Laboratory > Kinetics
Notes on Kinetics
Kinetics is the study of the rate of reactions, or how fast they occur under different conditions. It usually includes a study of the mechanisms of reactions, which is a look at how the reacting molecules break apart and then form the new molecules. This knowledge allows chemists to control reactions and/or design new or better ways to produce the desired products.
These notes are divided into six separate sections. They may be accessed in any order from this page or from within each section.
· Reaction Rates
· Reaction Order
· Activation Energy
· Catalysts
· Temperature Effects
· Reaction Mechanisms
Reaction Rates
What is meant by reaction rate?
Consider the reaction: 2N_{2}O_{5(g)} —–> 4NO_{2(g)} + O_{2(g)}
The rate is defined as the change in concentration of N_{2}O_{5} with respect to time or
rate = –DConc N_{2}O_{5 }/ Dt
where DConc N_{2}O_{5} = final conc N_{2}O_{5} – orig. conc. N_{2}O_{5}, and
Dt = final time – initial time
Reaction rate is always positive
Concentration is expressed in mol/L, whereas time can be in seconds, minutes, hours, days, etc.
Again, the rate of this reaction can be expressed as: rate = d(conc. N_{2}O_{5}) / dt
The instantaneous rate = – (slope of the tangent to conc. vs. time curve)
How are Reaction Rate and Concentration Related?
The rate decreases as the concentration of the starting materials (reactants) decreases.
plot of rate vs conc – if straight line, then it is said to be a first order reaction: in other words the rate of the reaction is directly proportional to the concentration.
What is the Rate Expression?
rate = k(conc N_{2}O_{5(g)})
this indicates that the rate of the reaction for 2N_{2}O_{5(g)} —–> 4NO_{2(g)} + O_{2(g)} depends upon the concentration of reactant
“k” is called the rate constant. It is independent of the other quantities in the equation.
It is dependent on:
1. the nature of the reaction
“fast” reactions typically have large rate constants
if k is small, the reaction is slow at ordinary concentration
2. the temperature
ordinarily, k increases with temperature
k = rate / conc. N_{2}O_{5(g)}
example: k = .056 mol/L^{.}min / .160 mol/L = 0.35/min
Reaction Order
What is meant by Order of the Reaction?
aA_{(g)} —–> products
rate = k(conc A)^{m}
The power(m) to which the concentration of A is raised in the rate expression describes the order of the reaction.
· If “m” = 0, the reaction is zero order
· If “m” = 1, it is first order
· If “m” = 2, it is second order
Conditions:
· m = 0: zero order – rate is independent of the concentration of reactant. Doubling concentration has noeffect on rate.
· m = 1: first order – rate is directly proportional to the concentration of the reactant. Doubling the concentration increases the rate by a factor of 2.
· m = 2: second order – the rate is <PROPORTIONAL< b>to the square of the concentration of the reactant. Doubling the concentration increases the rate by a factor of 4.
Given the following date for the reaction CH_{3}CHO —-> CH_{4} + CO
conc CH_{3}CHO (mol/L) | 0.10 | 0.20 | 0.30 | 0.40 |
rate (mol/L^{.}) | 0.085 | 0.34 | 0.76 | 1.4 |
Order of a reaction must be determined experimentally. It cannot be deduced from the coefficients of a balanced equation.
Two-reactant reaction:
aA_{(g)} + bB_{(g)} —-> products
rate = k(conc A)^{m}(conc B)^{n} refer to “m” as “the order of the reaction with respect to A” and “n” as “the order of the reaction with respect to B” the overall order of the reaction is the sum of the exponents, m + n.
example: CO + NO_{2} —–> CO_{2} + NO
rate = k(conc CO)(conc NO_{2})
This reaction is :
· first order with respect to CO, m = 1
· first order with respect to NO_{2}, m = 1
· second order overall, m + n = 2
Two reactant reactions are more difficult to monitor. The best way is to hold initial concentration of one reactant constant while varying the concentration of the other reactant.
example: 2H_{2} + 2NO —-> N_{2} + 2H_{2}O
rate = k(conc H_{2})^{m}(conc NO)^{n}
m = 1, as conc H_{2} goes from 0.1 to 0.2, the rate doubles from 0.1 to 0.2
n = 2, as conc NO goes from 0.1 to 0.2, the rate goes from 0.1 to 0.4
The rate equation is therefore: rate = k(conc H_{2})(conc NO)^{2}
the overall order for the reaction, m + n = 3, is third order.
