Arrhenius equation and activation energyEdexcel A-Level Chemistry: Subtopic test

1. 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.

   (a)

   What is the gradient of the straight line obtained by plotting ln k against 1/T?

   [1 mark]

   • AEₐ
   • BEₐ/R
   • C−Eₐ/R
   • Dln A

   (b)

   A catalyst is added and the experiment repeated. How does the gradient of the line change?

   [1 mark]

   • AIt becomes steeper (more negative)
   • BIt becomes less steep (smaller in magnitude)
   • CIt does not change
   • DIt becomes positive

   (c)

   Explain, in terms of particles, why the rate constant increases when the temperature increases.

   [2 marks]

   Total for question 1: 4 marks

2. 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.

   (a)

   Which pair of quantities, plotted on the axes, gives a straight line from which Eₐ can be found?

   [1 mark]

   • Aln k against 1/T, with T in kelvin
   • Bk against T, with T in kelvin
   • Cln k against T, with T in degrees Celsius
   • D1/k against 1/T, with T in kelvin

   (b)

   What is the activation energy of the reaction?

   [1 mark]

   • A0.87 kJ mol⁻¹
   • B866 kJ mol⁻¹
   • C−59.8 kJ mol⁻¹
   • D59.8 kJ mol⁻¹

   (c)

   Explain why each rate constant must be measured at a constant temperature, and why temperatures must be converted to kelvin before plotting.

   [2 marks]

   Total for question 2: 4 marks

3. 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.

   (a)

   Calculate the gradient of the line of ln k against 1/T between 300 K and 330 K.

   [3 marks]

   (b)

   Use your gradient to calculate the activation energy in kJ mol⁻¹. Then state, with a reason, how the gradient would differ for a reaction with a lower activation energy.

   [4 marks]

   Total for question 3: 7 marks

4. 4
   A student investigates the effect of temperature on the reaction between sodium thiosulfate and hydrochloric acid: Na₂S₂O₃(aq) + 2HCl(aq) → 2NaCl(aq) + S(s) + SO₂(g) + H₂O(l). The student mixes fixed volumes of the two solutions in a flask standing on a cross, and times how long it takes for the cross to disappear as the sulfur forms. The times taken are 100 s at 20 °C (293 K), 48 s at 30 °C (303 K), 24 s at 40 °C (313 K) and 12 s at 50 °C (323 K). The Arrhenius equation is ln k = ln A − Eₐ/RT and R = 8.31 J K⁻¹ mol⁻¹.

   (a)

   Describe how the student should carry out the investigation, and explain how the results are processed to find the activation energy.

   [6 marks]

   (b)

   Use the results at 20 °C and 50 °C to calculate the activation energy in kJ mol⁻¹, assuming that ln(1/t) can be used in place of ln k. Justify this assumption.

   [6 marks]

   Total for question 4: 12 marks

End of questions