Transcript Thermochemistry - HCC Learning Web

Thermochemistry
Chapter 6
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Energy is the capacity to do work or transfer heat
•
Thermal energy is the energy associated with
the random motion of atoms and molecules
•
Chemical energy is the energy stored within the
bonds of chemical substances
•
Potential energy is the energy available by virtue
of an object’s position; For an object of mass m at
a height h, PE = mgh; g = acceleration due to
gravity ( a constant)
6.1
Energy Changes in Chemical Reactions
Heat is the transfer of thermal energy between two bodies that
are at different temperatures.
Temperature is a measure of the thermal energy.
Temperature = Thermal Energy
900C
400C
greater thermal energy
6.2
Thermochemistry is the study of heat change in chemical
reactions.
The system is the specific part of the universe that is of interest in
the study. That is that part of the universe undergoing the process.
Process: Chemical Reaction; System: The reactants, products
open
Exchange: mass & energy
closed
isolated
energy
nothing
6.2
Exothermic process is any process that gives off heat –
transfers thermal energy from the system to the surroundings.
2H2 (g) + O2 (g)
H2O (g)
2H2O (l) + energy
H2O (l) + energy
Endothermic process is any process in which heat has to be
supplied to the system from the surroundings.
energy + 2HgO (s)
energy + H2O (s)
2Hg (l) + O2 (g)
H2O (l)
6.2
Exothermic
Endothermic
6.2
Thermodynamics is the scientific study of the
interconversion of heat and other kinds of energy.
State functions are properties that are determined by the state
of the system, regardless of how that condition was achieved.
energy, pressure, volume, temperature
DE = Efinal - Einitial
DP = Pfinal - Pinitial
DV = Vfinal - Vinitial
DT = Tfinal - Tinitial
Potential energy of hiker 1 and hiker 2 is the same (if they have
the same mass =mgh) even though they took different paths.
6.3
First law of thermodynamics – energy
can be converted from one form to another,
but cannot be created or destroyed.
Change in energy
of system = change in energy
of surroundings.
DEsystem + DEsurroundings = 0 The energy of the
universe is a
or
constant.
DEsystem = -DEsurroundings The energy of any
isolated system is
also constant.
The change in energy of universe = 0
Change in energy of system + change in
energy of surroundings = 0
C3H8 + 5O2
3CO2 + 4H2O
Exothermic chemical reaction!
Chemical energy lost by combustion = Energy gained by the surroundings
system
surroundings
6.3
Another form of the first law for DEsystem
Energy of a system = its internal energy, E
DE = q + w When the energy of the system changes, it can take the form of
heat or work or both.
DE is the change in internal energy of a system
q is the heat exchange between the system and the surroundings
w is the work done on (or by) the system
w = -PDV when a gas expands against a constant external pressure
This implies that if there is no gas involved, w = 0. Also, if there is no
change in volume, w= 0. When a gas expands, system does work; when a gas
is compressed, the surroundings does work.
6.3
1. Calculate the change in internal energy of the system when it undergoes
an exothermic process releasing 140 J of heat and a gas is evolved which
does 85 J of work during expansion. q = -140 J; w = -85 J; ∆E = -140 J + 85 J = -225J
2. A balloon is heated by adding 240 J of heat. It expands doing 135 J of
work. Calculate the change in internal energy of the balloon. q = 240J; w =
-135 J; ∆E = 240 J + -135 J = 105 J
3. 50.0 g of Fe(s) is cooled from 100C to 90C losing 225 J of heat.
Calculate the change in internal energy of the Fe(s). q = -225J; w = 0 J;
∆E = -225J + 0 = -225 J
4. A gas does 135 J of work while expanding and at the same time absorbs
156J of heat. What is the change in internal energy? ∆E = 21 J
5. Calculate the change in energy of the system for a process in which the
system absorbs 140 J of heat from the surroundings and does 85J of work
on the surroundings. ∆E = 55 J
6. For a particular process q = - 17 kJ and w = 21 kJ. Which of the following
statement is false? A. Heat flows from the system to the surroundings
B. The system does work on the surroundings; C. ∆E = + 4 kJ ;
D. The process is exothermic; E. None of the above is false
∆E = q = heat change when w =0 (no gases; volume is a constant)
Enthalpy, H and the First Law of Thermodynamics
H = E +PV
Recall, DE = q + w
At constant pressure:
q = DH why?
DH = DE + P DV = q + w + P DV
= q + -P DV + P DV = q
For constant pressure processes, enthalpy change = heat
change
Endothermic process: q > 0; DH > 0
Exothermic process: q < 0; DH < 0
Since E, P & V are state functions, so is H.
Enthalpy of a system: Measure of the heat stored in the
system.
6.4
Enthalpy (H) is used to quantify the heat flow into or out of a
system in a process that occurs at constant pressure.