Reactant Concentration and Time
for aA_{(g)} —-> products, the rate = k(conc A)
example: 2N_{2}O_{5} —-> 4NO_{2} + O_{2}
rate = k(conc N_{2}O_{5})
log_{10}(conc N_{2}O_{5}) = log_{10}(conc N_{2}O_{5})_{0} – kt / 2.30
k = rate constant
t = time
0 = original conc N_{2}O_{5} at t = 0
plotting log_{10} conc N_{2}O_{5} vs time gives a straight line
using the equation of a line, y = a + bx
· y = log_{10} conc N_{2}O_{5}
· x = time, t
· y intercept, a = log_{10}(conc N_{2}O_{5})_{0}
· slope, b = –k
the slope can also be determined from dy/dx. Given the values of -0.60 / 4.0 min, the slope would be = -0.15/min.
Therefore, substituting -0.15/min into –kt/2.30 and solving for k gives k = 2.30(0.15/min) = 0.35/min
Another way of writing this is log_{10}(X / X_{0}) = kt/2.30
· X_{0} = original concentration
· X = concentration of x at time t
· k = first order rate constant
Sample Problem 1: Calculate the concentration of N_{2}O_{5} after 4.0 min starting with a concentration of 0.160 mol/L.
Solution:
log(0.160/x) = (0.35)(4.0)/2.30 = 0.61
log 0.160 – log x = 0.61
-0.80 – log X = 0.61
-log x = 1.41
log x = -1.41
x = 0.040 mol/L
Sample Problem 2: How much time is required for the concentration to drop from 0.160 to 0.100 mol/L?
Solution:
t = (2.30/k)log(X_{0}/X)= (2.30/0.35)log(0.160/0.100)
t = (2.30)(0.20)/0.35 = 1.3 min
Sample Problem 3: At what time will half of the sample be decomposed?
Solution:
X = X_{0}/2, therefore X_{0} = 2X and X/X_{0} = 2
t = (2.30/0.35)log 2 = (2.30)(0.30)/0.35 = 2.0 min
Sample Problem 4: How long will it take for the concentration to drop from 0.160 to 0.080 M?
Solution:
t = (2.30/k)log(X_{0}/X) = (2.30/0.35) log(0.160/0.080)
t = (2.30)(0.30)/0.35 = 2.0 min
This is important!!
The time required for one half of a reactant to decompose via a first order reaction has a fixed value, independent of concentration. This is called the half-life and has the expression
t_{1/2} = 0.693/k
k is the rate constant for the first order reaction
half-life is inversely proportional to the rate constant k
if k is large, half-life is short; a slow reaction with a small k will have a relatively long half-life.
What About Reactions With Other Integral Orders?
Most common processes in gas phase reactions are second order. Zero order reactions are less common.
An example of a zero order reaction is:
2HI_{(g)} —-> H_{2(g)} + I_{2(g)}
rate = k(conc HI)^{0} = k
The reaction occurrs at a constant rate independent of concentration of HI.
Third order reactions are rare. One example of a third order reaction is:
2H<_{2} + 2NO —-> N_{2} + 2 H_{2}O
The following table is a summary of important information related to 0, 1st and 2nd order reactions:
Order | ||||
rate | conc-time | half-life | linear plot | |
0 | rate = k | X_{0} – X = k | X_{0}/2k | X vs t |
1 | rate = kX | log X_{0}/X = kt/2.30 | 0.693/k | log X vs t |
2 | rate = kX^{2} | 1/x – 1/X_{0} = kt | 1/X_{0}k | 1/X vs t |
To decide if a reaction is 0, 1st or 2nd order, list X, log_{10}X, and 1/X in a table, then plot each against time. The one with the linear plot is the order of the reaction.
Catalysts
What are Catalysts?
Certain substances, called catalysts, can increase the rate of a reaction without being consumed by it.
An example is the decomposition of H_{2}O_{2}:
2H_{2}O_{2} —-> 2H_{2}O + O_{2} – this reaction is ordinarily very slow, however, add NaI_{(aq)} to the H_{2}O_{2} and the reaction becomes very fast.
Catalysts can be described as being homogeneous or heterogeneous.
· Homogeneous: reaction takes place within a single phase, as in the example above with the H_{2}O_{2}.
· Heterogeneous: reaction involves two different phases, as in the example that follows.
