Chapter 6 Thermochemistry. Final Exam. ( May7, 2014 Wednesday) Instructional Complex 421 10:15 AM....
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Transcript of Chapter 6 Thermochemistry. Final Exam. ( May7, 2014 Wednesday) Instructional Complex 421 10:15 AM....
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
Two forms of energy
Kinetic energy
Potential energy
Kinetic energy: energy from motion
2
2
1mvEk
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).
Ei
Ef
∆E = Ef − Ei
E: State FunctionsSystem: a battery
State functions: E, P, V, T, do not depend on path.
W, Q: not state functions, depend on process.
W = −P∆V
PV work:
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)?
Example 6.4, page 245
Enthalpy: H = E + PV
Only PV work, constant P ∆H = Qp
∆H = Qp
∆H > 0 ↔ endothermic
∆H < 0 ↔ exothermic
Recall ….
EnthalpyEnthalpy
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
Example 6.6, page 250
Q = c m ∆T
c: specific heat (capacity)
m: mass
∆T: change of temperature
g
°C
J·°C−1·g−1
same sign as Q
Q = c m ∆T
C = c m
C: heat capacity, J ·°C−1
Q = C ∆T
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 6.2, page 242
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.)
Example 6.3, page 243
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
Read example 6.5,page 248
A constant-volumeCalorimeter
∆Erxn = −Qcal
= −Ccal ΔT
2H2(g) + O2(g) 2H2O(g) ∆H = −483.6 kJ
∆H = Hproducts − Hreactants
1. It is important to specify the state of each species in a thermochemical reaction.
∆H: enthalpy (change) of reaction, heat of reaction
2. ∆H of the reverse reaction is the negative of the original reaction.
3. ∆H depends on how the reaction is written.
Characteristics of enthalpy change of a reaction
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.
Example 6.7, page 252
HA
HB
∆HAB = HB − HA
HC
∆HAB = ∆HAC + ∆HCB
∆HAC = HC − HA
∆HCB = HB − HC
H1 = H2 + H3
Hess’s Law: ∆H for the overall reaction is equal to the sum ofthe 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 COto 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 combustionof 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 respectiveenthalpy 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
Sea level
Atlanta, GA
Reno, NV
1000 ft
4500 ft
3500 ft
Standard0 ft
H = H (P, T, phase), phase = s, l, g
Standard state
P = 1 atmT = temperature of interest, often 25 °Cstate = most stable form
The standard enthalpy of formation of a pure substance,∆H°f, is the change in enthalpy for the reaction that formsone mole of the substance from its elements, with allsubstances in their standard states.
“Sea level” : free elements at standard state
Calculate the standard enthalpy change for the following reaction
4NH3(g) + 5O2(g) 4NO (g) + 6H2O(g)
Example 6.11, page 261
ΔHrxn° = Σnp ΔHf°(products) − Σnr ΔHf° (reactants)
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)
For practice 6.11, page 261