Pseudo-1st-order reactionsTable of contentsUnder 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. IntroductionA 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 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 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: Reaction: 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.
Thus:
If we measure the the Half-Life in a Pseudo-1st Order reactionHalf-life refers to the time required to decrease the concentration of a reactant by half, so we must solve for t. Here, 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
The
Questions
Solutions1. 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 References
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