1. Chemical kinetics is primarily concerned with:
ⓐ. the equilibrium composition of a reacting system
ⓑ. reaction speed and the factors affecting it
ⓒ. the heat released or absorbed during every reaction
ⓓ. the electronic structures of isolated atoms
Correct Answer: reaction speed and the factors affecting it
Explanation: Chemical kinetics studies how rapidly chemical reactions occur and how their rates respond to changing conditions. It examines factors such as concentration, temperature, catalysts, and the nature of reactants. Equilibrium studies the final balance between forward and reverse processes, whereas kinetics focuses on how quickly the system changes. A reaction may be thermodynamically favourable but proceed extremely slowly. Rate and feasibility are related to different questions and should not be treated as interchangeable ideas.
2. Which pair places the relatively fast process first and the relatively slow process second under ordinary conditions?
ⓐ. Rusting of iron; burning of paper
ⓑ. Food spoilage; acid-base neutralisation
ⓒ. Burning of cooking gas; rusting of iron
ⓓ. Fermentation; formation of an ionic precipitate
Correct Answer: Burning of cooking gas; rusting of iron
Explanation: Burning of cooking gas occurs rapidly once ignition has taken place, whereas visible rusting usually develops over a much longer period. Reaction speed is compared by observing the time required for a noticeable chemical change. Acid-base neutralisation and many precipitation reactions are generally fast in solution. Food spoilage and fermentation are comparatively slower processes under ordinary conditions. A familiar example helps illustrate rate, but the precise speed still depends on the stated conditions.
3. Two mixtures contain the same reactants in the same proportions but are maintained at different temperatures. Their reaction speeds:
ⓐ. must be equal because the chemical equation is unchanged
ⓑ. depend only on the total masses of the reactants
ⓒ. become equal because the same products are obtained
ⓓ. may differ because reaction conditions influence rate
Correct Answer: may differ because reaction conditions influence rate
Explanation: A balanced chemical equation identifies the reactants and products but does not fix how rapidly the reaction occurs. Temperature can change the fraction of reacting particles able to undergo successful reaction events. Thus, two chemically identical mixtures may react at different speeds under different conditions. Obtaining the same products does not require the reactions to take the same time. Kinetic comparisons are meaningful only when the relevant experimental conditions are stated.
4. The symbols commonly used in chemical kinetics are represented appropriately by:
ⓐ. \(r\) for temperature, \(k\) for time, \(t\) for rate, and \(T\) for concentration
ⓑ. \(r\) for rate, \(k\) for rate constant, \(t\) for time, and \(T\) for temperature
ⓒ. \(r\) for amount, \(k\) for concentration, \(t\) for temperature, and \(T\) for rate
ⓓ. \(r\) for volume, \(k\) for pressure, \(t\) for concentration, and \(T\) for time
Correct Answer: \(r\) for rate, \(k\) for rate constant, \(t\) for time, and \(T\) for temperature
Explanation: The symbol \(r\) is commonly used for reaction rate, while \(k\) represents the rate constant in a rate law. Time is denoted by \(t\), and absolute temperature is written as \(T\). These symbols acquire precise meanings when they appear in kinetic equations. The rate \(r\) and the rate constant \(k\) are not identical quantities, even though both are connected through a rate law. Keeping \(t\) and \(T\) distinct also prevents confusion between time and temperature.
5. Match each quantity in Column I with its usual unit or symbol in Column II.
| Column I | Column II |
|---|
| P. Concentration | 1. \(K\) |
| Q. Time | 2. \(mol\,L^{-1}\) |
| R. Absolute temperature | 3. \(s\) |
| S. Reaction rate | 4. \(mol\,L^{-1}\,s^{-1}\) |
ⓐ. P-2, Q-3, R-1, S-4
ⓑ. P-3, Q-2, R-1, S-4
ⓒ. P-2, Q-1, R-3, S-4
ⓓ. P-4, Q-3, R-1, S-2
Correct Answer: P-2, Q-3, R-1, S-4
Explanation: Concentration is commonly measured in \(mol\,L^{-1}\), so P matches 2. Time may be expressed in seconds, giving Q-3, while absolute temperature is measured in kelvin, giving R-1. Rate combines concentration change with time and therefore commonly has the unit \(mol\,L^{-1}\,s^{-1}\). The unit of rate contains both a concentration part and an inverse-time part. It should not be confused with the unit of the rate constant, whose form can depend on reaction order.
