pseudo frst order reaction
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PSEUDO FIRST ORDER REACTION :
⇒the reaction that has order is equal to one but , it is bimolecular in nature is called pseudo first order reaction .
⇒ such type reactions takes place if the amount of either reactant is larger .
⇒ hydrolysis of cane sugar is an example of pseudo first order reaction .
⇒ C₁₂H₂₂O₁₁ + H₂O -------------> C₆H₁₂O₆ + C₆H₁₂O₆
here , rate = K [ C₁₂H₂₂O₁₁]
this is one of the example for pseudo first order reaction .
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hope it helps...!!!
PSEUDO FIRST ORDER REACTION :
⇒the reaction that has order is equal to one but , it is bimolecular in nature is called pseudo first order reaction .
⇒ such type reactions takes place if the amount of either reactant is larger .
⇒ hydrolysis of cane sugar is an example of pseudo first order reaction .
⇒ C₁₂H₂₂O₁₁ + H₂O -------------> C₆H₁₂O₆ + C₆H₁₂O₆
here , rate = K [ C₁₂H₂₂O₁₁]
this is one of the example for pseudo first order reaction .
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hope it helps...!!!
Answered by
1
Under certain conditions, the 2nd order kinetics can be well approximated as first order kinetics. These Pseudo-1st-order reactions greatly simplify quantifying the reaction dynamics.
Introduction
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.
In second order reactions with two reactant species, the rate of disappearance of AA is
−d[A]dt=k[A][B](1.1)(1.1)−d[A]dt=k[A][B]
The integrated form is
1[B]0−[A]0ln[B][A]0[B]0[A]=kt(1.2)(1.2)1[B]0−[A]0ln[B][A]0[B]0[A]=kt
when [B]0>[A]0[B]0>[A]0.
When [B]0>>[A]0[B]0>>[A]0, then [B]0≈[B][B]0≈[B] and Equation 1.21.2becomes
1[B]0−[A]0ln[B][A]0[B]0[A]≈1[B]ln[A]0[A]=kt(1.3)(1.3)1[B]0−[A]0ln[B][A]0[B]0[A]≈1[B]ln[A]0[A]=kt
or
[A]=[A]oe−[B]kt(1.4)(1.4)[A]=[A]oe−[B]kt
This functional form of the decay kinetics is similar ot the first order kinetics and the system is said to operate under pseudo-first order kinetics. 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, BB, would have a significantly low concentration. We can then assume that reactant AA concentration effectively 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, k′k′, with the reactant with assumed constant concentration, AA, to create a new rate constant (k=k′[A]k=k′[A]) that will be used in the new rate equation,
Rate=k′[B](1.5)(1.5)Rate=k′[B]
as the new rate constant so we can treat the 2nd order reaction as a 1st order reaction.
One 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 5 M HCl into a 55 M ocean, the concentration of the mixture would be closer or equal to that of the ocean because there is so much water physically compared to the HCl and also because 55 M is relatively larger compared to 5 M.
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 pseudost-order reaction equation as:
or 
where
[A]o[A]o is the initial concentration of AA,
[B]o[B]o is the initial concentration of BB,
kk is the pseudo-1st-order reaction rate constant,
k′k′ is the 2nd order reaction rate constant, and
[A][A] is the concentration of A at time tt.
By using natural log to both sides of the pseudo-1st-order equation we get:
 or 
1st order reaction mechanisms and additional information can be found in this link.
EXAMPLE 1
Consider the reaction between cyanide and bromopropane:
CH3CH2CH2Braq+CN−(l)→CH3CH2CH2CNaq+Br−(aq)(1.6)(1.6)CH3CH2CH2Braq+CN(l)−→CH3CH2CH2CNaq+Br(aq)−
The reaction involves two reactants, so it is a 2nd order reaction:
rate=k[CH3CH2CH2Braq][CN−(l)](1.7)(1.7)rate=k[CH3CH2CH2Braq][CN(l)−]
In this case, the concentration of cyanide equals 100 M, 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 kk and [CN−][CN−] at different concentrations of [CN−][CN−] (still in excess compared to [CH3CH2CH2Br][CH3CH2CH2Br] and plot out a graph of kk vs. [CN−][CN−] the slope will equal k′k′.
Introduction
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.
In second order reactions with two reactant species, the rate of disappearance of AA is
−d[A]dt=k[A][B](1.1)(1.1)−d[A]dt=k[A][B]
The integrated form is
1[B]0−[A]0ln[B][A]0[B]0[A]=kt(1.2)(1.2)1[B]0−[A]0ln[B][A]0[B]0[A]=kt
when [B]0>[A]0[B]0>[A]0.
When [B]0>>[A]0[B]0>>[A]0, then [B]0≈[B][B]0≈[B] and Equation 1.21.2becomes
1[B]0−[A]0ln[B][A]0[B]0[A]≈1[B]ln[A]0[A]=kt(1.3)(1.3)1[B]0−[A]0ln[B][A]0[B]0[A]≈1[B]ln[A]0[A]=kt
or
[A]=[A]oe−[B]kt(1.4)(1.4)[A]=[A]oe−[B]kt
This functional form of the decay kinetics is similar ot the first order kinetics and the system is said to operate under pseudo-first order kinetics. 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, BB, would have a significantly low concentration. We can then assume that reactant AA concentration effectively 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, k′k′, with the reactant with assumed constant concentration, AA, to create a new rate constant (k=k′[A]k=k′[A]) that will be used in the new rate equation,
Rate=k′[B](1.5)(1.5)Rate=k′[B]
as the new rate constant so we can treat the 2nd order reaction as a 1st order reaction.
One 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 5 M HCl into a 55 M ocean, the concentration of the mixture would be closer or equal to that of the ocean because there is so much water physically compared to the HCl and also because 55 M is relatively larger compared to 5 M.
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 pseudost-order reaction equation as:
or 
where
[A]o[A]o is the initial concentration of AA,
[B]o[B]o is the initial concentration of BB,
kk is the pseudo-1st-order reaction rate constant,
k′k′ is the 2nd order reaction rate constant, and
[A][A] is the concentration of A at time tt.
By using natural log to both sides of the pseudo-1st-order equation we get:
 or 
1st order reaction mechanisms and additional information can be found in this link.
EXAMPLE 1
Consider the reaction between cyanide and bromopropane:
CH3CH2CH2Braq+CN−(l)→CH3CH2CH2CNaq+Br−(aq)(1.6)(1.6)CH3CH2CH2Braq+CN(l)−→CH3CH2CH2CNaq+Br(aq)−
The reaction involves two reactants, so it is a 2nd order reaction:
rate=k[CH3CH2CH2Braq][CN−(l)](1.7)(1.7)rate=k[CH3CH2CH2Braq][CN(l)−]
In this case, the concentration of cyanide equals 100 M, 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 kk and [CN−][CN−] at different concentrations of [CN−][CN−] (still in excess compared to [CH3CH2CH2Br][CH3CH2CH2Br] and plot out a graph of kk vs. [CN−][CN−] the slope will equal k′k′.
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