A person's wound was exposed to some bacteria and then bacteria growth started to happen at the same place. The wound was later treated with some antibacterial medicine and the rate of bacterial decay (r) was found to be proportional with the square of the existing number of bacteria at any instance. Which of the following set of graphs correctly represents the 'before' and 'after' situation of the application of the medicine? [Given: mathrmN = textNo. of bacteria, textt = texttime, bacterial growth follows mathrmI^mathrmst order kinetics.]

Solution & Explanation

### Related Formula textBefore: fracdNdt = k_1 N implies N = N_0 e^k_1 t textAfter: -fracdNdt = k_2 N^2 implies frac1N - frac1N_0 = k_2 t ### Core Logic Let's analyze the kinetics for the two stages: 1. **Before applying medicine**: - Bacterial growth follows 1^textst order kinetics: fracdNdt = k_1 N. - Integrating this yields: N(t) = N_0 e^k_1 t. - The graph of fracNN_0 vs t is a rising exponential curve starting from 1 (since at t=0, fracNN_0 = 1). 2. **After applying medicine**: - Reductive rate is proportional to the square of existing bacteria (2^textnd order decay): -fracdNdt = k_2 N^2 implies fracdNN^2 = -k_2 dt - Integrating this yields: -frac1N = -k_2 t + C implies frac1N = k_2 t + frac1N_0 N(t) = fracN_01 + N_0 k_2 t - A plot of N vs t or fracNN_0 vs t for this decay is a hyperbolic curve decreasing gradually. - Option B correctly matches the exponential growth before medicine and the hyperbolic decay after medicine. ### Pattern Recognition First-order growth is an exponential curve (N_0 e^kt), while second-order decay behaves as a rational hyperbolic relationship (1 / (1 + bt)). Option B shows the exact transition from an exponential rise to a hyperbolic decay curve. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

More Chemical Kinetics Previous-Year Questions — Page 6

Q83 jee_main_2024_29_jan_morning Arrhenius Equation and Activation Energy
For a reaction taking place in three steps at same temperature, overall rate constant mathrmK = fracmathrmK_1mathrmK_2mathrmK_3 . If mathrmEa_1 , mathrmEa_2 and mathrmEa_3 are 40, 50 and 60 kJ/mol respectively, the overall Ea is ______ kJ/mol.
Numerical Answer. Answer: 30 to 30

Solution

### Related Formula K = A e^-E_a/RT ### Core Logic Given the relationship between the rate constants: K = fracK_1 cdot K_2K_3 Substituting the Arrhenius equation for each rate constant: A cdot e^-E_a/RT = fracA_1 cdot e^-E_a1/RT cdot A_2 cdot e^-E_a2/RTA_3 cdot e^-E_a3/RT Combining the exponential terms using rules of exponents: A cdot e^-E_a/RT = left(fracA_1 cdot A_2A_3right) cdot e^frac-(E_a1 + E_a2 - E_a3)RT ### Step 1: Equating Activation Energies By comparing the powers of e on both sides, the overall activation energy E_a is related to the individual steps as follows: E_a = E_a1 + E_a2 - E_a3 Substitute the given values (E_a1 = 40, E_a2 = 50, E_a3 = 60 kJ/mol): E_a = 40 + 50 - 60 E_a = 90 - 60 E_a = 30 text kJ/mol ### Pattern Recognition When rate constants are multiplied or divided (K = K_1^a K_2^b / K_3^c), the corresponding overall activation energy follows the linear combination of the exponents: E_a = a E_a1 + b E_a2 - c E_a3. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics
Q82 jee_main_2024_30_january_evening Rate of Chemical Reaction
mathrmNO_2 required for a reaction is produced by decomposition of mathrmN_2mathrmO_5 in mathrmCCl_4 as by equation 2mathrmN_2mathrmO_5(g) rightarrow 4mathrmNO_2(g) + mathrmO_2(g) The initial concentration of mathrmN_2mathrmO_5 is 3 mathrmmol mathrmL^-1 and it is 2.75 mathrmmol mathrmL^-1 after 30 minutes. The rate of formation of mathrmNO_2 is x times 10^-3 mathrmmol mathrmL^-1 mathrmmin^-1, value of x is
Numerical Answer. Answer: 17 to 17

