Exams › JEE Main › Chemistry
For a first-order reaction, the term 1/4 may be defined as the time required for the concentration of a reactant to decrease to three-fourths of its initial value. If the rate constant is k, this 1/4 is expressed as:
- 0.75/k
- 0.69/k
- 0.29/k
- 0.10/k
Correct answer: 0.29/k
Solution
For a first-order reaction, t = (2.303/k) log([R0]/[R]). For [R] = (3/4)[R0]: t = (2.303/k) log(4/3) = (2.303/k)(0.1249) = 0.29/k.
Related JEE Main Chemistry questions
- For the reaction A → products, the half-life is 120 min when the initial concentration of A is 8.0 × 10⁻² M. If the starting concentration is changed to 4.0 × 10⁻² M, the half-life increases to 240 min. What is the reaction order?
- For a first-order process A → B, if the rate constant is k and the starting concentration of A is 0.5 M, what is the half-life?
- For the elementary gas-phase reaction A(g) + 2B(g) → C(g) + D(g), an experiment starts with initial partial pressures p_A = 0.60 atm and p_B = 0.80 atm. If the partial pressure of C becomes 0.20 atm, then the reaction rate compared with the initial rate is
- From the experimental data below for the reaction A + B → products, identify the rate law that fits the observations.
Exp. 1: [A] = 0.012, [B] = 0.035, initial rate = 0.1
Exp. 2: [A] = 0.024, [B] = 0.070, initial rate = 0.8
Exp. 3: [A] = 0.024, [B] = 0.035, initial rate = 0.1
Exp. 4: [A] = 0.012, [B] = 0.070, initial rate = 0.8
- A radioactive substance is accidentally spread on the floor of a room. If its half-life is 30 days and the initial activity is 10 times the allowed safe level, after how many days will it become safe to enter the room?
- For the reversible reaction A2 + B2 ⇌ 2AB, the activation energies are 180 kJ mol⁻¹ for the forward direction and 200 kJ mol⁻¹ for the reverse direction. If a catalyst reduces the activation energy of both directions by 100 kJ mol⁻¹, what will be the enthalpy change of the reaction A2 + B2 ⇌ 2AB in the catalyzed pathway (in kJ mol⁻¹)?
⚔️ Practice JEE Main Chemistry free + battle 1v1 →