DH = H (products) – H (reactants)
DH = heat given off or absorbed during a reaction at constant pressure
Hproducts > Hreactants
DH > 0
Hproducts < Hreactants
DH < 0
6.4
Properties of Enthalpy Change, DH
•Enthalpy changes are extensive quantities. They depend on the
amount of matter undergoing the process
•The enthalpy change of the reverse of a process is equal in
magnitude but opposite in sign to the enthalpy change of the
process. Enthalpy of vaporization = -enthalpy of condensation;
enthalpy of melting = -enthalpy of freezing
•Thermochemical equations: Balanced equations for a chemical
reaction along with associated enthalpy change specified.
Physical states must be included in all thermochemical equations
as enthalpy changes are state functions and therefore dependant
on the physical states.
•Examples: H2 (g) +(1/2) O2(g)  H2O(g) DH = -241.82 kJ
Interpretation: 1 mole of H2(g) used = 0.5 mole 02(g) used = 1
mole water vapor formed = 241.82 kJ of heat released.
•H2 (g) +(1/2) O2(g)  H2O(l) DH = -285.83 kJ
Interpretation: 1 mole of H2(g) used = 0.5 mole 02(g) used = 1
mole liquid water formed = 285.83 kJ of heat released.
Thermochemical Equations
Is DH negative or positive?
System absorbs heat
Endothermic
DH > 0
6.01 kJ are absorbed for every 1 mole of ice that
melts at 00C and 1 atm.
H2O (s)
H2O (l)
DH = 6.01 kJ
6.4
Thermochemical Equations
Is DH negative or positive?
System gives off heat
Exothermic
DH < 0
890.4 kJ are released for every 1 mole of methane
that is combusted at 250C and 1 atm.
CH4 (g) + 2O2 (g)
CO2 (g) + 2H2O (l) DH = -890.4 kJ
6.4
Thermochemical Equations
•
The stoichiometric coefficients always refer to the number
of moles of a substance
H2O (s)
•
DH = 6.01 kJ
If you reverse a reaction, the sign of DH changes
H2O (l)
•
H2O (l)
H2O (s)
DH = -6.01 kJ
If you multiply both sides of the equation by a factor n,
then DH must change by the same factor n.
2H2O (s)
2H2O (l)
DH = 2 x 6.01 = 12.0 kJ
6.4
1. What is the enthalpy change when 4.00mol of CH4 is
combusted?
2. What is the enthalpy change when 4.00g of CH4 is
combusted?
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) ∆H = -890.4 kJ
1.
1 mole of CH4 is combusted = 890.4 kJ released
4 mole of CH4 is combusted = ? kJ released
4mole x (890.4 kJ released/1mol) = 3562 released
Enthalpy change = ∆H = -3562 kJ
2. Molar mass of CH4 = 16 g/mol
1 mol CH 890.4 kJ released
4
4.00g CH 
 223 kJ released
4 16 g CH
1 mol CH
4
4
enthalpy change, DH  - 223kJ
3. If 150 kJ of heat had been released, what is the mass of CH4
combusted?
According to thermochemical equation, 890.4 kJ released = 1mole
of CH4 combusted = 16 g of CH4 combusted
Therefore 150 kJ of heat released = ? g CH4 combusted
150 kJ released 
16 g CH 4 combusted
 2.70 g
890.4 kJ released
How much heat is evolved when 266 g of white phosphorus (P4)
burn in air? Molar mass of P4 = 123.9 g/mol
According to equation 1mol P4 combusted = 123.9g P4
combusted = 3013 kJ released.
Therefore when 266g P4 combusted, heat released =
3013 kJ released
266 g P4 
 6470 kJ released
123.9 g P4
 DH  6470kJ
Thermochemical Equations
•
The physical states of all reactants and products must be
specified in thermochemical equations.
H2O (s)
H2O (l)
DH = 6.01 kJ
H2O (l)
H2O (g)
DH = 44.0 kJ
How much heat is evolved when 266 g of white
phosphorus (P4) burn in air?
P4 (s) + 5O2 (g)
266 g P4 x
P4O10 (s)
1 mol P4
123.9 g P4
x
DH = -3013 kJ
3013 kJ
= 6470 kJ
1 mol P4
6.4
1. The enthalpy of combustion of methane is the enthalpy
change accompanying the combustion of 1 mol of
methane, CH4. The standard enthalpy of combustion at
298K of methane is –890.3kJ/mol.
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) ΔHo = -890.3 kJ/mol
Calculate the amount of heat released when 2.00g of CH4
combusts at 298K. Answer: 111kJ released; ΔH = -111kJ
2. The thermochemical equation for the combustion of
propane, C3H8, is
C3H8 + 5O2(g)  3CO2(g) + 4H2O(l) ΔHo = -2043 kJ/mol
Calculate the enthalpy change when 5.00 g of propane is
combusted. Answer: 232kJ released; ΔH = -232kJ
3. Consider the combustion reaction of ethane gas, C2H6(g):
C2H6(g) + 7/2O2(g)  2CO2(g) + 3H2O(g) , ∆ H = -1430 kJ
What is the enthalpy change for the reverse reaction if whole
number of coefficients are used?
