C H5
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THERMOCHEMISTRY Chapter 5
Transcript of C H5
THERMOCHEMISTRY
Chapter 5
Chapter 5
WARNING
There is a high risk of me assuming you know or understand things that you do not. Stop me as
necessary.
Chapter 5
The Nature of Energy
o Section 5.1 is really physicso Let’s look at those parts that are
germane to chemistry
Chapter 5
o Kinetic energy• Ek= ½ mv2
• Kinetic energy depends on mass and velocity
• Temperature is a measure of the average kinetic energy of the molecules in a sample
• Question: why does an iceberg have more energy, then a bucket of molten iron?
The Nature of Energy
Chapter 5
o Potential energy• Usually associated with positional Ep;
the rock on top of the hill• In chemistry, we care more about
chemical and electrostatic potential energies
• Chemical energies will be covered later in the chapter
The Nature of Energy
Chapter 5
o Electrical potential energy Eep
o Electrical potential energy is proportional to the charges involved divided by distance
o We can make it equal by multiplying by a proportionality constant:
Eep =k Q1Q2/do K = 8.99 x 109 J m/ C2 (joule meters per
coulomb squared)o 1J=1Kg m2/ s2
The Nature of Energy
Chapter 5
o Calorie: a measure of heato Capitalization is IMPORTANT:
• 1c =4.184 J• 1C = 1000c• Food wrappers use C
o One c is the energy required to heat 1g of water one Kelvin
o Yes, in thermodynamics K and R must be used instead of C and F
The Nature of Energy
Chapter 5
o Transferring energy: work & heato Work w = Force x distance
o Heat is energy transferred between objects of different temperatures (you DO NOT want to know the equations for heat transfer yet)
The Nature of Energy
Chapter 5
o Stoichiometry is accounting for mass (atoms in must equal atoms out)
o Thermodynamics is accounting for energy
Thermodynamics
Chapter 5
o System & Surroundings - The system is what we are looking at, the surroundings are the rest of the universe
o If work or energy are POSITIVE, work or heat go INTO the system, NEGATIVE work or heat go OUT
Thermodynamics
Chapter 5
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Chapter 5
o Internal Energy – the sum of ALL types of energy contained in a system
o In thermodynamics, we look at the changes in energy, ∆E
o ∆E = Efinal – Estart
o If ∆E is negative, energy left the system
Thermodynamics
Chapter 5
o The first law of thermodynamics equates energy, heat and work• ∆E = ∆q(heat) + ∆w(work)
o In English the change in energy of a system is the sum of the change in heat and the work done
o It means “You can only get out what you put in”
o TANSTAAFL (There aint no such thing as a free lunch)
Thermodynamics
Chapter 5
o ∆E of reactions: ∆Erxn
o Some reactions liberate heat, others require heat to go• HClaq+ NaOHaq H2O+ NaCl+ HEAT
• This is exothermic (heat exits)
• HgS(s) + HEAT Hg(l) + S9s)
• This is endothermic
Thermodynamics
Chapter 5
o State functions (and I DON’T mean dinner with Barak)
o State functions ONLY depend on what the state system is in NOW. I’m in front of the room. That’s a state system. It doesn’t mater how I got here, right now, I’m here.
Thermodynamics
Chapter 5
o The distance I traveled to get here is NOT a state function. I could have went right from the door to here. I could have come in, put my stuff down on the first desk, went to the bathroom (washed my hands for 20 sec), came back, and walked to the front.
Thermodynamics
Chapter 5
o Pressure – Volume worko When a system changes volume
and does worko w = -P ∆V (negative since if the
volume decreases, the work is done to the system
Enthalpy
Chapter 5
o In this chapter, the pressure always stays the same , so the work done id through changing volume w = -P ∆V
o It’s negative since if the piston expands, the work is done to the surroundings
Enthalpy
Chapter 5
o Enthalpy (H) is from the Greek word enthalpein meaning to warm
o It equals the heat flow when P-V work is the only work done
o H = E + PV or ∆H = ∆ E + P ∆ V
Enthalpy
Chapter 5
o If E = q + w, then we can put it in the enthalpy reaction as:
∆H = ∆E – w∆H = ∆(q +w) -w
∆H = ∆qp
Enthalpy
Chapter 5
o ∆H is the change in enthalpy from start to final ∆H=Hfinish-Hstart
o In a chemical reaction, the start are the reactants and the end is the products, so:
∆Hrxn =Hproducts-Hreactants
Enthalpy of Reaction
Chapter 5
HCl + NH3 → NH4Clo HNH3 is -80.29; o HNH4Cl is -314.4o HHCl is -92.30 o ΔH = Hproducts - Hreactants o ΔHproducts = -314.4 o ΔHreactants = -92.30 + (-80.29) = -172.59 o ΔH = -314.4 - 172.59 = 141.8
Enthalpy of Reaction
Chapter 5
HCl + NH3 → NH4Cl + 141.8 kJ
o Heats of reactions and enthalpies of compounds at specified temperatures & pressures are found in tables
o For any class problems, they will be given to you.
