Chemical Reactions.ppt

Introductory Chemistry B CH4751
Lecture Notes 11-20
Dr. Erzeng Xue
CH4751 Lecture Notes 11-20 (Erzeng Xue)

2
Chemical Reaction - Observation
Reaction (1) CH4 + 2O2 CO2 + 2H2O
Reaction (2) CH4 + CO2 2CO + 2H2
When carrying out these reactions we found that
 at 400K (123°C), the reaction (1) will proceed and reaction (2) will not
 at 1000K(723°C), reactions (1) & (2) both proceed; but
 rxn (1) can go complete (until CH4 or CO2 consumed completely)
 rxn (2) won’t complete (with a feed CH4=CO2=1 & CO=H2=0, max. conv.=63% at 1000K)
 The reaction (1) will give out heat, but the reaction (2) will require heat.
Why?
Chemical Reactions
CH4751 Lecture Notes 11 (Erzeng Xue)

3
Chemical Reaction Thermodynamics
 Each molecule contains certain types and quantity of chemical energy
 There is always energy change In a chemical reaction because of
 breaking / reformation of chemical bonds
 out-giving or in-taking heat
 There are different energies associated with a substances & a reaction
(A systematic study of various forms of energy & their changes is called
Thermodynamics)
We will learn some of these energies
 The meanings
 How to get values / do simple calculate
 How to use them as a tool to study chemical reactions
Chemical Reactions

4
 The heat of formation, H, (also called Enthalpy of Formation or Enthalpy)
 H is an energy associated with heat
 H is specific for each substance and is dependent of temperature & pressure
e.g. at 1000K: H°CH4=-89, H°O2=0, H°CO2=-394, H°H2O=-241, H°CO=-111, H°H2=0 (kJ/mol)
(H values for various substances can be found in physical chemistry/Chem Eng handbooks)
 In a reaction we are interested in the enthalpy change, DH, which is calculated using
For Rxn(1) CH4 + 2O2 CO2 + 2H2O DH°1000=-801 kJ/mol
Rxn (2) CH4 + CO2 2CO + 2H2 DH°1000=+260k J/mol
 The meaning
 When DH<0, a reaction releases heat  reaction is exothermic, as in rxn (1)
 When DH>0, a reaction requires heat  reaction is endothermic, as in rxn (2)
Chemical Reactions
refers to standard pressure (1 atm.)
Temperature
reac
T
,
i
i
prod
T
,
i
i
T )
H
v
(
)
H
v
(
H 0
0
0

 

D

5
 The Gibbs Free Energy, G, (also called Free Energy)
G is a thermodynamic function related to a reaction. It is a function of H, T & S (entropy)
G is specific for each substance & is a function of H, T & S (entropy)
e.g. at 1000K: G°CH4=-+30, G°O2=0, G°CO2=-395, G°H2O=-192, G°CO=-200, G°H2=0 (kJ/mol)
(G values for various substances can be found in physical chemistry/Chem Eng handbooks)
 The Gibbs Free energy change, DG, in a reaction can be calculated using
For Rxn(1) CH4 + 2O2 CO2 + 2H2O DG°400 DG°1000=-801 kJ/mol
Rxn (2) CH4 + CO2 2CO + 2H2 DG°400=+145, DG°1000=-24 kJ/mol
 Use of DG - Rxn(1) DG <0 at 400 & 1000K-spontaneous, Rxn(2) DG <0 at 400K, will not proceed
Chemical Reactions
reac
T
,
i
i
prod
T
,
i
i
T )
G
v
(
)
G
v
(
G 0
0
0

 

D
reaction can proceed (but we don’t know how fast it will be!)
reaction at equilibrium (no further change possible-‘dead’ state)
reaction will NOT proceed (or can proceed backward!)







D


D


D
0
0
0
T
T
T
G
G
G
for a reaction at
constant T, P,

6
Example of DH°T calculation
•
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) CH4 + CO2  2CO + 2H2
Coeff. 1 2 1 2 1 1 2 2
H°400( -77 0 -393 -242 -77 -393 -110 0 kJ/mol
H°1000 -89 0 -394 -248 -89 -394 -111 0 kJ/mol
Equation to use
DH°400 =[1x(-393)+2x(-242)]-[1x(-77)+2x(0)]= -800 kJ/mol
DH°1000 =[1x(-394)+2x(-248)]-[1x(-89)+2x(0)]= -801 kJ/mol
Reaction (1) DH°400 =[2x(-110)+2x(0)]-[1x(-77)+1x(-393)]=+250 kJ/mol
Reaction (2) DH°1000=[2x(-111)+2x(0)]-[1x(-89)+1x(-393)]= +260 kJ/mol
Note: The heat of formation of single element gases (O2, H2, N2 etc) is defined as zero.
Chemical Reactions
reac
T
,
i
i
prod
T
,
i
i
T )
H
v
(
)
H
v
(
H 0
0
0

 

D

7
Example of DH°T calculation
•
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) CH4 + CO2  2CO + 2H2
Coeff. 1 2 1 2 1 1 2 2
G°400( -42 0 -394 -224 -42 -394 -146 0 kJ/mol
G°1000 +19 0 -396 -193 +19 -396 -200 0 kJ/mol
Equation to use
DG°400 =[1x(-394)+2x(-224)]-[1x(-42)+2x(0)]= -800 kJ/mol
DG°1000 =[1x(-396)+2x(-193)]-[1x(+19)+2x(0)]= -801 kJ/mol
Reaction (1) DG°400 =[2x(-146)+2x(0)]-[1x(-42)+1x(-394)]=+144 kJ/mol
Reaction (2) DG°1000=[2x(-200)+2x(0)]-[1x(19)+1x(-396)]= -23 kJ/mol
Note: The Gibbs Free Energy of single element gas (O2, H2, N2 etc) is defined as zero.
Chemical Reactions
reac
T
,
i
i
prod
T
,
i
i
T )
G
v
(
)
G
v
(
G 0
0
0

 

D

8
 The values of DG°T and DH °T
Equations
In both cases G° and H° values for the reactants and products have to be those at the
reaction temperature, indicated by the subscript.
 For common substances, G° and H° values are given as a function of T in handbooks - okay
 For some less common substances, you may only find values at 298K, G°298 and H°298
How do you convert values of G°298 and H°298 to those of G°T and H°T ?
Here is the equations you can use to calculate the values of G°T and H°T from G°298 and H°298
in which, S°T is the entropy and Cp is the heat capacity at constant pressure
Chemical Reactions
reac
T
,
i
i
prod
T
,
i
i
T )
G
v
(
)
G
v
(
G 0
0
0

 

D
reac
T
,
i
i
prod
T
,
i
i
T )
H
v
(
)
H
v
(
H 0
0
0

 

D
298
0
298
0

 D

D

 i
,
change
phase
T
i
,
p
,
i
T
,
i H
dT
C
H
H
j
phase
T
i
,
p
T
,
i
T
,
i
T
,
i
T
,
i
T
Q
T
dT
C
S
S
S
T
H
G 
D


 298
0
298
0
0
0
0
where
-

9
 Summary
 Will a reaction proceed in the direction specified?
 Check DG°T value of the reaction. The DG°T value of a reaction can be calculated by
The G° values of reactants / products can be found in literature. Remember
 Is a reaction exothermic or endothermic?
 Check DH°T value of the reaction. The DH°T value of a reaction can be calculated by
The H° values of reactants / products can be found in literature
Chemical Reactions
reac
T
,
i
i
prod
T
,
i
i
T )
G
v
(
)
G
v
(
G 0
0
0

 

D
reac
T
,
i
i
prod
T
,
i
i
T )
H
v
(
)
H
v
(
H 0
0
0

 

D
reaction can proceed (but we don’t know how fast it will be!)
reaction at equilibrium (no further change possible-‘dead’ state)
reaction will NOT proceed (or can proceed backward!)