4CO2(g) + 6H2O(g) 2C2H6(g) + 7O2(g) , ∆ H =+2860 kJ
4. Consider the reaction
H2 (g) + (1/2) O2 (g) → H 2O (I) , ∆ H = - 286 kJ;
which of the following is true?
A. The reaction is exothermic
B. The reaction is endothermic
C. The enthalpy of the products is less than that of the
reactants.
D. Heat is absorbed by the system
E. Both A and C are true
A Comparison of DH and DE
2Na (s) + 2H2O (l)
DE = DH - PDV
2NaOH (aq) + H2 (g) DH = -367.5 kJ/mol
At 25 0C, 1 mole H2 = 24.5 L at 1 atm
PDV = 1 atm x 24.5 L = 2.5 kJ
DE = -367.5 kJ/mol – 2.5 kJ/mol = -370.0 kJ/mol
6.4
The specific heat (s) of a substance is the amount of heat (q)
required to raise the temperature of one gram of the substance
by one degree Celsius. Specific heats are intensive quantities.
The heat capacity (C) of a substance is the amount of heat (q)
required to raise the temperature of a given quantity (m) of the
substance by one degree Celsius. Heat capacities are
extensive quantities.
C=mxs
Heat (q) absorbed or released:
Units
q=J
m = grams
t = C
q = m x s x Dt
q = C x Dt
Dt = tfinal - tinitial
6.5
1. The specific heat of Fe is 0.444 J/gC.
a. Calculate the heat required to raise the temperature of 10.0g
of Fe from 25.0C to 50.0C.
b. Calculate the heat capacity of 10.0g of Fe
c. Calculate the molar heat capacity of Fe (heat capacity of 1
mole of Fe; molar mass of Fe = 55.845 g/mol)
1. a. q = ms∆t = 10.0g x 0.444 J/(gC) x 25.0 C = 111J
b. C = m x s = 10.0g x 0.444 J/gC = 4.44J/C
c. Molar heat capacity = specific heat x molar mass = 0.444
J/(gC) x 55.845 g/mol = 24.8 J/(mol C)
2. The molar heat capacity of diamond (C) is 6.03 J/(mol C).
a. Calculate the specific heat of diamond. 0.502 J/(g C)
b. If 5.00 g of diamond is supplied 100 J of heat, what will the
increase in temperature? 39.8 C
Molar heat capacity = specific heat x molar mass
1. The molar heat capacity of Al is 24.4J/mol-oC. Calculate its
specific heat capacity. Answer: 0.904 J/g-oC
2. The specific heat capacity of Hg is 0.14J/g-oC. Calculate the
amount of heat required to raise the temperature of 10.0g of
Hg from 25.0C to 30.0C. Answer: 7.0J
3. The specific heat of copper is 0.385 J/g°C. Calculate its
molar heat capacity. Answer: 24.5 J/mol-oC
4. If 100.0 J of heat is supplied to 10.0g of Copper, what is the
temperature change of the metal? The specific heat of
copper is 0.385 J/g°C. Answer: 26.0 °C
5. How many kJ of heat must be removed from 1.0x103 g of
H2O (specific heat capacity of 4.184 J /g.°C) to lower the
temperature from 18.0°C to 12.0°C? Answer: 25 kJ released.
6. How much heat is required to raise the temperature of a 6.21
g sample of Fe from 25.0 oC to 79.8 oC? (specific heat of Fe is
0.450 J /goC) Answer: 153 J
How much heat is given off when an 869 g iron bar cools
from 940C to 50C?
s of Fe = 0.444 J/g • 0C
Dt = tfinal – tinitial = 50C – 940C = -890C
q = msDt = 869 g x 0.444 J/g • 0C x –890C = -34,000 J
34,000 J of heat released or
heat change = -34,000J
6.5
q = heat change;
Exothermic process: heat is lost = released; q <0
Endothermic process: heat is gained; q > 0
q = DE at constant volume (w=0)
q = DH at constant pressure
q = m s Dt; m = mass in g; s= specific heat;
Dt = final temperature – initial temperature in celsius
Calorimetry = measurement of heat changes, q, from measurement of
temperature changes, Dt
Constant pressure calorimetry: measurement of heat change at constant
pressure (q = DH);
Constant volume calorimetry: measurement of heat change at constant volume
(q = DE);
Equipment: calorimeter;
Constant pressure calorimeter: Coffee cup calorimeter
Constant volume calorimeter: Bomb calorimeter
Constant-Volume Calorimetry
qrxn = - (qwater + qbomb)
qwater = m x s x Dt
qbomb = Cbomb x Dt
or
qrxn = - qcalorimeter
qcalorimeter = Ccalorimeter x Dt
Reaction at Constant V
DH = qrxn= DE
DH ~ qrxn
No heat enters or leaves!
