CHAPTER 5: ENERGETICS
· The whole universe want to have the smallest (lowest)
possible energy and the highest disorde
kinetic
*
En Ep potential
(speed) (stored inside species)
I Heat-energy (goes from higher to temperature
lower
Temperature a measure
-
of energy
System and surrounding
~ chemical reaction
(Hy
· + 02 > -
CO2 + 2420 combustion exo
↓
in air
·
NaOH + HC > Nal + H20 neutralisation
-
exo
(aq) (aq) (aq) (aq)
Topic : Exothermic (exo) and endothermic (endo reactions
Exo Endo
I
R P
Rectant >Product
>
-
↑ PP
-heat (can belight) heat is absorbed
>
is evolved
-
surrounding is
getting warm
-
surrounding gets colde
& & R- stable b
↓ I
P-more stable b more
they possess less ener
gY they possess less
energy
E(H) E(H) P
kJ kJ
R +Ho R IWH H>O
Hi P
progress progress
~
machish reaction
STB
-
H -
the enthalpy change is the difference between products and
reactants
-H =
P- R
, Topic Simple calculations
:
- He comb- when
energy released I mole of a substance is
burned completely under standart conditions (25 C)
·
1 Work out the oh when 29 of ethand is burned completely
- Homb =
- 1367kJ for 1 mol of CHON
n
=4
=
0 0434 mol
2
.
Imo CHOH--1367k5
0 on34 mol CHsOH XkJ
X = 59 . 3kJ
2 Work out the oh when octane burned completely
100g of is
*
-H comb = -
5470 1 -- 5470k5
100
12 01 x8 +
--Xk]
. 1 0148
.
X = -
4786k]
.
2 10% ethand II) -
100 100
Mr CcHjOH This
Mr
a) 90 % octane
2 17 . 7 88 .
m=
1kg
#) 1--1367 1 --
5470
1) 1009 9009 2 . 17 -
X 7 88.
-
X
X= -
46070kj
Melting Freezing I
Lendo Condensation -
exo
Boiling
b) (l >
-
(g) >
-
burns
Topic : The enthalpy change of solution and neutralization
NaOh 100cm HeO m=2 31g NaCH a = m +
c + ot
↓
.
9-heat
initial temperature 20 :
m-mass of substance heated
water c-specific heat capacity
final temperature : 26%
-T-change in temperature
/
, m = m(Hc0) + m (NaOH) = 102 .
31g
c =
4 18 .
Jgtk"- important
- T =
6 C Pexo This second step is known as the hydration
energy and can be defined as the enthalpy
q = mxc+ -T =
25655 change when 1 mole of gaseous ions
dissolves in sufficient water to give an
Enthalpy change of solution infinitely dilute solution.
1) NaOH , Na + or
(aq)
laq)
H -
the energy released or attached when I md
of a compound is dissolved completely into "infinite dilution
Mr (NaOH) =
nOgmol" 2 .
31g NaCh 25655
nog NaOH XJ
X= 14415 6 J mo+.
=44 4 .
kJ mol" (exothermic)
2) or
you can use the formula -H =
kJmolt
I
6 . a) MgCl Ig -
- T =
21 5-> 29 1
. . exo
Fimiting
reactant
H20-50mls
51 x 1 18 x
76 =
16205
9
= .
.
g &
Jund" 154 3k5mol
- =
I
154286
- -
=
+ = -
. 0105
0 .
24 31 + 35
.
. 45x2
Enthalpy change of neutralization -
is the enthalpy
change when I md H O formed when an acid (H) reacts
, is
with an alkali (out under standart conditions
100 cm3 of KOH (1moldmi) was mixed with 100cm of He
11 md dris) the temperature rised by 6 82% Work out
.
suneut.
· The whole universe want to have the smallest (lowest)
possible energy and the highest disorde
kinetic
*
En Ep potential
(speed) (stored inside species)
I Heat-energy (goes from higher to temperature
lower
Temperature a measure
-
of energy
System and surrounding
~ chemical reaction
(Hy
· + 02 > -
CO2 + 2420 combustion exo
↓
in air
·
NaOH + HC > Nal + H20 neutralisation
-
exo
(aq) (aq) (aq) (aq)
Topic : Exothermic (exo) and endothermic (endo reactions
Exo Endo
I
R P
Rectant >Product
>
-
↑ PP
-heat (can belight) heat is absorbed
>
is evolved
-
surrounding is
getting warm
-
surrounding gets colde
& & R- stable b
↓ I
P-more stable b more
they possess less ener
gY they possess less
energy
E(H) E(H) P
kJ kJ
R +Ho R IWH H>O
Hi P
progress progress
~
machish reaction
STB
-
H -
the enthalpy change is the difference between products and
reactants
-H =
P- R
, Topic Simple calculations
:
- He comb- when
energy released I mole of a substance is
burned completely under standart conditions (25 C)
·
1 Work out the oh when 29 of ethand is burned completely
- Homb =
- 1367kJ for 1 mol of CHON
n
=4
=
0 0434 mol
2
.
Imo CHOH--1367k5
0 on34 mol CHsOH XkJ
X = 59 . 3kJ
2 Work out the oh when octane burned completely
100g of is
*
-H comb = -
5470 1 -- 5470k5
100
12 01 x8 +
--Xk]
. 1 0148
.
X = -
4786k]
.
2 10% ethand II) -
100 100
Mr CcHjOH This
Mr
a) 90 % octane
2 17 . 7 88 .
m=
1kg
#) 1--1367 1 --
5470
1) 1009 9009 2 . 17 -
X 7 88.
-
X
X= -
46070kj
Melting Freezing I
Lendo Condensation -
exo
Boiling
b) (l >
-
(g) >
-
burns
Topic : The enthalpy change of solution and neutralization
NaOh 100cm HeO m=2 31g NaCH a = m +
c + ot
↓
.
9-heat
initial temperature 20 :
m-mass of substance heated
water c-specific heat capacity
final temperature : 26%
-T-change in temperature
/
, m = m(Hc0) + m (NaOH) = 102 .
31g
c =
4 18 .
Jgtk"- important
- T =
6 C Pexo This second step is known as the hydration
energy and can be defined as the enthalpy
q = mxc+ -T =
25655 change when 1 mole of gaseous ions
dissolves in sufficient water to give an
Enthalpy change of solution infinitely dilute solution.
1) NaOH , Na + or
(aq)
laq)
H -
the energy released or attached when I md
of a compound is dissolved completely into "infinite dilution
Mr (NaOH) =
nOgmol" 2 .
31g NaCh 25655
nog NaOH XJ
X= 14415 6 J mo+.
=44 4 .
kJ mol" (exothermic)
2) or
you can use the formula -H =
kJmolt
I
6 . a) MgCl Ig -
- T =
21 5-> 29 1
. . exo
Fimiting
reactant
H20-50mls
51 x 1 18 x
76 =
16205
9
= .
.
g &
Jund" 154 3k5mol
- =
I
154286
- -
=
+ = -
. 0105
0 .
24 31 + 35
.
. 45x2
Enthalpy change of neutralization -
is the enthalpy
change when I md H O formed when an acid (H) reacts
, is
with an alkali (out under standart conditions
100 cm3 of KOH (1moldmi) was mixed with 100cm of He
11 md dris) the temperature rised by 6 82% Work out
.
suneut.
