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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

System: part of universe selected for the thermodynamic study.

Surrounding: the part of the universe which surround the system.

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, E final

< E initial cellular respiration of glucose

Burning fossil fuels is an exothermic reaction

Endothermic

If

E > 0, E final

> E initial

Photosynthesis is an endothermic reaction

(requires energy input from sun)

Kinetic energy

(E

K

)

What is Energy?

Internal Energy E

Potential energy (E

P

)

Energy due to motion

Energy due to position

(stored energy)

Specific HEAT

J

°g

-1

C

-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/ o

C) = heat supplied (J) temperature ( o

C)

Heat Capacity = heat required to raise temp. of an object by 1 o

C

Specific heat capacity is the quantity of energy required to change the temperature of a 1g sample of something by 1 o

C

Specific Heat

Capacity (Cs)

J / o

C / g

=

=

Heat capacity

Mass

J / o

C g

Specific Heat Capacity

How much energy is transferred due to T difference?

The heat

(q) related to

“lost” or “gained” is a) sample mass b) 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

H

2

O

Ethylene glycol

Spec. Heat (J/ g•K)

4.184

2.39

Al glass

0.897

0.84

Aluminum

Specific Heat Capacity

If 25.0 g of Al cool from 310 o

C to 37 o

C, 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 o

C to 37 o

C, how many joules of heat energy are lost by the Al?

heat gain/lose = q = (sp. ht.)(mass)(∆T) where ∆T = T final

- T initial 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

H

2

O by 1.0 o

C.

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!

ة ظوُف ْحَم

heat transfer in

(endothermic), +q heat transfer out

(exothermic), -q

SYSTEM

∆E = q + w w transfer in

(+w) w transfer out

(-w)

1 st

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, CO

2

, H

2

O

Burning H

2 in air, body reactions, dissolving metals in acid, mixing acid and water, sugar dehydration

Endothermic Examples

Electrolysis (breaking water down into H

2

O

2 by running electricity in it) and

Photosynthesis, pasteurization, canning vegetables

2 H

2

+ O

2

2H

2

O +

4 g + 32 g

energy

36 g 136 600 cal

2H

2

O + energy

2H

2

+ O

36 g 136 600 cal

2

4g + 32 g

Changes in Internal Energy

If

E > 0, E final

> E initial

Therefore, the system

absorbed energy from the surroundings.

This energy change is called endergonic .

Changes in Internal Energy

If

E < 0, E final

< E initial

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

H

2

+ ½ O

2

→ H

2

O 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

H

2

H

2

(g) + ½ O

2

(g) + ½ O

2

(g)→ H

2

O (L) ΔH = -68.32 Kcal

(g)→ H

2

O (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

For solid and liquid

H =

E

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

or

R = 1.987 cal. (=2 cal.)

= 8.314 joules

Example 1:

Calculate

E for the following reaction:

2 CO

(g)

+ O

2(g)

→ 2 CO

2(g)

Where

H = - 135272 cal. At 25

°C

Solution:

H =

E +

nRT

n = 2 -3 = -1

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:

H

2

O

(l)

↔ H

2

O

(vap)

Sp. heat of vaporization = 540 cal/g. 1 mole H

2

O = 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 C

10

H

8 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

C

10

H

8(s)

+ 12 O

2(g)

→ 10CO

2(g)

+ 4H

2

O

(l)

6.4g at constant volume

→ -123KJ

128g at constant volume

E

E = = = -2460KJ = - 2460 x 10

6.4

3

J

H =

E +

nRT

= - 2460 x 10

3

+ (- 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

C

2

H

4

+ 3 O

2

= 2CO

2

(g) + 2H

2

O (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

-3 x -2 x 290] =-333.3 kcal

Given that

ΔE=-332. 19 kcal.

T= 273+17= 290k

R=2cal=2xlO

-3 kcals.

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 O

2

(g)

CO

2

(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 + O

2(g)

CO

2(g)

+ H

2

O

(l)

The thermochemical equation must be balanced firstly.

Example 4:

H for the combustion of liquid heptane C

CO

and H

2

O

(l)

7

H

Solution:

C

7

 n = 7-11 = -4

H

16(l)

+ 11O

2(g)

From thermochemical equation

7CO

2(g)

+ 8H

2

O

H =

E =

E +

= - 1151 x 10

3 nRT

H - A nRT = - 1151 x 10

-3

- (-2344) cal./mol.

