9.6 Collision theory  (Page 6/11)

In terms of collision theory, to which of the following is the rate of a chemical reaction proportional?

(a) the change in free energy per second

(b) the change in temperature per second

(c) the number of collisions per second

(d) the number of product molecules

Hydrogen iodide, HI, decomposes in the gas phase to produce hydrogen, H ₂ , and iodine, I ₂ . The value of the rate constant, k , for the reaction was measured at several different temperatures and the data are shown here:

What is the value of the activation energy (in kJ/mol) for this reaction?

The element Co exists in two oxidation states, Co(II) and Co(III), and the ions form many complexes. The rate at which one of the complexes of Co(III) was reduced by Fe(II) in water was measured. Determine the activation energy of the reaction from the following data:

The hydrolysis of the sugar sucrose to the sugars glucose and fructose,
${\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}+{\text{H}}_2\text{O}⟶{\text{C}}_6{\text{H}}_{12}{\text{O}}_6+{\text{C}}_6{\text{H}}_{12}{\text{O}}_6$

follows a first-order rate equation for the disappearance of sucrose: Rate = k [C ₁₂ H ₂₂ O ₁₁ ] (The products of the reaction, glucose and fructose, have the same molecular formulas but differ in the arrangement of the atoms in their molecules.)

(a) In neutral solution, k = 2.1 $\times$ 10 ⁻¹¹ s ⁻¹ at 27 °C and 8.5 $\times$ 10 ⁻¹¹ s ⁻¹ at 37 °C. Determine the activation energy, the frequency factor, and the rate constant for this equation at 47 °C (assuming the kinetics remain consistent with the Arrhenius equation at this temperature).

(b) When a solution of sucrose with an initial concentration of 0.150 M reaches equilibrium, the concentration of sucrose is 1.65 $\times$ 10 ⁻⁷ M . How long will it take the solution to reach equilibrium at 27 °C in the absence of a catalyst? Because the concentration of sucrose at equilibrium is so low, assume that the reaction is irreversible.

(c) Why does assuming that the reaction is irreversible simplify the calculation in part (b)?

Use the PhET Reactions&Rates interactive simulation to simulate a system. On the “Single collision” tab of the simulation applet, enable the “Energy view” by clicking the “+” icon. Select the first $A+BC⟶AB+C$ reaction (A is yellow, B is purple, and C is navy blue). Using the “straight shot” default option, try launching the A atom with varying amounts of energy. What changes when the Total Energy line at launch is below the transition state of the Potential Energy line? Why? What happens when it is above the transition state? Why?

Use the PhET Reactions&Rates interactive simulation to simulate a system. On the “Single collision” tab of the simulation applet, enable the “Energy view” by clicking the “+” icon. Select the first $A+BC⟶AB+C$ reaction (A is yellow, B is purple, and C is navy blue). Using the “angled shot” option, try launching the A atom with varying angles, but with more Total energy than the transition state. What happens when the A atom hits the BC molecule from different directions? Why?