Transcript File

THERMOCHEMISTRY- A
By
Dr. Hisham E Abdellatef
2011-2012
THERMOCHEMISTRY
The study of heat released or required by chemical
reactions
or heat changes caused by chemical reactions
•Surrounding: the part of the
universe which surround the
system.
System: part of universe
selected for the thermodynamic
study.
HEAT
HEAT
HEAT
HEAT
Heat of Reaction
• Heat of reaction: the change in
energy which accompanies a chemical
reaction.  H
• Endothermic reaction: the reaction which
is supplied by heat (absorb heat and  H is
+ ve).
• Exothermic reaction: the reaction which is
accompanied with evolution of heat
(evolve heat and  H is - ve).
System and Surroundings
In chemical reactions, heat is often transferred
from the “system” to its “surroundings,” or vice
versa.
• The substance or mixture of substances under
study in which a change occurs is called the
thermodynamic system (or simply the
system.)
• The surroundings are everything outside of
the thermodynamic system.
Exothermic
If E < 0, Efinal < Einitial
cellular respiration of glucose
Burning fossil fuels is
an exothermic
reaction
Endothermic
If E > 0, Efinal > Einitial
Photosynthesis is an
endothermic reaction
(requires energy input
from sun)
What is Energy?
Internal Energy E
Kinetic energy
(EK)
Energy due to
motion
Potential
energy (EP)
Energy due to position
(stored energy)
Specific HEAT J°g-1C-1
• The amount of heat which required to
raise the temperature of 1gof
substance by 1°C.
Molar heat capacity
• The amount of heat which required to
raise the temperature of one mole of
substance by 1°C
molar heat capacity = specific heat x M.wt.
• Heat of reaction:
• The change in energy which accompanies a
chemical reaction.
• Heat of combustion:
• The change in enthalpy (heat evolved) when
one mole of the substance is completely
burned in presence of excess oxygen.
• Heat of formation:
• The change in enthalpy (heat evolved) when
one mole of the substance is formed from its
elements their standard state. (t = 25 °C and
P = 1 atm).
HEAT CAPACITY
• J°C-1 Specific Heat x mass
• The amount of heat which required to
raise the temperature of the substance by
1°C.
Which has the larger heat capacity?
How do we relate change in temp. to the energy
transferred?
Heat capacity (J/oC) = heat supplied (J)
temperature (oC)
Heat Capacity = heat required to raise
temp. of an object by 1oC
Specific heat capacity is the quantity of energy
required to change the temperature of a 1g
sample of something by 1oC
Specific Heat
Capacity (Cs)
J / oC / g
=
Heat capacity
Mass
=
J / oC
g
Specific Heat Capacity
How much energy is transferred
due to T difference?
The heat (q) “lost” or “gained” is
related to
a)
b)
sample mass
change in T and
c)
specific heat capacity
Specific heat capacity =
heat lost or gained by substance (J)
(mass, g)(T change, K)
Specific Heat Capacity
Substance
Spec. Heat (J/g•K)
H2O
4.184
Ethylene glycol
2.39
Al
0.897
glass
0.84
Aluminum
Specific Heat Capacity
If 25.0 g of Al cool
from 310 oC to 37 oC,
how many joules of
heat energy are lost
by the Al?
Specific heat capacity =
heat lost or gained by substance (J)
(mass, g)(T change, K)
Specific Heat Capacity
If 25.0 g of Al cool from 310 oC to 37 oC, how many
joules of heat energy are lost by the Al?
heat gain/lose = q = (sp. ht.)(mass)(∆T)
where ∆T = Tfinal - Tinitial
q = (0.897 J/g•K)(25.0 g)(37 - 310)K
q = - 6120 J
Notice that the negative sign on q signals
heat “lost by” or transferred OUT of Al.
UNITS OF ENERGY
1 calorie = heat required to
raise temp. of 1.00 g of
H2O by 1.0 oC.
1000 cal = 1 kilocalorie = 1
kcal
But we use the unit called
the JOULE
1 cal = 4.184 joules
James Joule
1818-1889
FIRST LAW OF
THERMODYNAMICS
heat energy transferred
∆E = q + w
work done
by the
system
energy
change
Energy is conserved!