- (- 4x2x293)

= -1151 x 10

3

+ 2344

= -1148.656 x10

3 cal/mo

(i)

into

Example 5:

The heat of combustion of benzoic acid C

CO and H

2

O

6

H

5

COOH, into at constant pressure is - 78.2

k J/mole at

27°C calculate heat of combustion at constant volume?

Solution: C

6

H

5

H = - 78.2

COOH

(S) x 10

3

J

 n = 7-7.5 = -0.5

+ 7.5 O

2(g)

7CO

2(g)

+ 3H

2

O

H =

E +

 nRT

E =

H -

 nRT

E = - 73200 - (-0.5 x 8.3 x (27 + 273)) = -81935 J

(i)

• Try to solve: 3.2 g of naphthalene

C

10

H

O

2 solid when burnt in excess gas into CO

2(g) and H

2

O 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)

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)

H

2(g)

+ ½ O

2(g)

C

(graphite)

+ 2H

2(g)

H

2

O

(l)

CH

4(g)

H° f

H o

= -285.8 KJ f

= -74.9 KJ

H f o superscript o

means standard state

= 25 o

C and 1 atm pressure subscript f means formation from most stable elements

Example 6

:

Calculate the

H

° of the reaction

Fe

2

O

3(s)

+ 3CO

(g)

CO

(g)

→ 2Fe

(s)

= - 110.5KJ/mol

+ 3CO

2(g) given that:

H

° f

H

° f

Fe

2

O

3

CO

2(g)

= - 393.5 KJ/mol

= -822.2 KJ/mol

H

° f

Solution:

H o

= (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:

B

2

H

6(g)

+ 6 H

2

O

(l)

H° = - 493.4 KJ

2H

3

BO

3(s)

+ 6 H

2(g)

H° f

 

H° of H

3

BO f of H

2

O

3(S)

(l) is - 1088.7 KJ/mol, and is - 285.9 KJ/mol

Calculate the standard enthalpy of formation of

B

2

H

6

Solution:

= 2

H° f

H

3

B

2

H

6

- 493.4 KJ = 2x- 1088.7 KJ- [1x

285.9KJ]

BO

3

- [

H° f

+ 6

H° f f

H

2

O]

B

2

H

6

+ 6X -

= - 2177.4KJ - [

H° f

B

2

- 493.4KJ =- 462.0KJ -

H

6

H° f

B

2

H

6

= + 31.4KJ/mol

- 1715.4KJ]

H o f

B

2

H

6

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

ΔE

2

-

ΔE

1

=

Δ E

2

-

ΔE

1

= Δ Cv (T

2

-T

1

)

(3)

KIRCHOFF'S EQUATIONS.

2. At constant pressure

ΔH=H

2

-H

1

And finely …. ΔH

2

- ΔH

1

= ΔCp (T

2

-T

1

) (6)

Example 3 – 3

The heat of reaction

½ H

2

+ ½ Cl

2 =

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 H

2

+ 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

ΔH

2

-

ΔH

1

= ΔCp (T

2

-T

1

)

ΔH2 - (-22.1) = (-0.46 x 10

-3

) x 50= -21.123

T2-T1= (350-300)K=50K

Example 3 – 4

Calculate the standard heat of formation of

propane (C

(g) and H

1

2

3

H

8

) if its heat of combustion is

1

. The heats of formation of CO

O (L) are - 393.5 and -285.8 KJ molrespectively.

2

C

3

H

8

(g) + 5 O

2

C (s) + O

2

(g)

→ 3CO

2

(g)

→ CO

2

(g)

(g) + 4 H

2

O (L) Δ H c

= 2220.2 KJ

ΔHf=- 393.5 KJ

H

2

(g)+½O

2

(g)

→H

2

O (L) ΔHf=-285.8KJ

We should manipulate these equations in a way so as to get the required equation

3 C (s) + 4 H

2

(g)

→ C

3

H

8

(g)

ΔH=?

