101. A reaction mixture contains one strongly coloured reactant, while all other reactants and products are nearly colourless at the selected wavelength. Continuous absorbance measurement is especially suitable because:
ⓐ. absorbance directly gives the balanced-equation coefficients
ⓑ. absorbance tracks the coloured reactant concentration
ⓒ. colour guarantees that the reaction is zero order
ⓓ. absorbance is independent of concentration
Correct Answer: absorbance tracks the coloured reactant concentration
Explanation: Selecting a wavelength absorbed mainly by one species makes the measured signal comparatively specific to that species. As the coloured reactant is consumed, its absorbance changes in a way that can be related to its concentration. Continuous recording produces many closely spaced data points without repeatedly disturbing the mixture. The method does not determine reaction order merely from the presence of colour. Order must still be established from the kinetic relationship between rate and concentration or from appropriate integrated analysis.
102. A complete list of common factors that can influence reaction rate includes:
ⓐ. concentration, gas pressure, temperature, catalyst, physical state, reaction enthalpy, and light
ⓑ. concentration, gas pressure, equilibrium constant, entropy, surface area, electrical charge, and light
ⓒ. temperature, catalyst, physical state, product yield, stoichiometric coefficients, reaction medium, and light
ⓓ. concentration, pressure, temperature, catalyst, physical state, surface area, medium, reactant nature, and light
Correct Answer: concentration, pressure, temperature, catalyst, physical state, surface area, medium, reactant nature, and light
Explanation: Reaction rate can respond to several chemical and physical conditions. Concentration and gaseous pressure affect how frequently reacting particles encounter one another, while temperature changes the energetic distribution of particles. Catalysts provide an alternative pathway, and surface area matters especially in heterogeneous reactions. Solvent, physical state, reactant nature, and absorbed light can also influence particular systems. No single factor has the same importance in every reaction, so the chemical context must be stated.
103. A process has a negative Gibbs energy change under certain conditions but proceeds extremely slowly. This observation shows that:
ⓐ. thermodynamic feasibility does not guarantee a rapid reaction
ⓑ. every spontaneous reaction must be instantaneous
ⓒ. a slow reaction must have a positive Gibbs energy change
ⓓ. reaction rate and equilibrium tendency are identical quantities
Correct Answer: thermodynamic feasibility does not guarantee a rapid reaction
Explanation: Thermodynamic feasibility indicates whether a process has a favourable overall driving tendency under specified conditions. Kinetics determines how quickly the process can proceed along an available pathway. A reaction may be favourable yet remain slow because it faces a substantial activation barrier. Conversely, observing a rapid process does not by itself provide the equilibrium constant or Gibbs energy change. Rate and feasibility answer different chemical questions and must be evaluated separately.
104. Increasing the concentration of a reactant often increases the reaction rate because:
ⓐ. the activation energy of the reaction necessarily becomes zero
ⓑ. more particles per unit volume create more possible encounters
ⓒ. the equilibrium constant always becomes larger
ⓓ. every collision becomes chemically effective
Correct Answer: more particles per unit volume create more possible encounters
Explanation: A higher concentration places more reactant particles within a given volume. This generally increases the frequency with which reacting particles encounter one another. If a suitable fraction of those encounters is effective, the reaction rate rises. Increasing concentration does not guarantee that every collision has sufficient energy and proper orientation. The exact rate change is determined by the experimentally observed rate law rather than by a universal direct-proportionality rule.