D


D


D
0
0
0
T
T
T
G
G
G
for a reaction at
constant T, P,

10
Chemical Reaction Equilibrium
 Until now we assume reaction A + B C + D goes to complete
 Meaning a reaction only stops when either A or B is consumed completely
 Experimental observations
 Some reactions will ‘cease’ without complete consumption of limiting reactant
 Without altering reaction conditions (T, P, [ ] etc.) the ratio of conc. remains constant
 After a change (T, P, [ ] etc.), the ratio of conc’s changes to another constant value
Example 1: NH3(aq) + H2O(l) D NH4
+(l) + OH-(aq)
Follow concentrations of each component with time at constant T and P,
at t
 After changing T, a new constant ratio is established
 If [NH3] is reduced the amount [NH4] decrease accordingly
while the ratio above remains constant
 We say the reaction has reached equilibrium state
Chemical Reactions
constant
O]
][H
[NH
]
][OH
[NH
2
3
4



t
[NH4]1
[NH3]1
[NH3]2
[NH4]2

11
 Experimental observations
Example 2: 2NO(g) + O2(g) D 2NO2(g)
Again at constant T and P, when t
 Reaction equilibrium is achieved when t
 We say the reaction has reached equilibrium state
 The ratio of product concentration to reactants, with the stoichiometry coefficient as the
power index, is called reaction quotient
 When reaction quotient = constant value  reaction reaches equilibrium
 The value of reaction quotient at equilibrium, is called equilibrium constant, Keq
 At equilibrium, reactants may or may not be consumed completely
e.g. A feed gas mixture: NO=500ppm, O2=10%, N2=89.95%, achieves equilibria at the following T’s
Temperature / °C 50 325 500
NO remaining at equil / ppm 0 96.5 390
NO conversion at equil / % 100 80.3 22
Chemical Reactions
t
NO2
NO
O2
constant
2
2
2
2

O
NO
NO
P
P
P

12
 Important concepts of reaction equilibrium 2NO(g) + O2(g) D 2NO2(g)
 Is the reaction between reactants still going on?
YES. Reaction goes forward as well as reverses.
At equilibrium: Rforward = Rreverse
though there is no NET change of all conc.’s
 The equilibrium constant, Keq, has a meaning of
Keq=Rforward / Rreverse
 Changing T causes both Rforward & Rreverse to change, leading to a new Keq.
 If Keq >>1, which means Rforward >> Rreverse, the reaction tends to go forward
If Keq <<1, which means Rforward << Rreverse, the reaction tends to go backward
 Will there be an equilibrium constant for reactions that go complete?
YES. There is Keq for all reactions (at a constant P) once T and conc’s are fixed.
For a reaction that tends to go complete, Rforward >> Rreverse (Keq >>1).
Chemical Reactions
2
2
2
2
O
NO
NO
p
P
P
P
K 
t
NO2
NO
O2

13
 Equilibrium constant and Gibbs Free Energy
For reaction vAA + vBB D vCC + vDD
Remember: The value of DG° determines the direction of reaction
No more change possible reaction in equilibrium DG°= 0
 Is the DG value related to the equilibrium constant?
YES. DG° and Keq are related by the equation below (calculate one from the other)
DG°T= - RTln(Keq)
Chemical Reactions
reaction is spontaneous
reaction at equilibrium (no further change possible)
reverse reaction is spontaneous







D


D


D
0
0
0
T
T
T
G
G
G
for a reaction at
constant T, P,

14
 Equilibrium constant - for different type of rxns
General form: vAA + vBB D vCC + vDD
 gas phase 2NO(g) + O2(g) D 2NO2(g)
 gas-solid phase CaCO3(s) D CaO (s)+CO2(g)
 liquid phase NH3(aq)+H2O(l) D NH4
+(l)+OH-(aq)
 liquid-solid Cu(OH)2(s) D Cu2+(aq)+2OH- (aq)
 gas-liquid NH3(g)+H2O(l) D NH4OH(aq)
Chemical Reactions
)
(
]
[
[A]
[D]
[C]
T
f
P
P
P
P
B
K B
A
D
C
B
A
D
C
v
B
v
A
v
D
v
C
v
v
v
v
eq 


for liquid phase rxn
for gas phase rxn
2
2
2
2
O
NO
NO
p
P
P
P
K 
O]
][H
[NH
]
][OH
[NH
2
3
4



c
K
2
CO
p P
K 
2
2
]
][OH
[Cu 


c
K
3
1 NH
p P
/
K 
 For reactions that have gas components, we normally use pressure to represent the conc’s
 For reactions involves gas+liquid or gas+solid, only gas terms appear in the Keq expression

15
 Le Chatelier’s Principle
“When a system in equilibrium is subjected to an external stress, the system will
establish a new equilibrium, when possible, so as to minimise the external stress”
Stresses: Changes in [ ], temperature or pressure
Example: N2(g) + 3H2(g) D 2NH3(g) + heat (exothermic)
a) Effect of ([ ]). Increasing [ ] of substance shifts equil. in the direction of the long arrow
N2 + 3H2 2NH3 + heat
b) Effect of heat. Addition or removal of heat at constant temperature
Addition of heat: N2 + 3H2 2NH3 + heat
c) Effect of Pressure. (only affects reactions that have volume change before & after).
Increase in pressure: N2 + 3H2 2NH3 + heat
4 volumes  2 volumes
Factors Affecting Reaction Equilibrium
Chemical Reactions
&
or
const
2
2
3
3
2
2
2
2
3
H
N
NH
NH
H
N
3
H
N
2
NH











P
P
P
P
P
P
P
P
P
Kp

16
 Write the equilibrium expression for the following reactions and
determine the units for Keq:
1) 2O3(g) D 3O2(g)
2) Ag+(aq) + 2NH3(aq) D Ag(NH3)2
+(aq)
3) 2NaN3(s) D 3Na(s) + 3N2(g)
4) 2Na(s) + Cl2(g) D 2NaCl(s)
5) 2NaCl(s) D 2Na(s) + Cl2(g)
6) N2(g) + 3H2(g) D 2NH3(g)
Note 1. The Keq expression depends on how the rxn equation is written (compare rxns 4&5).
2). The unit of Keq depends on the way how the rxn eqn is written & the unit used of each.
Chemical Reactions
][atm]
[
[atm]
[atm]
2
3
2
3
3
2

 ,
P
P
K
O
O
p
]
][atm
[ 3
3
2

 ,
P
K N
p
][1/atm]
[
1
2

 ,
P
K
Cl
p
][atm]
[
2

 ,
P
K Cl
p
]
][1/atm
[
]
[atm][atm
]
[atm
]
[ 2
3
2
3
H
N
2
NH
2
2
3


 ,
P
P
P
Kp
]
/l
[mol
]
[
]
/l
l
[mol/l][mo
[mol/l]
]
[
]
][NH
[Ag
]
)
[Ag(NH 2
2
2
2
2
3
2
3


 

,
Kc

17
1. Analysis shows that a mixture of N2 (2.46 atm), H2 (7.38 atm) and NH3 (0.116 atm) at
472°C in reaction (N2(g) + 3H2(g) D 2NH3(g)) is in equilibrium state.
Calculate: 1) Keq; 2) DG°; 3). The total pressure. 4) Will the rxn be push to the product by
decreasing the reaction pressure? Give reason why? 5) Will the removal of NH3 from
reaction mixture promote the product formation? Explain why.
1)
2) DG°T= - RTln(Keq)=-8.314x(472+273)ln(2.79x10-5)=65 kJ/mol
3) Ptotal= PN2+PH2+ PNH3=2.46+7.38+0.116=9.956 atm
4) A decrease reaction P favours the reverse rxn because Volreactant > Volproduct.
5) Yes. As Keq=P2
NH3/(PN2xP3
H2)=constant, the removal of NH3 will reduce PNH3, to
compensate the change, more N2 and H2 will be converted to NH3 in order to keep
the same Keq (reaction quotient).
Chemical Reactions
2
5
3
2
3
H
N
2
NH
/atm
10
79
2
.38)
7
.46)(
2
(
)
166
0
(
2
2
3 



 .
.
P
P
P
Kp

18
Chemical Reaction
 What we know about a chemical reaction so far
 Reaction equation - quantitative representation of a chemical reaction
 Reaction stoichiometry coefficients and balancing reaction equations
 We can judge if a give chemical reaction would proceed in the direction specified
 A reaction has a tendency if DG of a reaction is smaller than zero
 We can decide if a given reaction gives out or takes up heat
 If DH < 0, reaction is exothermic; if DH > 0, reaction is endothermic
 For reactions that are feasible, to what extent they will complete
 Chemical reaction equilibrium, equilibrium constant
 Now we know a rxn would proceed in the direction specified, questions are:
- how fast that reaction is going to proceed under give conditions?
- how to quantitatively describe the rate of a reaction and make comparison?
- how can we explain that some reactions occur faster than others?
e.g. 2NO(g) + O2(g) = 2NO2(g) (slow); 2NaN3 = 2Na + 3N2 (very fast)
Chemical Reactions

19
Chemical Reaction Kinetics
 Chemical reaction kinetics study the rate of chemical reactions
 Definition of chemical reaction rate
 The number of moles of a reactant converted (consumed) in a reaction per unit time
for a reaction A + B  C + D (mol/s)
 When we say rate we always refer to ONE of the components in the reaction
 The minus sign refers to that the concentration of reactant decreases in reaction
 Reaction rate equation (or kinetic equation)
 Many forms exist
 The most common one
where k reaction rate constant
A0 pre-exponential factor
Ea reaction activation energy
a,b,l,d reaction orders with respect to A, B, C, D, respectively
R gas constant
T reaction temperature in Kelvin scale
Chemical Reactions
[A]
or
[A]
[A]
1
2
1
2
dt
d
r
t
t
r A
A 















