Consider the following statements related to temperature dependence of rate constants. Identify the correct statements, A. The Arrhenius equation holds true only for an elementary homogenous reaction. B. The unit of A is same as that of k in Arrhenius equation. C. At a given temperature, a low activation energy means a fast reaction. D. A and Ea as used in Arrhenius equation depend on temperature. E. When Ea >> RT, A and Ea become interdependent. Choose the correct answer from the options given below:

Solution & Explanation

### Related Formula Arrhenius equation is given by: k = A e^-fracE_aR T where: - k is the rate constant - A is the pre-exponential factor (frequency factor) - E_a is the activation energy ### Core Logic Evaluate each statement: - **A**: Arrhenius equation is an empirical relation that works well for both elementary and complex homogeneous reactions rightarrow *Incorrect*. - **B**: Since the exponential term e^-E_a/RT is dimensionless, the pre-exponential factor A has the exact same unit as the rate constant k rightarrow *Correct*. - **C**: For low E_a, the term e^-E_a/RT is large, giving a high rate constant k and a fast reaction rightarrow *Correct*. - **D**: A and E_a are assumed to be independent of temperature over a narrow range rightarrow *Incorrect*. - **E**: A and E_a remain independent parameters of the system, not interdependent rightarrow *Incorrect*. ### Step 1: Select correct statements Statements B and C are correct, matching Option (3). ### Pattern Recognition The exponential factor e^-E_a/RT represents the fraction of collisions with energy greater than the activation barrier. As E_a decreases, this fraction grows exponentially, explaining why low-activation pathways (like catalyzed reactions) run much faster. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

Reference Study Guides

Chemical Kinetics

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

More Chemical Kinetics Questions — jee_main_2025_03_april_evening

Practice all Chemical Kinetics previous-year questions →

Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)