Download Ch. 5 --Thermochemistry (I)

Ch. 5 --Thermochemistry (I)
• In this chapter we will examine the
relationships between chemical reactions
and energy changes involving heat.
Ch. 5 --Thermochemistry (I)
•Energy: “The capacity to do work.”
–In Physics, there are 2 main types of energy…
•Kinetic (energy of motion) = ½ mv2
•Potential (energy of position due to gravity)= mgh
–In Chemistry, we usually concern ourselves
with the heat energy gained or lost during
chemical reactions.
Energy Units
• SI Unit for energy is the joule, J:
1 Joule=1 kg m2/s2
– We sometimes use the calorie instead of the
joule:
1 cal = 4.184 J (exactly)
– A nutritional Calorie: 1 Cal = 1000 cal = 1 kcal
(So your 1 Calorie soft drink is really 1000 “real” calories!)
Systems and Surroundings
• For thermochemical reactions, we will need to define the
following parts of the reaction…
– System: part of the reaction we are interested in.
– Surroundings: the rest of the universe.
• Example: If we are interested in the reaction between
hydrogen and oxygen in a cylinder, then the H2 and O2 in
the cylinder form a system, and everything else is the
surroundings.
Transferring Energy
There are 2 ways to transfer energy in or out of a
system…
1) Work…applying a force to an object and moving
it a distance (W=Fxd)
2) Heat…the energy transferred from hotter objects
to colder objects
•
Energy is therefore capacity to do work OR
transfer heat.
We will concern ourselves with the transfer of heat!
The 1st Law of Thermodynamics
• Energy cannot be created or destroyed.
– The first law of thermodynamics is really just the law
of conservation of energy.
– The energy of (system + surroundings) is constant.
– Thus any energy transferred from a system must be
transferred to the surroundings (and vice versa).
• Internal Energy: total energy of a system.
– You cannot measure “absolute internal energy”
so we measure the change in energy.
Change in internal energy…∆E= Ef - Ei
The 1st Law of Thermodynamics
• When a system undergoes a physical or
chemical change, the change in internal
energy is given by the heat added to or
absorbed by the system plus the work done
on or by the system:
∆E= q + w
(E= energy, q=heat, w=work)
Sign Convention for q and w
heat goes in…q is (+).
heat goes out…q is (−).
work goes in…w is (+).
work goes out…w is (−).
Endothermic and Exothermic Processes
• An endothermic process is one that absorbs heat from the
surroundings. (+q)
An endothermic reaction feels cold.
Example--an “instant” ice pack
• An exothermic process is one that transfers heat to the
surroundings. (-q)
An exothermic reaction feels hot.
Example--burning paper
State Functions
• A state function depends only on the initial
and final states of a system.
– Example: The altitude difference between Denver and Chicago
does not depend on whether you fly or drive, only on the
elevation of the two cities above sea level.
– Similarly, the internal energy of 50 g of H2O(l) at 25 ºC does
not depend on whether we cool 50 g of H2O(l) from 100 ºC to
25 ºC or heat 50 g of H2O(l) at 0 ºC to 25 ºC.
State Functions
• A state function does
not depend on how
the internal energy is
used.
– Example: A battery in a
flashlight can be discharged by
producing heat and light. The
same battery in a toy car
produces heat and work. The
change in internal energy of
the battery is the same in both
cases.
State Functions
• ∆E is a state function.
Enthalpy
•Enthalpy, H:
-Enthalpy is heat transferred between the system and
surroundings carried out under constant pressure. (Open
containers are under constant pressure!)
-Enthalpy is also a state function.
∆H is (+) for endothermic reactions…just like q!
∆H is (−) for exothermic reactions…just like q!
Enthalpy
 Chemical reactions can absorb or release heat.
 They also have the ability to do work.
 For example, when a gas is produced, the gas can
be used to push a piston, doing work.
 Zn(s) + 2H+(aq)  Zn2+(aq) + H2(g)
 The work performed by the above reaction is
called pressure-volume work.
 When the pressure is constant,
W = - P∆V
Chapter 5
Enthalpy(state function), H: Heat
transferred between the system and
surroundings carried out under constant
pressure
H  E  PV
If the process occurs at constant pressure
H   E  PV 
 E  PV
Chapter 5
Since we know that
w   PV
We can write
H  E  PV
 qP  w
From equations 5.9 (pg. 165) and equation 5.5 (pg.
159)
∆H = qp
Chapter 5
Enthalpies of Reactions
•Again, we can only measure the change in enthalpy, ∆H.
∆H = Hfinal – Hinitial
•For a reaction…
∆Hrxn = H(products) – H(reactants)
•The enthalpy change that accompanies a reaction is called the
enthalpy of reaction or heat of reaction (∆Hrxn).
•Consider the equation for the production of water:
2H2(g) + O2(g) -----> 2H2O(g) ∆Hrxn = –483.6 kJ
-The equation tells us that 483.6 kJ of energy are released to the
surroundings when water is formed & the reaction would feel hot.
-These equations are called thermochemical equations.
Enthalpies of Reactions
•Enthalpy is an “extensive property” which means it depends on the
amount of reactant you start with. (“Intensive properties” do not
depend on the quantity of the substance, i.e. – density.)
+ 2O2(g)  CO2(g) + 2H2O(g)
2CH4(g) + 4O2(g)  2CO2(g) + 4H2O(g)
- Example: CH4(g)
∆H = -802 kJ
∆H = -1604 kJ
•If you reverse the reaction, change the sign on ∆H.