‫َم ْحفُوظة‬
heat transfer in
(endothermic), +q
heat transfer out
(exothermic), -q
SYSTEM
∆E = q + w
w transfer in
(+w)
w transfer out
(-w)
• 1st Law of Thermodynamics:
• Energy can neither created not destroyed,
only transformed from one form to another
Exothermic
Endothermic
6.2
Enthalpy Diagrams
• Values of H are measured experimentally.
• Negative values indicate exothermic reactions.
• Positive values indicate endothermic reactions.
A decrease in enthalpy
during the reaction; H
is negative.
An increase in
enthalpy during the
reaction; H is
positive.
Exothermic Examples
• Oxidation – wooden splint burning (giving off
light, heat, CO2, H2O
•
•
•
•
•
Burning H2 in air,
body reactions,
dissolving metals in acid,
mixing acid and water,
sugar dehydration
Endothermic Examples
• Electrolysis (breaking water down into H2 and
O2 by running electricity in it)
• Photosynthesis, pasteurization, canning
vegetables
• 2 H2 + O2  2H2O + energy
•
4 g + 32 g  36 g 136 600 cal
• 2H2O + energy  2H2 + O2
•
36 g 136 600 cal  4g + 32 g
Changes in Internal Energy
• If E > 0, Efinal > Einitial
–Therefore, the system
absorbed energy from
the surroundings.
–This energy change is
called endergonic.
Changes in Internal Energy
• If E < 0, Efinal < Einitial
–Therefore, the system
released energy to the
surroundings.
–This energy change is
called exergonic.
Thermochemical equations
• factors which affects the quantity of heat
evolved or absorbed during a physical or a
chemical transformation.
1. Amount of the reactants and products
2. Physical state of the reactants and
products
3. Temperature
4. Pressure
Thermochemical equations
It must essentially:
1.be balanced;
2.give the value of ΔE or ΔH corresponding
to the quantities of substances given by the
equation;
3.mention the physical states of the
reactants and products . The physical
states are represented by the symbols (s),
(L), (g) and (aq) for solid, liquid, gas and
aqueous states respectively.
Example of thermochemical
equation
H2 + ½ O2 → H2O
ΔH = -68.32 Kcal
1 mole of hydrogen reacts with 0.5 mole of oxygen, one
mole of water is formed and 68.32 Kcal of heats
evolved at constant pressure.
But, not specify whether water is in the form of steam
or liquid
H2 (g) + ½ O2 (g)→ H2O (L)
ΔH = -68.32 Kcal
H2 (g) + ½ O2 (g)→ H2O (g)
ΔH = -57.80 Kcal.
Effect of temperature ?????
Standard enthalpy change
ΔH°.
• The heat change at
– 298 K and
–one atmosphere pressure
is called the standard heat change
or standard enthalpy change. It
is denoted by ΔH°.
Enthalpy (H) of the reaction
(Comes from Greek for “heat inside”)
• the sum of internal energy and the product
of this pressure and volume.
H = E + PV
» E is the internal energy,
» P is the pressure and
» V is the volume of the system.
• It is also called heat content.
• ΔH = H product – H reactants = Hp – Hr
Calculation of ΔH from ΔE
• When the system changes at constant pressure, the change
in enthalpy, H, is
H = (E + PV)
H = E + PV
At constant pressure and temperature
The enthalpy of a chemical is measured in kilojoules per mole
(kJmol-1).
H = E + PV
H = E
For solid
and liquid
• At constant volume
Liquids and solids are neglected in calculation of n
n = moles of gaseous products - moles of gaseous reactants.
• In case of gases
H = E + PV
(I)
ΔV =Δn x V
• Δn = no of moles of products - no of moles of reactants
PxΔV = PVxΔn
(II)
• But PV = RT (for one mole of gas)
• Putting RT in place of PV in equation (II) we get
PΔV = RTΔn
• Substituting the value of P AV in equation (I) we get
ΔH = ΔE + Δn RT
R = 1.987 cal. (=2 cal.)
or
= 8.314 joules
Example 1:
• Calculate  E for the following reaction:
2 CO(g) + O2(g) → 2 CO2(g)
Where  H = - 135272 cal. At 25°C
Solution:
 n = 2 -3 = -1
H = E + nRT
T(K) = 25 + 273
- 135272 cal. = E + (- 1x2x298)
E = - 135272 + 596 = - 134676 cal.
Example 2:
Calculate E and H for vaporization of water at 100°C
and 1 atm., the specific heat of vaporization of water
under these conditions is 540 cal/g.