6. Which role of a balanced chemical equation is most relevant as a prerequisite for studying reaction rates?
ⓐ. It directly supplies the numerical value of the rate constant
ⓑ. It proves that every permitted reaction will occur rapidly
ⓒ. It gives stoichiometric relations among reactants and products
ⓓ. It determines the temperature required for the reaction
Correct Answer: It gives stoichiometric relations among reactants and products
Explanation: A balanced equation shows the relative stoichiometric amounts in which reactants are consumed and products are formed. These coefficients later help relate concentration changes of different species during a reaction. The equation alone does not provide the rate constant or the experimental rate law. It also does not show whether a reaction will be fast under a particular set of conditions. Stoichiometry describes proportional chemical change, while kinetics adds the time dependence of that change.
7. A reactant concentration may be reported in ______.
ⓐ. \(s^{-1}\)
ⓑ. \(K\,s^{-1}\)
ⓒ. \(L\,mol^{-1}\)
ⓓ. \(mol\,L^{-1}\)
Correct Answer: \(mol\,L^{-1}\)
Explanation: Concentration expresses the amount of a substance present per unit volume. When amount is measured in moles and volume in litres, the unit is \(mol\,L^{-1}\). The unit \(s^{-1}\) contains only an inverse-time term and is not a concentration unit. The unit \(L\,mol^{-1}\) is the reciprocal of molar concentration. Recognising the denominator is essential when concentration is later combined with time to express reaction rate.
8. A reaction is observed to occur very rapidly. The observation alone establishes that:
ⓐ. its rate is high under the stated conditions
ⓑ. its equilibrium constant must be extremely large
ⓒ. its enthalpy change must be negative
ⓓ. its products must be more stable than all possible reactants
Correct Answer: its rate is high under the stated conditions
Explanation: A rapid visible change provides information about how quickly the reaction proceeds under the stated conditions. It does not by itself determine the equilibrium constant, reaction enthalpy, or complete thermodynamic stability. Kinetic measurements concern the time scale of the process. Thermodynamic quantities concern energetic feasibility and equilibrium tendency. A reaction can be fast or slow regardless of whether its enthalpy change is positive or negative.
9. Consider the following statements about basic kinetic notation.
Statement I: \(t\) commonly represents time.
Statement II: \(T\) commonly represents reactant concentration.
Statement III: \(k\) commonly represents the rate constant.
The valid statements are:
ⓐ. I and II only
ⓑ. I and III only
ⓒ. II and III only
ⓓ. I, II and III
Correct Answer: I and III only
Explanation: The symbol \(t\) is conventionally used for time in kinetic relations. The capital symbol \(T\) represents absolute temperature rather than reactant concentration. A concentration is usually written using the chemical species in square brackets, such as \([R]\). The symbol \(k\) denotes the rate constant for a specified rate law under stated conditions. Confusing \(t\) with \(T\) can lead to using time and temperature in the wrong parts of later equations.
10. In the concentration expression used as a foundation for chemical kinetics, concentration represents:
ⓐ. amount of substance multiplied by the volume occupied
ⓑ. amount of substance present per unit volume
ⓒ. total mass of all substances divided by time
ⓓ. reaction temperature per mole of reactant
Correct Answer: amount of substance present per unit volume
Explanation: Concentration connects the amount of a chemical species with the volume in which that amount is present. Molar concentration is commonly expressed as moles per litre of solution. Multiplying amount by volume would not describe how densely the particles are distributed. Time and temperature are separate kinetic variables and are not part of the basic concentration definition. Concentration-time data later show how the composition of a reacting mixture changes as the reaction proceeds.