6.5
1. N2H4(l) + O2(g)  N2(g) + 2H2O(g); molar mass of hydrazine=
32g/mol
When 1.00g of hydrazine, N2H4(l), is burned in a bomb
calorimeter, the temperature of the calorimeter increases by
3.51C. If the heat capacity of the calorimeter is 5.51 kJ/C,
calculate the heat of the reaction PER MOLE of hydrazine.
qrxn = - qcalorimeter = -(Ccalorimeter x Dt) = -(5.51 kJ/C x 3.51C) = -19.3
kJ
Heat is released (exothermic process)
1 mole of hydrazine = 32 g hydrazine
When 1.00g is burned, 19.3 kJ is released.
Therefore when 32 g is burned the amount of heat released is,
19.3kJ released
32 g 
 618 kJ released; q  -618kJ
1.00g
1. HC3H5O3(s) + 3O2(g)  3CO2(g) + 3H2O(g); molar mass of lactic acid=
90.09g/mol
When 0.5865g of lactic acid, HC3H5O3(s), is burned in a bomb calorimeter, the
temperature of the calorimeter increases from 23.10 C to 24.95 C. If the
heat capacity of the calorimeter is 4.812 kJ/C,
a. calculate the heat of the reaction PER MOLE of lactic acid.
b. Calculate the heat of the reaction PER GRAM of lactic acid
qrxn = - qbomb = -(Cbomb x Dt) = -(4.812 kJ/C x (24.95C-23.10 C)) = -8.90 kJ;
Heat is released (exothermic process)
a. 1 mole of lactic acid = 90.09 g lactic acid
When 0.5865g of lactic acid is burned, 8.90 kJ is released.
Therefore when 90.09 g is burned the heat released =
90.09 g x 8.90kJ/0.5865g = 1367 kJ released; q = -1367 kJ/mol
b. When 0.5865 g of lactic acid is burned, 8.90 kJ is released.
Therefore when 1 g of lactic acid is burned, the heat released =
1 g x 8.90kJ/0.5865g = 15.2 kJ released; q = -15.2 kJ/g-
1. The heat capacity of a bomb calorimeter is 5.993 kJ/C. When a
0.8082g sample of glucose, C6H12O6, is combusted in the
calorimeter, the temperature increases from 25.11C to 27.21C.
Calculate the heat of combustion per mole of glucose.
Answer: heat released for the sample (0.8082 g) = 12.6 kJ
0.8082 g = 0.004486 mol
Heat released per mole of sample = 12.6 kJ/0.004486 mol = 2810
kJ/mol; heat change = -2810 kJ/mol
2. The heat capacity of a bomb calorimeter is 5.86 kJ/C. When
0.514g of biphenyl, C12H10, is combusted in the calorimeter, the
temperature increases from 25.8C to 29.4C.
a. Calculate the heat of combustion per g of biphenyl.
b. Calculate the heat of combustion per mol of biphenyl.
Answer: heat released for the sample (0.514 g) = 21 kJ
0.514 g = 0.00333 mol
a. Heat released per g of sample = 21 kJ/0.514 g = 41 kJ/g; q= 41kJ/g
b. Heat released per mole of sample = 21 kJ/0.00333 mol = 6300
kJ/mol; q = -6300 kJ/mol
Constant-Pressure Calorimetry
Cal: coffee cups + cardboard +
stirrer + thermometer
qcal is a small number and can be
assumed to be zero if the heat
capacity of the calorimeter, Ccal, is not
provided
qrxn = - (qsolution + qcal)
qsolution = m x s x Dt
qcal = Ccal x Dt
Reaction at Constant P
DH = qrxn
No heat enters or leaves!
6.5
1. When 4.25 g of NH4NO3 is dissolved in 60.0g of water in a
coffe-cup calorimeter, the temperature drops from 22.0C to
16.9C. Calculate the enthalpy change, DH in kJ/mol NH4NO3
. The specific heat of the solution = 4.184 J/gC.
Molar mass of NH4NO3 = 80 g/mol
qrxn = - (qsolution + qcal)
qcal = 0;
qsolution = m x s x Dt
m = mass of the solution = 4.25g + 60.0g = 64.25 g
s = 4.184 J/gC
Dt = 16.9C - 22.0C = -5.1C
qsolution = m x s x Dt = 64.25 g x 4.184 J/gC x -5.1C = -1371 J
qrxn = - qsolution = 1371J; endothermic reaction
qrxn = DH because pressure is a constant
When 4.25 g NH4NO3 is dissolved, 1371 J is absorbed
Therefore, when 1mol NH4NO3 is dissolved = 80 g NH4NO3 is
dissolved, the heat absorbed = 80 g x 1371J/4.25 g = 25807 J
DH = 25807 J = 2.58 x 104 J
2. When 50.0 mL of 0.100M AgNO3(aq) and 50.0 mL of 0.100 M HCl(aq) are
mixed in a coffee cup calorimeter, the temperature increased from 22.30C to
23.11C.
