[2]7a. Account for the observation that the rate equation for a chemical process cannot be obtained directly from the balanced chemical equation for the process.



The balanced chemical equation is a summary of the reactants and products and does not show the steps by which the reactants are converted into products. Steps are often needed to avoid the improbability of termolecular (or greater) collisions being required.

The rate of a chemical process depends on the rate of the slowest step in the step-wise process. This slow step may not contain all the species present in the overall equation.





7b. Unimolecular and bimolecular steps are often proposed in step wise mechanisms. Termolecular processes are proposed only rarely.

[1](i) What is meant by a step wise mechanism?



This is a detailed look at which reactants come together in collisions for the reaction to proceed. Steps are often needed to avoid termolecular (and higher) collisions being postulated.





[1](ii) Define the molecularity of a step in a step wise mechanism.



The molecularity is given by the number of molecules colliding in the particular step of the step-wise mechanism.



[2](iii) Suggest a reason for the observation made above.



Steps are often needed to avoid termolecular (and higher) collisions being postulated.





[6]8.Consider the following chemical reaction:

2 NO(g) + O2(g) 2 NO2(g)



An initial rates experiment was conducted with the following results:



Experiment # Initial [NO], mol/L Initial [O2], mol/L Initial rate, mol/L.s
1 0.0111 0.00433 0.0222
2 0.0220 0.00433 0.0902
3 0.0220 0.00900 0.180


Assume that the rate law has the form: rate = k [NO]x [O2]y

and obtain values for x, y and k.



The rate equation is:

rate = 4.2 x 104 L2.mol-2.s-1 [NO]2[O2]





9. A step wise reaction mechanism is proposed as follows:



step 1 2 NO(g) N2O2 fast

step 2 N2O2(g) + H2 H2O + N2O slow

step 3 N2O + H2 N2 + H2O fast



[2]a. What is the overall equation for this process?



This is obtained by adding together the three equations such that intermediates disappear:



2 NO + 2 H2 --> N2 + 2 H2O



[2]b. What intermediates (if any) are formed according to this mechanism?



Two intermediates are formed: N2O2 and N2O.



[1]c. Is there a catalyst involved in this process? If so, what is it?



There is no catalyst in this reaction.





[3]d. What is the rate equation for this mechanism in terms of experimentally measurable quantities.



The rate of a multistep process is the rate of the slowest step; In this case:



rate = k [N2O2] [H2]



However, the N2O2 is an proposed intermediate whose concentration is not directly measurable. So we must make an approximation, that the first fast step is actually at equilibrium, and write an equilibrium expression:



K = [N2O2] / [NO]2

from which



[N2O2] = K [NO]2



Substituting this expression for the [N2O2] in the rate equation gives:



rate = kK [NO]2 [H2]







[2]e. Assuming that the product energy is lower than the reactant energy (that is DH is negative, the reaction is exothermic), sketch an energy diagram for this three-step process.









10. The rate constant for the gas-phase decomposition of N2O5,

2 N2O5 2 N2O4 + O2,

has the following temperature dependence:

T (K) 338 318 298
k (s-1) 4.9 x 10-3 5.0 x 10-4 3.5 x 10-5


[2]a. Suggest a rate equation for this process and justify your suggestion.



Rate = k [N2O5]

(because of the units for k, it must be a first order reaction!)



[4]b. Calculate a value for the activation energy of this decomposition.



Use two of the sets of data given to create simultaneous equations which can be solved:



ln k = ln A - Ea / RT



Equation 1: ln 4.9 x 10-3 = ln A - Ea / (8.314*338)

Equation 2: ln 5.0 x 10-4 = ln A - Ea / (8.314*318)



Solving for Ea gives 144 kJ





[2]c. Calculate a value for the rate constant at 0C.



Again, use two sets of data to find k273



Or alternately, find ln A in the above, and use it:



From above, ln A = 51.24

So k273 = 5.0 x 10-6





[4]11. Researchers have created artificial red blood cells. These artificial red blood cells are cleared from circulation in the body by a first order reaction with a half life of 6.00 hours. An accident victim has received the red blood cells at 8:20 pm and is rushed to hospital, arriving at 9:45 pm (on the same day!). What percentage of the artificial red blood cells remain in the body?

(For a first order reaction: ln { a / (a - x) } = kt.)



For this process, a = 100%, (a - x) = % left.



For k, use the half life given, since t = (ln 2) / k.



So k = 0.116 hr-1.



ln { 100 / %left} = 0.116 hr-1 * 1.42 hr



% left = 84.8







[4]12. U-238 decomposes to Pb-206 with a half life of 4.47 x 109 years. A rock sample is 55.5 mole% uranium and 44.5 mole% lead. How old is the rock sample? (Radioisotopes decompose following first order kinetics.)



For this question, at start, assume the amount of uranium is 100%, at end amount is 55.5%



Substitute into the integrated rate equation:



ln (100/55.5) = kt



From the half life, k = (ln 2)/ t = 1.55 x 10-10 y-1



From which t = 3.8 x 109 years