thermochemistry. energy the ability to do work or transfer heat.the ability to do work or transfer...
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THERMOCHEMISTRY
Energy
• The ability to do work or transfer heat.– Work: Energy used to cause an
object that has mass to move.– Heat: Energy used to cause the
temperature of an object to rise.
Definitions #1
Energy: The capacity to do work or produce heat
Potential Energy: Energy due to position or composition
Kinetic Energy: Energy due to the motion of the object
2
2
1mvKE
Definitions #2
Law of Conservation of Energy: Energy can neither be created nor destroyed, but can be converted between formsThe First Law of Thermodynamics: The total energy content of the universe is constant
E = q + wE = change in internal energy of a system
q = heat flowing into or out of the system
-q if energy is leaving to the surroundings+q if energy is entering from the surroundings
w = work done by, or on, the system
-w if work is done by the system on the surroundings
+w if work is done on the system by the surroundings
Work problemsChapter 5
• 5.25• 5.27 A and B• 5.31 All
Calorimetry
The amount of heat absorbed or released during a physical or chemical change can be measured…
…usually by the change in temperature of a known quantity of water1 calorie is the heat required to raise the temperature of 1 gram of water by 1 C1 BTU is the heat required to raise the temperature of 1 pound of water by 1 F
The Joule
The unit of heat used in modern thermochemistry is the Joule
2
2111
s
mkgmeternewtonjoule
1 joule = 4.184 calories
A Bomb Calorimete
r
A Cheaper
Calorimeter
Specific HeatThe amount of heat required to raise the temperature of one gram of substance by one degree Celsius.Substance Specific Heat (J/g·K)
Water (liquid) 4.18
Ethanol (liquid) 2.44
Water (solid) 2.06
Water (vapor) 1.87
Aluminum (solid) 0.897
Carbon (graphite,solid) 0.709
Iron (solid) 0.449
Copper (solid) 0.385
Mercury (liquid) 0.140
Lead (solid) 0.129
Gold (solid) 0.129
Calculations Involving Specific Heat
Tmsq
s = Specific Heat Capacity
q = Heat lost or gainedT = Temperature
change
OR
Tm
qs
Problems
• 5.53 a and B
State Functions depend ONLY on the present state of the system
ENERGY IS A STATE FUNCTION
A person standing at the top of Mt. Everest has the same potential energy whether they got there by hiking up, or by falling down from a plane!
WORK IS NOT A STATE FUNCTION
WHY NOT???
State Functions
Usually we have no way of knowing the internal energy of a system; finding that value is simply too complex a problem.
State Functions• However, we do know that the
internal energy of a system is independent of the path by which the system achieved that state.– In the system below, the water could
have reached room temperature from either direction.
State Functions
• Therefore, internal energy is a state function.
• It depends only on the present state of the system, not on the path by which the system arrived at that state.
• And so, E depends only on Einitial and Efinal.
State Functions
• However, q and w are not state functions.
• Whether the battery is shorted out or is discharged by running the fan, its E is the same.– But q and w are
different in the two cases.
Work
When a process occurs in an open container, commonly the only work done is a change in volume of a gas pushing on the surroundings (or being pushed on by the surroundings).
WorkWe can measure the work done by the gas if the reaction is done in a vessel that has been fitted with a piston.
w = −PV
Work, Pressure, and Volume
VPw Expansion Compression+V (increase)
-V (decrease)
-w results +w results
Esystem decreases
Work has been done by the system on the surroundings
Esystem increases
Work has been done on the system by the surroundings
Energy Change in Chemical Processes
Endothermic:
Exothermic:
Reactions in which energy flows into the system as the reaction proceeds.
Reactions in which energy flows out of the system as the reaction proceeds.
+ qsystem - qsurroundings
- qsystem + qsurroundings
Endothermic Reactions
Exothermic Reactions
Enthalpy• If a process takes place at constant pressure (as the
majority of processes we study do) and the only work done is this pressure-volume work, we can account for heat flow during the process by measuring the enthalpy of the system.
• Enthalpy is the internal energy plus the product of pressure and volume:
H = E + PV
At constant pressure and volume the change in enthalpy is the heat gained or lost
H = q
Enthalpies of Reaction
The change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants:
H = Hproducts − Hreactants
Enthalpies of Reaction
This quantity, H, is called the enthalpy of reaction, or the heat of reaction.
Problems
5.57 A and B
Hess’s Law“In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or a series of steps.”
Hess’s Law
Hess’s Law Example Problem
Calculate H for the combustion of methane, CH4:
CH4 + 2O2 CO2 + 2H2O Reaction Ho
C + 2H2 CH4 -74.80 kJ
C + O2 CO2 -393.50 kJ
H2 + ½ O2 H2O -285.83 kJ
Step #1: CH4 must appear on the reactant side, so we reverse reaction #1 and change the sign on H.
CH4 C + 2H2 +74.80 kJ
Hess’s Law Example Problem
Calculate H for the combustion of methane, CH4:
CH4 + 2O2 CO2 + 2H2O Reaction Ho
C + 2H2 CH4 -74.80 kJ
C + O2 CO2 -393.50 kJ
H2 + ½ O2 H2O -285.83 kJ
CH4 C + 2H2 +74.80 kJ
Step #2: Keep reaction #2 unchanged, because CO2 belongs on the product side
C + O2 CO2 -393.50 kJ
Hess’s Law Example Problem
Calculate H for the combustion of methane, CH4:
CH4 + 2O2 CO2 + 2H2O Reaction Ho
C + 2H2 CH4 -74.80 kJ
C + O2 CO2 -393.50 kJ
H2 + ½ O2 H2O -285.83 kJ
CH4 C + 2H2 +74.80 kJ C + O2 CO2 -393.50 kJ
Step #3: Multiply reaction #2 by 2
2H2 + O2 2 H2O -571.66 kJ
Hess’s Law Example Problem
Calculate H for the combustion of methane, CH4:
CH4 + 2O2 CO2 + 2H2O Reaction Ho
C + 2H2 CH4 -74.80 kJ
C + O2 CO2 -393.50 kJ
H2 + ½ O2 H2O -285.83 kJ
CH4 C + 2H2 +74.80 kJ C + O2 CO2 -393.50 kJ
2H2 + O2 2 H2O -571.66 kJ
Step #4: Sum up reaction and H
CH4 + 2O2 CO2 + 2H2O -890.36 kJ
Calculation of Heat of Reaction Calculate H for the combustion of
methane, CH4: CH4 + 2O2 CO2 + 2H2OHrxn = Hf(products) - Hf(reactants)
Substance Hf
CH4 -74.80 kJ
O2 0 kJ
CO2 -393.50 kJ
H2O -285.83 kJ
Hrxn = [-393.50kJ + 2(-285.83kJ)] – [-74.80kJ]
Hrxn = -890.36 kJ