Chapter 3 Pharmacokinetics §PK process in the body §Kinetic processes.
Chapter 3 Kinetic
description
Transcript of Chapter 3 Kinetic
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William L Masterton
Cecile N. Hurley http://academic.cengage.com/chemistry/masterton
Edward J. Neth University of Connecticut
PHYSICAL CHEMISTRY
Nor Azira Irma Muhammad
UiTM Perlis
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3.1 Rates of reaction
3.2 Factors affecting rates of reaction
3.3 Rate Law and order of reaction
3.4 Methods to determine order of reactions
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Chemical kinetics- a branch of chemistry that deals with the rate of chemical reaction
Many chemical reactions are fast reactions, especially those involved reactions between oppositely charged ions in aqueous solution (e.g, neutralisation, precipitation reaction)
Reaction involving organic compounds are usually slow reactions because they involved breaking and making covalent bonds
Reaction rate: Rate of reaction the change in concentration of a reactant with
time (M/s) Consider the following reaction: C4H9Cl + H2O C4H9OH + HCl The above reaction was conducted using 0.1 M butyl chloride
aqueous solution and its concentration was determined as a function of time
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Graph of concentration C4H9Cl vs time was plotted
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[C4
H9
Cl]
Time (s)
Rate of reaction = concentration time. = decrease in concentration of butyl chloride time taken = - d [C4H9Cl] dt The average rate = concentration (of the first 50 s) 50 s (value taken from table / expt)
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The instantaneous rate is a rate of change at particular instant in time
The instantaneous rate at time t = gradient (slope) of the tangent to the curve at time, t (specific time) The instantaneous rate at the start of the reaction (t = o) is called
the initial rate Initial rate = gradient (slope) of the tangent to the curve at time = 0
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concentration
time t
Instantaneous
rate
initial rate
time
concentration
example;
the instantaneous rate at t
= y2 - y1 mol dm-3s-1
x2 - x1 x2
x1
y1
y2
example;
the initial rate
= y2 - y1 mol dm-3s-1
x2 - 0 y1
y2
0 x2
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(1) Collision theory (2) Transition state theory 1) Collision theory Three ideas;
i. Molecules must collide to react ii. Molecules must posses a certain minimum kinetic energy,
called the activation energy, Ea to initiate the chemical reaction. (otherwise they are unable to react)
iii. Molecules must collide in the right orientation
The basis concept of collision is that, for the reaction between two particles to occur, an effective collision must take place
Not all collision are effective in producing reaction, but only small proportion of the molecules react after collision
Collision frequency of reaction = total number of collision between molecules per unit time at specific T 6
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The effect of concentration on the rate of reaction the reaction rate will increase if the concentration of one or
more reactant increase [ ] high, frequency of collision high, because more particles
present in the same volume and more likely to collide - probability of collision with sufficient energy for reaction to occur. So, rate of reaction increase.
The effect of temperature Rate of chemical reaction increase with T increase At higher T, Arhenius suggested colliding molecules can react only
if they have total kinetic energy equal or greater than Ea (activation energy)
Ea (activation energy) is a minimum amount of energy required to initiate a chemical reaction, which means that Ea is a minimum kinetic energy that molecules must possess in order for a chemical reaction to occur
Ea is shown on diagram called reaction profiles /energy profile - Energy versus progress of reaction
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H reactant
H product
Ea
energy energy
Progress of reaction Progress of reaction
Ea
reactants
reactants poducts
poducts
H = - ve H = + ve
a) Reaction profile of an exothermic reaction b) Reaction profile of an endothermic reaction
Ea = activation energy
H = difference in energies between reactant and product/ heat of reactions = H product H reactants
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2) The transition state theory This theory focus on what happens to the reactant molecules as
they change into products The presence of a transitory intermediate stage (that lies
between the reactants and the products) which is called the transition state, which involved hypothetical species is activated complex
e.g : H-H + Cl [H----H---Cl] H. + HCl reactants activated complex products The complex does not always changed to product. (if it does, it
means that the reaction is occuring / proceed to give the product)
The reaction progress on x- axis for diagram of reaction profile represents the extent of the reaction.
The reaction starts with the reactants on the left, progress through an activated complex (transition state), and ends with the products on the right.