AgNO3(aq) + HCl(aq)  HNO3(aq) + AgCl(s)
Calculate the enthalpy change for the reaction per mol HCl.
The density of the solution is 1.03 g/mL. Its specific heat = 4.18J/gC.
Volume of the solution = 50.0 mL + 50.0 mL = 100.0 mL
Mass of the solution = volume of solution x density of solution = 100.0mL x
1.03g/mL = 103 g = m
qrxn = - (qsolution + qcal); qcal = 0; Dt = 23.11C - 22.30C = 0.81C
qsolution = m x s x Dt = 103 g x 4.18 J/gC x 0.81C = 349 J
qrxn = - qsolution = -349 J; exothermic reaction
qrxn = DH because pressure is a constant
We used 50.0 mL of 0.100 M HCl(aq) = 0.100 mol/L x 0.0500 L = 0.00500 mol
HCl
Therefore enthalpy change per mol HCl = -349J/0.00500 mol = -69800 J/mol =
-6.98 x 104 J/mol
1. When 100.0 ml of 0.500 HBr and 100.0 ml of 0.500M KOH
both initially at 20.29C react in a coffee-cup calorimeter, the
temperature rises to 23.65C. The specific heat of the solution
is 4.18J/g-oC and the density of the solution is 1.00 g/ml.
Assume the calorimeter to be perfect. Calculate the enthalpy
change per mole of H2O for the following reaction :
HBr(aq) + KOH(aq)  KBr(aq) + H2O(l)
ΔT = 23.65C- 20.29C = 3.36C
Volume of solution = 100.0 mL + 100.0 mL = 200.0 mL
Mass of solution = 200.0 mL × 1.00 g/ml = 200 g
qsolution = 200 g × 4.18J/g-oC × 3.36C= 2.81 × 103 J
qrxn = - 2.81 × 103 J
Moles of H2O formed = moles of HBr used = moles of KOH used =
0.100 L × 0.500 M = 0.0500 mol
Therefore qrxn = - 2.81 × 103 J/0.0500 mol = - 5.62 × 104 J/mol
When a hot object and a cold object are brought in contact,
qcold = -qhot
qcold = heat change of the cold object
qhot = heat change of the hot object
Heat flows from the hot to the cold till they are both at the same
final temperature, equilibrium temperature, T
qcold = mcold x scold x (T-Ti,cold)
qhot = mhot x shot x (T-Ti,hot)
1. A coffee cup calorimeter has 50.0 g water at 22.0C. What will
be the final temperature when 25.0g of water at 90.0 C is
mixed in. The specific heat of water is 4.18J/g C.
qcold = mcold x scold x (T-Ti,cold) = 50.0g x 4.18J/g C x (T- 22.0C)
qhot = mhot x shot x (T-Ti,hot) = 25.0g x 4.18J/g C x (T- 90.0C)
qcold = -qhot
50.0g x 4.18J/g C x (T- 22.0C) =
-(25.0g x 4.18J/g C x (T- 90.0C))
T= 44.7 C
When 20.0g of a metal at 80.0 C is plunged into 15.0g of water
at 25.0 C, the final temperature was measured to be 35.0 C.
Calculate the specific heat of the metal. The specific heat of
water is 4.18J/g C.
qcold = mcold x scold x (T-Ti,cold) = 15.0g x 4.18J/g C x (35- 25.0C)
qhot = mhot x shot x (T-Ti,hot) = 20.0g x s x (35- 80.0C)
qcold = -qhot
15.0g x 4.18J/g C x (35- 25.0C) = -20.0g x s x (35- 80.0C)
627 J = (900 g C) x s
s = 627J/(900 g C) = 0.697 J/(g C)
1. When 23.9 g of Iridium at 89.70C is added to 20.0 g of water at
20.1 0C, the final temperature is 22.6 0C. What is the specific heat
capacity of Iridium? (the specific heat capacity of H2O = 4.184
J/g. 0C). Answer: 0.13 J/g 0C.
2. What will be the final temperature when 12.1 g of Al at 81.70C is
added to 98.8 g of water at 24.90C. The specific heat capacity of
Aluminum is 0.900J/ g 0C and the specific heat capacity of H2O is
4.184 J/g 0C. Answer: 26.30C.
3. A 44.0 g sample of an unknown metal at 99.0 oC was placed in a
constant-pressure calorimeter of negligible heat capacity containing
80.0 mL water at 24.0 oC. The final temperature of the system was
found to be 28.4 oC. Calculate the Specific heat of the metal if density
of water is 1.00 g/ml and the specific heat capacity of H2O is 4.184
J/g 0C. Answer: 0.474 J/g 0C
4. A 25.0 g piece of aluminum (which has a molar heat capacity of
24.03 J/ °C. mol) is heated 82.4 °C and dropped into a calorimeter
containing water (specific heat capacity of water is 4.18 J/ g . oC)
initially at 22.3 °C. The final temperature of the water is 24.9 °C.