11. A concentration-time graph is to be drawn for a reacting species. The conventional placement of variables is:
ⓐ. time on the horizontal axis and concentration on the vertical axis
ⓑ. concentration on the horizontal axis and time on the vertical axis
ⓒ. temperature on the horizontal axis and time on the vertical axis
ⓓ. rate constant on the horizontal axis and concentration on the vertical axis
Correct Answer: time on the horizontal axis and concentration on the vertical axis
Explanation: Time is the independent variable in a concentration-time record because observations are made at selected times. Concentration is the dependent variable because its measured value changes as the reaction proceeds. Thus, \(t\) is placed on the horizontal axis and concentration on the vertical axis. Reversing the axes would also reverse the mathematical interpretation of the slope. The chosen orientation prepares the graph for later analysis of concentration change with time.
12. A vessel contains \(0.50\,mol\) of a reactant in a total solution volume of \(250\,mL\). Its molar concentration is:
ⓐ. \(0.50\,mol\,L^{-1}\)
ⓑ. \(1.0\,mol\,L^{-1}\)
ⓒ. \(4.0\,mol\,L^{-1}\)
ⓓ. \(2.0\,mol\,L^{-1}\)
Correct Answer: \(2.0\,mol\,L^{-1}\)
Explanation: \( \textbf{Known data:} \) Amount of reactant \(=0.50\,mol\) and solution volume \(=250\,mL\).
\( \textbf{Required quantity:} \) Molar concentration in \(mol\,L^{-1}\).
\( \textbf{Concentration relation:} \)
\[
C=\frac{n}{V}
\]
\( \textbf{Volume conversion:} \)
\[
250\,mL=0.250\,L
\]
\( \textbf{Substitution:} \)
\[
C=\frac{0.50\,mol}{0.250\,L}
\]
\( \textbf{Calculation:} \)
\[
C=2.0\,mol\,L^{-1}
\]
\( \textbf{Unit interpretation:} \) The result means that each litre of this solution would contain \(2.0\,mol\) of the reactant at the same composition.
\( \textbf{Final answer:} \) The molar concentration is \(2.0\,mol\,L^{-1}\); using \(250\) directly without converting \(mL\) to \(L\) would give an invalid unit-based result.
13. For a forward reaction monitored from its initial mixture, the usual concentration trends are:
ⓐ. reactant concentration decreases while product concentration increases
ⓑ. both reactant and product concentrations always decrease
ⓒ. reactant concentration increases while product concentration decreases
ⓓ. both concentrations must remain constant until the reaction ends
Correct Answer: reactant concentration decreases while product concentration increases
Explanation: Reactant particles are consumed as a forward reaction proceeds, so their concentrations generally decrease. Products are formed during the same progress, so their concentrations generally increase from their initial values. The exact amounts of change may differ because of stoichiometric coefficients. Concentrations need not change linearly with time, and their rates of change can vary during the process. These opposite trends explain why reactant concentration graphs usually slope downward while product concentration graphs slope upward.
14. On a graph with concentration on the vertical axis and time on the horizontal axis, the slope between two points represents:
ⓐ. concentration change divided by the corresponding time change
ⓑ. time change divided by the corresponding concentration change
ⓒ. the product of concentration and elapsed time
ⓓ. the sum of initial and final concentrations
Correct Answer: concentration change divided by the corresponding time change
Explanation: The slope of any graph is the change in the vertical-axis quantity divided by the change in the horizontal-axis quantity. Here the vertical quantity is concentration and the horizontal quantity is time. The slope is therefore written as \(\frac{\Delta[\text{species}]}{\Delta t}\). A decreasing concentration gives a negative signed slope, while an increasing concentration gives a positive signed slope. Taking the reciprocal would describe time change per concentration change and would not match the selected graph orientation.
