Arrhenius equation and activation energyEdexcel A-Level Chemistry: Flashcards

Exam questions on Arrhenius equation and activation energy

1. The rate constant, k, of a reaction depends on temperature according to the Arrhenius equation, ln k = ln A − Eₐ/RT, where Eₐ is the activation energy, R is the gas constant (8.31 J K⁻¹ mol⁻¹), T is the temperature in kelvin and A is a constant. A chemist plots ln k against 1/T for the reaction and obtains a straight line.
   Explain, in terms of particles, why the rate constant increases when the temperature increases.2 marks
2. A student measured the rate constant, k, for the hydrolysis of an ester at five temperatures between 298 K and 328 K. The student plotted ln k against 1/T and obtained a straight line with a gradient of −7.2 × 10³ K. The Arrhenius equation is ln k = ln A − Eₐ/RT and R = 8.31 J K⁻¹ mol⁻¹.
   Explain why each rate constant must be measured at a constant temperature, and why temperatures must be converted to kelvin before plotting.2 marks
3. The rate constant of a first-order reaction was measured at four temperatures: k = 0.128 s⁻¹ at 300 K, 0.244 s⁻¹ at 310 K, 0.448 s⁻¹ at 320 K and 0.792 s⁻¹ at 330 K. The Arrhenius equation is ln k = ln A − Eₐ/RT, where R = 8.31 J K⁻¹ mol⁻¹. A straight line is obtained when ln k is plotted against 1/T.
   Calculate the gradient of the line of ln k against 1/T between 300 K and 330 K.3 marks