Download Units of Energy Energy in Thermochemistry Thermochemistry

Units of Energy
Energy in Thermochemistry
Like we saw with pressure, many different units are used
throughout the world for energy.
SI unit for energy
1J =
1 kg ⋅ m 2
s2
Joule (J)
calorie (cal)
erg (erg)
electron volts (eV)
British thermal unit (Btu)
100º C
Amount of
energy it takes
to raise one
gram of water
1ºC
4.184 Joules (J) = 1 calorie (cal)
1 kcal = 1000 calories = 1 Nutritional Calorie (Cal)
Thermochemistry
In thermochemistry, the
universe is divided into two
parts:
The system: The
physical process or
chemical reaction in
which we are interested.
We can define the
system anyway we like.
The surroundings:
Everything else in the
universe.
Heat (q): energy transferred from an object
with a higher temperature to an object with a
lower temperature
50º C
heat
Heat cannot be transferred in the opposite
direction!
Vroom...
Work (w): energy transferred
against a force
Example: Combustion reactions allow
your car to move...performing work
Energy flow between system and
surroundings
Surroundings
Surroundings
System
System
Surroundings
Surroundings
Universe
System +
Surroundings
copper
After I add 2.0kJ of heat energy to each sample, will they
still be the same temperature?
5.0 g Al
5.0 g Cu
25°C
25°C
Energy (both heat
and work) can be
transferred between
the system and
surrounds.
The sign “+” or “” tells us the
direction of
energy flow.
Surroundings
Surroundings
_
+
System
System
Surroundings
Surroundings
Calorimetry
The energy released from the reaction
(or whatever is in the calorimeter) is
gained by the water. So…
-qrxn = qwater
Notice that they have opposite signs, which tells us the
direction of transfer.
However, assuming no heat loss, the magnitudes are
the same. The amount of heat lost by the reaction is
the same amount gained by the water.
Example
Exothermic vs. Endothermic processes
99.8oC
• A 28.2 gram sample of nickel is heated to
and placed
in a coffee cup calorimeter containing 150.0 grams of water
at 23.5oC. After the metal cools, the final temperature of the
metal and water is 25.0oC.
Feels warm to
the touch
• Which substance absorbed heat?
q= -
• Which substance released heat?
Feels cool to the
touch
• Calculate the heat absorbed by the substance you indicated
above.
System:
Surroundings
Exothermic
Endothermic
chemical
reaction
or process
System
q=+
Surroundings
Enthalpies of Reaction
Surroundings
Heat and Enthalpy
• Determine if the following processes are
endothermic or exothermic…
So how are q and ∆H related?
At constant pressure:
– Combustion of methane
– Neutralization of HCl
– Boiling
– Melting
– CaCO3 (s) CaO (s) + CO2 (g)
qp = ∆H
So, using a calorimeter, the ∆H of a chemical
reaction can be calculated from the change in
heat in the calorimeter.
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Enthalpies of Physical/Chemical Changes
Enthalpy (∆H) describes the change in heat
from a chemical reaction or process.
∆H = Hproducts – Hreactants
Where do these
energies come from?
Energy is released if:
∑Energyreactants > ∑Energyproducts
Energy is absorbed if:
∑Energyreactants < ∑Energyproducts
Thermodynamic Standard State
∆H˚
The “˚” (knot)
denotes enthalpy
under standard
states
Standard States
– A gas at 1 atm
– An aqueous solution (1 M conc.) at a pressure of
1 atm
– Pure liquids and solids
– The most stable form of elements at 1 atm and
25oC (298 K)
Direction of reaction and ∆H
2H2O(g) → 2H2(g) + O2(g)
• Heat of fusion (∆Hfus): Amount of heat
required to melt (solid liquid)
• Heat of vaporization (∆Hvap): Amount of heat
required to evaporate (liquid gas)
• Heat of sublimation (∆Hsub): Amount of heat
required to sublime (solid gas)
Why are there no values for ∆Hfreezing,
∆Hcondendsation, or ∆Hdeposition?
