Transcript H - Gordon State College

Chapter 6
Thermochemistry
Final Exam
(
May7, 2014
Wednesday)
.
Instructional
Complex 421
10:15 AM. –12:15
PM
Study of energy and its transformation ― thermodynamics
Thermodynamics of chemical reactions ― thermochemistry
Energy ― capacity to do work (W) and transfer heat (Q)
unit: J
W and Q ― two ways to transfer energy
Work = Force x Distance
W=F•d
SI unit:
W ― J, F ― N, d ― m
J=N•m
Kinetic energy
Two forms of energy
Potential energy
Kinetic energy: energy from motion
1 2
Ek  mv
2
potential energy: energy from the interaction between objects.
depends on the objects and the relative distance.
BrNO + BrNO  2NO +Br2
exothermic reaction
CH4 + 2O2  2H2O + CO2
Exothermic Process
system
surroundings
universe
Energy of the universe is constant.
Or say energy of the universe is conserved.
First law of thermodynamics
∆E = Q + W
E: energy of the system, including Ek and Ep. Or called
internal energy of the system.
∆E: change of internal energy of the system.
∆E = Ef − Ei
System absorbs heat from surroundings, Q > 0.
System releases heat to surroundings, Q < 0.
System does work on surroundings, W < 0.
Surroundings do work on system, W > 0.
Calculate ΔE for a system undergoing an endothermic process
in which 15.6 kJ of heat flows and where 1.4 kJ of work is
done on the system.
State Function
A property of a system whose value is determined only by
specifying the system’s state, not on how the system arrived
at that state.
The state of a system is specified by parameters such as
temperature, pressure, concentration and physical states (solid,
liquid, or gas).
Ef
Ei
∆E = Ef − Ei
System: a battery
E: State Functions
State functions: E, P, V, T,
do not depend on path.
W, Q: not state functions, depend on process.
PV work:
W = −P∆V
Example 6.4, page 245
To inflate a balloon you must do pressure–volume
work on the surroundings. If you inflate a balloon
from a volume of 0.100 L to 1.85 L against an
external pressure of 1.00 atm, how much work is
done (in joules)?
Enthalpy: H = E + PV
Only PV work, constant P ∆H = Qp
∆H = Qp
∆H > 0 ↔ endothermic
∆H < 0 ↔ exothermic
Recall ….
Enthalpy
Example 6.6, page 250
Identify each process as endothermic or exothermic
and indicate the sign of H.
(a) sweat evaporating from skin
(b) water freezing in a freezer
(c) wood burning in a fire
Q = c m ∆T
c: specific heat (capacity) J·°C−1·g−1
m: mass
g
∆T: change of temperature
same sign as Q
°C
Q = c m ∆T
C=cm
C: heat capacity, J ·°C−1
Q = C ∆T
Example 6.2, page 242
Suppose you find a copper penny (minted
before 1982, when pennies were almost
entirely copper) in the snow. How much
heat is absorbed by the penny as it warms
from the temperature of the snow, which is
–8.0 C, to the temperature of your body,
37.0 C? Assume the penny is pure copper
and has a mass of 3.10 g.
Example 6.3, page 243
A 32.5-g cube of aluminum initially at 45.8 C is
submerged into 105.3 g of water at 15.4 C.
What is the final temperature of both substances
at thermal equilibrium? (Assume that the
aluminum and the water are thermally isolated
from everything else.)
A constant-pressure
coffee-cup Calorimeter
ΔHrxn = − Qsoln
= − c m ΔT
Read example 6.8, page 253
∆E = Q + W
PV work: W = − P∆V
PV work only, constant P: ∆H = Qp
PV work only, constant V:
∆E = Qv
A constant-volume
Calorimeter
∆Erxn = −Qcal
= −Ccal ΔT
Read example 6.5,
page 248
Characteristics of enthalpy change of a reaction
∆H = Hproducts − Hreactants
2H2(g) + O2(g)  2H2O(g)
∆H = −483.6 kJ
∆H: enthalpy (change) of reaction, heat of reaction
1. It is important to specify the state of each species in a
thermochemical reaction.
2. ∆H of the reverse reaction is the negative of the original
reaction.
3. ∆H depends on how the reaction is written.
Example 6.7, page 252
An LP gas tank in a home barbeque contains 13.2
kg of propane, C3H8. Calculate the heat (in kJ)
associated with the complete combustion of all of
the propane in the tank.
HB
HA
HC
∆HAB = HB − HA
∆HAC = HC − HA
∆HCB = HB − HC
∆HAB = ∆HAC + ∆HCB
H1 = H2 + H3
Hess’s Law: ∆H for the overall reaction is equal to the sum of
the enthalpy changes of each individual step.
The enthalpy of reaction for the combustion of C to CO2 is
−393.5 kJ/mol C, and the enthalpy for the combustion of CO
to CO2 is −283.0 kJ/mol CO:
C(s) + O2(g)  CO2(g)
∆H1 = −393.5 kJ
CO(g) + ½ O2(g)  CO2(g)
∆H2 = −283.0 kJ
Using these data, calculate the enthalpy for the combustion
of C to CO:
C(s) + ½ O2(g)  CO(g)
∆H = ?
Calculate the ∆H for the reaction
2C(s) + H2(g)  C2H2(g)
Given the following chemical equations and their respective
enthalpy changes:
C2H2(g) + 5/2 O2(g)  2CO2(g) + H2O(l) ∆H1 = −1299.6 kJ
C(s) + O2(g)  CO2(g)
∆H2 = −393.5 kJ
H2(g) + ½ O2(g)  H2O(l)
∆H3 = −285.8 kJ
Try example 6.9 and for practice 6.9, page 256-257
Reno, NV
3500 ft
4500 ft
Atlanta, GA
1000 ft
Sea level
Standard
0 ft
H = H (P, T, phase), phase = s, l, g
Standard state
P = 1 atm
T = temperature of interest, often 25 °C
state = most stable form
The standard enthalpy of formation of a pure substance,
∆H°f, is the change in enthalpy for the reaction that forms
one mole of the substance from its elements, with all
substances in their standard states.
“Sea level” : free elements at standard state
Example 6.11, page 261
Calculate the standard enthalpy change for the following reaction
4NH3(g) + 5O2(g)  4NO (g) + 6H2O(g)
ΔHrxn° = Σnp ΔHf°(products) − Σnr ΔHf° (reactants)
For practice 6.11, page 261
Calculate the standard enthalpy change for the following reaction
2Al(s) + Fe2O3(s)  Al2O3(s) + 2Fe(s)
Al(s) + ½ Fe2O3(s)  ½ Al2O3(s) + Fe(s)