Transcript Class 22 - Department of Chemistry

CHEMISTRY 161
Chapter 6
www.chem.hawaii.edu/Bil301/welcome.html
THERMODYNAMICS
HEAT
CHANGE
quantitative study of heat and energy changes of a system
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
the state (condition) of a system is defined by
T, p, n, V, E
the state (condition) of a system is defined by
T, p, n, V, E
STATE FUNCTIONS
properties which depend only on the initial
and final state, but not on the way how this
condition was achieved
ΔV = Vfinal – Vinitial
Δp = pfinal – pinitial
ΔT = Tfinal – Tinitial
ΔE = Efinal – Einitial
Energy is a STATE FUNCTION
ΔE = m g Δh
IT DOES NOT MATTER WHICH PATH YOU TAKE
Hess Law
ENTHALPY, H
CH4(g) + 2O2(g)
- 802 kJ
Reactants
- 890 kJ
CO2(g) + 2H2O(g)
- 88 kJ
CO2(g) + 2H2O(l)
Products
Zeroth Law of Thermodynamics
a system at thermodynamical equilibrium
has a constant temperature
heat is spontaneous transfer of thermal energy two bodies
at different temperatures T1 > T2
spontaneous
T1
T2
First Law of Thermodynamics
energy can be converted from one form to another,
but cannot be created or destroyed
CONSERVATION OF ENERGY
SURROUNDINGS
+
-
SYSTEM
THE TOTAL ENERGY OF THE UNIVERSE IS CONSTANT
ΔE = ΔEsystem + ΔEsurrounding = 0
First Law of Thermodynamics
ΔEsystem = ΔQ + ΔW
ΔQ heat change
ΔW work done
DQ > 0 ENDOTHERMIC
DQ < 0 EXOTHERMIC
?
mechanical work
ΔW = - p ΔV
M
ΔV < 0
the energy of gas goes up
M
ΔV > 0
the energy of gas goes down
First Law and Enthalpy
ΔEsystem = ΔQ - pΔV
1.constant pressure
2. ideal gas law
→
enthalpy change Δ H
→
pV=nRT
p ΔV = Δn R T
ΔEsystem = ΔQ – R T Δn
ΔQ = ΔEsystem + R T Δn = ΔH
ΔQ = ΔEsystem + R T Δn = ΔH
Δ n = nfinal – ninitial
Calculate the energy change of a system for the reaction
process at 1 atm and 25C
2 CO(g) + O2(g) → 2 CO2(g) ΔHo = -566.0 kJ
ΔEsystem = ΔH0 - R T Δn
ΔEsystem = -563.5 kJ
A gas is compressed in a cylinder from a volume
of 20 L to 2.0 L by a constant pressure of 10 atm.
Calculate the amount of work done on the system.
Calculate the amount of work done against an
atmospheric pressure of 1.00 atm when 500.0
g of zinc dissolves in excess acid at 30.0°C.
Zn(s) + 2H+(aq) → Zn2+(aq) + H2 (g)
The heat of solution of KCl is +17.2 kJ/mol and the
lattice energy of KCl(s) is 701.2 kJ/mol. Calculate
the total heat of hydration of 1 mol of gas phase
K+ ions and Cl– ions.
Homework
Chapter 6, problems