105. The rate law is \(r=k[A]^2[B]\). If \([A]\) is doubled while \([B]\) is reduced to one-half of its original value at constant temperature, the new rate is:
ⓐ. twice the original rate
ⓑ. one-half of the original rate
ⓒ. equal to the original rate
ⓓ. four times the original rate
Correct Answer: twice the original rate
Explanation: \( \textbf{Original rate law:} \)
\[
r_1=k[A]_1^2[B]_1
\]
\( \textbf{Changed concentrations:} \)
\[
[A]_2=2[A]_1,\qquad [B]_2=\frac{1}{2}[B]_1
\]
\( \textbf{New-rate expression:} \)
\[
r_2=k(2[A]_1)^2\left(\frac{1}{2}[B]_1\right)
\]
\( \textbf{Concentration factors:} \)
\[
(2)^2\times\frac{1}{2}=4\times\frac{1}{2}
\]
\( \textbf{Rate ratio:} \)
\[
\frac{r_2}{r_1}=2
\]
\( \textbf{Interpretation:} \) The fourfold effect produced by doubling \([A]\) is partly offset by halving \([B]\).
\( \textbf{Final answer:} \) The new rate is twice the original rate.
106. Consider the following statements about the effect of reactant concentration on rate.
Statement I: Doubling a reactant concentration doubles the rate only when the reaction is first order with respect to that reactant.
Statement II: A zero-order rate is independent of reactant concentration within the range where zero-order behaviour applies.
Statement III: The concentration effect must be obtained from the rate law of the reaction.
The valid statements are:
ⓐ. I, II and III
ⓑ. I and II only
ⓒ. II and III only
ⓓ. I and III only
Correct Answer: I, II and III
Explanation: If the rate contains the factor \([A]^1\), doubling \([A]\) doubles the rate while other conditions remain fixed. For a zero-order dependence, the concentration term is effectively \([A]^0=1\), so changing \([A]\) does not change the rate within that kinetic regime. Other orders produce different responses, such as a fourfold change for second order or a factor of \(\sqrt{2}\) for half order. The balanced equation alone cannot determine these exponents for an overall reaction. Concentration effects must therefore be interpreted through the experimentally established rate law.
107. A graph is drawn with reaction rate on the vertical axis and reactant concentration \([A]\) on the horizontal axis. The graph is a horizontal straight line over the measured range. This graph indicates:
ⓐ. first-order dependence on \(A\)
ⓑ. second-order dependence on \(A\)
ⓒ. zero-order dependence on \(A\)
ⓓ. inverse first-order dependence on \(A\)
Correct Answer: zero-order dependence on \(A\)
Explanation: A horizontal rate-versus-concentration graph shows that the rate does not change as \([A]\) changes. This behaviour corresponds to a concentration exponent of zero. The rate law with respect to \(A\) is therefore proportional to \([A]^0\), which equals unity. A first-order graph would be a straight line through the origin, while a second-order graph would curve upward when rate is plotted directly against concentration. The zero-order interpretation applies only over the experimental range in which the horizontal behaviour is observed.
108. Assertion: Increasing a reactant concentration does not always increase the reaction rate.
Reason: A reaction may be zero order with respect to that reactant under the stated conditions.
ⓐ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓑ. Both Assertion and Reason are true, and Reason explains Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: A zero-order rate law does not contain a concentration-dependent factor for the stated reactant. Its rate may be written as \(r=k[A]^0=k\). Increasing \([A]\) therefore leaves the rate unchanged while the zero-order conditions remain valid. This provides a direct exception to the claim that every concentration increase must raise the rate. The Reason supplies the kinetic condition that explains the Assertion rather than merely describing an unrelated possibility.
109. The rate law is \(r=k[A]^{1/2}[B]^0\). The concentration of \(A\) is increased from \(0.16\,mol\,L^{-1}\) to \(0.36\,mol\,L^{-1}\), while \([B]\) is tripled. The new rate is:
ⓐ. \(1.5\) times the original rate
ⓑ. \(2.25\) times the original rate
ⓒ. \(3.0\) times the original rate
ⓓ. \(4.5\) times the original rate
Correct Answer: \(1.5\) times the original rate
Explanation: \( \textbf{Rate law:} \)
\[
r=k[A]^{1/2}[B]^0
\]
\( \textbf{Effect of }B\textbf{:} \)
\[
[B]^0=1
\]
\( \textbf{Concentration ratio for }A\textbf{:} \)
\[
\frac{[A]_2}{[A]_1}=\frac{0.36}{0.16}=2.25
\]
\( \textbf{Rate-ratio relation:} \)
\[
\frac{r_2}{r_1}=\left(\frac{[A]_2}{[A]_1}\right)^{1/2}
\]
\( \textbf{Substitution:} \)
\[
\frac{r_2}{r_1}=(2.25)^{1/2}
\]
\( \textbf{Calculation:} \)
\[
\frac{r_2}{r_1}=1.5
\]
\( \textbf{Condition interpretation:} \) Tripling \([B]\) has no effect because the partial order with respect to \(B\) is zero.
