Download Chapter 8 Thermochemistry

Chapter 8 Thermochemistry
Thermochemistry refers to the study of the heat flow that accompanies chemical
reactions.
Heat is a particular form of energy that is transferred from a body at a high temperature
to one at a lower temperature when they are brought into contact with each other.
8.1 Principles of Heat Flow
The system is the part of the universe upon which attention is focused. It could be as
simple as a sample of water on a “hot plate.”
The surroundings, which exchange energy with the system, comprise in principle the
rest of the universe. It includes only those materials in close contact with the system.
ex) Figure 8.1 pg 209
The surroundings consist of
1.) The hot plate
2.) The beaker
3.) The air
A. State Properties
The state of a system is described by giving to composition, temperature, and
pressure.
ex) system
50.0 g of H20 at 50.0 0C and 1 atm
State Properties depend only on the state of the system, NOT upon the way the system
reached that state.
ΔX= X final – X initial
Heat flow is not a state property.
B. Direction and sign of heat flow
q is positive when heat flows into the system from the surroundings
endothermic- q>0 (melting of ice)
q is negative when heat flows out of the system into the surroundings
exothermic- q< 0 (Bunsen burner)
C. Magnitude of heat flow
The amount of energy (q) is cited in J or KJ. * 1 KJ = 10 3 J
The energy (heat) required to change the temperature of a substance depends on
1.) The amount of substance (grams)
2.) The temperature change
3.) Identity of the substance ( different substances respond differently to being
heated)
Specific Heat Capacity- the amount of energy required to change the temperature of one
gram of a substance by 10C. (Water 4.184 J/g 0C) (pg 211)
q= cmΔt
ex) how much heat is given off by a 50.0 g sample of Cu when it cools from 80.0 0C to
50.0 0C?
q=cm Δt
= (0.382 J/g 0C) (50.0 g) (-30.0 0C)
= -573 J
ex) Suppose 652 J of heat is added to 15.0 g of water, originally of 20.0 0C. Find final
temperature?
q=cm Δt
Δt=q/cm
=652J/ (4.184 J/g 0C) (15.0g)
=10.4 0C
tf =20.0 0C + 10.4 0C
=30.4 0C
8.2 Measurement of Heat Flow: Calorimeter
Calorimeter is a device used to measure heat flow. The walls are insulated so there is no
exchange of heat with the air outside. It follows that the only heat flow is between the
reaction system and the calorimeter, thus
q reaction = -q calorimeter
A. Coffee- Cup Calorimeter
q reaction = -M water x 4.184 J/g 0C x Δt
ex) When 1.00g of NH4NO3 dissolves in 50.0 g of water in a coffee- cup calorimeter the
following reaction occurs:
NH4NO3  NH4 + NO3And the temperature of the water drops from 25.00 0C to 23.32 0C. Assuming that all the
heat absorbed by the reaction comes from the water; calculate q for the reaction system.
ΔT= t final-t initial
= 23.32 -25.00
= -1.68 0C
Q reaction= -M (water) x 4.184 J/g 0C x Δt
=-50.0g x 4.184 J/g 0C x -1.68 0C
= 351 J
B. Bomb Calorimeter
The heat flow, q, for a reaction is calculated from the temperature change multiplied
by the heat capacity of a calorimeter, which is determined in a preliminary
experiment.
q reaction = -0Ccal x Δt
see example 8.3 page 214
HWK, pg 233- 234, 2, 4, 6, 8, 10, 12
8.3 Enthalpy
Enthalpy: a property that reflects its capacity to exchange heat (q) with the
surroundings; defined so that H=q for a constant pressure process.
In other words, at constant pressure, the heat flow for the reaction system is equal to
the difference in enthalpy (H) between products and reactants. Enthalpy is a type of
chemical energy or “heat content.”
Q reaction = ΔH = H products –H reactants
At constant pressure
1.) In an exothermic reaction, the products have a lower enthalpy then the reactants;
thus ΔH is negative, and heat is given off to the surroundings.
q = ΔH<O
H products < H reactants
2.) In an endothermic reaction, the products have a higher enthalpy then the reactants
so OH is positive, and heat is absorbed from the surroundings.
q= ΔH>O
H products > H reactants
8.4 Thermochemical equations
Thermochemical equations: a chemical equation that shows the enthalpy relation
between products and reactants.
The appropriate value and sign for ΔH will be found on the right.
