What if the second gas phase molecule in the collision can actually be a second

Question

What if the second gas phase molecule in the collision can actually be a second molecule of A- and this collision has a noticeably different activation energy, so that the rate constants at the same temperature would be noticeably different? This would introduce a different path for producing A*: What is the rate law expression for the generation of product, in this case? If all nonreactant molecules (M) are removed from the system, what does this rate law expression simplify to? (Consider the case in which A is CO2, and a container of dry ice is pumped down to high vacuum conditions. The dry ice would later sublime, filling the vessel with CO2 (g), with no other gas molecules present in any noticeable concentration.)

Explanation / Answer

a) Derrive rate law

-1/2d[A]/dt = k2[A]^2

d[A*]/dt = k-2[A*][A]

k2[A] = k-2[A*]

[A*] = k2/k-2[A]

Substitute in the equation below,

-d[A*]/dt = kf[A*]

rate = kf(k2/k-2)[A] will be the rate law for the reaction.

b) If all the non-reactant molecules (M) is removed from the system, the rate law becomes,

-d[A]/dt = rate = kf(k2/k-2)[CO2]^2

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