Title: Figure 12'27: Layers of ions surrounding charged colloidal particles'
1Average84.2 Standard Deviation13.4
2Average161.6 Standard Deviation24.9
3Reaction Rates
- Chemical kinetics is the study of reaction rates,
how reaction rates change under varying
conditions, and what molecular events occur
during the overall reaction.
- What variables affect reaction rate?
Surface area of a solid reactant or catalyst.
4Dependence of Rate on Concentration
- Consider the reaction of nitric oxide with
hydrogen according to the following equation.
- Thus, the reaction is second order in NO, first
order in H2, and third order overall.
5Dependence of Rate on Concentration
- Zero and negative orders are also possible.
- The concentration of a reactant with a zero-order
dependence has no effect on the rate of the
reaction.
- Although reaction orders frequently have whole
number values (particularly 1 and 2), they can be
fractional.
6Dependence 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.
7Dependence 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.
8(No Transcript)
9(No Transcript)
10A Problem to Consider
- Iodide ion is oxidized in acidic solution to
triiodide ion, I3- , by hydrogen peroxide.
- A series of four experiments was run at different
concentrations, and the initial rates of I3-
formation were determined. - From the following data, obtain the reaction
orders with respect to H2O2, I-, and H. - Calculate the numerical value of the rate
constant.
11A Problem to Consider
- Comparing Experiment 1 and Experiment 2, you see
that when the H2O2 concentration doubles (with
other concentrations constant), the rate doubles. - This implies a first-order dependence with
respect to H2O2.
12A Problem to Consider
- 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-.
13A Problem to Consider
- 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.
14A Problem to Consider
- The reaction orders with respect to H2O2, I-, and
H, are 1, 1, and 0, respectively.
15A Problem to Consider
- You can now calculate the rate constant by
substituting values from any of the experiments.
Using Experiment 1 you obtain
16A Problem to Consider
- You can now calculate the rate constant by
substituting values from any of the experiments.
Using Experiment 1 you obtain
17Change of Concentration with Time
- A rate law simply tells you how the rate of
reaction changes as reactant concentrations
change.
- A more useful mathematical relationship would
show how a reactant concentration changes over a
period of time.
18Change of Concentration with Time
- A rate law simply tells you how the rate of
reaction changes as reactant concentrations
change.
- Using calculus we can transform a rate law into a
mathematical relationship between concentration
and time.
- This provides a graphical method for determining
rate laws.
19Concentration-Time Equations
20Concentration-Time Equations
- Using calculus, you get the following equation.
- Here At is the concentration of reactant A at
time t, and Ao is the initial concentration. - The ratio At/Ao is the fraction of A
remaining at time t.
21A Problem to Consider
- The decomposition of N2O5 to NO2 and O2 is first
order with a rate constant of 4.8 x 10-4 s-1. If
the initial concentration of N2O5 is 1.65 x 10-2
mol/L, what is the concentration of N2O5 after
825 seconds?
22A Problem to Consider
- The decomposition of N2O5 to NO2 and O2 is first
order with a rate constant of 4.8 x 10-4 s-1. If
the initial concentration of N2O5 is 1.65 x 10-2
mol/L, what is the concentration of N2O5 after
825 seconds?
- Substituting the given information we obtain
23A Problem to Consider
- The decomposition of N2O5 to NO2 and O2 is first
order with a rate constant of 4.8 x 10-4 s-1. If
the initial concentration of N2O5 is 1.65 x 10-2
mol/L, what is the concentration of N2O5 after
825 seconds?
- Substituting the given information we obtain
24A Problem to Consider
- The decomposition of N2O5 to NO2 and O2 is first
order with a rate constant of 4.8 x 10-4s-1. If
the initial concentration of N2O5 is 1.65 x 10-2
mol/L, what is the concentration of N2O5 after
825 seconds?
- Taking the inverse natural log of both sides we
obtain
25A Problem to Consider
- The decomposition of N2O5 to NO2 and O2 is first
order with a rate constant of 4.8 x 10-4 s-1. If
the initial concentration of N2O5 is 1.65 x 10-2
mol/L, what is the concentration of N2O5 after
825 seconds?
- Solving for N2O5 at 825 s we obtain
26Concentration-Time Equations
27Concentration-Time Equations
- Using calculus, you get the following equation.
- Here At is the concentration of reactant A at
time t, and Ao is the initial concentration.
28Graphing Kinetic Data
- In addition to the method of initial rates, rate
laws can be deduced by graphical methods.
- If we rewrite the first-order concentration-time
equation in a slightly different form, it can be
identified as the equation of a straight line.
- This means if you plot lnA versus time, you
will get a straight line for a first-order
reaction. (see Figure 14.9)
29Figure 14.9 A plot of log R versus time.
30Graphing Kinetic Data
- In addition to the method of initial rates, rate
laws can be deduced by graphical methods.
- If we rewrite the second-order concentration-time
equation in a slightly different form, it can be
identified as the equation of a straight line.
y mx b
31Graphing Kinetic Data
- In addition to the method of initial rates, rate
laws can be deduced by graphical methods.
- If we rewrite the first-order concentration-time
equation in a slightly different form, it can be
identified as the equation of a straight line.
32Half-life
- The half-life of a reaction is the time required
for the reactant concentration to decrease to
one-half of its initial value.
- For a first-order reaction, the half-life is
independent of the initial concentration of
reactant.
33Half-life
- The half-life of a reaction is the time required
for the reactant concentration to decrease to
one-half of its initial value.
- Solving for t1/2 we obtain
- Figure 14.8 illustrates the half-life of a
first-order reaction.
34Figure 14.8 A graph illustrating that the
half-life of a first-order reaction is
independent of initial concentration.
Half life, t1/2, is the time it takes for the
R to decrease by 1/2.
This is exactly like radioactive decay.
35Half-life
- Sulfuryl chloride, SO2Cl2, decomposes in a
first-order reaction to SO2 and Cl2.
- At 320 oC, the rate constant is 2.2 x 10-5 s-1.
What is the half-life of SO2Cl2 vapor at this
temperature?
36A model of SO2CI2(g)
37Half-life
- Sulfuryl chloride, SO2Cl2, decomposes in a
first-order reaction to SO2 and Cl2.
- At 320 oC, the rate constant is 2.20 x 10-5 s-1.
What is the half-life of SO2Cl2 vapor at this
temperature?
- Substitute the value of k into the relationship
between k and t1/2.
38Half-life
- Sulfuryl chloride, SO2Cl2, decomposes in a
first-order reaction to SO2 and Cl2.
- At 320 oC, the rate constant is 2.20 x 10-5 s-1.
What is the half-life of SO2Cl2 vapor at this
temperature?
- Substitute the value of k into the relationship
between k and t1/2.
39Half-life
- For a second-order reaction, half-life depends on
the initial concentration and becomes larger as
time goes on.
- Each succeeding half-life is twice the length of
its predecessor.