Observe the following reactions at T(K) I. A rightarrow textproducts. II. 5Br^-(aq) + BrO_3^-(aq) + 6H^+(aq) rightarrow 3Br_2(aq) + 3H_2O(l) Both the reactions are started at 10.00 am. The rates of these reactions at 10.10 am are same. The value of -fracDelta[Br^-]Delta t at 10.10 am is 2 times 10^-4 text mol L^-1 text min^-1. The concentration of A at 10.10 am is 10^-2 text mol L^-1. What is the first order rate constant (in textmin^-1) of reaction I?

Solution & Explanation

### Related Formula textRate of reaction = frac-1nu_i fracd[Reactant]dt textRate of first order reaction = k[A] ### Core Logic At t = 10 minutes (10:10 text am): For reaction II: 5Br^- + BrO_3^- + 6H^+ rightarrow 3Br_2 + 3H_2O The overall rate of reaction II is expressed by dividing the rate of disappearance of Br^- by its stoichiometric coefficient: textRate_II = -frac15fracDelta[Br^-]Delta t Given -fracDelta[Br^-]Delta t = 2 times 10^-4 text mol L^-1 text min^-1, we have: textRate_II = frac15 times (2 times 10^-4) = 4 times 10^-5 text mol L^-1 text min^-1 ### Step 1: Equating Rates The problem states that the rates of both reactions are identical at this time. Therefore, textRate_I = textRate_II = 4 times 10^-5 text mol L^-1 text min^-1. For the first-order reaction I (A rightarrow textproducts): textRate_I = k[A] ### Step 2: Calculating Rate Constant Substitute the known values at t = 10 text min: 4 times 10^-5 = k times (10^-2) k = frac4 times 10^-510^-2 = 4 times 10^-3 text min^-1 ### Pattern Recognition Remember to divide the given rate of disappearance of a species by its stoichiometric coefficient to find the true, normalized "Rate of Reaction" before equating it to another reaction's rate. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

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More Chemical Kinetics Previous-Year Questions

Q71 jee_main_2026_21_jan_morning Arrhenius Equation
Pre-exponential factors of two different reactions of same order are identical. Let activation energy of first reaction exceeds the activation energy of second reaction by 20text kJ mol^-1. If k_1 and k_2 are the rate constants of first and second reaction respectively at 300 K, then ln frack_2k_1 will be ..... (nearest integer) [R=8.3text J K^-1text mol^-1]
Numerical Answer. Answer: 8 to 8

Solution

### Related Formula ln k = ln A - fracE_aRT ### Core Logic For Reaction 1: ln k_1 = ln A - fracE_1RT For Reaction 2: ln k_2 = ln A - fracE_2RT (Pre-exponential factor A is the same). Subtracting the first from the second: ln k_2 - ln k_1 = -fracE_2RT - left(-fracE_1RTright) ln left(frack_2k_1right) = fracE_1 - E_2RT Given that E_1 exceeds E_2 by 20text kJ mol^-1, E_1 - E_2 = 20000text J mol^-1. T = 300text K, R = 8.3text J K^-1text mol^-1. ln left(frack_2k_1right) = frac200008.3 times 300 = frac2008.3 times 3 = frac20024.9 ln left(frack_2k_1right) = 8.032 Rounding off to nearest integer gives 8. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics
Q58 jee_main_2026_21_jan_evening First Order Reactions and Rate Constant
Decomposition of A is a first order reaction at T(K) and is given by textA(textg) rightarrow textB(textg) + textC(textg). In a closed 1 L vessel, 1 bar textA(textg) is allowed to decompose at T(K). After 100 minutes, the total pressure was 1.5 bar. What is the rate constant (in textmin^-1) of the reaction? (log 2 = 0.3)
  • A. (1) \ 6.9 times 10^-1
  • B. (2) \ 6.9 times 10^-3
  • C. (3) \ 6.9 times 10^-2
  • D. (4) \ 6.9 times 10^-4