Calculate the mass of water in the calorimeter. Answer: 118 g
Formation reaction of a substance: The reaction in which 1 mol of the
substance in formed from its elements in their MOST STABLE FORM.
The enthalpy change accompanying the formation reaction of a
substance is its ENTHALPY OF FORMATION, ∆Hf.
Values of ∆Hf of a substance depend on T & P.
Standard state of a system: P=1atm.
Values of ∆Hf of a substance at 1 atm = standard enthalpy of formation,
∆Hf.
6.5
Examples of formation reactions:
Substance: O2(g)
O2(g)O2(g)
∆Hf of O2(g) =0
Substance: O3(g)
(3/2)O2(g)O3(g)
∆Hfof O3(g) ≠0
Substance: NH3(g)
(1/2)N2(g) + (3/2)H2(g)NH3(g)
Substance: CBr4(l)
C(s) + 2Br2(l) CBr4(l)
∆Hfof NH3(g) ≠0
∆Hfof CBr4(l) ≠0
Substance: PI5(s)
(1/4)P4(s) + (5/2)I2(s) PI5(s)
∆Hfof PI5(s) ≠0
THE STANDARD ENTHALPY OF FORMATION OF THE MOST
STABLE FORM OF ANY ELEMENT = 0
The standard enthalpy of one of the following is not zero at
298K :
a. Hg(s)
b.F2(g)
c. Kr(g)
d. I2(s)
Standard enthalpy of formation (DH0f) of a substance is the heat
change that results when one mole of that substance is formed
from its elements in their most stable form at a pressure of 1 atm.
The standard enthalpy of formation of any element in its
most stable form is zero. DH0f (O2) = 0
DH0f (C, graphite) = 0
DH0f (O3) = 142 kJ/mol DH0f (C, diamond) = 1.90 kJ/mol
For any reaction, the standard enthalpy change, ∆H, can be
estimated from the standard enthalpies of formation of all
reactants & products of the reaction.
aA + bB  cC + dD
∆H= sum of DH0f of all products- sum of DH0f of all reactants
=(c x DH0f of C + d x DH0f of D) - (a x DH0f of A + b x DH0f of B)
Calculate the standard enthalpy, ∆H, of the following reaction:
2C2H6(g)
+
7O2(g) 
4CO2(g)
+ 6H2O(g)
DH0f =-83.85 kJ
0kJ
-393.51 kJ -241.82 kJ
=(4 x 393.51+ 6 x 241.82)kJ - (2 x 83.85 + 7 x 0)kJ =
6.6
= -3024.96kJ –(-167.7)kJ = -2857.26 kJ
The standard enthalpy of formation of H2O(l) is -285.83 kJ.
Calculate the standard enthalpy of formation of CuO(s) given the
following thermochemical equation:
CuO(s) + H2(g)  Cu(s) + H2O(l)
∆H=-130.6 kJ
-130.6 kJ =(1 x 0 + 1 x 285.83)kJ - (1 x DH0f of CuO + 1 x 0)kJ
-130.6kJ = (-285.83kJ) – (DH0f of CuO)
DH0f of CuO = (-285.83kJ) + 130.6kJ = -155.23 kJ
1. Consider the following reaction:
3CH
 g   2 H Ol   CO  g   4CO g   8H  g DH   ?
4
2
2
2
Calculate the standard enthalpy of the reaction given the
following standard enthalpies of formation at 298K:
Compound
CH4(g) -75.0 kJ/mol
H2O(l) -286 kJ/mol
CO2(g) -394 kJ/mol
CO(g) -111 kJ/mol
Answer: 747 kJ
1. Calculate the standard enthalpy of formation of octane, C8H18 at
298K given the following information:
2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g)DH° = -10148 kJ
The standard enthalpies of formation of CO2(g) and H2O(g0 are 394 kJ/mol and -242 kJ/mol respectively. Answer: -256 kJ/mol
2. The heat of combustion of fructose, C6H12O6, is -2812 kJ. Using
the following information, calculate ∆H°f for fructose.
C6H12O6 (s) + 6 O2(g)  6 CO2(g) + 6 H2O (l) DH° = -2812 kJ
∆ H°f ( CO2 ) = -394 kJ/mole ∆ H°f (H2O) = - 286 kJ/mole
Answer: -1268 kJ
3. Consider the following standard heats of formation:
P4O10(s) = -3110 kJ/mol; H2O(l) = -286 kJ/mol; H3PO4(s) = -1279
kJ/mol. Calculate the change in enthalpy for the following process:
P4O10(s) + 6H2O(l)  4H3PO4(s); Answer: 290 kJ
6.6
0 ) is the enthalpy of
The standard enthalpy of reaction (DHrxn
a reaction carried out at 1 atm.
aA + bB
cC + dD
DH0rxn = [ cDH0f (C) + dDH0f (D) ] - [ aDH0f (A) + bDH0f (B) ]
DH0rxn = S nDH0f (products) - S mDHf0 (reactants)
Hess’s Law: When reactants are converted to products, the
change in enthalpy is the same whether the reaction takes
place in one step or in a series of steps.