\( \textbf{Final answer:} \) The rate becomes \(1.5\) times its original value.
110. Select the row in which the stated concentration change and rate change are consistent.
| Row | Order with respect to \(A\) | Change in \([A]\) | Predicted rate factor |
|---|
| P | \(1\) | Tripled | \(9\) |
| Q | \(2\) | Doubled | \(2\) |
| R | \(0\) | Quadrupled | \(4\) |
| S | \(\frac{1}{2}\) | Quadrupled | \(2\) |
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row S
Explanation: For an order \(m\), changing \([A]\) by a factor \(f\) changes the rate by \(f^m\). A first-order reaction would respond to tripling by a factor of \(3\), not \(9\). A second-order reaction would respond to doubling by a factor of \(2^2=4\), not \(2\). A zero-order rate would remain unchanged when the concentration is quadrupled. For half-order dependence, quadrupling the concentration changes the rate by \(4^{1/2}=2\), so Row S is internally consistent.
111. The rate law is \(r=k[A][B]^2\). A solution containing fixed amounts of \(A\) and \(B\) is suddenly diluted to twice its original volume, with temperature unchanged. Immediately after dilution, the rate becomes:
ⓐ. one-half of the original rate
ⓑ. one-fourth of the original rate
ⓒ. one-eighth of the original rate
ⓓ. one-sixteenth of the original rate
Correct Answer: one-eighth of the original rate
Explanation: \( \textbf{Volume change:} \) The volume is doubled while the amounts of both reactants remain fixed.
\( \textbf{Concentration effect:} \)
\[
[A]_2=\frac{1}{2}[A]_1,\qquad [B]_2=\frac{1}{2}[B]_1
\]
\( \textbf{Rate law:} \)
\[
r=k[A][B]^2
\]
\( \textbf{Rate ratio:} \)
\[
\frac{r_2}{r_1}
=\left(\frac{[A]_2}{[A]_1}\right)
\left(\frac{[B]_2}{[B]_1}\right)^2
\]
\( \textbf{Substitution:} \)
\[
\frac{r_2}{r_1}
=\left(\frac{1}{2}\right)
\left(\frac{1}{2}\right)^2
\]
\( \textbf{Calculation:} \)
\[
\frac{r_2}{r_1}=\frac{1}{2}\times\frac{1}{4}
=\frac{1}{8}
\]
\( \textbf{Order interpretation:} \) Dilution acts on both concentration factors, and the squared dependence on \(B\) produces the larger contribution.
\( \textbf{Final answer:} \) The immediate rate becomes one-eighth of its original value.
112. The claim “doubling the concentration of any reactant always doubles the reaction rate” is unreliable because:
ⓐ. concentration can never be changed without altering temperature
ⓑ. the rate response depends on its rate-law exponent
ⓒ. rate constants become zero when concentration is doubled
ⓓ. reaction rates are independent of all concentrations
Correct Answer: the rate response depends on its rate-law exponent
Explanation: If the rate law contains \([A]^m\), doubling \([A]\) changes the rate by the factor \(2^m\). The factor is \(2\) only when \(m=1\). For \(m=2\), the rate becomes four times larger, while for \(m=0\), it remains unchanged. Fractional orders produce non-integral factors such as \(\sqrt{2}\). The concentration response is therefore a kinetic property established by experiment rather than a universal doubling rule.