Solution:
H2O(l) ↔ H2O(vap)
Sp. heat of vaporization = 540 cal/g. 1 mole H2O = 18g.
H = molar heat of vaporization = 540 x 18 at constant
pressure
Endothermic reaction since H = + 9720 cal.
n = 1 - 0 =1
 H = E + nRT
9720 = E + 1 x 2 x 373  E = 8974 cal.
Example 3:
6.4g of naphthalene C10H8 when burned under constant volume gave 123 KJ at
20°C, calculate E and H.
Solution: at constant volume the heat of combustion is equal to E gave means
exothermic
C10H8(s) + 12 O2(g) → 10CO2(g) + 4H2O(l)
6.4g at constant volume → -123KJ
128g at constant volume →E
E = = 128 x - 123 = -2460KJ = - 2460 x 103 J
6.4
H = E + nRT
= - 2460 x 103 + (- 2 x 8.3 x 293)
= - 2460 X10-3 - 4863.8 J
= - 2464863 J
= - 2464.863 KJ
Example
• The heat of combustion of ethylene at 17° C and
at constant volume is -332.19 kcal. Calculate the
heat of combustion at constant pressure
considering water to be in liquid state (R = 2 cal.).
The chemical equation for the combustion of ethylene is
C2H4 + 3 O2 = 2CO2(g) + 2H2O (1)
1 mole 3 moles 2moles
negligible volume
No. of moles of the products = 2
No. of moles of the reactants = 4
Δn =(2-4) =-2
ΔH = ΔE + Δn RT
ΔH = -332.19 +[ 2 x I0-3x -2 x 290] =-333.3 kcal
Given that
ΔE=-332. 19 kcal.
T= 273+17= 290k
R=2cal=2xlO-3kcals.
Example
• The heat of combustion of carbon monoxide at constant
volume and at 17° C is -283.3 Kj. Calculate its heat of
combustion at constant pressure(R= 8.314 J degree-1
mole-1).
CO(g) + ½ O2(g) →CO2(g)
1 mole ½ mole
1 mole
No. of mles of products = 1
No. of moles of reactants =1.5
n = No. of moles of products - No. of moles of reactants
=1-1.5 =-0.5
Given that :
ΔE =-283.3 kJ
T = (273+17) = 290 K.
R = 8.314 J or
8.314x10-3 KJ
ΔH = ΔE + Δn x RT
ΔH= -283.3 + (-0.5x (8.314x10-3) x 290] = - 283.3-1.20 =-284.5 KJ
Heat of combustion of CO at constant pressure is -284.5 kJ.
Heat of combustion
• The change in enthalpy (heat evolved) when 1 mole of a
substance is completely burnt in presence of excess
oxygen.
Organic compound + O2(g) → CO2(g) + H2O(l)
• The thermochemical equation must be balanced firstly.
Example 4:
 H for the combustion of liquid heptane C7H16 into
CO2(g) and H2O(l) is - 1151 kcal/mole at 20°C calculate
 E.
Solution:
C7H16(l) + 11O2(g) → 7CO2(g) + 8H2O(i)
n = 7-11 = -4
From thermochemical equation
 H =  E +  nRT
E =  H - A nRT = - 1151 x 10-3 - (- 4x2x293)
= - 1151 x 103 - (-2344) cal./mol.
= -1151 x 103 + 2344
= -1148.656 x103 cal/mo
Example 5:
The heat of combustion of benzoic acid C6H5COOH, into
CO2(g) and H2O(i) at constant pressure is - 78.2kJ/mole at
27°C calculate heat of combustion at constant volume?
Solution: C6H5COOH(S) + 7.5 O2(g) → 7CO2(g) + 3H2O(i)
 H = - 78.2 x 103 J
n = 7-7.5 = -0.5
 H =  E + nRT  E =  H - nRT
 E = - 73200 - (-0.5 x 8.3 x (27 + 273)) = -81935 J
• Try to solve: 3.2 g of naphthalene
C10H8 solid when burnt in excess
O2 gas into CO2(g) and H2O(l) under
constant volume gives 1423 KJ at
20°C calculate  E and  H? (M.Wt
= 128).
Calculate ∆H of
reaction?
