Exams › JEE Main › Chemistry
For the non-stoichiometric reaction 2A + B → C + D, the following kinetic data were obtained in three separate experiments, all at 298 K.
Initial Concentration (A) | Initial Concentration (B) | Initial rate of formation of C (mol L⁻¹ s⁻¹)
0.1 M | 0.1 M | 1.2 × 10⁻³
0.1 M | 0.2 M | 1.2 × 10⁻³
0.2 M | 0.1 M | 2.4 × 10⁻³
The rate law for the formation of C is:
- dc/dt = k[A][B]
- dc/dt = k[A]²[B]
- dc/dt = k[A][B]²
- dc/dt = k[A]
Correct answer: dc/dt = k[A]
Solution
The rate of formation of C is independent of the concentration of B, as evidenced by the constant rate observed when the concentration of A is held constant while varying B. This indicates that the reaction rate depends solely on the concentration of A, confirming that the correct rate law is dc/dt = k[A].
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 →