113. Compressing a gaseous reaction mixture at constant temperature generally raises the concentrations of gaseous reactants because:
ⓐ. the molar masses of the reactants increase
ⓑ. the activation energy automatically decreases
ⓒ. the gas occupies a smaller volume
ⓓ. the gas constant becomes larger
Correct Answer: the gas occupies a smaller volume
Explanation: Gas concentration is the amount of gas divided by the volume it occupies. Compression decreases the volume while the gaseous amounts initially remain unchanged. The concentrations and partial pressures of the gaseous reactants therefore increase. A higher concentration can alter the reaction rate according to the gaseous-reactant powers in the rate law. Compression does not change molar mass, the gas constant, or necessarily the activation energy.
114. Raising the external gas pressure has no necessary direct effect on the rate of a reaction occurring only between dissolved non-gaseous reactants because:
ⓐ. solution reactions never depend on concentration
ⓑ. reactant concentrations in solution may barely change
ⓒ. pressure can affect only equilibrium and never kinetics
ⓓ. dissolved reactants always behave as zero-order species
Correct Answer: reactant concentrations in solution may barely change
Explanation: Pressure strongly influences gas concentrations because gases are readily compressed. Liquids and solutions are much less compressible under ordinary conditions. If no gaseous reactant participates and the dissolved concentrations remain essentially unchanged, an external pressure increase need not alter the rate appreciably. This does not mean that all solution reactions are pressure-independent under every condition. The absence of a universal pressure effect follows from whether pressure actually changes the kinetic variables appearing in the rate law.
115. For ideal gaseous reactants, evaluate the effects described below.
Statement I: At fixed temperature, the partial pressure of a gas is proportional to its molar concentration.
Statement II: The rate response to pressure depends on the powers of gaseous-reactant terms in the rate law.
Statement III: Doubling every reactant partial pressure does not always merely double the rate.
The valid statements are:
ⓐ. I and II only
ⓑ. II and III only
ⓒ. I and III only
ⓓ. I, II and III
Correct Answer: I, II and III
Explanation: For an ideal gas, \(P_i=c_iRT\), so partial pressure and molar concentration are proportional at fixed temperature. A gaseous rate law may therefore be written using concentration or pressure variables when the appropriate form is established. If the overall pressure dependence is greater than one, doubling all relevant partial pressures can increase the rate by more than a factor of two. A zero-order pressure dependence would produce no rate change. The pressure response must be calculated from the actual exponents rather than assumed from the total-pressure change alone.
116. An inert gas is added to a reacting ideal-gas mixture at constant temperature and constant volume. The total pressure rises, but the amounts and partial pressures of the reacting gases remain unchanged. The immediate reaction rate:
ⓐ. remains unchanged because the reactant partial pressures remain unchanged
ⓑ. doubles because the total pressure has doubled
ⓒ. becomes zero because the mixture has been diluted by an inert gas
ⓓ. increases in direct proportion to the inert-gas pressure
Correct Answer: remains unchanged because the reactant partial pressures remain unchanged
Explanation: At constant volume and temperature, the partial pressure of each reacting gas depends on its own amount through \(P_i=\frac{n_iRT}{V}\). Adding an inert gas changes neither \(n_i\) nor \(V\) for the reactants. Their partial pressures therefore remain unchanged even though the total pressure increases. A rate law involving only the reacting-gas partial pressures consequently gives the same immediate rate. Total pressure alone is not a sufficient kinetic variable when its increase comes only from an inert component.
117. For a gaseous reaction obeying \(r=k(P_A)^2\) at fixed temperature, a graph of rate against \(P_A\) should be:
ⓐ. a horizontal line because rate is independent of \(P_A\)
ⓑ. a descending straight line because rate falls linearly with \(P_A\)
ⓒ. a rising straight line through the origin because \(r\propto P_A\)
ⓓ. an upward curve through the origin, since \(r\propto P_A^2\)
Correct Answer: an upward curve through the origin, since \(r\propto P_A^2\)
Explanation: The rate depends on the square of the reactant partial pressure. When \(P_A\) is doubled, the rate changes by the factor \(2^2=4\). A direct plot of \(r\) against \(P_A\) therefore follows a quadratic rather than a linear relation. The curve passes through the origin because \(r=0\) when \(P_A=0\) in this rate law. A straight-line graph would instead be obtained by plotting \(r\) against \((P_A)^2\).