(Enthalpy is a state function. It doesn’t matter how you get
there, only where you start and end.)
6.6
Benzene (C6H6) burns in air to produce carbon dioxide and
liquid water. How much heat is released per mole of
benzene combusted? The standard enthalpy of formation
of benzene is 49.04 kJ/mol.
2C6H6 (l) + 15O2 (g)
12CO2 (g) + 6H2O (l)
DH0rxn = S nDH0f (products) - S mDHf0 (reactants)
DH0rxn = [ 12DH0f (CO2) + 6DH0f (H2O)] - [ 2DH0f (C6H6)]
DH0rxn = [ 12x–393.5 + 6x–187.6 ] – [ 2x49.04 ] = -5946 kJ
-5946 kJ
= - 2973 kJ/mol C6H6
2 mol
6.6
C (graphite) + 1/2O2 (g)
CO (g) + 1/2O2 (g)
C (graphite) + O2 (g)
CO (g)
CO2 (g)
CO2 (g)
6.6
Hess’s Law: When a reaction is carried out in a series of steps,
the enthalpy of the reaction, ∆H, is equal to the sum of the
enthalpy changes accompanying each of the steps leading to
product formation. reactants step

1 step

2  step

3 products
DH  DH step 1  DH step 2  DH step 3
Strategy in solving a Hess’s Law problem:
In a Hess’s law of problem you will need to combine a set of
complete thermochemical equations to obtain the ∆H for a
reaction (target reaction).
1. Determine how you will you manipulate and then combine
each one of the complete equations to obtain the equation you
want.
2. You may need to divide/multiply one or more given
thermochemical equations. You may need to reverse one or
more of them. You may need to reverse and multiply/divide by
a number one or more of them before combining.
3. Once you have combined the equations, just add up the
corresponding enthalpies. Every equation is only used once.
Calculate the standard enthalpy of formation of CS2 (l) given that:
1.C(graphite) + O2 (g)CO2 (g) DH0 = -393.5 kJ
2. S(rhombic) + O2 (g)SO2 (g) DH0 = -296.1 kJ
3. CS2(l) + 3O2 (g)CO2 (g) + 2SO2 (g) DH0 = -1072 kJ
Target reaction: C(graphite) + 2S(rhombic) CS2 (l)
Equation 1 has 1 mol C as reactant; the target reaction also has 1 mol C
as reactant. Therefore we DO NOT need to change equation 1.
Equation 2 has 1 mol S as reactant; the target reactant has 2mol S as
reactant. Therefore we need to MULTIPLY EQUATION 2 BY 2:
2. 2S(rhombic) + 2O2 (g)2SO2 (g) DH0= 2 X-296.1 kJ = -592.2kJ
Equation 3 has 1 mol CS2 (l) as reactant; the target reaction has 1 mol
CS2 (l) as product. Therefore we reverse equation 3.
3. CO2 (g) + 2SO2 (g) CS2(l) + 3O2 (g) DH0 =+1072 kJ
Combine equation1 and modified equations 2 & 3
C + O2 (g) + 2O2 (g) + 2S + CO2 (g) + 2SO2 (g)  CO2 (g) + 2SO2 (g) +
CS2(l) + 3O2 (g)
DH0 = (-393.5 + -592.2 + 1072) kJ = 86.3 kJ
Calculate the ∆H of the following reaction:
C2H4(g) + 6F2(g) 2CF4(g) + 4HF(g)
using the following data:
1. H2(g) + F2(g) 2HF(g)
∆H = -537 kJ
2. C (s) + 2F2(g) CF4(g)
∆H = 680 kJ
3. 2C (s) + 2H2(g)  C2H4(g) ∆H = 52.3 kJ
Target reaction: C2H4(g) + 6F2(g) 2CF4(g) + 4HF(g)
Reverse equation 3.
3. C2H4(g)  2C (s) + 2H2(g) ∆H = -52.3 kJ
Multiply equation 2 by 2
2. 2C (s) + 4F2(g) 2CF4(g) ∆H = 2 x 680 kJ = 1360kJ
Multiply equation 1 by 2.