RT
E
A
k
k
dt
d
r a
A
-
exp
in which
[D]
[C]
[B]
[A]
[A]
0
d
l
b
a
Arrhenius equation
Kinetic
parameters

20
 Meanings of kinetic parameters, k, A0, Ea, a,b,l,d
 Reaction rate constant, k
 It tells how fast a reaction can occur
 It is a constant dependent on temperature but independent of concentrations
 Pre-exponential factor, A0
 It refers to the frequency of collision between molecules, the higher frequency, the faster rxn
 Reaction activation energy, Ea
 It can be understood as the energy
barrier for a reaction to overcome
 The higher Ea value is, the more
difficult for a reaction to occur
 Reaction orders w.r.t. each component, a,b,l,d
 The magnitude of these values reflects the effectiveness of each component in the reaction
 The values of a,b,l,d can be positive or negative or zero
 The values of a,b,l,d can be integrals or fraction
 All these kinetic parameters have to be determined experimentally
Chemical Reactions







RT
E
A
k a
-
exp
0
reaction process
Ea
reactant
product
energy

21
 Factors affecting the rate of a chemical reaction
 Reaction temperature, T
 An increase T will lead to increasing k, thus reaction rate.
 The dependence of k on T is given by differentiating k expression,
the higher Ea value is, the more significant of the effect of increasing T on the reaction rate
Concentration of reactants / products
 The effect of increasing a concentration is positive if the respective order is positive
 The larger the value of order is the stronger the effect of increasing conc on the rate
 When order equals to zero, there is no effect of concentration on the rate.
The presence of a catalyst
 A catalyst can alter reaction rate (speeding up desired rxns or slowing down undesired rxns)
Note: we assume rate of mass transfer (to meet) is sufficient high comparing to rA.
Chemical Reactions
[D]
[C]
[B]
[A]
-
exp
0
d
l
b
a







RT
E
A
r a
A
2
0
0
ln
-
ln
ln
-
exp
RT
E
dt
k
d
RT
E
A
k
RT
E
A
k a
a
a













22
 The reaction rate and mass transfer rate
When we discuss the reaction rate, it only makes sense if there are sufficient number
reactant molecules can be delivered to the reaction site.
We say a reaction is kinetic control when the rate of mass transfer > the rate of rxn rA.
 This means that molecules being transported
to the reaction site are ‘queuing’ for reaction
If the mass transfer rate is slower than the reaction rate, the overall rate we observed
will be the rate of mass transfer, not the reaction rate - diffusion control
 This means that molecules are waiting to be delivered
before reacting ‘queuing’ for reaction
The concept of rate determining step (r.d.s.)
 The slowest step in a reaction process determine the overall rate of a reaction
Chemical Reactions
mass
transfer reaction
mass
transfer
reaction

23
 Catalysis and catalysts
Catalyst is a substance which can alter reaction rate without itself being
destroyed or consumed (many other definitions and this is one of them)
95% of chemical industries apply one or more catalysts in their processes
e.g. polymerisation, air/water depolution, ammonia synthesis, cracking heavy oil to
LPG, etc
A catalyst can be an acid, a base; can be a liquid or a solid. Most industrial catalysts
are metals, metal oxides or a mixture of them formulated & made in special ways
Use of catalysts in industry
 Speeding up desired reactions thus increase the process output
 Slowing down undesired reaction thus reduce the unwanted waste products
 Altering reaction route by changing the relative speed of certain steps in a reaction network
therefore realising certain products which would not be possible without catalysts.
 Allowing some rxns to occur under mild conditions e.g. working with heat sensitive materials
 Enzymes are catalysts that participate in bio-active processes
 etc
Chemical Reactions

24
 Calculation of reaction rate
Example: a gas phase reaction 2N2O5=4NO2+O2 occurs at 300°C. The concentrations of N2O5 found
in the reaction mixture at different time intervals are given below:
t h 0 1 2 3 5 7 9
[N2O5] mol/L 1.40 1.07 0.80 0.58 0.34 0.18 0.09
Calculate the rxn rates w.r.t. N2O5, NO2 & O2 1) betw. 0-1h; 2) betw. 3-5h; 3) average betw 0-9h.
N2O5 consumption rate NO2 formation rate O2 formation rate
Eqn’s to use:
1) 0-1 h
2) 3-5h
3) aver.
Chemical Reactions
   
1
2
1
2
2
1
1
2
1
2
2
4
1
2
1
2 [A]
[A]
;
[A]
[A]
;
[A]
[A]
2
2
5
2
t
t
r
t
t
r
t
t
r O
NO
O
N












    h
mol/L
165
0
0
1
40
1
07
1
0.66;
0
1
40
1
07
1
0.33;
0
1
40
1
07
1 2
1
2
4
2
2
5
2















 .
.
.
r
.
.
r
.
.
r O
NO
O
N
    h
mol/L
06
0
0
1
58
0
34
0
0.24;
0
1
58
0
34
0
0.12;
3
5
58
0
34
0 2
1
2
4
2
2
5
2















 .
.
.
r
.
.
r
.
.
r O
NO
O
N
    h
mol/L
073
0
0
1
40
1
09
0
0.291;
0
1
40
1
09
0
0.146;
0
9
40
1
09
0 2
1
2
4
2
2
5
2















 .
.
.
r
.
.
r
.
.
r O
NO
O
N

25
 More about reaction rate 2N2O5 = 4NO2 + O2
1) 0-1 h rN2O5=0.33 rNO2=0.66 rO2=0.165 mol/Lh
2) 3-5h rN2O5=0.12 rNO2=0.24 rO2=0.06 mol/Lh
3) aver. rN2O5=0.146 rNO2=0.291 rO2=0.073 mol/Lh
For the same reaction, the reaction rate expressed by different components varies
with their stoichiometry coefficients
 rN2O5 : rNO2 : rO2 = 2 : 4 : 1 or
 Given reaction rate for one of the rxn components you should be able to calc others.
 For the same reaction the reaction rate may vary with the time
 because of change of reactant conc’s with time & rate in general is proportional to [ ]’s.
 When rxn orders w.r.t. reactant > 0 (usually they are) the rxn rate rbeginning > rlater
As reaction rate is a function of temperature, the determination of reaction of reaction
rate must be done at a constant temperature (you may need to determine the rate at
a different T, or you may need to vary temperature to determine such as Ea
Don’t forget to put the correct unit to the reaction rate you determined
Chemical Reactions
1
4
2
2
2
5
2 O
NO
O
N r
r
r



26
Chemical Reaction Calculations
Question 1. How many grams of water are produced in the oxidation of 1.0g of glucose,
C6H12O6? Reaction equation: C6H12O6 + 6O2 = 6CO2 + 6H2O
Step 1: Use molar mass of glucose to convert g to moles
1 mole C6H12O6=6x12(C)+12x1(H)+6x16(O)=180g/mol
number of moles C6H12O6=1.0g x (1mol/180g)=5.55x10-3 mol
Step 2: Use balanced equation to determine no. of moles of H2O produced
1 mole C6H12O6 produces 6 moles H2O
the no. of moles of H2O produced: 5.55x10-3 moles C6H12O6x6=0.033 moles H2O
Step 3: Convert moles of H2O to grams using molar mass
1 mole of H2O=2x1(H)+1x16(O)=18 g/mol
Grams of H2O produced: 0.033 mol of H2Ox(18g/mol)x = 0.6g H2O (answer)
Note: You cannot use the weight directly in the calculation. It has to be converted to moles.
Chemical Reactions

27
Question 2. In a reactor one put 180g of glucose (C6H12O6) and 160g of O2. Can you
produce 108g of H2O. Why? What is the maximum amount of H2O which can
be produced? Reaction equation: C6H12O6 + 6O2 = 6CO2 + 6H2O
Step 1: Convert all components from grams to moles:
No.moles of C6H12O6=180g/190g/mol=1 mole of C6H12O6
No.moles of O2= 160/32g/mol=5 mole of O2
No.moles of H2O= 108/18g/mol=6 mole of O2
Step 2: Find out how much glucose AND O2 you need to produce 108g H2O.
To produce 108g which is 6 moles of H2O, you will need 1mole glucose AND 6 moles of O2.
Do we have enough glucose? - Yes. Do we have enough O2? - No.
Step 3: Every 6 molecules of O2 will burn 1 molecule of glucose, this will proceed UNTIL
one of the reactant consumed completely, in this case O2.
When all O2 is consumed the reaction will stop and the max. amount of H2O which can be
produced can be calculated from O2 available: 5moles of O2 gives 5moles (or 90g) of H2O
Note: When one of reactants is consumed completely the reaction will stop.
Chemical Reactions