Enthalpy of Reaction
Chapter 5
The Ruleso Enthalpy is EXTENSIVE it depends
on the amount of material. Enthalpies are usually given in kJ/mol
Enthalpy of Reaction
Chapter 5
o The enthalpy of a reaction is equal to the enthalpy of the reverse reaction, but the opposite sign.
H2 + F2 2HF + 542kJ
Positive means heat given off H2 + F2 - 542kJ 2HF
542 kJ is how much energy needed to break 2HF back to it’s components
Enthalpy of Reaction
Chapter 5
o The enthalpy of a reaction depends on the physical state of the reactants. It takes energy to evaporate liquids and melt solids, and energy is released from condensing gases and freezing liquids. This energy MUST be accounted for.
Enthalpy of Reaction
Chapter 5
Enthalpy of Reaction
C6H12O6(s)+ 6O2 6CO2(g) + 6H2O(l)
∆H =-2803 Kj/mol. What if the water is a gas?
∆H H2O(l) H2O(g) +88 Kj/mol
Chapter 5
o Measurement of heat flowo Requires the use of heat capacity and
specific heato Heat capacity (C)– temperature change
when a mass of substance absorbs a certain amount of heat (go in a sauna wearing a necklace)
o C is the amount of heat required to raise the object’s one kelvin
Calorimetry
Chapter 5
Calorimetry
o Specific heat is a molar quantity – the amount of heat required to one gram one Kelvin
Tm
qC p
*
Change eTemperatur x Mass
dTransferreHeat ofAmount Heat Specific
Chapter 5
o How J are required to raise the temperature of 250g of Fe from 25oC to 100oC?
25oC = 298K, 100oC = 373KCsFe= 0.45 J/gK
Cs= q/m∆T 0.45= q/250*75
q= 8437.5J
Calorimetry
Chapter 5
o Works for solutions too.o How J are required to raise the
temperature of 250g of 2M NaCl from 25oC to 100oC?
Cssol= 1.2 J/gK
Cs= q/m∆T 1.2= q/250*75
q= 22,500J
Calorimetry
Chapter 5
o Bomb calorimetry
Calorimetry
Chapter 5
o In a calorimeter, all the energy entering the system can be accounted for.
o If we subtract the energy inputs, the only other heat source is the burning sample.
o So the heat derived from the sample is the ∆Hrxn
Calorimetry
Chapter 5
o When a 1.000 g sample of the rocket fuel hydrazine, N2H4, is burned in a bomb calorimeter which contains 1200 g of water, the temperature rises from 24.62°C to 28.16°C. If the C for the bomb is 840 J/°C, calculate
o ∆Hrxn reaction for combustion of a one-gram sample
o ∆Hrxn for combustion of one mole of hydrazine in the bomb calorimeter
Calorimetry
Chapter 5
o Solution o For a bomb calorimeter, use this equation:
∆Hrxn = -(qwater + qbomb)o ∆Hrxn = -(4.18 J / g·°C x mwater x Δt + C x Δt)o ∆Hrxn = -(4.18 J / g·°C x mwater + C)Δt o where q is heat flow, m is mass in grams, and
Δt is the temperature change. Plugging in the values given in the problem:
o ∆Hrxn = -(4.18 J / g·°C x 1200 g + 840 J/°C)(3.54°C)
o ∆Hrxn = -20,700 J or -20.7 kJ
Calorimetry
Chapter 5
Hess’s Law
o If a reaction is carried out in a series of steps, the ∆Hrxn is the sum of all the intermediate ∆Hs\
o The works if you can artificially separate a reaction into a number of steps.
o There are tables of ∆Hrxn to help
Chapter 5
o Sometimes the tables of ∆Hrxn use fractions in the chemical equation.
o If you have a problem using Hess’s law, you have to look at the tables and try to break your reaction into steps that are listed there.
