Every Book Back multiple-choice question from Chemical Kinetics (12th Standard Chemistry, Samacheer Kalvi) — each with the correct option highlighted and a clear, worked explanation. Free to read in English and Tamil.
Q1
For a first order reaction \(A \rightarrow B\) the rate constant is \(x\) \(min^{-1}\). If the initial concentration of A is \(0.01M\), the concentration of A after one hour is given by the expression.
- A. \(0.01 e^{-x}\)
- B. \(1 \times 10^{-2} (e^{-60})^{-x}\)
- C. \(1 \times 10^{-2} e^{-60x}\)Correct
- D. none of these
Explanation. The final concentration of a first-order reaction is determined by multiplying the initial concentration by the exponential decay factor, which accounts for the rate constant and the elapsed time in minutes.
Q2
A zero order reaction \(X \rightarrow Product\), with an initial concentration \(0.02M\) has a half life of \(10 min\). if one starts with concentration \(0.04M\), then the half life is
- A. 10 s
- B. 5 min
- C. 20 minCorrect
- D. cannot be predicted using the given information
Explanation. For zero-order reactions, the half-life is directly proportional to the initial concentration of the reactant; therefore, doubling the initial concentration results in a corresponding doubling of the half-life period.
Q3
For a first order reaction \(A \rightarrow product\) with initial concentration \(x\) \(mol L^{-1}\), has a half life period of 2.5 hours. For the same reaction with initial concentration \(\frac{x}{2}\) \(mol L^{-1}\) the half life is
- A. \(2.5 \times 2\) hours
- B. \(2.5 / 2\) hours
- C. 2.5 hoursCorrect
- D. Without knowing the rate constant, \(t_{1/2}\) cannot be determined from the given data
Explanation. In a first-order chemical reaction, the time taken for the reactant concentration to reduce to half its initial value is constant and does not change with different starting concentrations.
Q4
For the reaction, \(2NH_3 \rightarrow N_2 + 3H_2\), if \(-\frac{d[NH_3]}{dt} = k_1[NH_3]\), \(\frac{d[N_2]}{dt} = k_2[NH_3]\), \(\frac{d[H_2]}{dt} = k_3[NH_3]\) then the relation between \(k_1, k_2\) and \(k_3\) is
- A. \(k_1 = k_2 = k_3\)
- B. \(k_1 = 3k_2 = 2k_3\)
- C. \(1.5k_1 = 3k_2 = k_3\)Correct
- D. \(2k_1 = k_2 = 3k_3\)
Explanation. The relationship between different rate constants is established by equating the overall rate of reaction, which involves dividing each species' rate of change by its respective stoichiometric coefficient.
Q5
The decomposition of phosphine (\(PH_3\)) on tungsten at low pressure is a first order reaction. It is because the (NEET)
- A. rate is proportional to the surface coverageCorrect
- B. rate is inversely proportional to the surface coverage
- C. rate is independent of the surface coverage
- D. rate of decomposition is slow
Explanation. Under conditions of low pressure, the rate of a catalyzed surface reaction depends on the fraction of the catalyst surface that is occupied by the reacting molecules.
Q6
For a reaction \(Rate = k[acetone]^{3/2}\) then unit of rate constant and rate of reaction respectively is
- A. \(mol L^{-1} s^{-1}, mol^{-1/2} L^{1/2} s^{-1}\)
- B. \(mol^{-1/2} L^{1/2} s^{-1}, mol L^{-1} s^{-1}\)Correct
- C. \(mol^{1/2} L^{-1/2} s^{-1}, mol L^{-1} s^{-1}\)
- D. \(mol L s^{-1}, mol^{1/2} L s^{-1}\)
Explanation. The overall rate of reaction always has fixed units of molarity per unit time, while the rate constant units are determined mathematically by the specific order of the reaction.
Q7
The addition of a catalyst during a chemical reaction alters which of the following quantities? (NEET)
- A. Enthalpy
- B. Activation energyCorrect
- C. Entropy
- D. Internal energy
Explanation. A catalyst accelerates a reaction by providing an alternative mechanism that possesses a lower energy barrier, thus allowing more molecules to react effectively at the same temperature.
Q8
Consider the following statements: (i) increase in concentration of the reactant increases the rate of a zero order reaction. (ii) rate constant k is equal to collision frequency A if \(E_a = 0\) (iii) rate constant k is equal to collision frequency A if \(E_a = \infty\) (iv) a plot of ln k vs T is a straight line. (v) a plot of ln k vs 1/T is a straight line with a positive slope. Correct statements are
- A. (ii) onlyCorrect
- B. (ii) and (iv)
- C. (ii) and (v)
- D. (i), (ii) and (v)
Explanation. The Arrhenius equation demonstrates that the rate constant becomes identical to the frequency factor only when the activation energy is zero, making the exponential term equal to unity.