28
Question 3. As in Question 2, one puts 180g of glucose (C6H12O6) and 160g of O2. When
O2 is completely consumed, what is the glucose left & what is the percentage
conversion of glucose? Reaction equation: C6H12O6 + 6O2 = 6CO2 + 6H2O
Step 1: Convert all components from grams to moles:
No.moles of C6H12O6=180g/190g/mol=1 mole of C6H12O6
No.moles of O2= 160/32g/mol=5 mole of O2
No.moles of H2O= 108/18g/mol=6 mole of O2
Step 2: Find out how much glucose left after all O2 has consumed.
The molar ratio of glucose and O2 in the reaction=1:6. For a consumption of 5 moles of O2,
the amount glucose reacted will be 1x5/6=5/6 moles or 0.83 moles, or 0.83x180=150g.
The amount glucose left over=1-0.83=0.167moles or 180-150=30g
Step 3: The percentage conversion of glucose at complete conversion of O2
(try the weight base)
Note: The conversion (%) calculated based on moles is the same as that based on weight .
Chemical Reactions
%
.
%
.
%
conversion 3
83
100
1
833
0
1
100
[A]
[A]
[A]
(%)
in
out
in








29
Question 4. The brown gas NO2 can form colorless gas N2O4, 2NO2 D N2O4. At 25°C the
concentrations of NO2 & N2O4 are 0.018 M & 0.055 M respectively when at
equilibrium. 1) Calculate the equilibrium constant Keq at 25°C. 2) If in another
equilibrium system of the same gases at the same temperature, the NO2
concentration is found to be 0.08 M, what is the concentration of N2O4?
Step 1: Determine the equilibrium constant
From equilibrium constant definition:
Step 2: When at equilibrium
Chemical Reactions
170
0.018
055
0
]
[NO
]
O
[N
2
2
2
4
2



.
Keq
M
088
1
170
0.08
]
O
[N
170
0.08
]
O
[N
]
[NO
]
O
[N 2
4
2
2
4
2
2
2
4
2
.
Keq 







30
Question 5. The rate constants of a reaction are determined to be 3x10-5 mol/L.h at
200°C and 4x10-4 mol/L.h at 250°C. Estimate the reaction activation energy.
Arrhenius eqn relates the rate constant to activation energy
Mthd.1: lnk=lnA0+(-Ea/R)(1/T), A plot of lnk against 1/T will produce a
straight line, the slope of which is -Ea/R. So that Ea=slope x R
Ea=-12750 x 8.314=106,000 J/mol = 106 kJ/mol
Mthd 2:
Let A0,1=A0,2
T1=273+200 K, T2=273+250 K, k1=3x10-5 & k2=4x10-4 mol/L.h, R=8.314 JK/mol
Chemical Reactions
RT
/
Ea
e
A
k 
 0







 







































2
1
2
0
1
0
2
0
1
0
2
1
2
0
1
0
2
1
2
0
2
1
0
1
ln
ln
ln
&
2
1
2
1
2
1
RT
E
RT
E
A
A
e
A
e
A
k
k
e
A
e
A
k
k
e
A
k
e
A
k
a
a
,
,
RT
/
E
,
RT
/
E
,
RT
/
E
,
RT
/
E
,
RT
/
E
,
RT
/
E
,
a
a
a
a
a
a
 

















 








 












2
1
2
1
2
1
2
1
2
1
2
1
2
1
ln
or
ln
k
k
T
T
T
RT
E
T
RT
T
T
E
RT
E
RT
E
k
k
a
a
a
a
  
   
kJ/mol
106
10
4
10
3
ln
250
273
200
273
250
273
200
273
314
8
ln 4
5
2
1
2
1
2
1



























 

.
k
k
T
T
T
RT
Ea
1/T
ln k
slope= -12750

31
Question 6. Two catalysts A & B are compared for their catalytic activity for reaction RP.
When A is present it takes 10s for R to change from 2 to 0.5 moles and when
B is present it takes 20s for R to decrease from 5 to 2 moles at the same
temperature and with the quantities of catalyst.
Which catalyst is more active for the reaction concerned?
Answer: The activity of two catalysts can be compared based on the average reaction
rate when A & B presence separately.
The A catalyst is more active for the reaction concerned.
Chemical Reactions
mol/s
10
0
20
5
-
3
]
[
-
]
[
mol/s
15
0
10
2
-
.5
0
]
[
-
]
[
1
2
1
2
.
t
R
R
r
.
t
R
R
r
B
B
,
B
,
B
,
R
A
A
,
A
,
A
,
R



D




D


32
Question 7. Verify that the rate constant of a reaction following second order rate law
rA=-k[A]2 can be determined from the slope of a line obtained by plotting 1/[A]t
against reaction time t, where [A]t is the concentration of A measured at time t.
Answer: Second order rate law: (1)
rearrange: (2)
Define boundary conditions: at t=0, [A]=[A]0 and at t=t, [A]=[A]t
integrate eqn (2), t from 0-t and [A] from [A]0 to [A]t
(3)
compare eqn (3) with linear eqn Y=aX+B, which is a straight line with slope a
Let
A plot of vs. t will give a straight line with slope=k.
Chemical Reactions
2
[A]
[A]
k
dt
d
rA 


kdt
d
k
dt
d




 2
2
[A]
[A]
[A]
[A]
 
0
t
0
t
t
0
[A]
[A] 2
2
[A]
1
[A]
1
0
[A]
1
[A]
1
[A]
[A]
1
-
[A]
[A] t
0


























 
 kt
t
k
dt
k
d
kdt
d
0
t [A]
1
and
[A]
1



 b
t
X
,
k
a
,
Y
t
[A]
1
t
slope=k
1/[A]

33
Question 8. Reaction RP follows the second order rate law rR=-k[R]2. Verify that the time
required for the reactant R to fall to a half of its initial value is t1/2=1/(k[R]).
Answer: Second order rate law: (1)
After integration of eqn (1) with the boundary conditions:
at t=0, [R]=[R]0 & at t=t1/2, [R]=[R]t1/2=0.5[R]0
Chemical Reactions
2
[R]
[R]
k
dt
d
rR 


0
2
1
0
0
0
0
2
1
0
0
2
1
0
2
1
0
0
t
[R]
1
[R]
1
[R]
2
1
[R]
1
0.5[R]
1
1
[R]
1
0.5[R]
1
[R]
1
0.5[R]
1
[R]
1
[R]
1
k
t
k
k
t
kt
kt
kt
/
/
/
/
































34
Homogeneous and Heterogeneous Reactions
 In a chemical reaction, the reactants can be in various physical states
 Homogeneous - All of reactants & products in the same phase & no phase boundary
 Heterogeneous - Involving multi-phases & phase boundary crossing
Chemical Reactions
Phase Type Example
gas - gas Homog. 2NO + O2 = 2NO2
gas - liquid Hetrog. CO2 + H2O = H2CO3
gas - solid Hetrog. O2 + Fe = Fe2O3
liquid - liquid (miscible)* Homog. NaOH + HCl = NaCl + H2O
liquid - solid Hetrog. CaO + H2O = Ca(OH)2
solid - solid Hetrog. CaCO3 = CaO + CO2
* When two immiscible liquid, such as oil and water is regarded as heterogeneous type.

35
Phase & Phase Change
 A substance can exist in different physical states
 Gas / vapour
 Liquid
 Solid
Note: Some other states may sometime mentioned.
such as liquid-crystal, super-critical state, gel. etc.
 The temperatures at which phase changes occur
vary with substances and circumstances (e.g. P)
 The energy required for phase change varies with
substances and type of phase change.
 The energy possessed by molecules of the same
substance at different state are different
Chemical Reactions
liqui
d
solid
sublimation
deposition
condensation
evaporation
melting
freezing
Energy
level
gas
Temperature
Energy (heat) added
ice
liquid
water
water
vapour
melting
evaporating

36
Reactions In Liquid Phase - Solution
 Solution = solute + solvent
Solute is a solid or a gas - solute is dissolved in solvent (e.g. NaCl + H2O, O2 + H2O)
Solute is another liquid - solute and solvent are miscible (e.g. C2H6O + H2O)
 solute in a solution can exist as molecules or ions, or both (such as weak acid)
 some solutes are dissolved in a solvent in any proportions (e.g. C2H6O in water); others are
only dissolved in a solvent in certain proportion - solubility limitation (e.g. NaCl or N2 in H2O)
 Concentration of a solution - the amount of solute in the solution
Molar concentration (molarity) - number of moles solute in ONE litre solution
 Molarity is the most commonly used concentration unit in chemistry
Weight percentage (wt.%) of solute in solution
definition: solute wt%=100% x (wt of solute)/(wt. of solute + wt. of solvent)
e.g. A solution contains 20g solute & 30g solvent
solute wt%=100% x 20/(20+30)=40wt% (the wt% of solvent =100%-40%=60%)
Chemical Reactions
solute - substance that is dissolved
solvent - dissolving medium

