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Summary Energetic IB chemistry HL

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  • September 6, 2024
  • 25
  • 2024/2025
  • Summary
  • Secondary school
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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.

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