15. On a reactant concentration-time graph, the concentration falls from \(0.80\,mol\,L^{-1}\) at \(0\,s\) to \(0.50\,mol\,L^{-1}\) at \(30\,s\). The signed slope of the line joining these points is:
ⓐ. \(+0.010\,mol\,L^{-1}\,s^{-1}\)
ⓑ. \(-0.010\,mol\,L^{-1}\,s^{-1}\)
ⓒ. \(-0.030\,mol\,L^{-1}\,s^{-1}\)
ⓓ. \(+0.030\,mol\,L^{-1}\,s^{-1}\)
Correct Answer: \(-0.010\,mol\,L^{-1}\,s^{-1}\)
Explanation: \( \textbf{Initial point:} \) At \(t_1=0\,s\), the concentration is \([R]_1=0.80\,mol\,L^{-1}\).
\( \textbf{Final point:} \) At \(t_2=30\,s\), the concentration is \([R]_2=0.50\,mol\,L^{-1}\).
\( \textbf{Slope relation:} \)
\[
\text{slope}=\frac{[R]_2-[R]_1}{t_2-t_1}
\]
\( \textbf{Concentration change:} \)
\[
\Delta[R]=0.50-0.80=-0.30\,mol\,L^{-1}
\]
\( \textbf{Time change:} \)
\[
\Delta t=30-0=30\,s
\]
\( \textbf{Substitution:} \)
\[
\text{slope}=\frac{-0.30\,mol\,L^{-1}}{30\,s}
\]
\( \textbf{Calculation:} \)
\[
\text{slope}=-0.010\,mol\,L^{-1}\,s^{-1}
\]
\( \textbf{Sign meaning:} \) The negative sign shows that the reactant concentration decreases as time increases.
\( \textbf{Final answer:} \) The signed slope is \(-0.010\,mol\,L^{-1}\,s^{-1}\); its magnitude is positive, but the graph slope itself remains negative.
16. A solution contains \(0.30\,mol\) of reactant in \(600\,mL\). Without changing the amount of reactant, the solution is concentrated to a final volume of \(300\,mL\). The final concentration is:
ⓐ. \(0.25\,mol\,L^{-1}\)
ⓑ. \(0.50\,mol\,L^{-1}\)
ⓒ. \(1.00\,mol\,L^{-1}\)
ⓓ. \(2.00\,mol\,L^{-1}\)
Correct Answer: \(1.00\,mol\,L^{-1}\)
Explanation: \( \textbf{Amount present:} \) The reactant amount remains \(0.30\,mol\).
\( \textbf{Final solution volume:} \)
\[
300\,mL=0.300\,L
\]
\( \textbf{Concentration relation:} \)
\[
C=\frac{n}{V}
\]
\( \textbf{Why this relation applies:} \) The required quantity is the amount of reactant per litre of the final solution.
\( \textbf{Substitution:} \)
\[
C=\frac{0.30\,mol}{0.300\,L}
\]
\( \textbf{Calculation:} \)
\[
C=1.00\,mol\,L^{-1}
\]
\( \textbf{Comparison with the initial state:} \) The initial concentration was \(\frac{0.30}{0.600}=0.50\,mol\,L^{-1}\).
\( \textbf{Volume effect:} \) Halving the volume while keeping the amount fixed doubles the concentration.
\( \textbf{Final answer:} \) The final concentration is \(1.00\,mol\,L^{-1}\).
17. Consider the following statements for a reaction progressing in the forward direction.
Statement I: Reactant concentrations generally decrease with time.
Statement II: Product concentrations generally increase with time.
Statement III: The stoichiometric coefficients help relate the amounts of different species consumed and formed.
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: Reactants are consumed during forward reaction progress, so their concentrations generally decrease. Products are formed, so their concentrations generally increase from their initial values. A balanced equation provides the stoichiometric ratios connecting the consumption and formation of different species. These ratios do not state how rapidly the process occurs, but they relate the concentration changes once reaction progress is known. Concentration trends and stoichiometric relationships must therefore be used together without treating the equation itself as a rate law.