Amount of reactant, and ∆H
In chemistry, we usually specify energy in terms of a
specific amount of reactant or product. For example:
2H2(g) + O2(g) → 2H2O(g)
∆H =-241.8 kJ
Which means if we react 2 mole of H2 with 1
mole of O2, 241.8kJ of heat energy will be
released.
This can be
read as:
But what if we react 4 moles of H2 with 2
moles of O2?
-241.8 kJ per 2
moles H2
∆H =-483.6 kJ
-241.8 kJ per 2
moles of H2O
-241.8 kJ per 1
mole O2
Thermochemical Equations and Stoichiometry
What about the reverse reaction of:
2H2(g) + O2(g) → 2H2O(g)
Enthalpies of Phase Changes
• 2Al (s) + Fe2O3 (s) 2Fe (s) + Al2O3 (s)
∆Ho = -852 kJ
∆H =-241.8 kJ
∆H =+241.8 kJ
All we do is change the sign. Instead of energy
being released (as in the first reaction), energy
is absorbed when breaking water molecules
apart.
• How much heat is released if 10.0 grams of
Fe2O3 reacts with excess Al?
– 10.0 g Fe2O3 x mol Fe2O3
– x mol Fe2O3 852 kJ / 1 mol Fe2O3
Hess’s Law
Hess’s Law
• If a compound cannot be directly synthesized
from its elements, we must use multiple
reactions to calculate the enthalpy of reaction
∆H is a state function…it doesn’t matter what
route we take to get there. Same change in
energy, ∆H.
These two
steps (solid →
liquid → gas)
should have the
same change in
energy as…
• Hess’s Law: change in enthalpy is the same
whether the reaction occurs in one step or in
a series of steps
• Look at direction of reaction and amount of
reactants/products
…this one step
(solid → gas)
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Hess’s Law
Hess’s Law
• Does the energy change in a reaction depend
on the number of steps in the reaction?
∆H = 53.00 kJ
• H2(g) + I2(s) → 2HI(g)
I2(s) → I2(g)
∆H = 62.44 kJ
• Values of enthalpy change
– For a reaction in the reverse direction, enthalpy is
numerically equal but opposite in sign
• Reverse direction, heat flow changes; endothermic
becomes exothermic (and vice versa)
– Proportional to the amount of reactant consumed
• H2(g) + I2(g) → 2HI(g)
• Twice as many moles = twice as much heat
– Depends on the physical state of the reactants
and products (i.e., elemental state)
∆H = ?
• How can we add the first two reactions to get
the third?
• Hess’s Law: ∆Hsum of steps = ∆H1 + ∆H2
21
Application of Hess’s Law
22
Hess’ Law Problem
• We can use known values of ∆Ho to calculate
unknown values for other reactions
• P4 (s) + 3 O2 (g) P4O6 (s)
∆H = -1640.1 kJ
• P4 (s) + 5 O2 (g) P4O10 (s) ∆H = -2940.1 kJ
• Given:
• CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g)
• H2O (l) H2O (g)
• What is ∆Ho for the following reaction?
P4O6 (s) + 2 O2 (g) P4O10 (s)
∆H = ?
∆H = -802 kJ
∆H = 88 kJ
• Find the enthalpy for the following reaction:
• CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H = ?
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Standard Enthalpies of Formation
Standard Enthalpies of Formation
• Can use measured enthalpies of formation to determine the
enthalpy of a reaction (use Appendix B in back of book)
• Standard enthalpy of formation (∆Hof): heat needed
to make 1 mole of a substance from its stable
elements in their standard states
∆Horxn = Σn∆Hof (products) – Σn∆Hof (reactants)
− Σ = sum
– n = number of moles (coefficients)
∆Hof = 0 for a stable (naturally occurring) element
• Which of these have ∆Hof = 0?
– CO(g), Cu(s), Br2(l), Cl(g), O2(g), O3(g), O2(s)
• Do the following equations represent standard
enthalpies of formation? Why or why not?