118. A gaseous reaction obeys \(r=kP_A(P_B)^2\). Its initial rate is \(1.00\times10^{-3}\,mol\,L^{-1}\,s^{-1}\). At constant temperature, the mixture is compressed from volume \(V\) to \(\frac{2V}{3}\), without changing its composition. The new initial rate is:
ⓐ. \(2.25\times10^{-3}\,mol\,L^{-1}\,s^{-1}\)
ⓑ. \(3.375\times10^{-3}\,mol\,L^{-1}\,s^{-1}\)
ⓒ. \(4.50\times10^{-3}\,mol\,L^{-1}\,s^{-1}\)
ⓓ. \(6.75\times10^{-3}\,mol\,L^{-1}\,s^{-1}\)
Correct Answer: \(3.375\times10^{-3}\,mol\,L^{-1}\,s^{-1}\)
Explanation: \( \textbf{Initial rate:} \)
\[
r_1=1.00\times10^{-3}\,mol\,L^{-1}\,s^{-1}
\]
\( \textbf{Volume change:} \)
\[
V_2=\frac{2}{3}V_1
\]
\( \textbf{Partial-pressure factor:} \) At constant temperature and unchanged gaseous amounts, pressure varies inversely with volume.
\[
\frac{P_2}{P_1}=\frac{V_1}{V_2}
=\frac{V_1}{\frac{2}{3}V_1}
=\frac{3}{2}
\]
\( \textbf{Rate law:} \)
\[
r=kP_A(P_B)^2
\]
\( \textbf{Overall pressure exponent:} \)
\[
1+2=3
\]
\( \textbf{Rate factor:} \)
\[
\frac{r_2}{r_1}
=\left(\frac{3}{2}\right)^3
\]
\[
\frac{r_2}{r_1}=\frac{27}{8}=3.375
\]
\( \textbf{New rate:} \)
\[
r_2=3.375\left(1.00\times10^{-3}\right)
\]
\[
r_2=3.375\times10^{-3}\,mol\,L^{-1}\,s^{-1}
\]
\( \textbf{Final answer:} \) The new initial rate is \(3.375\times10^{-3}\,mol\,L^{-1}\,s^{-1}\).
119. When the temperature of a reacting system is raised while concentrations and other conditions are kept unchanged, the reaction rate generally:
ⓐ. decreases because particles move farther apart
ⓑ. remains unchanged because the balanced equation is unchanged
ⓒ. increases because the rate constant usually increases
ⓓ. becomes independent of the nature of the reactants
Correct Answer: increases because the rate constant usually increases
Explanation: Raising temperature generally increases the rate constant \(k\) for reactions with ordinary positive activation energies. The rate law may retain the same concentration terms, but the larger value of \(k\) produces a greater rate at the same concentrations. The balanced chemical equation does not specify how rapidly the reaction occurs. Temperature also affects the energy distribution of reacting particles, increasing the fraction able to undergo successful reaction events. The exact increase depends on the reaction and the temperature range rather than on one universal factor.
120. Assertion: A reaction can proceed faster at a higher temperature even when the reactant concentrations are initially the same.
Reason: The rate constant of the reaction commonly increases with temperature.
ⓐ. Both Assertion and Reason are true, and Reason explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: A rate law has the general structure of a rate constant multiplied by concentration-dependent terms. If the initial concentrations are held fixed, the concentration factors remain the same in the two experiments. A higher temperature commonly produces a larger value of \(k\), so the calculated reaction rate increases. The Reason therefore directly explains the Assertion. The temperature effect should not be attributed to an automatic increase in reactant concentration when the mixture composition has not changed.