37
Reactions In Liquid Phase - Solution
 Some important notes on the chemical reaction in solution
 When a solute dissolve in solution, solute can either be present as molecules or ions
 as molecules. e.g. dissolving sugar in water - not electronic conductivity.
 as ions. e.g. dissolving salt (NaCl) in water - Na+ & Cl- both conduct electricity.
 Strong acids or base, when dissolved in H2O, form only ions in solution
 Weak acid or base, when dissolved in H2O, form mixture of molecules and ions (partial dissociatn)
 The solubility of some gas solutes, when dissolved in a solvent, depends on the
pressure of the solute gas above the solution
ideal solution: Pi=xiPi* or xi=Pi / Pi* in which
 Solvent molecules may form weak bond with solute molecules (e.g. Hydrogen-bond),
which may to a certain degree change the reactivity of solute.
 The presence of solute may affect certain properties of the resultant solution
e.g. boiling point elevation, freezing point depression, osmosis pressure, etc.
Chemical Reactions
Pi - vapour pressure of solute i
xi - mole fraction of solute i in solution
Pi* - equil. vapour pressure of pure solute i

38
Gas Phase Reactions
 Distinctive features of molecules in gas phase in relation to reaction
 Having high energy
 Sometime a liquid or even a solid substance is heat to gas phase to react
e.g. steam reforming hydrocarbons, heavy oil cracking,
 Moving freely within the space of reaction
 High mass transfer rate - General magnitude of mass transfer rate in solid, liquid and gas
solid 100 liquid 103 gas 105
 High heat transfer rate - This is very important for reactions involving heating/cooling
 In practice, many reactions in which the reactants are liquid or solid at normal temperature
are carried out at elevated temperatures in order to convert the reactants to gas
 Compressible therefore sensitive to the reaction pressure
 This has implication on the reactions involving the change of number of moles before and
after reaction.
 The main disadvantages of the gas phase reactions are
 Usually high volume (large reactor)
 not suitable for heat sensitive substances if heating to high temperature is required.
Chemical Reactions

39
Solid Phase Reactions
 Solid phase reactions are usually slow due to limited mobility of molecules
 When a solid reacts with another reactant which is liquid or gas, the
reaction starts from outer surface of the solid
 In industry if a reaction involves a solid reactant (at ordinary temperature)
what we usually do is
dissolving solid in solvent
heating it to above the melting point so that it takes part in reaction as a liquid
 Many catalytic reactions use solid catalysts
The reactant in this case can be a liquid or a gas or liquid-gas mixed phase
Solid catalysts are very easy to separate from liquid, gas or a liquid-gas mixture
It is easy to handle solid catalysts from practical point of view (loading, discharge etc)
Chemical Reactions

40
Reactions involve multi-phases
 Many reactions involve multi-phase in one reactor
 reactants and products can be presented as any combination of two or three phases.
 When multi-phases present in a rxn following issues become important:
 Relative rate of reactant and/or product molecules diffusion within each phase as well
as through phase boundaries must match the rate of reaction
 Solubility of solids and/or gases in liquid phase
 When a porous solid is involved the liquid/gas molecules transport within the pore
 Both pressures (for gas phase) and concentrations (for liquid) are interlinked in the
reaction network therefore these have to be considered systematically.
 When reaction involves heating or cooling, as most of reactions do, this has to be
dealt with by considering both mass transfer and heat transfer within and between
different phases.
 The main advantage of multi-phase reactions is the easiness for separation
Chemical Reactions

41
Example of Reactions involving multi-phases
 The long journey for reactant molecules
j. travel within gas phase
k. cross gas-liquid phase boundary
l. travel within liquid phase
m. cross liquid-solid phase boundary
n. reach outer surface of solid
o. travel with pore
p. reach reaction site
q. be adsorbed on the site and activated
r. react with other reactant molecules (either
adsorbed or approached from surface above
 Product molecules must follow the reverse
process to return to gas phase
 Heat transfer follows similar process
Chemical Reactions
j
r
gas phase
pore
porous
solid
liquid
phase
k
l
mn
o
p q
gas phase
reactant molecule

42
Acids and Bases
 Acids & Bases are one of the most important classes of chemicals
 Acids and bases have been know to human for a long time
 Acids taste sour (in fruit), change colour of certain dye
 Bases taste bitter and feel slippery (like in soap, lime water)
 Acids and bases are widely present in nature,
 especially in plants, electrolyte balance in life system cycle etc
 Acids and bases are widely used in industry for various purpose
 Dissolving chemicals, e.g. HF, aqua regia (HCl:HNO3=3:1)
 Reagents for producing various chemicals
 Catalysing various types of reactions
 Titration in volumetric analysis
 etc
Chemical Reactions

43
Acids and Bases - Definition
 Classical definition
Acids - Substances that, when dissolved in water, increase the concentration of H+ ions
e.g. HCl(g) H+(aq) + Cl-
(aq)
Note: H+, which is a proton only (no e- ), is actually bond with water molecule forming H3O+, the rxn is
HCl(g) + H2O (l)  H3O+(aq) + Cl
-
(aq)
For simplicity, we often use H+ instead of H3O+.
Bases - Substance that, when dissolved in water, increase the concentration of OH-
ions
e.g. NaOH OH
-
(aq) + Na+(aq)
NH3 + H2O  NH4
+ + OH
-
 Brønsted-Lowry definition
Acid is proton donor and Base is proton acceptor
(because H+ is a proton and OH
-
of a base reacts with H+ giving water)
Chemical Reactions
H2O
H2O

44
Conjugate Acid and Base Pairs
 An acid & a base always work together to transfer proton (donate-accept). A substance
can function as an acid only if another substance behaves simultaneously as a base.
 When an acid or a base is dissolved in water, ions are released - this process involves
proton transfer. To mark the process and link the ions with its original acid or base,
conjugate acid-base pairs are defined.
 Acid and conjugate base always appear in pair; likewise base and conjugate acid appear in pair
 When an acid losses proton (H+) it becomes the conjugate base of that acid (e.g. HX to X-)
when a base receives a proton (H+) it becomes the conjugate acid of that base (H2O to H3O+)
 If an acid dissolves in water, H2O is a base; if a base dissolves in water, H2O becomes an acid.
Chemical Reactions
remove H+
HX(aq) + H2O (l)  X
-
(aq) + H3O+(aq)
acid base conjugate base conjugate acid
add H+
remove H+
HCl(aq) + H2O(l)  Cl
-
(aq) + H3O+(aq)
acid base conjugate conjugate
base acid
add H+
add H+
NH3(aq) + H2O(l)  NH4
+
(aq) + OH-(aq)
base acid conjugate conjugate
acid base
remove H+

45
Strengths of Acids and Bases
 The strength of acids and bases
 The strength of an acid is the ability to donate proton,
or increase [H+] when acid is dissolved in water.
 likewise, the ability to accept proton, or [OH-],
determine the strength of a base
 Common acids and their relative strengths
Strong acids, paired with bases with negligible basicity
- Able to completely transfer their proton to water
- Their conjugate bases are the weakest, with negligible
tendency to accept proton
Weak acids, paired with week bases
- These acids are partially dissociated to ions
- Their conjugate bases are also weak, with limited ability of
accepting proton
 Acids with negligible acidity, paired with strong bases
- These class of acids, though carrying H, give out no [H+]
- Their conjugate bases, however, are strong bases
 Water can act as acid as well as base
Chemical Reactions
acid base
HCl Cl
-
H2SO4 HSO4
-
HNO3 NO3
-
H3O H2O
HSO4 SO4
2-
H3PO4 H2PO4
HF F
-
HC2H3O2 C2H3O2
-
H2CO3 HCO3
-
H2S HS
-
H2PO4 HPO4
2-
NH4 NH3
HCO3 CO3
2-
HPO4 PO4
3-
H2O OH
-
OH O
2-
H2 H
-
CH4 CH3
-
acid
strength
increase
base
strength
increase
negligible
weak
strong
negligible
weak
strong

46
Acid and Base Equilibrium
 The extent of ionisation of an acid or a base in water
 Some acids (or bases) ionise in water completely, leaving no molecules behind
 Other acids (or bases) ionise partially in water, forming an equilibrium between
molecules and ions
e.g. HF(aq) + H2O (l) D F
-
(aq) + H3O+(aq) (1)
NH3(g) + H2O (l) D NH4
+(aq) + OH
-
(aq) (2)
 The tendency of ionisation of an acid (or a base) varies with the type of acids, we
can use the concept of reaction equilibrium to indicate the degree of ionisation.
The ‘equilibrium constant’ used to describe the degree of ionisation of an acid is
called acid-dissociation constant, Ka, which is defined as
for equili. (1)
for equili. (2)
Chemical Reactions
[HF]
]
][H
[F
or
[HF]
]
O
][H
[F -
3
- 



 a
eq
a K
K
K
]
[NH
]
][OH
[NH
3
4



 eq
a K
K
ions
molecule
The higher the Ka value, the higher
ion conc., the higher acidity/basicity