2N_{2}O —-> 2N_{2} + O_{2} – this reaction is usually slow, but if the reactants are bought into contact with a metal such as gold, the reaction speeds up considerably.
A catalyst lowers the activation energy required for the reaction.
Consider the reaction 2N_{2}O —-> 2N_{2} + O_{2} again. The E_{a} = 250 kJ, however, using the Au catalyst, the E_{a} is lowered to 120 kJ.
The reduction in activation energy occurs because the catalyst provides an alternate pathway of lower energy for the reaction.
The N_{2}O is chemically adsorbed on the metal surface. A bond is formed between the O of the N_{2}O and an Au atom. This weakens the bond joining the O to the N, thus making it easier for the molecule to break apart.
Now, consider again the decomposition of the hydrogen peroxide. It appears to take place in a 2-step process:
1. H_{2}O_{2} + I^{–}_{(aq)} —-> H_{2}O + IO^{–}_{(aq)}
2. H_{2}O_{2} + IO^{–}_{(aq)} —-> H_{2}O + O_{2} + I^{–}_{(aq)}
The sum of these two reactions is then 2H_{2}O_{2} —->H_{2}O
The I^{–} are not consumed in the reaction, for every I^{–} usesd in the first step, an I^{–} is produced in the second.
A catalyst has no effecct on the relative energies of the reactants and products, nor on the equilibrium constant, K_{c}. It merely speeds up the reaction, thus allowing it to reach equilibrium more quickly.
So, What Are Enzymes?
Enzymes are organic catalysts (proteins) of high molecular mass. They allow reactions which occur slowly under ordinary conditions to occur readily in the body.
An important example is that of the metabolism of sugar. The reaction is difficult to bring about directly. It will usually turn into a charred, black mess if you try to burn it in a flame, however, in the body, sugar is metabolized at 37 ^{o}C in a series of biochemical reactions. The end products are CO_{2} and H_{2}O.
Each step is catalyzed by a particular enzyme adapted for that purpose. The reactant, called the substrate fits into a specific site on the enzyme surface and is held in place by intermolecular forces. There, it can react with another species.
Inhibitors are substances which diminish the activity of an enzyme. One way they do this is to occupy the space on the enzyme where the substrate is supposed to fit. These usually have a structure similar to that of the substrate they replace.
The structures of enzymes have been studied with x-ray crystallography.
Temperature Effects
How Are Reaction Rates Affected By Temperature?
The rates of most reactions increase with a rise in temperature.
example
· the use of a pressure cooker to cook foods faster
· the use of a refrigerator to store food and slow its spoilage
The general rule is than an increase in temperature by 10 ^{o}C doubles the reaction rate.
The Kinetic Theory can be used to explain the effect of temperature on reaction rates.
Raising the temperature increases the fraction of molecules having very high kinetic energies. These are the ones most likely to react when they collide. The higher the temperature, the larger the fraction of molecules that can provide the activation energy needed for reaction.
The rate constant, k, becomes larger as the temperature increases.
What is the relation between k and the temperature, T?
f = e^{-Ea/RT}, where f is the fraction of molecules having an energy equal to or greater than E_{a}.
· e is the base of natural logs
· R is the gas constant
· T is the temperature in K
If we assume that the rate constant, k, is directly proportional to “f”, then,
k = cf = ce^{-E/RT}, where c is a proportionality constant.
Now, if we do a bit on math on this equation we get:
ln k = ln c – E_{a}/RT
log_{10} k = log_{10} c – E_{a}/2.30 RT
Next, substitute R = 8.31 J/mol^{.}K and let log_{10} c = A
log_{10} k = A – E_{a} / (2.30)(8.31)T
The equation in this form is called the Arrhenius Equation.
If you look at it, you will notice that it is in the form y = a + bx, with y being log_{10} k and x = 1/T.
A plot of log_{10} k vs 1/T should be a straight line. E_{a} for a reaction can be obtained from the slope of this line.
slope = -E_{a} / (2.30)(8.31)
The value of k at a particular temperature can be calculated if it is known for some other temperature
example:
The E_{a} for a reaction is 9.32 x 10^{4} J. At 27 ^{o}C, k = 1.25 x 10^{-2} L/mol^{.}s. Calculate the value of k at 127 ^{o}C.
Solution: Use this form of the Arrhenius Equation
log_{10} (k_{2} / k_{1}) = (E_{a} / (2.30)(8.31))(T_{2} – T_{1}) / T_{2}T_{1} | | This form is sometimes called |