The activated complex is very unstable. The energy gap between energies of the reactants and the activated complex is called the Ea. 9
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The rate law is an equation that relates the rate of reaction to the concentration of the reactants. This rate law can only be determined by experiment
Example: aA + bB cC + dD
If the rate is to the concentration of A only and not B, then
Rate of reaction [A] [B]0
Rate of reaction = k [A][B]0
= k [A]
The reaction is said to be first order with respect to A and zero order with respect to B
The overall order is 1 + 0 = 1, that is first order
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Rate = k [A]m [B]n
(the expression is called rate law / rate equation) m = order of reaction with respect to A
n = order of reaction with respect to B
Overall order of reaction = m + n
The proportionality constant, k is called the rate constant.
k value depends on the specific reaction, the temperature of the reaction and the presence of a catalyst (if any).
The larger value of k, the faster a reaction proceeds. Once the k and order of reaction are known, the reaction rate can be predicted for any concentration of A and B
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The order of reaction (m + n) and the rate constant (k) can be determined by using:
I. The reaction rate method
II. The linear plots
III.The half-life method, t1/2
IV. Initial rate method
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The order of reaction can be found by plotting a reaction rate against the concentration
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k
rate rate rate
[A] [A] [A]
First order Zero order Second order
y= kx y= kx2 y= k
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Example: Bromine reacts with methanoic acid according to the equation: The table below shows the rates of reactions at specific concentrations
)(2)(2)()(2 aqHaqBraqHCOOHaqBr
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[Br2] mol dm-3 Rate (10-5 mol dm-3s-1)
8.0 2.75
7.0 2.40
5.0 1.70
4.0 1.35
2.0 0.70
Show that the reaction between bromine and methanoic acid is a first order by a) calculation b) A graphical method
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Solution:
a) Calculation (Refer to the table)
when the concentration of bromine is doubled, the rate of reaction is doubled.
Thus, the reaction is first order with respect bromine. so, rate [Br2].
b) By graph
0
0.5
1
1.5
2
2.5
3
0 2 4 6 8 10
[Br2]
Rate
of
reacti
on
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A sloping straight line graph is obtained, when the graph of rate against concentration is plotted. This shows that the reaction is first order with respect to bromine Rate [Br2]
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The order of reaction can be found by plotting a linear plots with respect to a given reactant
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ln [A]t
time
y
x
time time
x
y x
y
[A]t 1/ [A]t
Slope = y / x = - k /2.303
First order Zero order second order
Slope = y / x = k
Slope = y / x = - k
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A first order reaction with respect to a reactant A, is a reaction in which the rate of reaction is directly proportional to the concentration of A.
If the concentration of a reactant is doubled, the rate of reaction is also doubled.
The rate equation for a first order reaction is, rate reactions = k [A] where, k = rate constant [A] = concentration in mol dm-3
the unit of k is time-1 (that is s-1, min-1, h-1) Examples of first order reactions:
i. Radioactive decay
ii. Catalytic decomposition of hydrogen peroxide
iii. Thermal decomposition of dinitrogen pentoxide
UkeradioactivofrateHeThU
238
92
4
2
234
90
238
92
)()(2)(2 222 gasOliqOHaqOH
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24252 22 OONON
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For the reaction:
A products
Having the rate law :
where [A] t = concentration of A at time t
Rearrange the expression:
Ak
dt
Adreactionofrate
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ot
ot
t
o
A
A
AnlktAnl
ktAnlAnl
dtkA
Ad
tegratein
dtkA
Ad
t
O
:
ln [A]t
t
A straight line with -ve slope
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Rate = - d [A] = k[A]1
dt
d[A] = - k[A]1
dt Integrate:
1 d[A] = - k dt [A] ln [A]t = - kt + c , t = 0 c = ln [A]0 ln [A]t = - kt + ln [A]0
ln [A]t
t
A straight line with -ve slope
A B (reactant) (product)
y -mx c
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CH3CO-C2H5 + H2O CH3COOH + C2H5OH
For example, the hydrolysis of ethyl ethanoate is first order
respect to the ester and first order with respect to the water and acid.
