Under certain condictions, the 2nd order kinetics can be well approximated as first order kinetics. These Pseudo-1st-order reactions greatly simplify quantifying the reaction dynamics.
A 2nd order reaction can be challenging to follow mostly because the two reactants involved must be measured simultaneously. There can be additional complications because certain amounts of each reactant are required to determine the reaction rate, for example, which can make the cost of one's experiment high if one or both of the needed reactants are expensive. To avoid more complicated, expensive experiments and calculations, we can use the Pseudo-1st-order reaction, which involves treating a 2nd order reaction like a 1st order reaction.
A 2nd order reaction has the rate equation because it involves two reactants, A and B. In a pseudo-1st-order reaction, we can manipulate the initial concentrations of the reactants. One of the reactants, A, for example, would have a significantly high concentration, while the other reactant, B, would have a significantly low concentration. We can then assume that reactant A's concentration remains constant during the reaction because its consumption is so small that the change in concentration becomes negligible. Because of this assumption, we can multiply the reaction rate, , with the reactant with assumed constant concentration, A, to create a new rate constant ( = ). This new rate constant, , will be used in the new rate equation,, as the new rate constant so we can treat the 2nd order reaction as a 1st order reaction. Another way to create a pseudo-1st-order reaction is to manipulate the physical amounts of the reactants. For example, if one were to dump a liter of 5M HCl into a 55M ocean, the concentration of the mixture would be closer or equal to that of the ocean because there is so much water physcially compared to the HCl and also because 55M is relatively larger compared to 5M.
In theory, if we have an instance where there are more than two reactants involved in a reaction, all we would have to do is make the reaction appear like it is first order. If there were three reactants, for example, we would make two of the three reactants be in excess (whether in amount or in concentration) and then monitor the dependency of the third reactant.
We can also write the pesudo-1st-order reaction equation as:
where is the initial concentration of A, is the initial concentration of B, is the pseudo-1st-order reaction rate constant, is the 2nd order reaction rate constant, and is the concentration of A at time .
By using natural log to both sides of the pseudo-1st-order equation we get:
1st order reaction mechanisms and additional information can be found in this link.
Using cyanide and bromopropane as an example:
The reaction involves two reactants, so it is a 2nd order reaction:
In this case, the concentration of cyanide (CN-) equals 100M, which is significantly higher than the concentration of bromopropane.
Because the concentration of cyanide is significantly higher than the concentration of bromopropane, we can assume cyanide's concentration will stay constant in the reaction.
If we measure the the and at different concentrations of (still in excess compared to ) and plot out a graph of vs. the slope will equal .
Half-Life in a Pseudo-1st Order reaction
Half-life refers to the time required to decrease the concentration of a reactant by half, so we must solve for t. Here, will be the reactant in excess, and its concentration will stay constant. is the initial concentaration of A; thus the half-life concentration of A is .
The pseudo-1st-order reaction equation can be written as:
By using natural log to both sides of the pseudo-1st-order equation, we get:
Because the half-life concentration of A is 0.5[A]o at half-life time t1/2:
Recall that equals :
The cancels out:
1. Because [B] is in excess we mulitply 99M with 3.67M-1s-1
(99M)(3.67M-1s-1) = 363.33s-1
2. Because [A] is in excess we can mulitply the k' with [A]o to find k
(109M)(45M-1s-1) = 4905s-1
t1/2 = (ln0.5 / -k)
t1/2 = 1.41 X 10-4s
3. Use the equation [A] = [A]oe-k'[B]ot
(55M) = (99M)e-k'(1000M)(39s)
k' = 1.507 X 10-5M-1s-1
4. Use the equation [A] = [A]oe-k'[B]ot
[A] = (1)e-(0.6 M-1s-1)(45)
[A] = 1.88 X10-12M
5. Use the equation R = k'[A][B]
R = (0.1M-1s-1)(560M)(0.2M)
Rate = 11.2Ms-1
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