interval:}\) For \(30-90\ \mathrm{s}\), the duration is \(60\ \mathrm{s}\). \[ -\Delta[\mathrm R]_2=(2.0\times10^{-3})(60) \] \[ -\Delta[\mathrm R]_2=0.12\ \mathrm{mol\,L^{-1}} \] \(\textbf{Total reactant consumption:}\) \[ -\Delta[\mathrm R]_{\mathrm{total}}=0.18+0.12 \] \[ -\Delta[\mathrm R]_{\mathrm{total}}=0.30\ \mathrm{mol\,L^{-1}} \] \(\textbf{Convert to product formation:}\) For \(\mathrm R\rightarrow2\mathrm S\), two stoichiometric units of S are produced per unit of R consumed. \[ \Delta[\mathrm S]=2(0.30) \] \[ \Delta[\mathrm S]=0.60\ \mathrm{mol\,L^{-1}} \] The total elapsed time is \(90\ \mathrm{s}\). \[ r_{\mathrm{av,S}}=\frac{\Delta[\mathrm S]}{\Delta t} \] \[ r_{\mathrm{av,S}}=\frac{0.60}{90} \] \[ r_{\mathrm{av,S}}=6.666\ldots\times10^{-3}\ \mathrm{mol\,L^{-1}\,s^{-1}} \] The product-formation rate is positive, and its unit is concentration per unit time. A simple average of the two supplied rates would incorrectly assign equal weighting to intervals of 30 s and 60 s. \(\textbf{Final answer:}\) \[ \boxed{r_{\mathrm{av,S}}\approx6.67\times10^{-3}\ \mathrm{mol\,L^{-1}\,s^{-1}}} \]

Explanation: \(\textbf{Concentration dependence:}\) The experimentally determined rate law specifies how the reaction rate depends on the concentrations of A and B. A has a partial reaction order of \(1/2\), while B has a partial order of zero. Therefore, changing the concentration of B does not affect the rate under the given law. The rate ratio between the two mixtures is \[ \frac{r'}r=\left(\frac{[\mathrm A]'}{[\mathrm A]}\right)^{1/2}\left(\frac{[\mathrm B]'}{[\mathrm B]}\right)^0 \] \(\textbf{Substitution:}\) \[ \frac{r'}r=4^{1/2}\times2^0 \] \[ \frac{r'}r=2\times1=2 \] Thus, \[ r'=2r \] \(\textbf{Rate-constant check:}\) The rate constant \(k\) does not change merely because the reactant concentrations have changed at fixed temperature under the same kinetic conditions. Therefore, \[ k'=k \] A fractional reaction order does not mean the concentration itself must change fractionally. Instead, the rate responds according to the exponent appearing in the experimentally established rate law. The doubled concentration of B contributes a factor of unity because its partial order is zero. \(\textbf{Final answer:}\) \[ \boxed{\left(\frac{r'}r,\frac{k'}k\right)=(2,1)} \]

Explanation: The problem involves two successive changes in reactant concentrations. The first change alters A and B in opposite directions. The second change dilutes both concentrations simultaneously. Since the temperature and kinetic conditions remain unchanged, the rate constant is the same throughout. \(\textbf{Initial reaction rate:}\) \[ r_0=k[\mathrm A]_0^2[\mathrm B]_0 \] \(\textbf{Stage 1 — Prepare the second mixture:}\) The new concentrations are \[ [\mathrm A]_1=2[\mathrm A]_0 \] \[ [\mathrm B]_1=\frac14[\mathrm B]_0 \] Using the rate law, \[ r_1=k(2[\mathrm A]_0)^2\left(\frac14[\mathrm B]_0\right) \] \[ r_1=k(4[\mathrm A]_0^2)\left(\frac14[\mathrm B]_0\right) \] \[ r_1=k[\mathrm A]_0^2[\mathrm B]_0=r_0 \] The fourfold increase arising from the squared concentration of A exactly compensates for the fourfold reduction in the concentration of B. \(\textbf{Stage 2 — Dilute the second mixture:}\) Adding an equal volume of solvent doubles the solution volume. Because no reaction occurs during this dilution, the amounts of A and B remain unchanged while both their concentrations are halved. \[ [\mathrm A]_2=\frac12[\mathrm A]_1 \] \[ [\mathrm B]_2=\frac12[\mathrm B]_1 \] The rate after dilution is \[ r_2=k\left(\frac{[\mathrm A]_1}{2}\right)^2\left(\frac{[\mathrm B]_1}{2}\right) \] \[ r_2=\frac{1}{4}\times\frac12\,k[\mathrm A]_1^2[\mathrm B]_1 \] \[ r_2=\frac18r_1 \] Since \(r_1=r_0\), \[ r_2=\frac18r_0 \] \(\textbf{Consistency check:}\) The overall concentration of A returns to its original value, whereas the overall concentration of B becomes one-eighth of its original value. The final rate must therefore be one-eighth of the original rate, in agreement with the staged calculation. \(\textbf{Final answer:}\) \[ \boxed{r_2=\frac{r_0}{8}} \]