In the presence of excess H2O and H+, only a small fraction of acid
or water used in reaction The constant of acid and water hardly alter during the course of
the reaction So, the reaction appear to be zero order with respect to the acid
and water and the rate of hydrolysis is depend only on the concentration of ester
Rate of hydrolysis = k [ ester] Thus, acid hydrolysis of ester is first order The order of reaction which sometimes altered by the conditions
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O
[H+]
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A second order reaction is a reaction in which the reaction rate is proportional to the product of the concentrations two reactants
The rate equation for a second order reaction is, rate = k [A] [B] or rate= k [A]2
where, k = rate constant [A] = concentration in mol dm-3
the unit of k is dm3 mol-1 time-1 Examples of second order reactions:
i. The hydrolysis of iodomethane is second order
i. The thermal decomposition of hydrogen iodide is second
order
][][
)()()()(
3
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NaOHICHkhydrolysisofrate
aqNaIliqOHCHaqNaOHliqICH
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2
22
][
)()()(2
HIkiondecompositofrate
gasIgasHgasHI
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Rate of reaction = - (rate of disappearance of A) = k [A]2
ot
ot
o
t
Akt
A
Akt
A
ACt
cktA
dtkA
Ad
kdtA
Ad
Akdt
Ad
11
11
1,0
][
1
][
2
2
2
22
A straight line graph with a + ve slope
1/[A]
t
y mx c
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The rate of zero order reaction does not depend on the concentration of the reactants
For zero order reaction, the rate law is rate = k [A]0 = k where, k = rate constant [A] = concentration in mol dm-3
the unit of k is mol dm-3 time-1
Examples of zero order reactions:
i. Reaction between iodine and propanone
The reaction is slow, even at high temperature. However,
addition of dilute acid provides H+ ion which catalyze the reaction. The graph of the reaction rate against the iodine concentration is a straight line parallel to the horizontal axis. This means reaction rate remains constant even though the reactant, iodine is being used. Thus, the reaction is zero order with respect to iodine
)()()()( 2323 aqHIaqICOCHCHaqIaqCOOHCH
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-
o
o
o
o
AktA
AC
AAotat
CktA
dtkAd
kdtAd
kdtAd
kAkdt
Ad
,
][
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[A]
t
A straight line graph with -ve slope
y c mx
t
t
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The half-life, t1/2, of a reaction is the time required for the concentration of a reactant to decrease to half-life of its initial concentration
i. For the first order reaction the half life is independent of the initial concentration
Thus first half-life (t 1/2) = Second half-life (t1/2)
2
1
2
1
693.0
2
tk
ort
nlk
25
[A], concentration
Times (minutes)
12
10
8
6
4
2
t1/2 t1/2 t1/2
0 2 4 6
First order reaction
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1.5
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ii. For the second order reaction, the half life is inversely proportional to the initial concentration. Thus, second half life (t1/2) =2t1/2
2
1
10 tA
k
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t1/2
Second order reaction
[A]
times 2t1/2
5
2.5
10
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iii. For the zero order reaction, a sloping straight line is obtained
2
0
12
][
t
Ak
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Zero order reaction
[A]
times
-
nA
A
r
r
2
1
2
1
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Where: r1 and r2 = initial rates of experiment 1 and 2 respectively A1 ,A2 , B1 & B2 = initial concentration of A and B for experiment 1 and 2 respectively m = order of reaction with respect to A n = order of reaction with respect to B
r1 = k [A]
1 [B]
1
r2 = k [A]
2 [B]
2
m n
m n
m
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Example;
Reaction A + 2B C have been studying at 25C and the result are shown below.
