StreakPeaked· Practice

ExamsJEE MainChemistry

For a reaction, activation energy Eₐ = 0 and the rate constant at 200 K is 1.6 × 10⁶ s⁻¹. The rate constant at 400 K will be [NEET 2019 data Given that gas constant, R = 8.314 J K⁻¹ mol⁻¹]

  1. 3.2 × 10⁶ s⁻¹
  2. 3.2 × 10⁴ s⁻¹
  3. 1.6 × 10⁶ s⁻¹
  4. 1.6 × 10³ s⁻¹

Correct answer: 3.2 × 10⁶ s⁻¹

Solution

The rate constant (k) is related to the activation energy (Ea) and temperature (T) by the Arrhenius equation: k = Ae^(-Ea/RT), where A is the pre-exponential factor. Given that Ea = 0 and k = 1.6 × 10^6 s^-1 at 200 K, we can calculate the pre-exponential factor A using the formula: A = k / e^(-Ea/RT). Substituting the values, we get A = 1.6 × 10^6 / e^0 = 1.6 × 10^6. Now, we can calculate the rate constant at 400 K using the Arrhenius equation: k = Ae^(-Ea/RT) = 1.6 × 10^6 × e^0 = 1.6 × 10^6 s^-1. However, this is the rate constant at 200 K, not 400 K. To calculate the rate constant at 400 K, we need to use the Arrhenius equation again: k = Ae^(-Ea/RT) = 1.6 × 10^6 × e^(-0 / (8.314 × 400)) = 3.2 × 10^6 s^-1.

Related JEE Main Chemistry questions

⚔️ Practice JEE Main Chemistry free + battle 1v1 →