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CHEMICAL
KINETICS
Chemical Kinetics
- The area of chemistry that
concerns reaction rates.
Reaction Rates
- The questions posed in this chapter will be:
- How is the rate of a reaction measured?
- What conditions will affect the rate of a reaction?
- How do you express the relationship of rate to the variables affecting the rate?
- What happens on a molecular level during a chemical reaction?
Reaction Rate
•Change in concentration (conc) of a reactant or
product per unit time.
Rate =
conc of A at time 2 conc of A at time 1
2 1
t t
t t
[ ]
∆ ∆
A t
Definition of Reaction Rate
- The reaction rate is the increase in molar
concentration of a product of a reaction
per unit time.
- It can also be expressed as the decrease in
molar concentration of a reactant per unit
time.
1. Rate of formation of product = ∆[product]/∆t This is the average rate over time 2. Rate of reaction of reactant = - ∆[Reactant]/∆t This is the average rate over time
Definition of Reaction Rates
- Consider the gas-phase decomposition of
dintrogen pentoxide.
2 N 2 O 5 (g) → 4 NO 2 (g) + O 2 (g )
- If we denote molar concentrations using brackets, then the change in the molarity of O 2 would be represented as
where the symbol, ∆ (capital Greek delta), means “the change in.”
∆ [O 2 ]
- Figure 14.5 shows the increase in
concentration of O 2 during the decomposition
of N 2 O 5.
Definition of Reaction Rates
- Note that the rate decreases as the reaction
proceeds.
Figure 14.4 The instantaneous rate of reaction In the reaction
The concentration of O (^2) increases over time. You obtain the instantaneous rate from the slope of the tangent at the point of the curve corresponding to that time.
2 N 2 O 5 (g) → 4 NO 2 (g) + O 2 (g)
10
Figure 12.3: A plot of the concentration of N 2 O (^5) is a function of time for the reaction: 2N 2 O 5 ( soln ) º 4NO 2 ( soln ) + O 2 ( g ) (at 45ºC). Note that the reaction rate at [N 2 O 5 ] = 0.90 M is twice that at [N 2 O 5 ] =
0.45 M. • Because the amounts of products and
reactants are related by stoichiometry, any
substance in the reaction can be used to
express the rate.
Definition of Reaction Rates
t
[NO]
Rate ofdecompositionofN 2 O 5 - 2 5
∆
∆
- Note the negative sign. This results in a
positive rate as reactant concentrations
decrease.
Factors Affecting Reaction Rates
- Temperature at which a reaction occurs.
- Usually reactions speed up when the temperature increases.
- A good “rule of thumb” is that reactions approximately double in rate with a 10 o^ C rise in temperature. - Concentration of a catalyst. - A catalyst is a substance that increases the rate of a reaction without being consumed in the overall reaction. - The catalyst generally does not appear in the overall balanced chemical equation (although its presence may be indicated by writing its formula over the arrow).
Factors Affecting Reaction Rates
2 () 2 ()
( ) (^2 2 2) ( ) (^2) l g HBraq H Oaq → HO + O
- Concentration of a catalyst.
Factors Affecting Reaction Rates
2 (l) 2 (g )
HBr(aq) 2 H 2 O 2 (aq) → 2 HO + O
- Figure 14.2 shows the HBr catalyzed decomposition of H 2 O 2 to H 2 O and O 2.
- A catalyst speeds up reactions by reducing the “activation energy” needed for successful reaction.
- A catalyst may also provide an alternative mechanism, or pathway, that results in a faster rate.
Figure 14.2: Catalytic decomposition of hydrogen peroxide. Photo courtesy of James Scherer.
Effect of Surface Area of Solid or Catalyst (in case of solids) or Pressure (gases)
Increasing the surface area of a solid reactant or catalyst or the pressure of a gaseous reactant (without increasing the actual amount of the gas) usually increases the reaction rate.
In the case of the solid reactant or catalyst - there are more available reaction sites. In the case of the gas, the likelihood collisions resulting in reaction increases
Factors Affecting Reaction Rates Figure 12.8: The decomposition reaction: 2N 2 O( g ) º 2N 2 ( g ) + O 2 ( g ) takes place on a platinum surface.
Types of Rate Laws
- Differential Rate Law: expresses how rate
depends on concentration.
- Integrated Rate Law: expresses how
concentration depends on time.
Dependence of Rate on
Concentration
- 2 NO 2 (g) º 2 NO (g) + O 2 (g)
- When the concentration of nitrogen dioxide is doubled, the reaction rate doubles.
