Thermodynamics Introduction ThermoDynamics

Μηχανική

27 Οκτ 2013 (πριν από 4 χρόνια και 8 μήνες)

122 εμφανίσεις

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.