Heat of Formation :
(Enthalpies of Formation) H°f
• is the enthalpy change for the formation
of one mole of the substance from its
elements, at standard pressure (1 atm)
and a specified temperature (25°C)
H2(g) + ½ O2(g) → H2O(l)
C (graphite) + 2H2(g) → CH4(g)
 H°f = -285.8 KJ
 Ho f = -74.9 KJ
Hf
o
superscript o means standard state
= 25oC and 1 atm pressure
subscript f means formation from most stable elements
Example 6:
Calculate the  H° of the reaction
Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)
given that: H°f CO2(g) = - 393.5 KJ/mol
 H°f Fe2O3 = -822.2 KJ/mol
H°f CO(g) = - 110.5KJ/mol
Solution:  Ho = (products) - (reactants).
= [(3X-393.5) + (2x0)] - [(lx-822.2) + (3x-110.5)]J
= -1180.5 + 1153.7 = -26.8 KJ.
Example 7:
Give the following data:
B2H6(g) + 6 H2O(l) → 2H3BO3(s) + 6 H2(g)
 H° = - 493.4 KJ
 H°f of H3BO3(S) is - 1088.7 KJ/mol, and
  H°f of H2O(l) is - 285.9 KJ/mol
Calculate the standard enthalpy of formation of
B2H6
Solution:
= 2 H°f H3BO3 - [ H°f B2H6 + 6 H°f H2O]
- 493.4 KJ = 2x- 1088.7 KJ- [1x H°f B2H6 + 6X 285.9KJ]
= - 2177.4KJ - [H°f B2H6 - 1715.4KJ]
- 493.4KJ =- 462.0KJ -  HofB2H6
H°fB2H6 = + 31.4KJ/mol
The equations representing
the variation of heat
changes of reaction with
temperature are known as
Kirchoff's equations.
VARIATION OF HEAT (OR ENTHALPY)
OF REACTION WITH TEMPERATURE
KIRCHOFF'S EQUATIONS.
1. At constant volume,
internal energies
of the reactants
and products.
ΔE=E2-E1
Differentiating this equation with respect to temperature at constant
volume, we get
KIRCHOFF'S EQUATIONS.
heat capacities
But we have already seen that
heat capacities of
the products and
reactants
Integrating
between temperature T l and T2 , we have
ΔE2- ΔE1 =
Δ E2 -ΔE1 = Δ Cv (T2-T1)
(3)
KIRCHOFF'S EQUATIONS.
2. At constant pressure
ΔH=H2-H1
And finely …. ΔH2- ΔH1= ΔCp (T2-T1) (6)
Example 3 – 3
The heat of reaction
½ H2 + ½ Cl2 = HCl
at 27 C° is -22.1 Kcal.
Calculate the heat of reaction at 77°C. The molar heat
capacities at constant pressure at 27° C for hydrogen,
chlorine and HCl are 6.82, 7.70 and 6.80 cal mol-1,
respectively.
Here ½ H2 + ½Cl- → HCl Δ H = -22.1 Kcal
Δ Cp = Heat capacities of products - Heat capacities of reactants
= 6.80-[½(6.82) +½ (7.70)]
T2 = 273 + 77 =350 K
= 6.80 - 7.26 = -0.46 x 10-3 Kcal
T1 = 273 + 27 = 300 K
T2-T1= (350-300)K=50K
ΔH2- ΔH1= ΔCp (T2-T1)
ΔH2 - (-22.1) = (-0.46 x 10 -3) x 50= -21.123
Example 3 – 4
• Calculate the standard heat of formation of
propane (C3H8) if its heat of combustion is
2220.2 KJ mol-1. The heats of formation of CO2
(g) and H2O (L) are - 393.5 and -285.8 KJ mol1 respectively.
C3H8 (g) + 5 O2 (g) → 3CO2 (g) + 4 H2O (L)
Δ Hc = 2220.2 KJ
C (s) + O2 (g) → CO2(g)
ΔHf=- 393.5 KJ
H2(g)+½O2(g) →H2O (L)
ΔHf=-285.8KJ
We should manipulate these equations in a way so as to get the
required equation
3 C (s) + 4 H2 (g) → C3H8 (g)
ΔH=?
Multiplying equation (ii) by 3 and equation (iii) by 4 and adding up we
get
3C(s) + 3O2 (g) → 3 CO2 (g) ΔH = -1180.8 KJ
4H2 (g) +2O2(g) → 4H20(L)
ΔH = -1143.2 KJ
3C(s) + 4H2 (g) + 502(g) → 3CO2(g) +4H2O(L) ΔH= - 2323.7 KJ
subtracting equation (i) from equation (iv), we have
3C(s)+4H2(g) → C3H8(g)
Δ H= -103.5 KJ
Heat of transition
• the change in enthalpy which occurs when one
mole of an element changes from one allotropic
form to another.