Q9
In a reversible reaction, the enthalpy change and the activation energy in the forward direction are respectively \(-x kJ mol^{-1}\) and \(y kJ mol^{-1}\). Therefore, the energy of activation in the backward direction is
- A. \((y-x) kJ mol^{-1}\)
- B. \((x+y) J mol^{-1}\)
- C. \((x-y) kJ mol^{-1}\)
- D. \((x+y) \times 10^3 J mol^{-1}\)Correct
Explanation. In an exothermic reversible reaction, the activation energy for the reverse process is calculated by adding the magnitude of the enthalpy change to the forward activation energy.
Q10
What is the activation energy for a reaction if its rate doubles when the temperature is raised from 200K to 400K? (R = 8.314 JK\(^{-1}\)mol\(^{-1}\))
- A. 234.65 kJ mol\(^{-1}\) K\(^{-1}\)
- B. 434.65 kJ mol\(^{-1}\) K\(^{-1}\)
- C. 2.305 kJ mol\(^{-1}\)Correct
- D. 334.65 J mol\(^{-1}\) K\(^{-1}\)
Explanation. Using the Arrhenius equation for two temperatures, the activation energy is calculated from the ratio of rate constants. Since the rate doubles between 200K and 400K, substituting the values into the formula yields approximately 2.305 kJ/mol.
Q11
This reaction follows first order kinetics. The rate constant at particular temperature is \(2.303 \times 10^{-2}\) hour\(^{-1}\). The initial concentration of cyclopropane is 0.25 M. What will be the concentration of cyclopropane after 1806 minutes?
- A. 0.125MCorrect
- B. 0.215M
- C. 0.25 \times 2.303M
- D. 0.05M
Explanation. First, 1806 minutes is converted to 30.1 hours. Applying the first-order integrated rate law with the given rate constant and time shows the initial concentration of 0.25 M reduces by half to 0.125 M.
Q12
For a first order reaction, the rate constant is 6.909 min\(^{-1}\). The time taken for 75% conversion in minutes is
- A. \(\frac{3}{2} \log 2\)
- B. \(\frac{2}{3} \log 2\)Correct
- C. \(\frac{3}{2} \log \frac{3}{4}\)
- D. \(\frac{2}{3} \log \frac{4}{3}\)
Explanation. A 75% conversion leaves 25% of the reactant. In first-order kinetics, the time is calculated using \(t = (2.303/k) \log([A_0]/[A])\). Substituting \(k = 6.909\) and \([A_0]/[A] = 4\) simplifies the expression to \((2/3) \log 2\).
Q13
In a first order reaction x \(\rightarrow\) y ; if k is the rate constant and the initial concentration of the reactant x is 0.1M, then, the half life is
- A. \(\frac{\log 2}{k}\)
- B. \(\frac{0.693(0.1)}{k}\)
- C. \(\frac{\ln 2}{k}\)Correct
- D. none of these
Explanation. The half-life of a first-order reaction is independent of the initial reactant concentration. It is derived solely from the rate constant \(k\) using the standard relationship \(t_{1/2} = \ln 2 / k\).
Q14
Predict the rate law of the following reaction based on the data given below: 2A + B \(\rightarrow\) C + 3D
| Reaction | [A] (min) | [B] (min) | Rate (M s\(^{-1}\)) |
| :--- | :--- | :--- | :--- |
| 1 | 0.1 | 0.1 | x |
| 2 | 0.2 | 0.1 | 2x |
| 3 | 0.1 | 0.2 | 4x |
| 4 | 0.2 | 0.2 | 8x |
- A. rate = k[A]\(^2\)[B]
- B. rate = k[A][B]\(^2\)Correct
- C. rate = k[A][B]
- D. rate = k[A]\(^{1/2}\)[B]\(^{3/2}\)
Explanation. Comparing experiments 1 and 2 shows the rate doubles when [A] is doubled, indicating first-order for A. Comparing 1 and 3 shows the rate quadruples when [B] is doubled, indicating second-order for B. The rate law is \(k[A][B]^2\).
Q15
The rate constant of a reaction is \(5.8 \times 10^{-2}\) s\(^{-1}\). The order of the reaction is
- A. First orderCorrect
- B. zero order
- C. Second order
- D. Third order
Explanation. The reaction order can be determined from the units of the rate constant. A rate constant with units of s\(^{-1}\) corresponds to a first-order reaction, where the rate is proportional to concentration to the power of one.