Determine the a) rate law or rate of equation b) order of reaction c) rate constant, k
d) write the rate equation for the reaction
Exp [A]
(mol dm-3)
[B]
(mol dm-3)
Initial reaction rate
(mol dm-3s-1)
1 0.1 0.1 5.5 x 10-6
2 0.2 0.1 2.2 x 10-5
3 0.4 0.1 8.8 x 10-5
4 0.1 0.3 1.65 x 10-5
5 0.1 0.6 3.3 x 10-5
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To find order of reaction for A consider exp. 1 and 2, where [B] is keep constant that is [B] = 0.1
So, we have r1=5.5 x 10-6, r2 = 2.2 x 10-5 A1=0.1 A2=0.2
To find order of reaction for B consider exp. 4 and 5, where [A] is keep constant that is [A] = 0.1
So, we have r1=1.65 x 10-5, r2 = 3.3 x 10-5 B1=0.3 B2=0.6
A respect to with
order second isreaction theThus,
2
5.025.0
5.025.0
1.0
1.0
2.0
1.0
102.2
105.55
6
2
1
2
1
2
1
m
glmgl
x
x
B
B
A
A
r
r
m
nm
nm
a) So, rate of reaction or rate law = k [A]m[B]n= k [A]2[B]1 b) And order of reaction = m+ n = 2 +1 =3
B respect to with
orderfirst isreaction theThus,
1
5.05.0
5.05.0
6.0
3.0
1.0
1.0
103.3
1065.15
5
2
1
2
1
2
1
n
glngl
x
x
B
B
A
A
r
r
n
nm
nm
-
1623
323
136
2
105.5
1.01.0
105.5
][][
][][
2
sdmmolx
moldmmoldm
smoldmx
BA
reactionofratek
BAkreactionofrate
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c) Rate constant, k - can use exp 1, 2, 3,4 & 5 for calculation because the value of k is same for all experiments
d) Write the rate equation Rate = k [A]2[B] = 5.5 x 10-3[A]2[B]
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1. Titration method 2. Colour changes by visual inspection or by using colorimeter 3. Gas syringe for measuring the volume of gas released 4. Pressure changes 5. Conductivity changes for reactions in which the number of
ions in solution changes 6. Monitoring the rotation of plane polarised light
Example: Titration method
H2O2 + 2H+ + 2I- I2 + 2H2O Measuring the concentration of I2 using standard S2O3-
(thiosulphate). At measured time interval, samples of the reaction mixture are removed using pipette and titrated with S2O3-
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Is a series of simple steps that lead from the initial reactants to the final products of a reaction
An elementary reaction represent, at the molecular level, a single stage in the progress of the overall reaction
One of the main reasons for determining the order of a reaction is to see whether the experimentally determined overall order sheds any light on the detailed mechanism by which the reaction occurs
By knowing the reaction mechanism, a chemist can more effectively control the reaction or predict new reaction
Two requirements must be met: 1) The mechanism must account for the experimentally determined
rate 2) The mechanism must be consistent with the stoichiometry of the
overall/net reaction
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The exponent in the rate law for an elementary reaction are the same as the stoichiometric coefficients in the chemical equation for the reaction (this is not usually the case with the rate law of the overall reaction)
Reversible reaction One elementary reaction may be much slower than all the others
the rate determining step (1) A mechanism with a slow step followed by a fast step. Example: decomposition of hydrogen peroxide, H2O2
2H2O2 (aq) 2H20 (liq) + O2 (gas) In the decomposition of H202, the facts (determined by the
experiment) are: i. The rate of decomposition of H2O2 is first order in both H2O2
and I-(second order overall) ii. The reactant I- (iodide) is unchanged during the reaction (acts as
a catalyst) and therefore does not appear in the equation for the net reaction.(overall becomes first order)
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I-
-
The mechanism suggested, Slow step: H2O2 + I
- H20 + OI-
Fast step: H2O2 + OI- H2O + O2 + I
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overall/net: 2H2O2 2H20 + O2 I- = is a catalyst (recovered at the end of the reaction) OI- = is an intermediate The slow step determines the rate of the overall rate So, the slow step is the rate determining step We set the reaction rate to that of the slow step; rate = rate of the slow step = k [H2O2][I-] rate = k [H2O2] where k = k [I-] = constant (the [I-] remains constant through out the reaction)
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(2) A mechanism with a fast reversible step followed by a slow step The smog forming reaction: 2NO (gas) + O2 (gas) 2NO2 (gas) The rate law investigated experimentally: rate = k [NO]2 [O2] Suggested mechanism: fast step : slow step : N2O2 + O2 2NO2 overall : 2NO + O2 2NO2 The rate equation based on the rate determining step (slow step) rate = k2 [N2O2] [O2] We can state the experimentally determined rate law only in terms of
