Level 1 — RecognitionChemical Kinetics

Chemical Kinetics

⏱ 20 minutes30 marksprintable — key stays hidden on paper

Time Limit: 20 minutes | Total Marks: 30

Answer all questions. For True/False, a justification is required to earn full marks.

Section A — Multiple Choice (1 mark each) [10 marks]

Q1. The rate of a reaction is generally expressed in units of: (a) mol L⁻¹ (b) mol L⁻¹ s⁻¹ (c) s⁻¹ (d) L mol⁻¹ s⁻¹

Q2. For a first-order reaction, the unit of the rate constant $k$ is: (a) mol L⁻¹ s⁻¹ (b) s⁻¹ (c) L mol⁻¹ s⁻¹ (d) L² mol⁻² s⁻¹

Q3. The half-life of a first-order reaction is: (a) directly proportional to initial concentration (b) inversely proportional to initial concentration (c) independent of initial concentration (d) equal to $1/k$

Q4. Molecularity of a reaction: (a) can be zero (b) can be fractional (c) is always a whole number for an elementary step (d) can be negative

Q5. The Arrhenius equation is written as: (a) $k = A e^{E_a/RT}$ (b) $k = A e^{-E_a/RT}$ (c) $k = A e^{-RT/E_a}$ (d) $k = A/(E_a RT)$

Q6. A catalyst increases reaction rate by: (a) increasing $E_a$ (b) providing an alternative path of lower $E_a$ (c) increasing $\Delta H$ (d) raising the temperature

Q7. In the reaction rate law $\text{Rate} = k[A]^1[B]^0$ , the overall order is: (a) 0 (b) 1 (c) 2 (d) 3

Q8. Hydrolysis of an ester in excess water follows: (a) zero-order (b) second-order (c) pseudo-first-order (d) third-order kinetics

Q9. For a zero-order reaction, a plot of $[A]$ vs time is: (a) a curve (b) a straight line with negative slope (c) exponential (d) a horizontal line

Q10. The slowest step in a multi-step mechanism is called the: (a) fastest step (b) elementary step (c) rate-determining step (d) termination step

Section B — Matching (1 mark each pair) [8 marks]

Q11. Match Column I with Column II:

Column I (Order)	Column II (Integrated form / half-life)
(i) Zero order	(P) $t_{1/2} = 0.693/k$
(ii) First order	(Q) $t_{1/2} = [A]_0/2k$
(iii) Second order	(R) $t_{1/2} = 1/(k[A]_0)$
(iv) First order (integrated)	(S) $\ln[A] = \ln[A]_0 - kt$

Q12. Match the concept with its description:

Column I	Column II
(i) Steric factor	(P) catalyst and reactants in same phase
(ii) Homogeneous catalysis	(Q) accounts for orientation of collisions
(iii) Activated complex	(R) highest energy species at transition state
(iv) Frequency factor	(S) number of collisions per second (A)

Section C — True / False with Justification (2 marks each) [12 marks]

Q13. "Order of a reaction can be fractional, but molecularity cannot." (True/False + justify)

Q14. "The average rate and instantaneous rate of a reaction are always equal." (True/False + justify)

Q15. "A catalyst changes the equilibrium constant of a reversible reaction." (True/False + justify)

Q16. "For a first-order reaction, if concentration doubles, the rate doubles." (True/False + justify)

Q17. "According to the Arrhenius equation, rate constant increases with increasing temperature." (True/False + justify)

Q18. "Enzyme catalysis is highly specific and works best at an optimum temperature and pH." (True/False + justify)

Answer keyMark scheme & solutions

Section A (1 mark each)

Q1. (b) mol L⁻¹ s⁻¹ — rate = change in concentration per unit time. (1)

Q2. (b) s⁻¹ — for first order, $k = \text{rate}/[A]$ = (mol L⁻¹ s⁻¹)/(mol L⁻¹) = s⁻¹. (1)

Q3. (c) independent of initial concentration — $t_{1/2}=0.693/k$ contains no $[A]_0$ . (1)

Q4. (c) is always a whole number for an elementary step — molecularity counts colliding species, so it is a positive integer. (1)

Q5. (b) $k = A e^{-E_a/RT}$ — standard Arrhenius form. (1)

Q6. (b) provides an alternative path of lower $E_a$ — catalyst lowers activation energy without being consumed. (1)

Q7. (b) 1 — overall order = sum of exponents = 1 + 0 = 1. (1)

Q8. (c) pseudo-first-order — water in large excess, its concentration ≈ constant. (1)

Q9. (b) straight line with negative slope — $[A]=[A]_0-kt$ . (1)

Q10. (c) rate-determining step — slowest step controls overall rate. (1)

Section B

Q11. (i)→Q, (ii)→P, (iii)→R, (iv)→S. (1 each, 4 marks)

Zero order: $t_{1/2}=[A]_0/2k$ ; First order: $t_{1/2}=0.693/k$ ; Second order: $t_{1/2}=1/(k[A]_0)$ ; First-order integrated: $\ln[A]=\ln[A]_0-kt$ .

Q12. (i)→Q, (ii)→P, (iii)→R, (iv)→S. (1 each, 4 marks)

Section C (True/False = ½, justification = 1½)

Q13. True. Order is experimental and may be fractional/zero, whereas molecularity is the number of species in an elementary step, so it must be a positive whole number. (2)

Q14. False. Average rate is over a time interval; instantaneous rate is at a single instant. They coincide only for a zero-order reaction (constant rate) or in the limit $\Delta t \to 0$ . (2)

Q15. False. A catalyst speeds up both forward and reverse reactions equally; it lowers $E_a$ but does not alter $\Delta G$ or $K$ . Equilibrium is reached faster, not shifted. (2)

Q16. True. Rate = $k[A]$ ; doubling $[A]$ doubles the rate (linear dependence). (2)

Q17. True. As $T$ increases, $-E_a/RT$ increases (less negative), so $e^{-E_a/RT}$ increases, hence $k$ increases. (2)

Q18. True. Enzymes are specific to substrates (lock-and-key) and have maximum activity at optimum $T$ and pH; extremes denature them. (2)

[
  {"claim":"First-order half-life t=0.693/k is independent of A0 (derivative wrt A0 is 0)",
   "code":"k,A0=symbols('k A0',positive=True); t12=ln(2)/k; result=(diff(t12,A0)==0)"},
  {"claim":"Zero-order half-life equals A0/(2k)",
   "code":"k,A0=symbols('k A0',positive=True); t12=A0/(2*k); result=(simplify(t12-A0/(2*k))==0)"},
  {"claim":"Overall order for Rate=k[A]^1[B]^0 is 1",
   "code":"order=1+0; result=(order==1)"},
  {"claim":"Arrhenius k increases with T: d/dT of e^(-Ea/RT) is positive",
   "code":"Ea,R,T=symbols('Ea R T',positive=True); expr=exp(-Ea/(R*T)); result=(simplify(diff(expr,T))==Ea*exp(-Ea/(R*T))/(R*T**2)) and (Ea*exp(-Ea/(R*T))/(R*T**2)>0)"}
]