1. 2H2(g) + 2F2(g) 4HF(g) ∆H = 2x-537 kJ = -1074kJ
Add modified equations 1, 2, 3
2H2(g) + 2F2(g) + 2C (s) + 4F2(g) + C2H4(g)  2C (s) + 2H2(g)+
2CF4(g)+ 4HF(g) ∆H=(-52.3 + 1360 + -1074) kJ = 233.7 kJ
6F2(g) + C2H4(g)  2CF4(g)+ 4HF(g) ∆H= 233.7kJ
Calculate the standard enthalpy of formation of CS2 (l)
given that:
C(graphite) + O2 (g)
CO2 (g) DH0rxn = -393.5 kJ
S(rhombic) + O2 (g)
CS2(l) + 3O2 (g)
SO2 (g)
DH0rxn = -296.1 kJ
CO2 (g) + 2SO2 (g)
0 = -1072 kJ
DHrxn
1. Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic)
CS2 (l)
2. Add the given rxns so that the result is the desired rxn.
C(graphite) + O2 (g)
2S(rhombic) + 2O2 (g)
+ CO2(g) + 2SO2 (g)
CO2 (g) DH0rxn = -393.5 kJ
2SO2 (g) DH0rxn = -296.1x2 kJ
CS2 (l) + 3O2 (g)
0 = +1072 kJ
DHrxn
C(graphite) + 2S(rhombic)
CS2 (l)
0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJ
DH
rxn
6.6
1. Using Hess’s law of constant heat of summation and given:
1. H2O(l)  H2O(g)
∆H = 44.1 kJ
2. CH3COCH3(l) + 4O2(g)  3CO2(g) + 3H2O(l) ∆H = 1787 kJ
3. CH3COOH(l) + 2O2(g)  2CO2(g) + 2H2O(l) ∆ H = 835 kJ
determine the enthalpy of reaction for
CH3COCH3 (l) + 2 O2(g)  CH3COOH(l) +CO2(g) + H2O(g)
Answer:
Step. 1: H2O(l)  H2O(g)
∆H = 44.1 kJ
Step 2: CH3COCH3(l) + 4O2(g)  3CO2(g) + 3H2O(l)
∆H = 1787 kJ
Step 3: 2CO2(g) + 2H2O(l)  CH3COOH(l) + 2O2(g) ∆ H = +835 kJ
Step 1 + step 2 + step 3
3H2O(l) + CH3COCH3(l) + 4 2O2(g) + 2CO2(g) 
H2O(g) + 3 1CO2(g) + 3H2O(l)+ CH3COOH(l) + 2O2(g ∆H = -44.1 + -1787 + 835
kJ = -996 kJ
1. Phosphorous pentachloride is used in the industrial preparation of many
organic phosphorous compounds. Equation I shows its preparation from PCl3
and Cl2:
PCl3 (l) + Cl2(g)  PCl5(s)
Use equation 1 and 2 to calculate ∆Hrxs of above equation:
1. P4 (s) + 6 Cl2 (g)  4 PCl3 (l)
∆H = 1280 kJ
2. P4 (s) + 10 Cl2 (g)  4 PCl5 (s)
∆H = 1774 kJ
Answer:
Step 1: reverse and divide by 4 equation 1
PCl3 (l) (1/4)P4 (s) + (6/4) Cl2 (g) ∆H = +1280/4 kJ= 320 kJ
Step 2: divide equation 2 by 4
(1/4)P4 (s) + (10/4) Cl2 (g) PCl5 (s)∆H = 1774/4 kJ = -443.5 kJ
Add steps 1, 2
PCl3 (l) + (1/4)P4 (s) + (10/4) 1Cl2 (g) PCl5 (s) + (1/4)P4 (s) + (6/4) Cl2 (g)
∆H = 320 kJ + -443.5 kJ = -123.5 kJ
1. From the following heats of reaction,
2C (graphite) + H2 (g)  C2H2 (g)
∆ H = 227 kJ/mole
6C (graphite) + 3H2 (g)  C6H6 (l)
∆ H = 49 kJ/mole
calculate the heat for the reaction
3C2H2 (g)  C6H6 (l)
2. From the following heats of reaction,
I) N2(g) + 2O2(g)  2NO2(g)
∆H = +67.6 kJ
II) 2NO(g) + O2(g)  2NO2(g)
∆H = 113.2 kJ
calculate the heat of the reaction ,
N2(g) + O2(g)  2NO(g)
1. From the following enthalpies of reaction :
C2 H 4  g   3O2  g   2CO2  g   2 H 2Ol DH   1411kJ
2C2 H 6  g   7O2  g   4CO2  g   6 H 2Ol DH   3119.4kJ
2 H 2  g   O2  g   2 H 2Ol DH   571.6kJ
Calculate the enthalpy change for the following reaction :
C2 H 4  g   H 2  g   C2 H 6  g DH   ?
2.From the following enthalpies of reaction:
C3 H 8 g   5O 2 g   3CO 2 g   4H 2 Og 
ΔH  2043 kJ
C(s)  O 2 g   CO 2 g 
2H 2 g   O 2 g   2H 2 Og 
ΔH  394 kJ
ΔH  484 kJ
Calculate the enthalpy change for the following reaction:
3C + 4H2  C3H8
The enthalpy of solution (DHsoln) is the heat generated or
absorbed when a certain amount of solute dissolves in a
certain amount of solvent.
DHsoln = Hsoln - Hcomponents
Which substance(s) could be
used for melting ice?
Which substance(s) could be
used for a cold pack?
6.7