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Rate Laws The rate of a reaction can be expressed in a second
way. For the hydrolysis of acetyl chloride, we can write
That is:
nCOCl][CHrate 3
nCOCl][CHrate 3k
2
Rate Laws The rate of a reaction can be expressed in a second
way. For the hydrolysis of acetyl chloride, we can write
That is:
The exponent n is determined by experiment. k is called the rate constant. Note that this constant
does not depend on the concentration of acetyl chloride.
nCOCl][CHrate 3
nCOCl][CHrate 3k
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For a more general reaction such as: a A + b B c C + d D
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For a more general reaction such as: a A + b B c C + d D
The forward rate (left right) is given by:
y[B]x[A]rate k
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For a more general reaction such as: a A + b B c C + d D
The forward rate (left right) is given by:
Note that x and y are NOT necessarily related to the stoichiometric coefficients a and b in the above equation. They are numbers which are constants for a given reaction.
y[B]x[A]rate k
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Rate Law: An expression relating the rate of a reaction to the rate constant and the concentrations of the reactants raised to the appropriate powers.
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Rate Law: An expression relating the rate of a reaction to the rate constant and the concentrations of the reactants raised to the appropriate powers.
Reaction Order: The power to which the concentration of a reactant needs to be raised in the rate law expression. For a reaction with more than one species present, it is the sum of the reaction orders of the individual species in the rate law.
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Suppose in the preceding reaction that x = 1 and y = 2. The reaction is said to be first-order in species A and second-order in species B.
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Suppose in the preceding reaction that x = 1 and y = 2. The reaction is said to be first-order in species A and second-order in species B.
The reaction would be described as third-order (sum of the reaction orders of the individual species reacting).
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Important Point The exponents x and y that determine the order of
the reaction are not found from the stoichiometric coefficients in the overall balanced equation.
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Important Point The exponents x and y that determine the order of
the reaction are not found from the stoichiometric coefficients in the overall balanced equation.
These exponents must be determined experimentally.
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Some examples: 2 N2O5(g) 4 NO2(g) + O2(g)
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Some examples: 2 N2O5(g) 4 NO2(g) + O2(g)
The rate law is given by:
]O[Nt
]O[N2
rate 5252 k
1
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Some examples: 2 N2O5(g) 4 NO2(g) + O2(g)
The rate law is given by:
Note that the rate is NOT given by:
]O[Nt
]O[N2
rate 5252 k
1
252 ]O[Nrate k
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For the reaction H2(g) + I2(g) 2 HI(g)
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For the reaction H2(g) + I2(g) 2 HI(g)
The rate of reaction is:
]][I[Hrate 22k
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For the reaction H2(g) + I2(g) 2 HI(g)
The rate of reaction is:
It is just a coincidence that the exponents on the concentration terms in this rate law match the stoichiometric coefficients in the overall balanced equation for this example.
]][I[Hrate 22k
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For the reaction CHCl3(g) + Cl2(g) CCl4(g) + HCl(g)
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For the reaction CHCl3(g) + Cl2(g) CCl4(g) + HCl(g)
The rate of reaction is:
1/223 ]][Cl[CHClrate k
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For the reaction CHCl3(g) + Cl2(g) CCl4(g) + HCl(g)
The rate of reaction is:
The exponents are not required to be integers, though this is often the case.
1/223 ]][Cl[CHClrate k
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Integrated rate laws: Concentration changes over time
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Analysis of some simple classes of reaction by Order
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Analysis of some simple classes of reaction by Order
Zero-order reactions
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Analysis of some simple classes of reaction by Order
Zero-order reactions
The simplest class of reactions to treat are zero-order. In practical applications, this case does not arise very frequently, but it is simple to treat.
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For a zero-order reaction, the rate of reaction is independent of the concentrations of the reactants.
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For a zero-order reaction, the rate of reaction is independent of the concentrations of the reactants.
If the following reaction: A P follows zero-order kinetics,
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For a zero-order reaction, the rate of reaction is independent of the concentrations of the reactants.
If the following reaction: A P follows zero-order kinetics, then
kk
0[A]t
[A]rate
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For a zero-order reaction, the rate of reaction is independent of the concentrations of the reactants.
If the following reaction: A P follows zero-order kinetics, then
So the rate is a fixed constant k and does not change with time.
kk
0[A]t
[A]rate
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Notation: The concentration of reactant A at time t is denoted
as and the concentration of reactant A at the start of the reaction, taken as t = 0, is denoted by
, which is the initial concentration of A.
t[A]
0[A]
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Notation: The concentration of reactant A at time t is denoted
as and the concentration of reactant A at the start of the reaction, taken as t = 0, is denoted by
, which is the initial concentration of A.