Solution

### Related Formula k = frac2.303t logleft(fracP_02P_0 - P_texttotalright) or equivalent first-order expression. ### Core Logic For textA(textg) rightarrow textB(textg) + textC(textg): - Initial pressure: P_0 = 1 text bar - At time t = 100 text min, pressure of A remaining = 1 - P, pressures of B and C = P. - Total pressure P_texttotal = 1 - P + P + P = 1 + P = 1.5 text bar implies P = 0.5 text bar. Remaining pressure of A = 1 - 0.5 = 0.5 text bar. ### Step 1: Calculating Rate Constant k = frac1100 lnleft(frac10.5right) = frac0.693100 = 6.9 times 10^-3 text min^-1 ### Pattern Recognition Sees: gaseous phase first-order kinetics with total pressure data. Trap: Confusing partial pressure of reactant with total pressure in rate expressions. ### Chapter Mix Class 12 Chemistry: Chemical Kinetics
Q62 jee_main_2026_22_january_morning First Order Reaction Rates
A rightarrow products (First order reaction). Three sets of experiment were performed for a reaction under similar experimental conditions. Run 1 Rightarrow 100 mL of 10 M solution of reactant A Run 2 Rightarrow 200 mL of 10 M solution of reactant A Run 3 Rightarrow 100 mL of 10 M solution of reactant A + 100 mL of H_2O added. The correct variation of rate of reaction is
  • A. textRun 1 = Run 2 = Run 3
  • B. textRun 3 < Run 1 = Run 2
  • C. textRun 3 < Run 1 < Run 2
  • D. textRun 1 < Run 2 < Run 3

Solution

### Related Formula textRate = k[A] Where [A] is the molar concentration of reactant A. ### Core Logic For a first order reaction, the rate is directly proportional to the concentration of the reactant, not the total volume or the total number of moles. In Run 1: [A] = 10text M. textRate_1 = k(10). In Run 2: [A] = 10text M. textRate_2 = k(10). (Volume increased, but molarity is identical). In Run 3: 100 mL of 10 M solution is diluted with 100 mL water. New volume is 200 mL. M_1 V_1 = M_2 V_2 implies 10 times 100 = M_2 times 200 implies M_2 = 5text M. [A] = 5text M. textRate_3 = k(5). ### Step 1: Final Conclusion Therefore, textRate_3 < textRate_1 = textRate_2. ### Pattern Recognition Rate laws depend exclusively on molar concentration (M). Diluting the solution decreases rate, while just taking a larger volume of the same stock solution keeps the rate identical. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics
Q73 jee_main_2026_22_january_morning Arrhenius Equation
The temperature at which the rate constants of the given below two gaseous reactions become equal is ____ K. (Nearest integer). X rightarrow Y quad k_1 = 10^6e^frac-30000T P rightarrow Q quad k_2 = 10^4e^frac-24000T Given: ln 10 = 2.303
Numerical Answer. Answer: 1303 to 1303

Solution

### Related Formula k = A e^-E_a/RT ### Core Logic Equate the two rate constants: k_1 = k_2 10^6e^frac-30000T = 10^4e^frac-24000T Divide both sides by 10^4: 10^2e^frac-30000T = e^frac-24000T Divide both sides by e^frac-30000T: 100 = frace^frac-24000Te^frac-30000T 100 = e^frac6000T ### Step 1: Solve for T Take the natural logarithm (ln) on both sides: ln(100) = frac6000T 2 ln(10) = frac6000T Substitute ln 10 = 2.303: 2 times 2.303 = frac6000T 4.606 = frac6000T T = frac60004.606 = 1302.64text K ### Step 2: Rounding Nearest integer is 1303. ### Pattern Recognition Simple exponential equating. Group powers of 10 on one side and exponentials on the other, then apply natural log. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics
Q57 jee_main_2026_22_january_evening Arrhenius Equation and Temperature Dependence
Correct statements regarding Arrhenius equation among the following are: (A) Factor e^-E_a/RT corresponds to fraction of molecules having kinetic energy less than E_a. (B) At a given temperature, lower the E_a, faster is the reaction. (C) Increase in temperature by about 10^circtextC doubles the rate of reaction. (D) Plot of log k vs frac1T gives a straight line with textslope = -fracE_aR. Choose the correct answer from the options given below:
  • A. B and D only
  • B. A and B only
  • C. A and C only
  • D. B and C only

Solution

### Related Formula k = A e^-E_a/RT ln k = ln A - fracE_aRT implies log k = log A - fracE_a2.303 R T ### Core Logic Statement (A): BANNED - e^-E_a/RT represents fraction of molecules with energy ge E_a (not less). Statement (B): CORRECT - Lower activation energy E_a increases the rate constant k, speeding up the reaction. Statement (C): CORRECT - For most reactions, a 10^circtextC rise in temperature doubles the rate coefficient. Statement (D): INCORRECT - Plot of log k vs 1/T has slope equal to -fracE_a2.303 R (the factor 2.303 is missing). ### Pattern Recognition Sees: Arrhenius statements. Shortcut: Watch for missing 2.303 in log slope equation and 'less than' vs 'greater than' in exponential fraction definition. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

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