• P white → P red
• S monoclipic → S rhombic
ΔH= -1.028 kcal
ΔH=-0016 Kcal
Hess's law of constant heat summation
If a reaction is carried out in a series of
steps, ΔH for the reaction is the sum of
ΔH’s for each of the steps.
A →B + q1
ΔH1= q1
B → C + q2
ΔH2 = q2
C → Z + q3
ΔH3 = q3
A →z + Q1 ΔH1 = -Q1
The total evolution of heat = q1 +q2 +q3 =Q2
According to Hess's law Q1= Q2
We can find H(a) by
subtracting H(b) from H(c)
Hess’s Law
& Energy Level Diagrams
Forming CO2 can occur in a
single step or in a two steps.
∆Htotal is the same no matter
which path is followed.
Hess’s Law
& Energy Level Diagrams
Forming H2O can occur in a
single step or in a two
steps.
∆Htotal is the same no matter
which path is followed.
Illustrations of Hess's law
• (1) Burning of carbon to CO2
1st way:
C(S) + 02(g) → CO2 (g)
ΔH = -94.05kcal (-393.50
KJ)
2nd way:
C(s)+½O2(g) →CO(g)
ΔH =-26.42 kcal and
CO2 (g) + ½ O2 (g) → CO2 (g)
ΔH = -67.71 kcal
Overall change C (s) + O2 (g) →CO2 (g) ΔH = -94.13
kcal
Illustrations of Hess's law
(2) Formation of sodium by Hydroxide from Na
1st way:
2Na (s) + ½ O2 (g) → Na2O (s)
Na2O + H2O (L) →2 NaOH (aq)
2 Na (s) + H2O (L) + ½ O2 (g) → 2 NaOH (aq)
2nd way
2 Na (s) + 2 H2O (L) → 2 NaOH (aq) + H2 (g)
H2 (g) + ½ O2 (g) → H2O (g)
2 Na (s) + H2O (L) + ½ O2 (g) → 2 NaOH (aq)
ΔH = -100 kcal
Δ H = -56 kcal
ΔH = - 156 Kcal
ΔH = - 88 Kcal
ΔH = - 68.5 Kcal
ΔH = - 156 Kcal
Example 8:
Calculate the heat of hydrogenation of acetylene
1) C2H2(g) + 2 H2(g) → C2H6(g) H1 = ?
At 25°C and 1 atm., given that;
2) 2C2H2(g) + 5 O2(g) → 4CO2 + 2H2O(l) H2 = - 2602 KJ
3) 2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(I) H3 =-3123KJ
4) H2(g)+ ½O2(g) →H2O(l)
H4= -286KJ
Solution:
5) C2H2(g) + 5/2 O2(g) → 2CO2(g) + H2O(l)
2 ‫) علي‬2( ‫بقسمة‬
H5 = -2602/2 = -1301KJ
6) 2H2(g) + O2(g) → 2H2O(I)
2 ‫ في‬4 ‫بضرب‬
 H6 = 2 (-286KJ) = - 572 KJ
7) 2CO2(g) + 3H2O (l) → 2C2H6(g) + 7/2 O2(g)2 ‫) علي‬3( ‫عكس و قسمة‬
H7 = + 3123/2 = +1561KJ
Adding equation 5,6 and 7 gives
C2H2(g) + 2H2(g) + 7/2O2(g) + 2CO2 + 3H2O(I) →
2CO2(g) + 3H2O(|) + C2H6(g) + 7/2 O2(g)
Canceling elements that are the same on both
sides, we get
C2H2(g) + 2H2(g) → C2H6(g)
H1 =  H5 +H6 +  H7
= (-1301 KJ) + (-572 KJ) + (1561 KJ) = -312 KJ
Calorimetry
• We measure heat flow using calorimetry.
• A calorimeter is a device used to make this
measurement.
• A “coffee cup” calorimeter may be used for
measuring heat involving solutions.
A “bomb” calorimeter is used to
find heat of combustion; the
“bomb” contains oxygen and a
sample of the material to be
burned.
Bomb Calorimeter
Coffee-cup
calorimeter.