Solution

### Related Formula textRate of Reaction (ROR) = -frac12 fracDelta [mathrmN_2mathrmO_5]Delta t = frac14 fracDelta [mathrmNO_2]Delta t ### Core Logic First, find the rate of disappearance of mathrmN_2mathrmO_5. -fracDelta [mathrmN_2mathrmO_5]Delta t = - frac(2.75 - 3)30 = frac0.2530 mathrmmol \, L^-1 \, min^-1 Now, equate it to the general Rate of Reaction: textROR = -frac12 fracDelta [mathrmN_2mathrmO_5]Delta t = frac12 left(frac0.2530right) = frac0.12530 = frac1240 mathrmmol \, L^-1 \, min^-1 ### Step 1: Calculate the Rate of Formation of mathrmNO_2 Rate of formation of mathrmNO_2 = fracDelta [mathrmNO_2]Delta t = 4 times textROR = 4 times frac1240 = frac160 mathrmmol \, L^-1 \, min^-1 Convert this to scientific notation to find x: frac160 approx 0.01666 = 16.66 times 10^-3 mathrmmol \, L^-1 \, min^-1 Rounding to the nearest integer, we get x = 17. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics
Q81 jee_main_2024_30_jan_morning First Order Reactions
The rate of first order reaction is 0.04text mol L^-1s^-1 at 10 minutes and 0.03text mol L^-1s^-1 at 20 minutes after initiation. Half life of the reaction is ________ minutes. (Given log 2=0.3010, log 3=0.4771)
Numerical Answer. Answer: 24 to 24.1

Solution

### Related Formula Rate = k[A] [A] = [A]_0 e^-kt t_1/2 = fracln 2k ### Core Logic For a first order reaction, rate is directly proportional to concentration. R_1 = k[A]_10 = k[A]_0 e^-k(10 times 60) R_2 = k[A]_20 = k[A]_0 e^-k(20 times 60) ### Step 1: Setting up equations 0.04 = k[A]_0 e^-600k dots (1) 0.03 = k[A]_0 e^-1200k dots (2) ### Step 2: Solving for k Dividing equation (1) by (2): frac0.040.03 = frace^-600ke^-1200k frac43 = e^600k Take natural log on both sides: lnleft(frac43right) = 600k k = fracln(4/3)600 text s^-1 ### Step 3: Calculating half life t_1/2 = fracln 2k = fracln 2fracln(4/3)600 = frac600 ln 2ln 4 - ln 3 text seconds Convert to minutes by dividing by 60: t_1/2 = frac10 ln 2ln 4 - ln 3 text minutes Substitute log values (since ln x = 2.303 log x, the 2.303 cancels out): t_1/2 = 10 times fraclog 2log 4 - log 3 text minutes t_1/2 = 10 times frac0.30102(0.3010) - 0.4771 t_1/2 = 10 times frac0.30100.6020 - 0.4771 = 10 times frac0.30100.1249 t_1/2 = 24.099 approx 24 text minutes ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics
Q82 jee_main_2024_31_jan_evening First Order Kinetics
r = k[A] for a reaction, 50\% of A is decomposed in 120text minutes. The time taken for 90\% decomposition of A is ________ minutes.
Numerical Answer. Answer: 398.5 to 399.5

Solution

### Related Formula k = frac0.693t_1/2 t = frac2.303k logleft( fracaa - x right) ### Core Logic Since r = k[A], the reaction follows first-order kinetics. The half-life (50\% decomposition) is t_1/2 = 120text minutes. ### Step 1: Calculating for 90% decomposition For 90\% completion of the reaction, [A]_0 = 100 and [A]_t = 100 - 90 = 10. t = frac2.303k log frac10010 t = frac2.303left( frac0.693t_1/2 right) log (10) t = frac2.303 times 1200.693 times 1 t = 398.78text minutes Rounding off to the nearest integer, we get 399text minutes. ### Pattern Recognition For a first order reaction, t_90\% approx 3.32 times t_50\%. 120 times 3.32 = 398.4, so roughly 399. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics
Q73 jee_main_2024_31_jan_morning First Order Reactions
Integrated rate law equation for a first order gas phase reaction is given by (where P_i is initial pressure and P_t is total pressure at time t)
  • A. k = frac2.303t times log fracP_i(2P_i - P_t)
  • B. k = frac2.303t times log frac2P_i(2P_i - P_t)
  • C. k = frac2.303t times log frac(2P_i - P_t)P_i
  • D. k = frac2.303t times fracP_i(2P_i - P_t)

Solution

### Core Logic Consider a general gas phase reaction: A rightarrow B + C Initial (t=0): quad P_i quadquad 0 quadquad 0 At time t: quad P_i - x quadquad x quadquad x Total pressure at time t: P_t = (P_i - x) + x + x = P_i + x x = P_t - P_i Partial pressure of A at time t (P_A): P_A = P_i - x P_A = P_i - (P_t - P_i) = 2P_i - P_t For a first-order reaction: k = frac2.303t log fracP_0P_t Here, P_0 = P_i and the pressure of the reactant at time t is P_A. k = frac2.303t log fracP_i2P_i - P_t ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics
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