47
Quantifying the Strength of Acids and Bases
 [H+] and [OH-] are the measure of the strengths of acids and bases
 We know that an acid when dissolved in water releases [H+] and a base gives [OH-]
 We also know that the strengths of an acid or a base depend on the [H+] and [OH-]
 It comes naturally that [H+] & [OH-] are used to indicate the strengths of acids/bases
 The range of [H+] and [OH-]
 Dilute aqueous solutions at 25°C always give,
Kw=[H+][OH-]=1.0x10-14
 For an acid [H+]>[OH-], Kw=[H+][OH-]=1.0x10-14
 For a base [OH-]>[H+], Kw=[H+][OH-]=1.0x10-14
 For pure water, which is neutral
[H+]=[OH-]=1.0x10-7, Kw=[H+][OH-]=1.0x10-14
 pH scale
 For convenience the low value of [H+] and [OH-], we use the scale of log10 [H+]
define pH= -log10[H+] Scale: 1-14. Acid pH=0-7 [H+]>[OH-]; strong acids have low pH
Base pH=7-14 [OH-]>[H+]; strong bases have high pH
Note: When using [OH-] (which is less used), we have pOH= -log10[OH-] (=14-pH)
Chemical Reactions
water can act as an acid as well as a base
at equilibrium H2O D H+ + OH-
Equili. constant at 25 °C is found to be
Further examine other aqueous solution
the same relation holds
14
-
2
10
1.0
]
][OH
[H
O]
[H
]
][OH
[H



 



w
K
In pure water [H2O] is constant
Known [H+], [OH-] can be calculated by this eqn.

48
Calculation of pH
Example 1: Calculate pH of 0.05M HNO3 solution
HNO3 + H2O  H3O+ + NO3
-
HNO3 is a strong acid, HNO3 ionizes completely in water, i.e. [H3O+]= 0.05M
pH = - log10[0.05] = 1.3
Example 2: Calculate pH and pOH of 0.05M NaOH solution
NaOH + H2O  Na+ + OH-
NaOH is a strong base, NaOH ionizes completely in water, i.e. [OH-]=0.05M,
Kw = [H3O+][OH-] = 1 x10-14 M2
[H3O+] = 1 x10-14 M2 / [OH-] = 1 x10-14 M2 / 0.05 M = 2 x 10-13 M
pH = - log10[2 x 10-13] = 12.7
pOH = 14 - pH = 14 - 12.7 = 1.3
(why is this result the same as that of example1?)
Chemical Reactions

49
Calculation of pH
Example 3: What is the [OH-], in mol/L, in a solution whose pH is 9.72?
Known: pH = - log10[H3O+] = 9.72  [H3O+] = 1.9 x 10-10 (mol/L)
for any aqueous solution Kw = [H3O+][OH- ] = 1.0 x 10-14 (mol/L)2
 [OH- ] = Kw / [H3O+] = 11.0 x 10-14 (mol/L)2 / 1.9 x 10-10 (mol/L) = 5.3 x 10-5 (mol / L)
Example 4: The acid-dissociation constant, Ka, of hydrofluoric acid is 6.8x10-4. What is
the [H3O+] in a 2M HF solution? What is the pH of the solution?
HF(aq) + H2O (l) D F
-
(aq) + H3O+(aq)
initial 2 0 0
at equili. 2 - x x x
By definition
Solve the eqn for x ( = [H3O+])
pH = - log10[H3O+] = - log10(0.0365) = 1.44
Chemical Reactions
4
3
-
10
8
6
-
2
[HF]
]
O
][H
[F 






 .
x
x
x
K
K eq
a
M
0365
0
10
8
6
2
10
8
6 4
4
2
.
x
.
x
.
x 






 


50
Aqueous Equilibria and Some Applications
 In chemistry many aqueous systems involve equilibria
 Human body fluids are in electrolyte equilibria in order to function properly
 Electrolyte: aqueous solutions that contain ions
 Plants contain weak acids, which maintain right balance for plants to grow
 Many properties of a solution that has ions are affected by its equilibrium state.
 etc. (In a broad sense, harmony=balance=equilibria)
 Many phenomena in chemistry can be studied by means of equilibria.
We will look at:
 The behaviour of an equilibrated electrolyte solution when other ions are added
Applications
 Buffer effect
 Acid-base titration
 Solubility of ionic substances and the factors affecting it
Chemical Equilibria

51
The Common-Ion Effect from Equilibrium
 Considering the following two cases
Case 1. What is the pH of 0.3M acetic acid HC2H3O2 solution, (Ka=1.8x10-5)?
HC2H3O2(aq) D H+ (aq) + C2H3O2
-(aq)
initial 0.3 0 0
at equilibrium 0.3-x x x
By definition
Solve eqn for x
Chemical Equilibria
5
2
3
2
-
2
3
2
10
8
1
-
0.3
]
O
H
[HC
]
O
H
][C
[H 





 .
x
x
x
Ka
  64
2
10
3
2
-log
]
-log[H
pH
M
10
3
2
]
[H
3
3
.
.
.
x












Note: HC2H3O2 is a weak acid (Ka<<1) and the water solutn of HC2H3O2 is an electrolyte solution.

52
The Common-Ion Effect from Equilibrium
 (cont’d)
Case 2. What is the pH of solutn contains 0.3M acetic acid HC2H3O2 & 0.3M NaC2H3O2?
HC2H3O2(aq) D H+(aq) + C2H3O2
-(aq)
initial 0.3 0 0
at equilibrium 0.3-x x 0.3+x
By definition
Solve equ for x
 Compare cases 1 & 2:
The extent of ionisation of HC2H3O2 is reduced by the presence of NaC2H3O2 (which has
C2H3O2
- ion in common with HC2H3O2)
 This is called the Common-ion Effect. It works in many equilibrated electrolyte solutions
such as buffer solutions, solubility of ionic compounds etc.
Chemical Equilibria
  5
2
3
2
-
2
3
2
10
8
1
-
0.3
3
0
]
O
H
[HC
]
O
H
][C
[H 






 .
x
x
.
x
Ka
Note:
NaC2H3O2 ionises in water completely
NaC2H3O2(aq) D Na+(aq) + C2H3O2
-aq)
  74
4
10
8
1
-log
]
-log[H
pH
M
10
8
1
]
[H
5
5
.
.
.
x











 Note: The presence of NaC2H3O2 &
Na+ does not change Ka value
Note: C2H3O2
- is the conjugate base
of HC2H3O2

53
Buffered Solutions
 Behaviour of a solution containing a weak conjugate acid-base pair
equilibrium of weak acid HX(aq) D H+(aq) + X-(aq)
acid-dissociation constant
If a base, OH-, is added, OH-(aq) + HX(aq) D H2O(aq) + X-(aq)  [HX]  & [X-]
If an acid, H+, is added, H+(aq) + X-(aq) D HX(aq)  [X-]  & [HX]
When the addition of OH- or H+ is small compared to [HX] & [X-], the change to [HX] & [X-] is very
small, so does the ratio [HX] / [X-]  the [H+] thus pH will remain almost constant.
 A Buffered Solution (also called Buffer) contains a weak conjugate acid-base
pair. It can resist drastic change of pH upon the adding strong acid or base.
 Buffers solutions are widely used in biology and biochemistry because of the
need of maintaining certain pH for some reactions/process to occur properly.
Note: Buffer solutions can be made for all pH ranges. The amount of acid or base it can
neutralise before pH begins to change (called buffer capacity) depends on the [HX] & [X-].
Chemical Equilibria
As the HC2H3O2 and
C2H3O2
- pair in case 2
]
[X
[HX]
]
[H
[HX]
]
][X
[H
-
-
a
a K
K 

 


54
Solubility of Ionic Compounds
 Equilibrium between solid of ionic compound and its ions when dissolved
dissolving
CaCO3(s) D Ca2+(aq) + CO3
2-(aq)
precipitation
 Equilibrium constant:
solubility-product constant Ksp=[Ca2+] [CO3
2-]
 When adding another strong electrolyte Na2CO3, which dissociates completely in
water solution and contains common ions CO3
2-, into the above equilibrated solution
The above equilibrium will ‘shift’ to the left, meaning that the CaCO3 solubility 
Reason? - Common-ion effect (Ksp is constant, [CO3
2-] [Ca2+]  [CaCO3] )
 Addition of common ions alter the equilibrated solubility of an ionic compound.
(you may like to link this with the cases such as scale formation in kettle, kidney stone, etc.)
Chemical Equilibria
Note: The solid CaCO3 does not
appear in the expression