Consider: When one gram of NH4NO3, q reaction= 351 J
Thus when one mole (80.05 g) dissolves 0.351 KJ/g * 80.05 g = 28.1 KJ
ΔH=28.1 KJ
Thus: NH4NO3(s) ----- NH4 (aq) + NO3 (aq)
Consider: H2 (g) + Cl2 (g) ----- 2HCL (g)
When one gram of H2 reacts, q reaction= - 91.6 KJ
Thus, one mole (2.02 G) reacts
-91.6 KJ/g (2.02 g/mol) = -185 KJ/mol
H2 (g) + Cl2 (g) ----- 2HCL (g)
ΔH= -185 KJ
Remember
1.) Sign of ΔH indicates
endothermic ΔH is +
exothermic
ΔH is –
2.) Coefficients represent the number of moles
3.) The phases (s), (l), (g), (aq) of all species must be specified
4.) The value quoted for ΔH applies when products and reactants are at the same
temperature, usually 250C.
Rules of Thermochemistry
1.) The magnitude of ΔH is directly positioned to the amount of reactant or product.
Ex) boil samples of H20
H2(g) + Cl2(g)  2HCL(g)
ΔH= -185 KJ
-185 KJ/ 1 mol H2 or 1mol Cl2/ -185 KJ or -185 KJ/ 2 mol HCL or ect….
See examples 8.4 and 8.5 pg 217
Heat of Fusion: heat absorbed when a solid melts
Heat of Vaporization: heat absorbed when a liquid vaporizes
2.) ΔH for a reaction is equal in magnitude but opposite in sign to H for the reverse
reaction.
Ex) if 6.00 KJ of heat is absorbed when a mole of ice melts, then 6.00 KJ of heat
should be evolved when a mole of liquid water freezes.
See example 8.6 pg 218
3.) The value of ΔH for a reaction is the same whether it occurs in one step or in a
series of steps.
Hess’ Law: A relation stating that the heat flow in a reaction that is the sum of the
two other reactions is equal to the sum of the heat flows in these two reactions.
ΔH= ΔH1 + ΔH2+….
See example 8.7 pg 219
Hwk 18, 20, 22, 28, 32, 34
8.5 Enthalpies of Formation
ΔH can be calculated for any reaction, using quantities known as enthalpies of
formation.
Meaning = Δ H0f
ΔH0f , standard molar enthalpy of formation of a compound is equal to the
enthalpy change when one mole of the compound is formed at a constant pressure of
1 atm and a fixed temperature (250C) from the elements.
Ex) Ag (s, 250C) + ½ Cl2 (g, 250C, 1atm)------AgCl (s, 250C)
ΔH =-127.1 KJ
Thus, H0f AgCl(s) = -127.1 KJ/mol
See Table 8.3 page 221
Most enthalpies are exothermic thus negative quantities
Note:
ΔH0f BR2(l) = 0
ΔH0f O2 (g) = 0
Calculation at ΔH0
The standard enthalpy change, ΔH0 , for a given Thermochemical equation is equal to
the sum of the standard enthalpies of formation of the product compounds minus the
sum of the standard enthalpies of formation of the reactant compounds.
ΔH0 =
∑H0f products ---
∑H0f reactants
Coefficients must be taken into account
aA(g) + bB(g)------- cC(g) + dD(g)
∑ΔH0f products = c ΔH0f C(g) + d Δ H0f D(g)
∑ΔH0f reactants = a Δ H0f A(g) + b ΔH0f B(g)
Thus
ΔH0 = c ΔH0f C(g) + d ΔH0f D(g)-[ a ΔH0f A(g) + b ΔH0f B(g)]
ex) Calculate ΔH0 for the combustion of one mole of propane, C3H8, according to
the equation
C3H8 (g) + 5O2(g) 3CO2 (g) + 4H2O(l)
ΔH0 = 3 ΔH0fCO2(g) + 4 Δ H0f H2O(l) –( ΔH0f C3H8 (g) + 5 ΔH0f O2(g))
ΔH0= 3 mol (-393.5 KJ/mol) + 4mol (-285.8 KJ/mol) –( 1 mol(-103.8 kj/mol)+ 5mol(0))
ΔH0= -2219.9 KJ
ex) the combustion of benzene is
C6H6 + 15/2 O2(g)  6CO2(g) + 3H2O(l)
ΔH0=-3267.41 KJ
Use table 8.3 to obtain heats of formations for CO2 and H2O, calculate the heat of
formation of benzene, C6H6.