Explanation: The initial-rate method allows the concentration dependence of each reactant to be identified experimentally. The exponents \(m\) and \(n\) must be determined from experiments in which one reactant concentration changes while the other remains constant. The stoichiometric coefficients of the overall reaction, even if known, are not required for this determination. \(\textbf{Determine the partial order in A:}\) Compare Experiments 1 and 2. The concentration of B remains constant, while the concentration of A doubles. The observed rate decreases from \(0.020\) to \(0.010\ \mathrm{mol\,L^{-1}\,s^{-1}}\). Using the rate ratio, \[ \frac{r_2}{r_1}=\left(\frac{[\mathrm A]_2}{[\mathrm A]_1}\right)^m \] \[ \frac{0.010}{0.020}=\left(\frac{0.20}{0.10}\right)^m \] \[ \frac12=2^m \] Therefore, \[ m=-1 \] A negative partial order means that increasing the concentration of A decreases the rate under this experimentally established law. \(\textbf{Determine the partial order in B:}\) Compare Experiments 1 and 3. The concentration of A remains constant, while B doubles. The observed rate becomes four times its original value. \[ \frac{r_3}{r_1}=\left(\frac{[\mathrm B]_3}{[\mathrm B]_1}\right)^n \] \[ \frac{0.080}{0.020}=\left(\frac{0.20}{0.10}\right)^n \] \[ 4=2^n \] Therefore, \[ n=2 \] \(\textbf{Established rate law:}\) \[ r=k[\mathrm A]^{-1}[\mathrm B]^2 \] The overall order is \(m+n=-1+2=1\), but the individual exponents determine how the rate changes when both concentrations are altered. \(\textbf{Predict the rate in Experiment 4:}\) Compare Experiment 4 directly with Experiment 1. The concentration of A doubles, while the concentration of B becomes four times its original value. \[ \frac{r_4}{r_1}=\left(\frac{0.20}{0.10}\right)^{-1}\left(\frac{0.40}{0.10}\right)^2 \] \[ \frac{r_4}{r_1}=2^{-1}\times4^2 \] \[ \frac{r_4}{r_1}=\frac12\times16 \] \[ \frac{r_4}{r_1}=8 \] Thus, \[ r_4=8(0.020) \] \[ r_4=0.160\ \mathrm{mol\,L^{-1}\,s^{-1}} \] \(\textbf{Physical interpretation:}\) Doubling A reduces the rate by a factor of two because its partial order is negative one. Increasing B fourfold raises the rate by a factor of sixteen because its partial order is two. The combined effect is an eightfold increase. These factors multiply rather than add, and the rate constant remains unchanged because the temperature and kinetic conditions are fixed. \(\textbf{Final answer:}\) \[ \boxed{r_4=0.160\ \mathrm{mol\,L^{-1}\,s^{-1}}} \]

Explanation: The concentration of A decreases during the reaction, but the corresponding concentration of B must first be determined from the reaction stoichiometry. The experimentally established rate law contains a negative partial order with respect to A. Consequently, decreasing A can increase the rate, even though A is a reactant. However, B has a positive partial order, and its consumption tends to reduce the rate. The actual change depends on the combined effects. \(\textbf{Stoichiometric concentration change:}\) For the reaction \[ 2\mathrm A+\mathrm B\rightarrow\mathrm P \] two stoichiometric units of A are consumed for every unit of B consumed. The decrease in A is \[ -\Delta[\mathrm A]=0.80-0.40 \] \[ -\Delta[\mathrm A]=0.40\ \mathrm{mol\,L^{-1}} \] Therefore, the corresponding decrease in B is \[ -\Delta[\mathrm B]=\frac12(0.40) \] \[ -\Delta[\mathrm B]=0.20\ \mathrm{mol\,L^{-1}} \] The concentration of B at the later instant is \[ [\mathrm B]_t=0.50-0.20 \] \[ [\mathrm B]_t=0.30\ \mathrm{mol\,L^{-1}} \] \(\textbf{Apply the experimental rate law:}\) \[ r=k[\mathrm A]^{-2}[\mathrm B] \] At the two instants, \[ \frac{r_t}{r_0}=\left(\frac{[\mathrm A]_t}{[\mathrm A]_0}\right)^{-2}\left(\frac{[\mathrm B]_t}{[\mathrm B]_0}\right) \] Substitution gives \[ \frac{r_t}{r_0}=\left(\frac{0.40}{0.80}\right)^{-2}\left(\frac{0.30}{0.50}\right) \] \[ \frac{r_t}{r_0}=\left(\frac12\right)^{-2}\left(\frac35\right) \] \[ \frac{r_t}{r_0}=4\times0.60 \] \[ \frac{r_t}{r_0}=2.4 \] \(\textbf{Chemical interpretation:}\) The decrease in A increases the rate by a factor of four because its partial order is negative two. The simultaneous decrease in B multiplies the rate by \(0.60\). The resulting rate is therefore \(2.4\) times the initial rate. This illustrates that reactant consumption does not necessarily imply a decreasing instantaneous rate when an experimentally determined rate law contains negative concentration dependence. \(\textbf{Final answer:}\) \[ \boxed{\frac{r_t}{r_0}=2.4} \]