55
Solubility of Ionic Compounds
 Equilibrium between solid and its constituent ions containing OH-
dissolving
Mg(OH)2(s) D Mg2+(aq) + 2OH-(aq)
precipitation
 Many metal hydroxides are partially dissolved in solutn (or precipitated when formed)
solubility-product constant Ksp=[Mg2+] [OH-]2
 When adding an acid, H+, into the above equilibrated solution, a reduction of solution
pH occur due to the following reaction: H+ + OH- D H2O, OH- in the solution is
consumed thus reduced [OH-]  [Mg2+] [Mg(OH)2] .
 This is also a kind of common-ion effect but working in reverse direction.
 Due to consumption of one of ions in the equilibrated solid-ions solution, the
equilibrated solubility of an ionic compound is increased.
(you may like to link this with the cases such as kettle de-scaling, tooth decay etc.)
Chemical Equilibria
Note: the stoichiometric number raised
to power in Ksp expression

56
Titration
 A method to determine the concentration of a particular solute in a solution
 Titration reactions
 Acid base reactions (acid / base indicators, etc)
 Oxidation-reduction reactions (colour change, etc
 Precipitation (cloud appearance, etc)
 Standard solution - a solution with known concentration which is used to titrate.
 Equivalent point - The point at which stoichiometrically equivalent quantities are
brought together. It is a theoretical point of reaching stoichiometry
 End point - The point at which a pre-determined indication of reaching the equivalent
point effects. It is practically the point one stops adding standard solution
and is usually very close to the Equivalent point
 Usual means of indicating the arrival at the end point
 Colour (indicators or the colour change of the substance itself before and after equiv. point
 Conductivity if the quantities of ions is used as an indication
 Others such as precipitation formation …
Chemical Applications

57
Acid - Base Titration
 Typical titration curves of strong
& weak acids by Strong base
 pH change of weak acids is less
drastically as that of strong acids
because of equilibrium shifting
Chemical Applications
0 10 20 30 40 50 60
pH
14
12
10
8
6
4
2
0
0 20 40 60 80 100
mL NaOH
H3PO3 H2PO3
-
HPO3
2-
mL NaOH
pH
14
12
10
8
6
4
2
0
strong acid
Ka=10-2
Ka=10-4
Ka=10-6
Ka=10-10
Ka=10-8
Equivalent point
 Typical titration curve of polyprotic
acids by Strong base
 Different equilibria of ions with
different charges

58
Introductory to Organic Chemistry
 Organic chemistry
 A branch of chemistry devoted to the study of carbon-containing (organic) molecules
 All life forms on earth have organic molecules as their basic building blocks
 The most important carbon-containing molecules are hydrocarbons (HC’s)
 General characteristics of hydrocarbons (HC’s)
 Hydrocarbons (HC’s) - molecules contains mainly carbon (C), hydrogen (H)
 Bonds and general structures
 The bonds of HC’s are mainly covalent, formed between C and C (C-C, C=C or CC), H (C-
H), O (C-O or C=O) and others such as N (C-N).
 The C-C bonds forms the backbone or skeleton of HC’s and H’s are at surface
 Functional groups (FG’s)
 Many FG’s attached to C-C skeleton give HC’s various unique function and properties.
 Stability
 All HC’s can burn in oxygen easily giving heat.
 C-C single bonds are most stable (- > = > ), C-H & C-FG’s are easy to break
Chemistry for Life

59
 There are four types of HC’s (based on the kinds of C-C bonds)
 Alkanes (C-C)
 contain only C-C bonds in this group of molecules, also called saturated HC’s
 Alkenes (C=C)
 contain C=C bonds
 Alkynes (CC)
 contain CC bonds unsaturated HC’s
 Aromatics
 carbon atoms are connected in a planar ring structure
 usually possess special odour
Chemistry for Life

60
 Alkanes
 Most stable HC’s
 Its molecular formula can generally be written as CnH2n+2, where n is the number of
carbon atoms in molecule.
 Most common alkanes
Name Molecular Condensed Lewis structure Boiling
formula structure formulary point
methane CH4 CH4 -161°C
ethane C2H6 CH3CH3 or CH3-CH3 -89°C
propane C3H8 CH3CH2CH3 -44°C
butane C4H10 CH3CH2CH2CH3 -0.5°C
pentane C5H12 CH3CH2CH2CH2CH3 36°C
Chemistry for Life
H
H
H-C-H
H
H
H-C-C-H
H
H H
H
H-C-C-C-H
H
H
H
H
H
H
H-C-C-C-C-H
H
H
H
H
H
H H
H
H-C-C-C-C-C-H
H
H
H
H
H
H
H
H

61
 Some general points about HC’s
 Straight-chain HC’s: All carbon atoms are joined in a non-branched chain
C-C C-C-C -C-C-C-C- C-C-C-C-C=C-C-C-C
 Branched-chain HC’s C
-C-C-C- -C-C-C C-C-C-C-C=C-C-C C-C-C-C-C=C
C C C C C
 Structural isomers
compounds that have the same C C
molecules formulas but with C-C-C-C-C=C C-C-C-C-C=C
different bonding arrangement C C C C
 There is a system way of naming HC’s
to differentiate the isomers with different structures
 We can sometimes write only carbon atoms to show the structure & bonds, as
shown above
Chemistry for Life
All of these are isomers of C9H18

62
 Alkenes
 Less stable than alkanes
 Its molecular formula can generally be written as CnH2n, where n is the number of
carbon atoms in a molecule.
 The double bond C=C can locate between any two C (n equal to or larger than 2).
 There can be more than one double bonds in an alkene
 Isomers exists when n equal or larger than 4.
Common alkenes
ethene or ethylene CH2=CH2 or CH2CH2 (ethene is a plant hormone)
propene or propylene CH3-CH=CH2 or CH2CHCH2 (play key role in fruit ripening)
Chemistry for Life

63
 Alkynes
 Least stable - very active
 It molecular formula can generally be written as CnH2n-2, where n is the number of
carbon atoms in molecule.
 The triple bond CC can locate between any two C (n equal to or larger than 2).
 There can be more than one triple bonds in an alkyne
 Isomers exists when n equal or larger than 4.
Common alkenes
acetylene CH CH or CHCH very active. Burn in oxygen - oxiacetylene torch
with flame temperature 3200K.
Very important intermediates in chemical industry
Chemistry for Life

64
 Aromatic HC’s
 More stable than alkenes and alkynes, though there are unsaturated bonds.
 Most common aromatic HC is benzene
 Examples of other aromatic compounds
Chemistry for Life
H
H
or
C
C
C
C
C
C
H
H H
H
CH3
Tuluene
CH3
CH3
H3C
H3C
H3C
H
H
Cholesterol
HO

65
 Some common functional groups
Alcohols (name suffix -ol)
Soluble in H2O; use in food, medicine, cholesterol is an alcohol
Ethers (name suffix ether)
Mostly used as solvent
Aldehydes (name suffix -al)
Flavour (vanilla, cinnamon etc)
Ketones (name suffix -one)
Such as acetone used extensively as solvent
Carboxylic acids (name suffix -oic acid)
Sour veg, fruits; application in polymers, fibres, paints
Esters (name suffix -oate)
very pleasant odour (fruits)
Amines and Amides (name suffix -amide)
Are key functional group in protein structure
Chemistry for Life
R-OH
R-O-R’
O
R-C-H
O
R-C-R’
O
R-C-OH
O
R-C-O-R’
O
R-C-N-R’

66
Introductory to Biochemistry
 Some general remarks on Biochemistry
 Biochemistry - Biological chemistry which studies living species in chemical means
 Looking at the compositions and structure of biochemical molecules, and try to understand
their functions at molecular level.
 Looking at the properties these molecules especially from biological point of view
 The change processes of these molecules in relation to its role in life cycle
 Making use of our knowledge for human benefits
 General observations on biochemical molecules
 The molecules are generally very large, molecular wt in the range of 1,000’s -1,000,000’s
 They are generally very complicated in structure, yet they contains mainly C, O, H as their
main building blocks and some other atoms such N, P, S in their functional groups
 The specific ways of these molecules are structured make them specific functions in
biological processes.
 All life processes of mammals and other animals on the Earth require energy (processes of
bio-molecule synthesis are endothermic in large), which, ultimately coming from the Sun,
are obtained indirectly through plant photosynthesis.
Chemistry for Life

67
Introductory to Biochemistry - Proteins
 Proteins and Amino acids
 Proteins are big molecules present in living cells (~50% dry wt of our body)
animal tissues, skin, hair, nails, muscles
 All proteins are chemically similar
 All proteins are composed of the same building blacks - a-amino acids
 a-amino acids are linked by amide groups, which is formed by reaction between
-C-O- and H-N- groups of 2 amino acids after dropping a H2O, into proteins
Chemistry for Life
H
R
+H3N-C-C-O-
O
H
R
H2N-C-C-OH
O
or
H
H
O
H
R
+H3N-C-C-O-
O
H
e.g. + = + H2O
H
R’
+H3N-C-C-O-
O H
R
+H3N-C-C-N-C-C-O-
O O
H
R’
amide group

