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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