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Slides on Enthalpy and its Function, Slides of Chemistry

Enthalpy is the sum of the internal energy and the product of pressure and volume and its general formula is H=U+PV.

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Chapter 2. The First Law.
P.27
2.5(a) Enthalpy
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Chapter 2. The First Law.

P.

2.5(a) Enthalpy

Justification 2.

The relation

H = q

p

-^

For a general infinitesimal change in the state of the system,

U

changes to

U +

d U

,^ p

changes to

p +

d

p , and

V

changes to

V +

d

V , so

H

changes from

U + pV

to

  • •^

Recognize

U

pV

=

H

on the right,

H

changes to

  • • •^

Now substitute d

U

= d

q^

  • d

w^

into this expression,

  • •^

If the system is in mechanical equilibrium with its surroundings at a pressure

p

and does only expansion work, we can write d

w^

= −

p d

V^

and obtain

  • •^

Now impose the condition that the heating occurs at constant pressure bywriting d

p^

= 0. Then

-^

dH=dq (at constant p, no additional work)

Example 2.

Relating

H and

U

-^

The internal energy change when 1.0 mol CaCO

in the form of calcite 3

converts to aragonite is +0.21 kJ. Calculate the difference between theenthalpy change and the change in internal energy when the pressure is 1.0bar given that the densities of the solids are 2.71 g cm

and 2.93 g cm

−3,

respectively.

-^

Answer:

-^

The volume of 1.0 mol CaCO

(100 g) as aragonite is 34 cm 3

3 , and that of 1.

mol CaCO

as calcite is 37 cm 3

  1. Therefore,
  • •^

(1 Pa m

3 = 1 J). Hence,

  • •^

only 0.1% of

U

. Usually ignore the difference between H and U of

condensed phases, except at very high p, when

pV

no longer negligible

-^

Self Test 2.

Calculate the difference between

H

and

U

when 1.0 mol Sn(s,

grey) of density 5.75 g cm

changes to Sn(s, white) of density 7.31 g cm

at

10.0 bar. At 298 K,

H = +

2.1 kJ.

-^

Correct Answer:

H −

U = −

4.4 J

Enthalpy of perfect gas

•^

Illustration 2.

The relation between

H

and

U for gas-phase reactions

  • In the reaction: 2 H

(g) + O 2

(g) 2

2 H

O(l), 2

•^

n

g^

3 mol. At 298 K, when

RT

= 2.5 kJ

mol

, the enthalpy and internal energy

changes in the system

Answer

  • The enthalpy change is• • (1 A V s = 1 J ). 0.798 g of water is (0.

g)/(18.02 g mol

) = (0.798/18.02) mol H

O, 2

the enthalpy of vaporization per mole ofH

O is 2

• •^

H

O(l) 2

H

O(g) , 2

n

g^

= +1 mol,

Test 2.3 (p.44)

  • The molar enthalpy of vaporization of

benzene at its boiling point (353.25 K) is30.8 kJ mol

. What is the molar internal

energy change? For how long would thesame 12 V source need to supply a 0.50 Acurrent in order to vaporize a 10 g sample?

-^

Correct Answer:

+27.9 kJ mol

, 660 s

Chapter 2. The First Law.

P.

2.5(c) The variation of enthalpy with temperature

Example 2.

Evaluating an increase in enthalpy with temperature

(p.45)

  • What is the change in molar enthalpy of N

2

when it is heated from 25

°C to 100

°C?

( Table 2-2)

Chapter 2. The First Law.

P.

2.6 Adiabatic changes

Adiabatic processes

-^

Consider a stage in a reversible adiabaticexpansion when the pressure inside and out is

p

.

The work done when the gas expands by d

V

is

d

w

= −

p

d

V

; however, for a

perfect gas

, d

U

=

C

dV

T

.

-^

Therefore,because for an

adiabatic

change (d

q

= 0)

d

U

= d

w

  • d

q

= d

w

, we can equate these two

expressions for d

U

and write

  • •^

We are dealing with a

perfect gas

, so we can

replace

p

by

nRT

/ V

and obtain

Adiabatic processes

  • Because ln(

x

/ y

) = −ln(

y

/ x

), this expression

rearranges to

  • • With

c

C

/ V

nR

we obtain (because ln

x

a^

a

ln

x

  • • which implies that (

T

/f

T

)i c^

V

/i

V

) and,f

upon rearrangement,

Adiabatic processes

(4)

-^

The initial and final states of a perfect gas satisfy theperfect gas law regardless of how the change of statetakes place, so use

pV

=

nRT,

  • •^

However, we have just shown that

  • •^

where we use the definition of the heat capacity ratiowhere

γ

=

C

p ,m

/ C

V ,m

and the fact that, for a perfect

gas,

C

p ,m

C

V ,m

=

R

(

Sec.2.

). Then we combine the

two expressions, to obtain

  • •^

which rearranges to

p

V i

γi

=

p

V f

γf

Illustration 2.

Work of adiabatic expansion

-^

Consider the adiabatic, reversible expansion of 0.020mol Ar, initially at 25

°C, from 0.50 dm

3

to 1.00 dm

The molar heat capacity of argon at constant volume is12.48 J K

mol

, so

c

= 1.501. From

  • •^

It follows that

T = −

110 K, from

  • •^

Note that temperature change is independent of theamount of gas but the work is not.

-^

Test 2.

Calculate the final temperature, the work done,

and the change of internal energy when ammonia isused in a reversible adiabatic expansion from 0.50 dm

3

to 2.00 dm

3 , the other initial conditions being the same.

-^

Correct Answer:

195 K, −56 J, −56 J

Illustration 2.

The pressure change

accompanying adiabatic expansion

  • When a sample of argon (for which

at 100 kPa expands reversibly andadiabatically to twice its initial volume thefinal pressure will be

  • • • For an isothermal doubling of volume, the

final pressure would be 50 kPa.