Half-life of a Reaction
The half-life, t1/2, of a reaction is the amount of time it will take the concentration, [A], of a reactant to reduce to half its initial concentration, [Ao]. That is to say,
@ t1/2, [A] = [Ao]/2 ...……(i)
The half-life of the various orders of reaction can be calculated by substituting the above equation into the respective integrated rate laws.
(a) Half-life of a Zeroth-Order Reaction
Recall that the zeroth-order integrated rate law is given by:
[A] = [Ao] - kt …………(ii)
[Ao] - [A] = kt
@ t1/2, [A] = [Ao]/2
kt1/2 = [Ao] - ([Ao]/2)
kt1/2 = (2[Ao] - [Ao])/2
kt1/2 = [Ao]/2
t1/2 = [Ao]/2k …………(iii)
Equation (iii) above, gives the half-life for a zeroth-order reaction, which can be obtained by finding the product of half the initial reactant concentration and the reciprocal of the reaction rate constant ([Ao]/2 × 1/k).
(b) Half-life of a First-Order Reaction
Recall that the first-order integrated rate law is given by:
ln [A] = ln [Ao] - kt …………(iv)
kt = ln [Ao] - ln [A]
@ t1/2, [A] = [Ao]/2
kt1/2 = ln [Ao] - ln ([Ao]/2)
kt1/2 = ln ([Ao]/([Ao]/2))
kt1/2 = ln 2
t1/2 = ln 2/k …………(v)
Equation (v) gives the half-life for a first-order reaction, which is independent of the initial concentration of the reactant. It can be obtained by dividing the natural log of 2 (ln 2 = 0.6932) by the reaction rate constant.
Equation (v) is applied when calculating the half-life of most radioactive substances. This is because nuclear or radioactive reactions are generally first-ordered.
(c) Half-life of a Second-Order Reaction
Recall that the integrated rate law for a second-order reaction is given by:
1/[A] = 1/[Ao] + kt …………(vi)
kt = 1/[A] - 1/[Ao]
@ t1/2, [A] = [Ao]/2
kt1/2 = 1/([Ao]/2) - 1/[Ao]
kt1/2 = 2/[Ao] - 1/[Ao]
kt1/2 = 1/[Ao]
t1/2 = 1/[Ao]k …………(vii)
Equation (vii) gives the half-life for a second-order reaction, which can be obtained by finding the reciprocal of the product of the initial concentration of the reactant and the reaction rate constant (1/([Ao] × k)).
Do These:
Question 1
Using the information provided in Example 1 of the Chemical Kinetics (Part II), calculate the half-life of the 1-chloroethane (C2H5Cl) for experiments 1 and 4.
Question 2
Determine the half-life of the 1-bromopropane (C3H7Br) in Example 3 of the Chemical Kinetics (Part II)
Question 3
Calculate the half-life of the peroxydisulfate ion (S2O8--) using the information provided in Example 4 of the Chemical Kinetics (Part II).
Mechanism of Reactions
The mechanism of a chemical reaction is the detailed description of the elementary steps involved in the given reaction. Some reactions proceed directly from the reactants to the products stage, without passing through an intermediate stage. Such reactions are known as one-step reactions, whereas those that pass through the intermediate stage are known as multi-step reactions. An intermediate is actually a very reactive or unstable species produced in one stage and consumed in another stage of a multi-step reaction.
Example of a multi-step reaction is the decomposition of ozone molecules to oxygen gas, according to the equations:
O3(g) ----> O2(g) + O. …………(a)
O3(g) + O. ----> 2O2(g) ……...(b)
2O3(g) ----> 3O2(g) …………….(c)
The mechanism of the two-step reaction is as explained below:
The ozone molecule undergoes a photolytic cleavage (the breaking of bond by the action of UV light) in (a) to produce a highly unstable oxygen atom radical, O. as the intermediate species. The radical is consumed in (b) when it attacks an ozone molecule to produce two molecules of oxygen gas. The addition of (a) and (b) gives the overall reaction (c).
In a multi-step reaction, the elementary step with the slowest rate of reaction is known as the rate determining step of the overall reaction. Remember, a chain, they say, is as strong as its weakest link. Which means that no matter how strong a chain looks, its strength is limited by strength of its weakest link.
The same analogy applies here, because no matter how fast the other reaction steps may be, the overall rate of the reaction is limited by the rate of slowest reaction step. This explains why the rate law of the overall reaction is determined by the stoichiometry of the rate determining step and not the stoichiometry of the overall reaction.
It is important to note that we can propose two or more mechanisms for a particular reaction. However, the reasonable one will be the mechanism that is consistent with the experimentally determined kinetic data for the reaction.