Explanation: The two systems have the same overall stoichiometric coefficients, but these coefficients do not determine their experimentally observed rate laws. System I is second order with respect to A and zero order with respect to B. System II is first order with respect to both A and B. Therefore, the same concentration changes produce different rate responses. The rate constants remain unchanged within their respective systems because only the initial concentrations are being altered. \(\textbf{Rate change in System I:}\) \[ r_1=k_1[\mathrm A]^2 \] When the concentration of A becomes half its original value, \[ \frac{r_1'}{r_0}=\left(\frac12\right)^2 \] \[ r_1'=\frac{r_0}{4} \] Increasing B has no effect on System I because B does not appear in its rate law. \(\textbf{Rate change in System II:}\) \[ r_2=k_2[\mathrm A][\mathrm B] \] The concentration factors are \(1/2\) for A and \(3\) for B. \[ \frac{r_2'}{r_0}=\frac12\times3 \] \[ r_2'=\frac{3r_0}{2} \] \(\textbf{Compare the altered rates:}\) \[ \frac{r_2'}{r_1'}=\frac{3r_0/2}{r_0/4} \] \[ \frac{r_2'}{r_1'}=\frac32\times4 \] \[ \frac{r_2'}{r_1'}=6 \] The equality of the original rates does not require the two systems to respond identically to later concentration changes. The difference arises from their distinct experimental concentration dependencies, not from the shared stoichiometric equation. \(\textbf{Final answer:}\) \[ \boxed{\frac{r_2'}{r_1'}=6} \]

Explanation: The observed rate change results from two different influences. One is the change in reactant concentrations, and the other is the temperature-dependent change in the rate constant. The two contributions must be separated before the rate-constant ratio can be obtained. \(\textbf{Rate at the first temperature:}\) \[ r_1=k_1[\mathrm A]_1^2[\mathrm B]_1 \] \(\textbf{Rate under the altered conditions:}\) The new concentrations are \[ [\mathrm A]_2=\frac12[\mathrm A]_1 \] \[ [\mathrm B]_2=2[\mathrm B]_1 \] Therefore, \[ r_2=k_2\left(\frac{[\mathrm A]_1}{2}\right)^2(2[\mathrm B]_1) \] Simplifying, \[ r_2=k_2\left(\frac14[\mathrm A]_1^2\right)(2[\mathrm B]_1) \] \[ r_2=\frac{k_2}{2}[\mathrm A]_1^2[\mathrm B]_1 \] \(\textbf{Separate temperature and concentration effects:}\) Dividing the second rate by the first, \[ \frac{r_2}{r_1}=\frac{k_2}{k_1}\times\frac12 \] The observed rate ratio is \(1.5\). \[ 1.5=\frac{k_2}{k_1}\times\frac12 \] Hence, \[ \frac{k_2}{k_1}=1.5\times2 \] \[ \frac{k_2}{k_1}=3 \] \(\textbf{Consistency check:}\) If the temperature had remained unchanged, the concentration changes alone would have reduced the rate to half its original value. The observed rate instead increases to \(1.5\) times its original value. The temperature-dependent rate constant must therefore have increased by a factor of three. The final result is dimensionless because the same rate constant is compared at two temperatures. \(\textbf{Final answer:}\) \[ \boxed{\frac{k_2}{k_1}=3} \]

Explanation: The Assertion is false because the frequently used approximate rule concerning a temperature rise of 10 K is not an exact universal mathematical law. For many reactions, the rate constant increases substantially when temperature rises, often by a factor of approximately two or three over a 10 K interval near ordinary temperatures. However, the actual multiplier depends on the particular reaction and the temperature range. An approximate doubling between 298 K and 308 K does not guarantee an exact doubling between 308 K and 318 K. The Reason is true because the Arrhenius equation relates the rate constant to absolute temperature and activation energy. Its mathematical form is \[ k=Ae^{-E_a/(RT)}. \] Here, \(A\) is the frequency factor, \(E_a\) is the activation energy, and \(R\) is the gas constant. For a temperature-independent activation energy and frequency factor over the interval, the ratio of rate constants depends on the difference between the reciprocals of the two absolute temperatures. Equal temperature increases do not correspond to equal changes in reciprocal temperature. Consequently, successive increases of 10 K need not produce exactly identical rate multipliers. The Reason identifies the scientific basis for rejecting the exact universal doubling claim in the Assertion. The important distinction is between an approximate empirical observation and the quantitative temperature dependence described by the Arrhenius equation.