68
 Structure of proteins - 4 levels
 Primary structure - amino acids sequence
 Secondary structure - a-helix
 Tertiary structure - folded individual peptide
 Quaternary structure - aggregation of 2 or more peptides
Chemistry for Life

69
 Enzymes
 Enzyme is one of the most important classes of proteins. Each enzyme is capable
of catalysing very specific reactions with living organisms
 More about amino acids
 There are many different amino acids, the difference being the R groups
 Our body requires 20 amino acids
 Our body can synthesise 10 of these 20
 The other 10 (called essential amino acids) must be ingested.
Chemistry for Life
H
R
H2N-C-C-OH
O
e.g.
H
H
H2N-C-C-OH
O H
CH3
H2N-C-C-OH
O
Glycine Alanine

70
Introductory to Biochemistry - Carbohydrates
 Carbohydrates (hydrate of carbon)
 It has general formula of Cx(H2O)y
 It is a form of ‘sugar’ used for store energy by plants
 Carbohydrates can be divided according to the number of units of ‘basic sugar’
 Monoaccharides - containing a single unit of sugar (that cannot be broken by a acid)
The most important monoaccharides are glucose and fructose
 Glucose - the most abundant carbohydrate, C6(H2O)6 or C6H11OH (aldehyde sugar)
 Fructose - present in most of fruit, C6(H2O)6 or C6H11OH (ketone sugar)
 Diaccharides - composed of two monoaccharides
The most most important diaccharide is sucrose and lactose
 Sucrose (table sugar) - is a sugar composed of 1 glucose + 1 fructose.
The invert sugar (sweeter than sucrose) is made from hydrolysis of sucrose converting part of
glucose to fructose. The sugars made from sugar beets and canes are the same.
 Lactose (milk sugar) - is a sugar composed of 1 glucose + 1 galactose
 Polyaccharides - composed more than two units of monoaccarides
Chemistry for Life

71
Introductory to Biochemistry - Carbohydrates
 Some of most important polyaccharides
 Starch
• Many crops contains mainly starch (corn, potatoes, wheat, rice etc). It is a major way
plants store their energy
• It consists of mainly glucose. due to different way in which glucose units are joined
together, starch may be unbranched or branched in structure
• Hydrolysis of starch, catalysed by enzyme within our digestive system, gives glucose
 Glycogen
• These type of polyaccharides can be synthesised within our body and stored in liver and
muscles. It services as immediate energy source of our body.
 Cellulose
• It forms major structural unit of plants (e.g. wood 50%, cotton fibres 100% cellulose).
• usually unbranched chain of glucose units with average molecular weight 500,000 amu.
• The enzymes within our body which help hydrolysis starch cannot digest cellulose.
Chemistry for Life

72
Introductory to Biochemistry - Nucleic Acids
 Functions of nucleic acids
 Nucleic acids are chemical carriers of an organism’s genetic information
 They are also chemical controller of cell development through controlling protein
synthesis
 Composition of nucleic acids
 The nucleic acids are bio-polymers (molecules are linked together through
polymerisation reaction like the formation of proteins from amino acids)
 The basic building blocks of nucleic acids are called necleotides.
 nucleotides are formed from the following units:
1. a phosphoric acid molecule, H3PO4
2. a five-carbon sugar
3. a nitrogen-containing organic base
 Type of Nucleic acids
 DNA
 RNA
Chemistry for Life

73
Introductory to Biochemistry - DNA & RNA
 DNA - Deoxyribonucleic Acids
 DNA has huge molecular weight ranging from 6~16 million amu
 DNA is found primarily in the nucleus of living cell
 DNA stores the genetic information of the cell & controls the production of protein
 RNA - Ribonucleic Acids
 RNA has smaller molecular weight ranging from 20,000~40,000 amu
 RNA is mostly found outside the nucleus in the cytoplasma (substance around cell
membrane)
 RNA carries the information stored by DNA out of nucleus of cell into cytoplasma,
where proteins are synthesised based on the instruction delivered by RNA
 The differences in composition of DNA and RNA
 The only difference between DNA and RNA
is the 5-carbon sugar units in the nucleotides.
DNA has deoxyribose
NRA has ribose
Chemistry for Life
HOCH2
l
C
l
H
O
H
l
C
l
OH
H
l
C
l
OH
H
l
C
l
OH
HOCH2
l
C
l
H
O
H
l
C
l
OH
H
l
C
l
H
H
l
C
l
OH
ribose deoxyribose

74
Introduction to Spectroscopes
 Atoms & molecules can have different
states each having a specific E level
 Ground state
The normal state (the lowest possible E level)
 Excited state
When absorbing electromeganetic radiation
(e.m.r.). Atoms/molecules rise their E levels
 Atoms/molecules at excited state are not stable
& tends to return to their ground state.
 In the process of returning their ground states,
the energy gained earlier is release in a form of
e.m.r. (photons).
 Absorption/Emission spectra
 E.m.r. has energy determined by its frequency
 E.m.r. can be absorbed and emitted by atoms/
molecules contain specific Information of A/M.
 These e.m.r.’s can be recorded and analysed
- the base of many spectroscopic techniques.
Analytical Techniques
n = 1
n = 2
n = 3,
etc.
D
Energy
n=1
n=2
n=3
n=4
1s
2s
2p
3s
3p
4s
3d
4p
4d
4f
S0
v1
v2
v3
v4
S2
v1
v2
v3
v4
T1
v1
v2
v3
v4
Atomic level
Molecular level

75
Introduction to
UV-Visible Absorption Spectroscopy
 The electromnetic radiations
X-Ray UV Visible IR Microwave
200nm 400nm 800nm
Wavelength (nm)

76
Introduction to UV-Visible Absorption Spectroscopy
 UV Absorption Spectrometer
Sample
90C
Detector
UV Light Source
Monochromator Monochromator
Emit fluorescent light
as energy decreases
Ground state
Antibonding
Antibonding
Nonbonding
Bonding
Bonding
Energy s
p
ss’
pp’
ns’
n
np’
Electron's molecular energy levels
s’
p’
UV Spectrometer Applications
Protein
Amino Acids (aromatic)
Pantothenic Acid
Glucose Determination
Enzyme Activity (Hexokinase)
Visible Spectrometer Applications
 Niacin
 Pyridoxine
 Vitamin B12
 Metal Determination (Fe)
 Fat-quality Determination (TBA)
 Enzyme Activity (glucose oxidase)

77
Introduction to UV-Visible Absorption Spectroscopy
UV Absorption Spectra Visible Absorption Spectra

78
Introduction to Mass Spectrometry
 The working principle
1. Sample molecules are ionised in
the ion source section
2. The ions with different masses
and charges travel through a
magnet field at different speeds,
arriving to detector at different
time scales
3. The charges of ions are
converted to electricity current,
the intensity of which is then the
measure of the concentration of
the molecules
 The measurement results are
directly linked to the atomic
mass of molecules
 Very useful in detecting organic
molecules & in isotopic tracing
analysis
Schematic diagramme of a single-focusing mass
spectrometr with an electron-impact ion source
lon source mass analyser detector
heated filament
produce electrons
beam which
collide and ionise
sample molecules
Ions with different mass & e-
changes travel at different
speeds under the magnetic
field, reaching the detector
at different time.
The charges carried
by the ions are
converted to electricy
current and detected

79
Introduction to Mass Spectrometry
Mass spectrum of chlorine gas
m/e ratio Corresponding ion
35 35Cl+
37 35Cl+
70 35Cl+ - 35Cl+
72 35Cl+ - 37Cl+
74 37Cl+ - 37Cl+

80
Introduction to Chromatography
 The working principle of Chromatography
The level of Interaction (adsorption) betw.
packing material & sample A, B differs,
resulting in different speeds of travel of A &
B in a media (paper, column etc.)
 Usually sample to be analysed is
injected into a carrier (gas or liquid)
 Carrier is usually inert (does not react
with packing materials)
 The components in sample, being
separated after chromatography, are
analysed by TCD or mass spec.
 Types of chromatography
LC - Liquid (carrier & A,B) Chromatography
GC - Gas (carrier & A,B) Chromatography
HPLC - High Pressure Liquid Chromatography
t
gas or
liquid
sample (A+B)
injection A B
effluent
column packing, P (stationary phase)
t
c
t
c
c
A
B
A
B
t = to
t = ti
t = te
effluent
Assuming P ‘likes’ A (A stay with P longer)
carrier
(C)
effluent
effluent
v

81
Introduction to Chromatography
Typical GC setup GC chromatograph

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