Twitter: @gmtacademy
WhatsApp: 07034776117
Facebook: www.facebook.com/greatermindstutors
The half-life, t1/2, of a reaction is the amount of time it will take the concentration, [A], of a reactant to reduce to half its initial concentration, [Ao]. That is to say,
@ t1/2, [A] = [Ao]/2 ...……(i)
The half-life of the various orders of reaction can be calculated by substituting the above equation into the respective integrated rate laws.
(a) Half-life of a Zeroth-Order Reaction
Recall that the zeroth-order integrated rate law is given by:
[A] = [Ao] - kt …………(ii)
[Ao] - [A] = kt
@ t1/2, [A] = [Ao]/2
kt1/2 = [Ao] - ([Ao]/2)
kt1/2 = (2[Ao] - [Ao])/2
kt1/2 = [Ao]/2
t1/2 = [Ao]/2k …………(iii)
Equation (iii) above, gives the half-life for a zeroth-order reaction, which can be obtained by finding the product of half the initial reactant concentration and the reciprocal of the reaction rate constant ([Ao]/2 × 1/k).
(b) Half-life of a First-Order Reaction
Recall that the first-order integrated rate law is given by:
ln [A] = ln [Ao] - kt …………(iv)
kt = ln [Ao] - ln [A]
@ t1/2, [A] = [Ao]/2
kt1/2 = ln [Ao] - ln ([Ao]/2)
kt1/2 = ln ([Ao]/([Ao]/2))
kt1/2 = ln 2
t1/2 = ln 2/k …………(v)
Equation (v) gives the half-life for a first-order reaction, which is independent of the initial concentration of the reactant. It can be obtained by dividing the natural log of 2 (ln 2 = 0.6932) by the reaction rate constant.
Equation (v) is applied when calculating the half-life of most radioactive substances. This is because nuclear or radioactive reactions are generally first-ordered.
(c) Half-life of a Second-Order Reaction
Recall that the integrated rate law for a second-order reaction is given by:
1/[A] = 1/[Ao] + kt …………(vi)
kt = 1/[A] - 1/[Ao]
@ t1/2, [A] = [Ao]/2
kt1/2 = 1/([Ao]/2) - 1/[Ao]
kt1/2 = 2/[Ao] - 1/[Ao]
kt1/2 = 1/[Ao]
t1/2 = 1/[Ao]k …………(vii)
Equation (vii) gives the half-life for a second-order reaction, which can be obtained by finding the reciprocal of the product of the initial concentration of the reactant and the reaction rate constant (1/([Ao] × k)).
Do These:
Question 1
Using the information provided in Example 1 of the Chemical Kinetics (Part II), calculate the half-life of the 1-chloroethane (C2H5Cl) for experiments 1 and 4.
Question 2
Determine the half-life of the 1-bromopropane (C3H7Br) in Example 3 of the Chemical Kinetics (Part II)
Question 3
Calculate the half-life of the peroxydisulfate ion (S2O8--) using the information provided in Example 4 of the Chemical Kinetics (Part II).
Mechanism of Reactions
The mechanism of a chemical reaction is the detailed description of the elementary steps involved in the given reaction. Some reactions proceed directly from the reactants to the products stage, without passing through an intermediate stage. Such reactions are known as one-step reactions, whereas those that pass through the intermediate stage are known as multi-step reactions. An intermediate is actually a very reactive or unstable species produced in one stage and consumed in another stage of a multi-step reaction.
Example of a multi-step reaction is the decomposition of ozone molecules to oxygen gas, according to the equations:
O3(g) ----> O2(g) + O. …………(a)
O3(g) + O. ----> 2O2(g) ……...(b)
2O3(g) ----> 3O2(g) …………….(c)
The mechanism of the two-step reaction is as explained below:
The ozone molecule undergoes a photolytic cleavage (the breaking of bond by the action of UV light) in (a) to produce a highly unstable oxygen atom radical, O. as the intermediate species. The radical is consumed in (b) when it attacks an ozone molecule to produce two molecules of oxygen gas. The addition of (a) and (b) gives the overall reaction (c).
In a multi-step reaction, the elementary step with the slowest rate of reaction is known as the rate determining step of the overall reaction. Remember, a chain, they say, is as strong as its weakest link. Which means that no matter how strong a chain looks, its strength is limited by strength of its weakest link.
The same analogy applies here, because no matter how fast the other reaction steps may be, the overall rate of the reaction is limited by the rate of slowest reaction step. This explains why the rate law of the overall reaction is determined by the stoichiometry of the rate determining step and not the stoichiometry of the overall reaction.
It is important to note that we can propose two or more mechanisms for a particular reaction. However, the reasonable one will be the mechanism that is consistent with the experimentally determined kinetic data for the reaction.
Twitter: @gmtacademy
WhatsApp: 07034776117
Facebook: www.facebook.com/greatermindstutors
Comments
Post a Comment