Q16
For the reaction N\(_2\)O\(_5\)(g) \(\rightarrow\) 2NO\(_2\)(g) + 1/2 O\(_2\)(g), the value of rate of disappearance of N\(_2\)O\(_5\) is given as \(6.5 \times 10^{-2}\) mol L\(^{-1}\) s\(^{-1}\). The rate of formation of NO\(_2\) and O\(_2\) is given respectively as
- A. \(3.25 \times 10^{-2}\) mol L\(^{-1}\) s\(^{-1}\) and \(1.3 \times 10^{-1}\) mol L\(^{-1}\) s\(^{-1}\)
- B. \(1.3 \times 10^{-2}\) mol L\(^{-1}\) s\(^{-1}\) and \(3.25 \times 10^{-2}\) mol L\(^{-1}\) s\(^{-1}\)
- C. \(1.3 \times 10^{-1}\) mol L\(^{-1}\) s\(^{-1}\) and \(3.25 \times 10^{-2}\) mol L\(^{-1}\) s\(^{-1}\)Correct
- D. None of these
Explanation. Based on stoichiometry, the rate of formation of NO\(_2\) is twice the disappearance of N\(_2\)O\(_5\) (\(2 \times 6.5 \times 10^{-2}\)), and the rate of O\(_2\) formation is half (\(0.5 \times 6.5 \times 10^{-2}\)).
Q17
During the decomposition of H\(_2\)O\(_2\) to give dioxygen, 48 g O\(_2\) is formed per minute at certain point of time. The rate of formation of water at this point is
- A. 0.75 mol min\(^{-1}\)
- B. 1.5 mol min\(^{-1}\)
- C. 2.25 mol min\(^{-1}\)
- D. 3.0 mol min\(^{-1}\)Correct
Explanation. 48 g of O\(_2\) equals 1.5 moles (48/32). In the balanced equation \(2H_2O_2 \rightarrow 2H_2O + O_2\), two moles of water are produced for every mole of oxygen, leading to a rate of 3.0 mol/min.
Q18
If the initial concentration of the reactant is doubled, the time for half reaction is also doubled. Then the order of the reaction is
- A. ZeroCorrect
- B. one
- C. Fraction
- D. none
Explanation. In zero-order reactions, the half-life is directly proportional to the initial concentration (\(t_{1/2} = [A_0]/2k\)). This means that if the starting concentration is doubled, the time taken for half the reaction to complete also doubles.
Q19
In a homogeneous reaction A \(\rightarrow\) B + C + D, the initial pressure was P\(_0\) and after time t it was P. expression for rate constant in terms of P\(_0\), P and t will be
- A. \(k = \frac{2.303}{t} \log \frac{2P_0}{3P_0 - P}\)Correct
- B. \(k = \frac{2.303}{t} \log \frac{2P_0}{P_0 - P}\)
- C. \(k = \frac{2.303}{t} \log \frac{3P_0 - P}{2P_0}\)
- D. \(k = \frac{2.303}{t} \log \frac{2P_0}{3P_0 - P^2}\)
Explanation. Calculating the partial pressure of reactant A at time \(t\) from the total system pressure \(P\) gives \((3P_0 - P)/2\). Substituting this into the first-order rate constant formula produces the correct logarithmic expression.
Q20
If 75% of a first order reaction was completed in 60 minutes, 50% of the same reaction under the same conditions would be completed in
- A. 20 minutes
- B. 30 minutesCorrect
- C. 35 minutes
- D. 75 minutes
Explanation. For any first-order reaction, the time required for 75% completion is twice the half-life (\(t_{50\%}\)). Dividing 60 minutes by 2 yields a half-life (50% completion time) of 30 minutes.
Q21
The half life period of a radioactive element is 140 days. After 560 days, 1 g of element will be reduced to
- A. \(\frac{1}{2}\) g
- B. \(\frac{1}{4}\) g
- C. \(\frac{1}{8}\) g
- D. \(\frac{1}{16}\) gCorrect
Explanation. A period of 560 days corresponds to exactly 4 half-lives (560/140). After 4 half-lives, the remaining mass is calculated as \((1/2)^4\) times the original mass, resulting in \(1/16\) g.
Q22
The correct difference between first and second order reactions is that (NEET)
- A. A first order reaction can be catalysed; a second order reaction cannot be catalysed.
- B. The half life of a first order reaction does not depend on [A\(_0\)]; the half life of a second order reaction does depend on [A\(_0\)].Correct
- C. The rate of a first order reaction does not depend on reactant concentrations; the rate of a second order reaction does depend on reactant concentrations.
- D. The rate of a first order reaction is independent of temperature jumps.
Explanation. A distinguishing feature of first-order reactions is that their half-life (\(0.693/k\)) is independent of the initial concentration. In second-order reactions, the half-life is inversely proportional to the initial concentration.
Q23
If a first order reaction is 93.75% complete in 2 hours, the half life of the reaction is
- A. 15 minutes
- B. 45 minutes
- C. 30 minutesCorrect
- D. 60 minutes
Explanation. 93.75% completion means only 6.25% (\(1/16\)) of the reactant remains, representing 4 half-lives (\(2^4=16\)). Dividing the total time of 120 minutes (2 hours) by 4 gives a half-life of 30 minutes.