- We need a mathematical expression to relate the rate of the reaction to the concentrations of the reactants.. Rate = - )[NO 2 ]/ )t =?
Figure 12.2: Representation of the reaction: 2NO 2 ( g ) º 2NO( g ) + O 2 ( g ). (a) The reaction at the very beginning (t = 0). (b) and (c) As time passes, NO 2 is converted to NO and O 2.
Figure 12.1: Starting with a flask of nitrogen dioxide at 300°C, the concentrations of nitrogen dioxide, nitric oxide, and oxygen are plotted versus time.
Dependence of Rate on
Concentration
- Determining the Rate Law.
- One method for determining the order of a reaction with respect to each reactant is the method of initial rates.
- It involves running the experiment multiple times, each time varying the concentration of only one reactant and measuring its initial rate.
- The resulting change in rate indicates the order with respect to that reactant.
Dependence of Rate on
Concentration
- Determining the Rate Law.
- If doubling the concentration of a reactant has a doubling effect on the rate, then one would deduce it was a first-order dependence.
- If doubling the concentration had a quadrupling effect on the rate, one would deduce it was a second-order dependence.
- A doubling of concentration that results in an eight-fold increase in the rate would be a third- order dependence.
Rate = k[A] m
A Problem to Consider
- Iodide ion is oxidized in acidic solution to triiodide ion, I 3 -^ , by hydrogen peroxide.
- A series of four experiments was run at different concentrations, and the initial rates of I 3 -^ formation were determined.
- From the following data, obtain the reaction orders with respect to H 2 O 2 , I -^ , and H +.
- Calculate the numerical value of the rate constant.
H (^) 2 O 2 ( aq )+ 3 I ( aq )+ 2 H ( aq )→ I 3 ( aq )+ 2 H 2 O ( l ) − + −
A Problem to Consider
Exp. 4 0.010 0.010 0.00100 1.15 x 10 -
Exp. 3 0.010 0.020 0.00050 2.30 x 10 -
Exp. 2 0.020 0.010 0.00050 2.30 x 10 -
Exp. 1 0.010 0.010 0.00050 1.15 x 10 -
H 2 O 2 I - H + Initial Rate [mol/(L.^ s)]
Initial Concentrations (mol/L)
- Comparing Experiment 1 and Experiment 2, you see that when the H 2 O 2 concentration doubles (with other concentrations constant), the rate doubles.
- This implies a first-order dependence with respect to H 2 O 2.
A Problem to Consider
Exp. 4 0.010 0.010 0.00100 1.15 x 10 -
Exp. 3 0.010 0.020 0.00050 2.30 x 10 -
Exp. 2 0.020 0.010 0.00050 2.30 x 10 -
Exp. 1 0.010 0.010 0.00050 1.15 x 10 -
H 2 O 2 I - H + Initial Rate [mol/(L.^ s)]
Initial Concentrations (mol/L)
- Comparing Experiment 1 and Experiment 3, you see that when the I-^ concentration doubles (with other concentrations constant), the rate doubles.
- This implies a first-order dependence with respect to I-^.
A Problem to Consider
Exp. 4 0.010 0.010 0.00100 1.15 x 10 -
Exp. 3 0.010 0.020 0.00050 2.30 x 10 -
Exp. 2 0.020 0.010 0.00050 2.30 x 10 -
Exp. 1 0.010 0.010 0.00050 1.15 x 10 -
H 2 O 2 I - H + Initial Rate [mol/(L.^ s)]
Initial Concentrations (mol/L)
- Comparing Experiment 1 and Experiment 4, you see that when the H+^ concentration doubles (with other concentrations constant), the rate is unchanged.
- This implies a zero-order dependence with respect to H+^.
A Problem to Consider
Exp. 4 0.010 0.010 0.00100 1.15 x 10 -
Exp. 3 0.010 0.020 0.00050 2.30 x 10 -
Exp. 2 0.020 0.010 0.00050 2.30 x 10 -
Exp. 1 0.010 0.010 0.00050 1.15 x 10 -
H 2 O 2 I - H + Initial Rate [mol/(L.^ s)]
Initial Concentrations (mol/L)
- Because [H+] 0 = 1, the rate law is:
Rate k[H 2 O 2 ][I ] − =
- The reaction orders with respect to H 2 O 2 , I -^ , and H +, are 1 , 1 , and 0 , respectively.