substances in the net equation. So, we must cancell out/ eliminate N2O2
2NOk1 N2O2k -1
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k2
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Fast step:
Forward rate = reverse rate equilibrium Rate formation of N2O2 = rate of disappearance of N2O2 k1 [NO] 2 = k-1[N2O2] [N2O2] = k1 [NO]2 k-1 Substitute [N2O2] in the rate- law for the rate-determining
step rate = k2 [N2O2] [O2] = k2 ( k1 [NO]2 ) [O2] k-1 = k2k1 [NO]2 [O2] k-1 rate = k [NO]2[O2] So, the proposed mechanism is consistent with the observed
rate law
2NOk1 N2O2k -1
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Example of Question; For the reaction: H2 (gas) + I2 (gas) 2HI (gas) A proposed mechanism is fast step : slow step : 2I- + H2 2HI What is a) The net equation based on this mechanism b) The order of reaction according to this mechanism Answer: a) Net equation: I2 + H2 2HI b) Rate law for the rate determining step: rate = k2 [I-]2 [H2] 1 to eliminate I-, we assume rapid equilibrium in the first step rate forward = rate of the reverse reaction k1 [I2] = k-1[I-]2 [I-]2 = k1 [I2] k-1 substitute in equation 1 rate = k2k1 [I2] [H2] = k [I2] [H2] k-1
I2k1
2I-
k-1
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k2
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Arrhenius conducted research on the effects of temperature on reaction rates
For most reactions, the increase in rate with increasing temperature is not linear but exponential
Arrhenius equation is useful in analysing the effect of temperature on the rate constant and
the reaction rate it allow us to determine the activation energy, Ea, if the value of
the rate constants are known at different temperatures
ntconstaratek
(K)etemperaturT
)Jmol(8.31ntconstagasR
energyactivationEa
collision)of(frequencyntconstaArrheniusA:where
Aek
11
RT
Ea
K
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2.303RT
EaAglkgl
RT
EaAlnknl
Aek RTEa
40
k
T (k)
lg k
1/ T
Gradient = -Ea / 2.303 R
At the temperature high, the rate constant k also high and therefore the reaction rate also increases
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Example: in the presence of platinum as a catalyst, hydrogen iodide
decomposes to form hydrogen and iodine. The activation energy for this reaction is 58 KJmol-1. Calculate the ratio of the rate constant at 30C and 20C. Comment on your answer.
Answer: Let the rate constants at 30C and 20C be represented by k1
and k2, respectively
2.20.1
2.2
1052.4
109.9
11
11
2
1
2027331.8
100058
3027331.8
100058
2
1
11
1
11
1
k
k
k
k
KKJmol
Jmol
KKJmol
Jmol
e
e
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The calculation shows that for an increase of 10oC, the initial reaction rate increases by approximately twofold
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Exercise:
Calculate activation energy, Ea for the reaction:
2HI H2 + I2 from the k at (i) 250C = 2.5 x 10-5s-1
(ii) 550C = 1.0 x 10-3 s-1
R = 8.314 Jmol-1K-1
Answer: 9.99 x 104 J/mol
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ln k1 = ln A1 - Ea 1 - 1
k2 A2 8.314 T1 T2
ln 2.5 x 10-5 s-1 = ln A1 - Ea 1 - 1
1.0 x 10-3 s-1 A2 8.314 298 328
-3.688 = -Ea (3.069 x 10-4)
8.314
Ea = - 3.688 x 8.314
- 3.069 x 10-4
= 9.99 x104 J/mol
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Answer: 9.99 x 104 J/mol
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A catalyst is a a substance that alters the rate of chemical reaction without itself being chemically changed at the end of the reaction
Catalyst (positive catalyst)- speed up the reaction Inhibitor (negative catalyst)- slow down the reaction Only a small amount of catalyst is needed to achieve a large/big
increase in reaction rate Catalyst lower the activation energy of the reaction
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Ea catalysed uncatalysed
energy
Progress of reaction
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Catalyst are often highly specific a catalyst for one reaction is not necessarily a catalyst for another reaction
Catalyst may be poisoned/ becomes ineffective by small amounts of substances such as lead, As or CN-
Homogenous catalyst is a catalyst that exists in the same phase as the reactants. e.g acid catalyst of ester
CH3COOCH3 + H2O+ CH3COOH + CH3OH
Heterogenous catalysis occurs when the catalyst and the reactants are in different phases. e.g. the transition metals, Al2O3, SiO3, V2O5
Autocatalysis occurs when one of the products of the reaction is a catalyst for the reaction
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H+