Question

You are studying the kinetics of the reaction:

                                                                        H2(g) + F2(g) → 2HF(g)

You are attempting to determine a mechanism for the reaction. You run 2 separate reaction trials keeping one reactant at a much higher pressure than the other reactant. The lower pressure reactant begins at 1.000 atm. OOPS! You forget to record which reactant was at the higher pressure. The data you collected for the first trial are:

Pressure of HF(atm)                             Time(min)

0                                                                     0

0.300                                                           30.0

0.600                                                           65.8       

0.900                                                           110.4

1.200                                                           169.1

1.500                                                           255.9

When you ran the second trial( in which the higher pressure reactant was much higher than the first trial) you determine the that the value of the pseudo-rate constants are equal. Another student in your lab is studying the same reaction. She has found that the reaction is 40 times faster at 55oC than at 35oC. You also know from the energy level diagram that the mechanism has 3 steps and the first step has the highest activation energy. You look up the bond energies of the reactants and find H-H = 432 kJ/mol ;    F-F = 154 kJ/mol; H-F = 565 kJ/mol

a- Sketch a qualitative energy level diagram for the mechanism

- Identify Ea , position of activated complexes and position of intermediates on your graph

b- Develop a reasonable mechanism for the reaction

c- Which reactant was limiting in the trials (aka the low pressure reactant)  

Answer 1

a. See your text : “You know from the energy level diagram that the mechanism has 3 steps and the first step has the highest activation energy “. In a Energy vs. Reaction coordinate graph simply draw a reaction pathway having 3 succesive activated complexes, the first one having the highest activation energy (i.e., the first step is the rate limiting step).

You can calculate the activation energy Ea for this first reaction step (rate limiting step) using the equation:

Ln(k2/k1) = (Ea/R)(1/T1 – 1/T2)

Ln (40) = (Ea/8.314)(1/308 – 1/328)

3.69 = (Ea/8.314)(0.00325 – 0.00305)       

3.69 = 0.00020(Ea/8.314)      

Ea = 153 400 J/mol = 153 kJ/mol

This value is close to the bond energy of F-F.

You may conclude that the reaction in the first step (rate determining step) is

       F-F = 2F·

The rate law is

r = dC_HF/2dt = k C_F2

where k is the pseudo-rate constant.

One may increase the pressure of H₂ without an observable effect on the reaction rate. You may conclude that the pressure of H₂ was different in the two trials.

b. F-F    = 2F·                        (slow, rate determining step)

      F· + H-H = HF + H·         (fast, not rate determining step)

      H· + F·   = HF                   (fast, not rate determining step)

      F₂ + H₂ = 2HF

c. The rate law is

r = dC_HF/2dt = k C_F2

where k is the pseudo-rate constant.

One may increase the pressure of H₂ without an observable effect on the reaction rate. You may conclude that the pressure of H₂ was different in the two trials.

      F₂ was the rate limiting reactant (the low pressure reactant) .