Multiplying equation (ii) by 3 and equation (iii) by 4 and adding up we get

3C(s) + 3O

2

(g) → 3 CO

2

4H

2

(g) ΔH = -1180.8 KJ

(g) +2O

2

(g) → 4H

2

0(L) ΔH = -1143.2 KJ

3C(s) + 4H

2

(g) + 50

2

(g) → 3CO

2

(g) +4H

2

O(L) ΔH= - 2323.7 KJ subtracting equation (i) from equation (iv), we have

3C(s)+4H

2

(g) → C

3

H

8

(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

S white

→ P red 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

B

→ C + q2

C

→ Z + q3

Δ

H

1

= q1

Δ

H

2

= q2

Δ

H

3

= q

3

A →z + Q

1

ΔH

1

= -Q

1

The total evolution of heat = q1 +q2 +q3 =Q2

According to Hess's law Q1= Q2

We can find

H

subtracting

H

(a)

(b) by from

H

(c)

Hess’s Law

& Energy Level Diagrams

Forming CO

2 can occur in a single step or in a two steps.

∆H total is the same no matter which path is followed.

Hess’s Law

& Energy Level Diagrams

Forming H

2

O can occur in a single step or in a two steps.

∆H total is the same no matter which path is followed.

Illustrations of Hess's law

(1) Burning of carbon to CO

2

1 st way:

C(S) + 0

2

KJ)

(g) → CO

2

(g) ΔH =

-94.05kcal (-393.50

2 nd way:

C(s)+½O

2

(g) →CO(g) ΔH =-26.42 kcal and

CO

2

(g) + ½ O

2

(g) → CO

2

(g) ΔH = -67.71 kcal

Overall change C (s) + O kcal

2

(g) →CO

2

(g) ΔH =

-94.13

Illustrations of Hess's law

(2) Formation of sodium by Hydroxide from Na

1 st way:

2

Na (s) + ½ O

2

(g)

→ Na

2

O (s) ΔH = -100 kcal

Na

2

O + H

2

O (L)

→2 NaOH (aq) Δ H = -56 kcal

2 Na (s) + H

2

O (L) + ½ O

2

(g)

→ 2 NaOH (aq) ΔH = - 156 Kcal

2 nd way

2 Na (s) + 2 H

2

O (L)

→ 2 NaOH (aq) + H

2

H

2

(g) + ½ O

2

(g)

→ H

2

(g) ΔH = - 88 Kcal

O (g) ΔH = - 68.5 Kcal

2 Na (s) + H

2

O (L) + ½ O

2

(g)

→ 2 NaOH (aq) ΔH = - 156 Kcal

Example 8:

Calculate the heat of hydrogenation of acetylene

1) C

2

H

2(g)

+ 2 H

2(g)

C

2

H

6(g)

At 25°C and 1 atm., given that;

H

1

= ?

2) 2C

3) 2C

2

2

4) H

2(g)

H

H

2(g)

6(g)

+ 5 O

2(g)

+ 7O

+ ½O

2(g)

2(g)

H

2

O

(l)

4CO

4CO

2

2(g)

+ 2H

2

O

(l)

+ 6H

2

O

(I )

H

2

H

3

H

4

= - 2602 KJ

=-3123KJ

= -286KJ

Solution:

5) C

2

H

2(g)

+ 5/2 O

2(g)

H

5

2CO

2(g)

+ H

2

O

(l)

= -2602/2 = -1301KJ

2 يلع ) 2 ( ةمسقب

6) 2H

2(g)

+ O

2(g)

2H

2

H

6

O

(I)

2

= 2 (-286KJ) = - 572 KJ

يف 4 برضب

7) 2CO

2(g)

+ 3H

2

O

(l)

2C

2

H

6(g)

+ 7/2 O

2(g)

2 يلع

H7 = + 3123/2 = +1561KJ

) 3 ( ةمسق و سكع

Adding equation 5,6 and 7 gives

C

2

H

2(g)

2CO

2(g)

+ 2H

2(g)

+ 3H

2

O

+ 7/2O

2(g)

(|)

+ C

2

H

6(g)

+ 2CO

2

+ 3H

+ 7/2 O

2(g)

2

O

(I)

Canceling elements that are the same on both sides, we get

C

2

H

2(g)

+ 2H

2(g)

C

2

H

6(g)



H

1

=

H

5

+

H

6

+

H

7

= (-1301 KJ) + (-572 KJ) + (1561 KJ) = -312 KJ

Calorimetry

We measure heat flow using calorimetry .

A calorimeter measurement.

is a device used to make this

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.

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