Hess’s Law
Chapter 5
o Use the thermochemical equations shown below to determine the enthalpy for the reaction:2H2O(l)+NH3(g)=>NO2(g)+7/2H2(g)
o 2NH3(g)=>N2(g) + 3H2(g) H=138KJ o 1/2N2(g)+ 2H20(l)=>NO2(g)+2H2(g)
H=255KJ
Hess’s Law
Chapter 5
Hess’s Law
2H2O(l)+NH3(g)=>NO2(g)+7/2H2(g)
2NH3(g)=>N2(g) + 3H2(g) H=138KJ
We only have 1 NH3(g), let’s divide the equation in half:
NH3(g)=> ½ N2(g) + 3/2 H2(g) H=69KJ
½ N2(g)+ 2H20(l)=>NO2(g)+2H2(g) H=255KJ
Chapter 5
NH3(g)=> ½ N2(g) + 3/2 H2(g) H=69KJEq. 2:½ N2(g)+ 2H20(l)=>NO2(g)+2H2(g)
H=255KJ o Now add eq. 1 to eq. 2:o NH3(g)+½ N2(g)+ 2H20(l) ½ N2(g) +
3/2 H2(g) NO2(g)+2H2(g) 69KJ+255Kj
Hess’s Law
Chapter 5
NH3(g)+½ N2(g)+ 2H20(l) ½ N2(g) + 3/2 H2(g) NO2(g)+2H2(g) 69KJ+255Kj
Let’s cancel the N2s and add the H2s:
2H2O(l)+NH3(g)=>NO2(g)+7/2H2(g)
o Since the two steps complete the Rxn., we add the ∆Hs
Hess’s Law
Chapter 5
NH3(g)=> ½ N2(g) + 3/2 H2(g) H=69KJ
½ N2(g)+ 2H20(l)=>NO2(g)+2H2(g)
H=255KJ∆Hrxn= 69KJ+ 255KJ = 324KJ
Hess’s Law
Chapter 5
o It is a good thing to know how much energy is liberated when compounds are formed from their constituent elements
o This is called the enthalpy of formation ∆Hf
Enthalpy of Formation
Chapter 5
o To compare enthalpies, we need to evaluate the reactions at the same environmental conditions
o The is called the standard stateo The standard state is defined as 298K
and 1 atmosphere (101.3 KPa)o We denote the fact the enthalpy is
measured at standard conditions with a superscript ∆Ho
f
Enthalpy of Formation
Chapter 5
o Elements in their LOWEST ENERGY STATE have enthalpies of 0
o The lowest energy state for oxygen is O2 NOT O. Energy must be added (495 Kj/mol) to break the molecules, os the enthalpy is not 0
Enthalpy of Formation
Chapter 5
o Negative ∆Hof indicate that heat is
given off during formationo Positive ∆Ho
f indicate energy must be added during formation
o Compounds that require energy to be formed are those that will give off energy when decomposing (fuels)
Enthalpy of Formation
Chapter 5
o You can use ∆Hof to determine ∆Hrxn
o From the following heats of reaction2 SO2(g) + O2(g) 2 SO3(g)
∆H = – 196 kJ (a)2 S(s) + 3 O2(g) 2 SO3(g)
∆H = – 790 kJ (b)calculate the heat of reaction for
S(s) + O2(g) SO2(g) (c)
Enthalpy of Formation
Chapter 5
Enthalpy of Formation
Eq a 2 SO2(g) + O2(g) 2 SO3(g) has the 2 SO2(g) on the left. We can re-write the equation as the reverse reaction if we change the sign of the ∆H
2 SO3(g) 2 SO2(g) + O2(g)
∆H= +196KJ
Chapter 5
(a) 2 SO3(g) 2 SO2(g) + O2(g)
Eq b has the SO3 as well:
(b) 2 S(s) + 3 O2(g) 2 SO3(g)
Lets combine a & b (b first)2 S(s) + 3 O2(g) 2 SO3(g) (b)
2 SO3(g) 2 SO2(g) + O2(g) (a)
Cancel out the 2SO3s and one of the O2s gives
2S(s) + 2O2(g) 2SO2(g)
Enthalpy of Formation
Chapter 5
2S(s) + 2O2(g) 2SO2(g)
Eq c was S(s) + O2(g) SO2(g)
So that’s HALF of –a+b. Adding those ∆H and dividing by 2:
½ (196 + – 790 kJ) = -297KJ
Enthalpy of Formation
Chapter 5
o They are the same thing, the only difference is one goes into a gas tank, the other in a mouth.
o Determining the energy available from food or fuel should be easy for you now, we’ve done it for the past 20 slides.
Food & Fuels
