Thermodynamics Introduction ThermoDynamics

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27 Οκτ 2013 (πριν από 3 χρόνια και 9 μήνες)

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Thermodynamics Introduction


Thermo
: heat


Dynamics
: change


-

thermodynamics: study of heat changes

-

thermochemistry: study of the heat changes of chemical reactions


system surroundings


-

w
hat you observe/ are interested in


-

everything else


first law of thermodynamics: energy cannot be created or destroyed


∆E
system

=
-
∆E
surroundings

also written as:

∆E
universe

= 0 or ∆E
universe
= ∆E
system

+ ∆E
surroundings


Thermal energy: average of kinetic energy

-

three types of kinetic motion in gases: vibrational, translational, rotational energy


E
k

= E
tra
ns

+ E
vib

+ E
rat


Temperature: a measure of average kinetic energy


Heat: “q” or “Q”, change in thermal energy


Reactions:

-

releases heat, Q
rxn

is negative, Q
surroundings

is positive


exothermic

-

absorbs heat, Q
rxn

is positive, Q
surroundings

is negative


e
ndothermic


Enthalpy:

-

thermal energy

-

same as “q”



Example Question:

What is the molar heat of solution of NaOH?


Surroundings





System

VH
2
O= 100.0mL




mNaOH= 2.13g

mH
2
0= 100.0g

Ti(H
2
O)= 20.0
o
C

Tf(H
2
O)= 25.1
o
C

∆T= Tf


Ti


= 5.0
o
C



Analysis:


q= mc∆t


q: heat (J)

m: mass (g)

c: specific heat capacity (J/g
.
c)

∆t: temperature (
o
C or K)


q= mc∆t


q
system
=
-
q
surroundings


Calculating enthalpy of the system:

q
surroundings

= m
surroudings
c
surrouding
s∆t
surroundi
ngs



= (100.0g)(4.18J/g c) (5.1
o
C)



= 2 131.8 J


q
system
=
-
q
surroundings



Therefore
, q
system
=
-

2 131.8 J



Calculating moles of NaOH:

m
NaOH

= 2. 13g

M
NaOH

= 39.99714 g/ mol

n
NaOH
= m/M


= 0.05325


Using ratios to solve:

Q
system

/ n
system

= Q
mol
/ n
mol

-

2 131.8 J / n
system

= 0.05325mol / 1 mol

n
system
=
-

40 033.80282 J/mol x 10
-
3



=
-

40 KJ/mol


Therefore, the heat of the solution is


40 KJ/mol.