-Example: CO2(g) + 2H2O(g)  CH4(g) + 2O2(g)
∆H = +802 kJ
•Enthalpy change for a reaction depends on the state of the reactants
and products.
-Example: CH4(g) + 2O2(g) ----> CO2(g) + 2H2O(l) ∆H = -890 kJ
If water vapor is the product, ∆H = -802 kJ because…
2H2O(l) -------> 2H2O(g)
∆H = +88 kJ
Calorimeters
•Calorimeters measure heat flow. It measures changes in
water temperature after a reaction is performed.
(Constant Pressure)
Bomb Calorimeter
Usually studies combustion (Constant Volume)
Measuring Heat, q
•Heat capacity is the amount of energy required to raise
the temperature of an object by 1 ºC.
•Molar heat capacity, (C) is the heat capacity of 1 mol
of a substance.
Cp=∆H/∆T
•Specific heat, (c), or specific heat capacity is the heat
capacity of 1 g of a substance.
• q = (grams of substance) x (specific heat) x ∆T.
q=mc∆T
c (water) = 4.184 J/g ºC or 1.0 cal/g ºC
Measuring Heat
•Practice Problems: How many joules of heat will raise the
temperature of exactly 50 g of water at 25 ºC to 75 ºC
q=mc∆T
q = (50 g) x (4.18J/g ºC) x (50 ºC) = 10450 J
*Note: q(solution) = − q(rxn)
•What is the molar heat capacity for water?
C = (4.18 J/g ºC) x (18.0 g/mole) = 75.2 J/mol ºC
So…
C = cxM
Hess’s Law
• If a reaction is carried out in a series of steps, ∆H for the reaction
is the sum of ∆H for each of the steps.
• The total change in enthalpy is independent of the number of
steps.
• Total ∆H is also independent of the nature of the path.
CH4(g) + 2O2(g) -----> CO2(g) + 2H2O(g)
∆H = –802 kJ
2H2O(g) ------> 2H2O(l)
∆H = –88 kJ
____________________________________________________
(Total) CH4(g) + 2O2(g) ------> CO2(g) + 2H2O(l)
∆H = –890 kJ
(Add)
• Hess’s Law provides a useful means of calculating energy
changes that are difficult to measure…(in this case it’s getting
liquid water to form instead of water vapor.)
Enthalpies of Formation
•If a compound is formed from its constituent elements, then the
enthalpy change for the reaction is called the enthalpy of
formation, ∆Hf .
Example: C(s) + O2(g)  CO2(g)
∆Hf = −393.5 kJ
Standard state (standard conditions) refer to the substance at:
1 atm and 25ºC (298 K)
(We will assume that reactants and products are both at 25 ºC unless otherwise stated.)
•Standard enthalpy, ∆Hº, is the enthalpy measured when
everything is in its standard state.
•Standard enthalpy of formation of a compound, ∆Hºf , is the
enthalpy change for the formation of 1 mole of compound with all
substances in their standard states.
Enthalpies of Formation
•If there is more than one state for a substance under
standard conditions, the more stable one is used.
-Example: When dealing with carbon we use graphite because
graphite is more stable than diamond or C60.
•By definition, the standard enthalpy of formation of the
most stable form of an element is zero. Why?
-Because there is no formation reaction needed when the element
is already in its standard state.
∆Hºf C(graphite) = zero
∆Hºf O2(g) = zero
∆Hºf Br2(l) = zero
Using ∆Hºf to calculate ∆Hºrxn
• In
general:
∆Hºrxn =  n∆Hºf (products) –  m∆Hºf (reactants)
Where n and m are the coefficients from the balanced
chemical equation.
Let’s do one for practice!!!
Using ∆Hºf to calculate ∆Hºrxn
•Example: Use Hess’s Law to calculate ∆Hºrxn for…
C3H8(g) + 5O2(g) -------> 3CO2(g) + 4H2O(l)
• Step 1, the products…Form 3CO2 and 4H2O from their elements:
3C(s) + 3O2(g) ------>3CO2(g)
∆H1 = 3∆Hºf[CO2(g)]…tripled it!
4H2(g) + 2O2(g) ------> 4H2O(l)
∆H2 = 4∆Hºf[H2O(l)]…quadrupled it!
• Step 2, the reactants…(Note: O2 has no enthalpy of formation since
it is in the elemental state, so we concern ourselves with C3H8.)
3C(s) + 4H2(g) ------> C3H8(g)
∆H3 = ∆Hºf [C3H8(g)]
• Step 3, Now look up the values on p.177 and do the math!
∆Hºrxn = ∆Hºf (products) – ∆Hºf (reactants)
∆Hºrxn = [3(–393.5 kJ) + 4(–285.8 kJ)] – [(– 103.85 kJ)]
∆Hºrxn = –1180.5 –1143.2 + 103.85 = –2220 kJ
Heat of Fusion & Heat of Vaporization
• Molar heat of fusion (∆Hfus) : the amount of energy required to take
1 mole of a solid to the liquid state.
Example: H2O(s)  H2O(l)
∆Hfus = 6.01 kJ
-Heat of fusion is usually greater for ionic solids than molecular
solids since ionic solids are more strongly held together.
For ice to water … ∆Hfus= 6.01 kJ/mol
• Molar heat of vaporization (∆H vap): the amount of energy required
to take 1 mole of a liquid to the gaseous state.
Example: H2O(l)  H2O(g) ∆Hvap = 40.67 kJ
For water to steam … ∆Hvap= 40.67 kJ/mol
These values are mainly used as conversion factors!