For a given reaction mathrmRrightarrow mathrmP,mathrmt_1 / 2 is related to [mathrmA]_0 as given in table :
[A]_0 / mathrmmol\,L^-1t_1/2 / mathrmmin
0.100200
0.025100
Given: log 2 = 0.30 Which of the following is true? A. The order of the reaction is frac12 . B. If [mathrmA]_0 is 1mathrmM , then mathrmt_1/2 is 200sqrt10 min C. The order of the reaction changes to 1 if the concentration of reactant changes from 0.100 mathrmM to 0.500 mathrmM . D. t_1 / 2 is 800 mathrm~min for [mathrmA]_0 = 1.6 mathrmM Choose the correct answer from the options given below:

Solution & Explanation

### Related Formula The dependence of half-life on initial concentration is given by: t_1/2 propto frac1[A]_0^n-1 ### Step 1: Finding the reaction order (n) Using the values provided: frac(t_1/2)_1(t_1/2)_2 = left( frac[A]_0,2[A]_0,1 right)^n-1 frac200100 = left( frac0.0250.100 right)^n-1 Rightarrow 2 = left( frac14 right)^n-1 2 = 2^-2(n-1) Rightarrow 1 = -2n + 2 Rightarrow n = frac12 Hence, statement A is correct. ### Step 2: Checking half-life at other concentrations Since n = frac12, t_1/2 propto sqrt[A]_0. - For [A]_0 = 1\,mathrmM: frac200t_1/2 = sqrtfrac0.11 Rightarrow t_1/2 = 200sqrt10\,mathrmmin Hence, statement B is correct. - For [A]_0 = 1.6\,mathrmM: frac200t_1/2 = sqrtfrac0.11.6 = sqrtfrac116 = frac14 Rightarrow t_1/2 = 800\,mathrmmin Hence, statement D is correct. ### Pattern Recognition Sees: Half-life reducing as initial concentration decreases. Trap: Assuming all reactions are first or zero order without calculations. Shortcut: Reduction of [A]_0 by 4 causes reduction of t_1/2 by 2 rightarrow indicates a square root dependence (t_1/2 propto sqrtA_0), which implies n = 0.5. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

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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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