ΔH0 = 6 ΔH0fCO2(g) + 3 ΔH0f H2O(g)- ΔH0f C6H6(l)
-3267.41 KJ= 6 mol (-393.5 KJ/mol) + 3mol (-285.8 KJ/mol) –( 1 mol Δ H0f C6H6(l))
ΔH0f C6H6(l) = 49.0 KJ/mol
Enthalpies of Formation of Ions in Solution
The enthalpies of formation of ions in water solution is done exactly the same way as
above with the exception that ΔH0f H+= 0. Table 8.4 will also be utilized. (pg 224)
Ex) For the reaction that occurs when HCL is added to water, H0 is found to be -74.9
KJ. Find ΔH0f Cl- (aq)
HCl(g)  H+(aq) + Cl-(aq)
ΔH0= -74.9 KJ
ΔH0= ΔH0f H+ (aq) +
ΔH0f Cl-(aq)- ΔH0f HCl(g)
-74.9 KJ = 0 + ΔH0f Cl-(aq) - (1 mol (-92.3 KJ/mol)
-74.9 KJ = ΔH0f Cl-(aq)+ 92.3 KJ
ΔH0f Cl-(aq)= -167.2 KJ
ex) When hydrochloric acid is added to a solution of sodium carbonate, carbon
dioxide gas is formed. The equation for the reaction is
2 H+(aq) + CO3-2(aq)  CO2(g) + H2O(l)
Calculate
ΔH0 for this Thermochemical equation.
ΔH0f= ΔH0f CO2(g) + Δ H0f H2O(l)-[2 ΔH0f H+(aq) + ΔH0f CO3-2(aq)]
ΔH0= 1 mol (-393.5 KJ/mol) + 1mol (-285.8 KJ/mol) –1 mol(-677.1g/mol))
ΔH0= -2.2 KJ
Hwk pg 236 Que 38, 42, 46, 48
8.6 Bond Energy
Bond energy is defined as ΔH when one mole of bonds is broken in the gaseous
state.
H2(aq)------2H(g)
ΔH=436KJ
Cl2(g)
ΔH= 243 KJ
Bond energies are listed in Table 8.5 page 225
When chemical bonds are broken, bond energy is +
When chemical bonds are formed, bond energy is –
H(aq) + Cl(g) ----- HCL(g)
ΔH= -431 KJ
If the bonds in the reactants are stranger than those in the products the reaction is
endothermic.
If the strong bonds are formed at the expense of weak bonds the reaction is
exothermic.
Multiple bonds have a greater number of bonding electrons thus a greater bond
energy.
Bond energy C-C= 347 KJ/mol
Bond energy C= C = 612 KJ/mol
Bond energy C≡ C = 820 KJ/mol
8.7 The First Law of Thermodynamics
Thermochemistry: study of heat flow
Thermodynamics: deals with all kinds of energy effects in all kinds of process; deals
with heat and work (w)
Work includes all forms of energy except heat.
Remember energy (E) can either be neither created nor destroyed.
E system = - ΔE surroundings
1st Law of Thermodynamics
In any process, the total change in energy of the system, E, is equal to the sum of the
heat absorbed q, and the work, w, done on the system.
ΔE=q + w
q and w are positive when heat or work enters a system from the surroundings.
q and w are negative when heat or work leaves a system to the surroundings.
ex) Calculate E of a gas for a process in which the gas
a.) Absorbs 20J of heat and does 12J of work by expanding.
b.) Evolves 30J of heat and has 52J of work done on it as it contracts.
a.) q= +20J
w=-12J
ΔE= 20J -12J
=8J
b.) q=-30J
w= +52J
ΔE=-30J+52J
=22J
Expansion Work
When a chemical reaction is carried out directly in the laboratory, only one type of
work is involved: work expansion (or contraction) done by a gas against a restraining
pressure (atmosphere pressure)
w= -P ΔV= - Δng x RT
Where Δng is the change in the number of moles of gas in the process and R= 8.31
J/mol x K
The quantities q and w are not stable properties. Their value depends upon the path
by which a process is carried out. Thus at
1.) Constant volume
*no change in volume, thus, no expansion work associated with constant volume
wv=0
subscript v indicates constant volume
ΔE=qv +wv
ΔE= qv
2.) Constant pressure
*the heat flow for a constant pressure process is
wp= - ΔngRT
qp= ΔE-wp
qp= ΔE + ΔngRT
thus
qp= qv + ΔngRT
ex) Calculate the differences in KJ between qp and qv for the decomposition of one
mole of ammonium chloride at 25 0C .
NH4CL(s)-----NH3(g) + HCL(g)
ΔngRT
Δng= 2mol -0= 2mol
ΔngRT= (2mol)(8.31 J/mol x K)(298 K)
=4.95 x 103 J
=4.95 KJ
Hwk pg 236- 237 38, 42, 46, 48, 56, 58