From the expression we can write
t[A]
0[A]
k
t
[A]
t[A][A] 0t k
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Recall that the equation of a straight line is
mxby
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Recall that the equation of a straight line is
So comparison with , indicates that a plot of versus t will yield a straight line with a slope = -k.
mxby
t[A][A] 0t kt[A]
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Recall that the equation of a straight line is
So comparison with , indicates that a plot of versus t will yield a straight line with a slope = -k.
mxby
t[A][A] 0t kt[A]
t[A]
t
kslope
0
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A plot of the rate versus time would look like:
rate
t0
[A]of tindependen
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First-order reactions From a practical standpoint, this is an important
case, and occurs commonly.
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First-order reactions From a practical standpoint, this is an important
case, and occurs commonly. If the following reaction: A P follows first-order kinetics, then
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First-order reactions From a practical standpoint, this is an important
case, and occurs commonly. If the following reaction: A P follows first-order kinetics, then
[A]t
[A]rate k
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First-order reactions From a practical standpoint, this is an important
case, and occurs commonly. If the following reaction: A P follows first-order kinetics, then
The equation can be solved (with a slight modification and using calculus).
[A]t
[A]rate k
[A]t
[A] k
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The result is:
t[A][A]ln
t
0 k
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The result is:
Because of the following property of logs:
t[A][A]ln
t
0 k
XYlnln(Y)ln(X)Y
Xln
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The result is:
Because of the following property of logs:
then the above equation can be written as
t[A][A]ln
t
0 k
XYlnln(Y)ln(X)Y
Xln
t[A][A]ln
0
t k
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The preceding equation can be put in the form (by taking antilogs of both sides of the equation):
t[A][A]
0
t ke
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The preceding equation can be put in the form (by taking antilogs of both sides of the equation):
that is:
t[A][A]
0
t ke
t[A][A] 0tke
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The preceding equation can be put in the form (by taking antilogs of both sides of the equation):
that is:
You can convert the preceding equation back into by taking ln of both sides of the equation.
t[A][A]
0
t ke
t[A][A] 0tke
t[A][A]ln
0
t k
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Note that the equation
is that of a straight line if we identify y with the term
and x with the time t (the constant b would be
zero).
t[A][A]ln
t
0 k
t
0[A][A]ln
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Note that the equation
is that of a straight line if we identify y with the term
and x with the time t (the constant b would be
zero).
t[A][A]ln
t
0 k
t
0[A][A]ln
t0
t
0[A][A]ln
kslope
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An alternative plot can be made: Rewriting
as
so the following plot can be made:
t[A][A]ln
t
0 k
t[A]ln)([A]ln 0t k )(
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An alternative plot can be made: Rewriting
as
so the following plot can be made:
t[A][A]ln
t
0 k
t0
kslope
t[A]ln)([A]ln 0t k )(
)( t[A]ln
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A plot of versus t will show the following appearance:
t0
t[A]
t[A]
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A plot of versus t will show the following appearance:
This is an exponential fall-off of the concentration with time.
t0
t[A]
t[A]
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Second-order reactions
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Second-order reactions We will examine two cases: Both cases arise fairly commonly.
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Second-order reactions We will examine two cases: Both cases arise fairly commonly. First case – a single reactant species.
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Second-order reactions We will examine two cases: Both cases arise fairly commonly. First case – a single reactant species.
If the following reaction: A P follows second-order kinetics,
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Second-order reactions We will examine two cases: Both cases arise fairly commonly. First case – a single reactant species.
If the following reaction: A P follows second-order kinetics, then
2[A]t
[A]rate k
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Second-order reactions We will examine two cases: Both cases arise fairly commonly. First case – a single reactant species.
If the following reaction: A P follows second-order kinetics, then
This equation can be solved (using calculus) for to yield the following result.
2[A]t
[A]rate k
t[A]
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t[A][A] 0t
k 11
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This is of the form of a straight line if you identify y with
and x with the time t.
t[A][A] 0t
k 11
t[A]1
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This is of the form of a straight line if you identify y with
and x with the time t. The constant b would be
and the slope = k.
t[A][A] 0t
k 11
t[A]1
0[A]1
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t0
kslope t[A]1
0[A]1
intercept