18. Match each graph feature in Column I with its kinetic interpretation in Column II.
| Column I | Column II |
|---|
| P. Downward concentration-time curve | 1. Product concentration increasing |
| Q. Upward concentration-time curve | 2. Reactant concentration decreasing |
| R. Steeper change over the same time interval | 3. Greater magnitude of concentration change per unit time |
| S. Horizontal-axis variable | 4. Time |
ⓐ. P-1, Q-2, R-4, S-3
ⓑ. P-2, Q-3, R-1, S-4
ⓒ. P-3, Q-1, R-2, S-4
ⓓ. P-2, Q-1, R-3, S-4
Correct Answer: P-2, Q-1, R-3, S-4
Explanation: A downward concentration curve indicates that the plotted species is being consumed, so P matches reactant concentration decreasing. An upward curve indicates formation, so Q matches product concentration increasing. Over the same time interval, a steeper concentration change has a greater magnitude of change per unit time. Time is conventionally placed on the horizontal axis in a concentration-time graph. The sign of the slope distinguishes increase from decrease, while its magnitude reflects how rapidly the concentration changes.
19. In a constant-volume vessel, the reaction \(2A\rightarrow B\) causes the concentration of \(A\) to decrease from \(0.90\,mol\,L^{-1}\) to \(0.50\,mol\,L^{-1}\). Assuming no product was initially present, the concentration of \(B\) formed is:
ⓐ. \(0.10\,mol\,L^{-1}\)
ⓑ. \(0.40\,mol\,L^{-1}\)
ⓒ. \(0.20\,mol\,L^{-1}\)
ⓓ. \(0.80\,mol\,L^{-1}\)
Correct Answer: \(0.20\,mol\,L^{-1}\)
Explanation: \( \textbf{Reaction relation:} \)
\[
2A\rightarrow B
\]
\( \textbf{Initial concentration of }A\textbf{:} \)
\[
[A]_0=0.90\,mol\,L^{-1}
\]
\( \textbf{Final concentration of }A\textbf{:} \)
\[
[A]_t=0.50\,mol\,L^{-1}
\]
\( \textbf{Concentration of }A\textbf{ consumed:} \)
\[
0.90-0.50=0.40\,mol\,L^{-1}
\]
\( \textbf{Stoichiometric interpretation:} \) Every \(2\) concentration units of \(A\) consumed form \(1\) concentration unit of \(B\) at constant volume.
\( \textbf{Product concentration formed:} \)
\[
\Delta[B]=\frac{0.40}{2}=0.20\,mol\,L^{-1}
\]
\( \textbf{Initial-product condition:} \) Since no \(B\) was initially present, its final concentration equals the concentration formed.
\( \textbf{Coefficient check:} \) Using \(0.40\,mol\,L^{-1}\) directly for \(B\) would ignore the \(2:1\) stoichiometric ratio.
\( \textbf{Final answer:} \) The concentration of \(B\) formed is \(0.20\,mol\,L^{-1}\).
20. Assertion: For the reaction \(3X\rightarrow 2Y\), the numerical decrease in \([X]\) need not equal the numerical increase in \([Y]\).
Reason: The concentration changes of \(X\) and \(Y\) are related through their stoichiometric coefficients.
ⓐ. 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: The equation \(3X\rightarrow 2Y\) shows that \(3\) stoichiometric units of \(X\) are consumed for every \(2\) units of \(Y\) formed. Consequently, the raw concentration changes of the two species are not generally equal. Under constant-volume conditions, their magnitudes satisfy \(\frac{-\Delta[X]}{3}=\frac{\Delta[Y]}{2}\). The Reason gives the exact stoichiometric basis for the Assertion. Equal concentration changes would apply only to a suitable \(1:1\) stoichiometric relationship.
