Chemistry - A Molecular Science

Example 9.5

Predict the sign of

ΔS

o for each of the following processes:

a) H

(g) + I 2

(g) 2

→

2HI(g)

There are 2 moles of gas on each side of the

equation, so the number of moles of gas

does not change. Thus, we predict that

ΔS

o ~ 0.

Tabulated value:

SΔ

o = +21J

-1.K

b) 3H

(g) + N 2

(g) 2

→

2NH

(g) 3

SΔ

o < 0 because the number of moles of gas decreases (4 mol

→

2 mol).

Tabulated value:

SΔ

o = -199 J

.K

-1^

c) Ag(s) + NaCl(s)

→

AgCl(s) + Na(s)

SΔ

o ~ 0 because there are no gases involved.

All reactants and products are in the same

state, so we cannot predict the sign, but we

can predict that the magnitude is small.

Tabulated value:

SΔ

o = +33 J

-1.K

. Note that this is comparable to Part a where there was

no change in the number of moles of gas but small compared to Part b where there was a change in the number of moles of gases.

d) H

O(l) 2

→

H

O(g) 2

SΔ

o > 0 because a gas is produced.

Tabulated value:

ΔS

o = +118 J

.K

-1.

Example 9.6

Indicate which process in each pair increases the entropy of the system more:

a) adding 10 J of heat to neon at 300 K or adding 50 J of heat to neon at 300 K

The heat is added at the same

temperature, so the greater entropy change occurs when

the greater amount of heat is

added. Consequently, adding 50 J results in a greater

entropy change than adding 10 J at 300 K.

b) adding 10 J of heat to neon at 300 K or adding 10 J of heat to neon at 800K

The amount of heat added is the same, so the greater entropy change occurs when the heat is added at the lower tem

perature. Consequently, adding 10 J results in a greater

entropy change at 300 K.

c) adding 50 J of heat to neon at 800 K or adding 10 J of heat to neon at 300 K

ΔS = q/T. (50/800) > (10/300), so adding 50 J at

800 K causes a larger entropy change.

d) melting 1 g of neon at its melting point or evaporating 1 g of neon at its boiling point

ΔS

fusion

= S

liquid

• S

solid

~ 0

because no gases are involved.

ΔS

evaporation

= S

gas

• S

liquid

>> 0

because gas is produced.

Chapter 9 Reaction Energetics

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Carolina

State

University