– 2Ag (l) + Cl2 (g) 2AgCl (s)
– Ca (s) + F2 (g) CaF2 (s)
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Heats of Formation
Group Work
∆Horxn
= Σ n∆Hof,products - Σ n∆Hof,reactants
Calculate values of ∆Ho for the following
• Use Standard Heat of Formation values to calculate the
enthalpy of reaction for:
• C6H12O6(s) 2 C2H5OH(l) + 2 CO2(g)
•
rxns:
• 1) CaCO3 (s) CaO (s) + CO2 (g)
• 2) 2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
∆Hof (C6H12O6(s)) = -1260.0 kJ/mol
∆Hof (C2H5OH(l)) = -277.7 kJ/mol
∆Hof (CO2(g)) = -393.5 kJ/mol
∆H˚ of (in kJ/mol):
CaCO3 (-1207.1)
CaO (-635.5)
CO2 (-393.5)
C6H6 (49.0)
CO2 (-393.5)
H2O (-285.8)
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Bond Dissociation Energies
Bond Energy
• Bond dissociation energy (BDE). Energy needed to
break one 1 mol of a specific bond (eg. H-H, C-H, C=C,
C-C, etc.)
• Bond dissociation energy varies somewhat from one
molecule to another, or even within one molecule, so in
most cases, we have use an average bond energy (D).
•
•
•
•
•
For example:
H-OH
H-O
H-OOH
Average =
502 kJ/mol
427 kJ/mol
431 kJ/mol
453.3 kJ/mol for O-H
• Note: there are certain bonds where the BDE is not and average.
These are bonds that only exist within diatomic molecules (eg. H-H,
H-Br, Cl-Cl, etc.)
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Bond Dissociation Energies
∆Horxn
Enthalpy of Reaction
= ΣBDE (reactants) + - ΣBDE (products)
endothermic
energy input
• Use bond energies to calculate the enthalpy change for
the following reaction:
exothermic
energy released
NΞN = 945 kJ/mol
• N2(g) + 3H2(g) → 2NH3(g)
• Or: ∆Horxn = ΣBDE (reactants) – ΣBDE (products)
H-H = 436 kJ/mol
N-H = 390 kJ/mol
∆Hrxn = [BEN ≡ N + 3BEH-H] – [6BEN-H]
∆Hrxn = [945 + 3(436)] – [6(390)] = -87 kJ
ΣBDE(react) > ΣBDE(prod) endothermic
ΣBDE(react) < ΣBDE(prod) exothermic
• measured value = -92.2 kJ
• We use only when heats of formation are not available, since
bond energies are average values for gaseous molecules
• Why are the calculated and measured values different?
31
Enthalpy of Reaction
32
Practice Problems
• Which amount of energy is higher?
– Energy that is released when 55g of Al is cooled
from 5˚C to -25 ˚C
– Energy that is absorbed when 64g of Cu is heated
from 298K to 315K?
• Use bond energies to calculate the enthalpy change for
the decomposition of nitrogen trichloride:
NCl3 (g) N2 (g) + Cl2 (g) (hint: balance first!!)
∆Hrxn?
N-Cl = 200 kJ/mol
N≡N = 945 kJ/mol
Cl-Cl = 243 kJ/mol
• 8750 J of heat are applied to a 170 g sample of
metal, causing a 56oC increase in its
temperature. What is the specific heat of the metal?
Which metal is it?
33
Practice Problems
Practice Problems
• Practice: Worked Example 8.10, page 320….
• Identify how to set up the following problems:
• Calculate the ∆Ho of reaction for:
– C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l)
∆Hof C3H8(g): -103.95 kJ/mol; ∆Hof CO2(g): -393.5 kJ/mol;
∆Hof H2O(l): -285.8 kJ/mol
• 8750 J of heat are applied to a 170 g sample of metal,
causing a 56oC increase in its temperature. What is the
specific heat of the metal? Which metal is it?
• C2H4(g ) + 6F2(g) 2CF4(g) + 4HF(g) ∆Ho = ?
– H2 (g) + F2 (g) 2HF (g) ∆Ho = -537 kJ
– C (s) + 2F2 (g) CF4 (g) ∆Ho = -680 kJ
– 2C (s) + 2H2 (g) C2H4 (g)∆Ho = 52.3 kJ
• Use average bond energies to determine the enthalpy of
the following reaction (from Table 7.1).
– CH4 (g) + Cl2 (g) CH3Cl (g) + HCl (g)
– (BEC-Cl = 328 kJ/mol)