A Problem to Consider
Exp. 4 0.010 0.010 0.00100 1.15 x 10 -
Exp. 3 0.010 0.020 0.00050 2.30 x 10 -
Exp. 2 0.020 0.010 0.00050 2.30 x 10 -
Exp. 1 0.010 0.010 0.00050 1.15 x 10 -
H 2 O 2 I - H + Initial Rate [mol/(L.^ s)]
Initial Concentrations (mol/L)
- You can now calculate the rate constant by substituting values from any of the experiments. Using Experiment 1 you obtain:
2 10 /( )
010 0. 010 /
(^1510612) L mol s molL
s k = × ⋅ ×
× = −
− −
A Problem to Consider
Exp. 4 0.010 0.010 0.00100 1.15 x 10 -
Exp. 3 0.010 0.020 0.00050 2.30 x 10 -
Exp. 2 0.020 0.010 0.00050 2.30 x 10 -
Exp. 1 0.010 0.010 0.00050 1.15 x 10 -
H 2 O 2 I - H + Initial Rate [mol/(L.^ s)]
Initial Concentrations (mol/L)
- You can now calculate the rate constant by substituting values from any of the experiments. Using Experiment 1 you obtain:
L
0. 010 mol L
k 0. 010 mol Ls
1. 15 106 mol = × × ⋅
× −
Figure 12.4: A plot of ln[N 2 O 5 ]
versus time.
A Problem to Consider
- The decomposition of N 2 O 5 to NO 2 and O 2 is first order with a rate constant of 4.8 x 10-4^ s-1^. If the initial concentration of N 2 O 5 is 1.65 x 10-2^ mol/L, what is the concentration of N 2 O 5 after 825 seconds?
- The first-order time-concentration equation for this reaction would be: - kt [NO ]
[NO ] ln 2 5 o
2 5 t
A Problem to Consider
- The decomposition of N 2 O 5 to NO 2 and O 2 is first order with a rate constant of 4.8 x 10-4^ s-1^. If the initial concentration of N 2 O 5 is 1.65 x 10-2^ mol/L, what is the concentration of N 2 O 5 after 825 seconds?
- Substituting the given information we obtain: - (4.80 10 s ) ( 825 s)
1. 65 10 mol/L
[NO]
ln 2 25 t = × -^4 -^1 ×
×
−
A Problem to Consider
- The decomposition of N 2 O 5 to NO 2 and O 2 is first order with a rate constant of 4.8 x 10-4^ s-1^. If the initial concentration of N 2 O 5 is 1.65 x 10-2^ mol/L, what is the concentration of N 2 O 5 after 825 seconds?
- Substituting the given information we obtain:
- 65 10 /
[ ] ln (^2) 2 5 = ×
− mol L
N O t
A Problem to Consider
- The decomposition of N 2 O 5 to NO 2 and O 2 is first order with a rate constant of 4.8 x 10-4^ s-1^. If the initial concentration of N 2 O 5 is 1.65 x 10-2^ mol/L, what is the concentration of N 2 O 5 after 825 seconds?
- Taking the inverse natural log of both sides we obtain:
1. 65 10 mol/L
[N O] -0.
2
2 5 t = e =
× −
A Problem to Consider
- The decomposition of N 2 O 5 to NO 2 and O 2 is first order with a rate constant of 4.8 x 10-4^ s-1^. If the initial concentration of N 2 O 5 is 1.65 x 10-2^ mol/L, what is the concentration of N 2 O 5 after 825 seconds?
- Solving for [N 2 O 5 ] at 825 s we obtain:
[N 2 O 5 ] = (1.65 × 10 -^2 mol/L) × ( 0. 673 ) = 0. 0111 mol/ L
Figure 12.5: A plot of [N 2 O 5 ]
versus time for the
decomposition reaction of N 2 O 5.
Half-Life of a First-Order
Reaction
- t 1/2 = half-life of the reaction
- k = rate constant
•For a first-order reaction, the half-life does
not depend on concentration.
t
k
1/2 =^
Second-Order Rate Law
- For a A → products in a second-order
reaction,
Rate =
A
A
∆ t
k^2
A
A o
= kt
Concentration-Time Equations
- Second-Order Rate Law
- You could write the rate law in the form
2 k[A] t
[A] Rate = ∆
∆ = −
Concentration-Time Equations
- Second-Order Rate Law
- Using calculus, you get the following equation.
t [A] o
1 kt [A]
1 = +
- Here [A] (^) t is the concentration of reactant A at time t , and [A] (^) o is the initial concentration.
Figure 12.6: (a) A plot of ln[C 4 H 6 ] versus t. (b) A plot of 1n[C 4 H 6 ] versus t.
