Find the change in entropy when 87.3 g of water vapor condense, given that water’s heat of

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# Find the change in entropy when 87.3 g of water vapor condense, given that water’s heat of - PowerPoint PPT Presentation

i.e., heat (energy). temperature. = –77.9 . For a system in which heat is transferred at constant temperature…. -- T in K. -- common unit for entropy . Find the change in entropy when 87.3 g of water vapor condense, given that water’s heat of vaporization is 5.99 kJ/mol. 373.15 K.

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i.e., heat (energy)

temperature

= –77.9

For a system in which heat is

transferred at constant temperature…

-- T in K

-- common unit for entropy 

Find the change in entropy when 87.3 g of water

vapor condense, given that water’s heat of

vaporization is 5.99 kJ/mol.

373.15 K

= –0.0779

Third Law of Thermodynamics:

The entropy of a pure,

crystalline substance at

absolute zero is…

ZERO.

--

that would be a state

of perfect order

(impossible)

fewer bonds;

fewer restrictions on motion of atoms;

more degrees of freedom;

more possible microstates

2 NH3

N2 + 3 H2

Entropy increases when:

1. the number of gas particles increases

+DS

2. liquids or solutions

are formed from solids

3. gases are formed

from liquids or solids

2 mol HCl(g) @ 300 K or 4 mol HCl(g) @ 300 K

(same volume)

1 mol HCl(g) @ 300 K or 1 mol Ar(g) @ 300 K

(same volume)

2 mol HCl(g) @ 300 K or 2 mol HCl(g) @ 300 K

(in a 10-L vessel)

(in a 5-L vessel)

Which has the greater entropy?

1 mol O2(g) @ 300 K or 1 mol O2(g) @ 500 K

(same volume)

1 mol KCl(s) @ 300 K or 1 mol HCl(g) @ 300 K

actually, at 1 bar = 105 Pa = 0.987 atm

w/increasing molar mass

w/increasing # of atoms in formula

Calculating Entropy Changes

standard molar entropies, So: molar entropy values

of substances in their standard states

(i.e., pure substances at ~1 atm)

So values typically…

--

are NOT zero

--

--

In a chemical reaction…

DSo = SnSoP – SmSoR

(n and m are the coeff. for each substance)

tabulated values of

So in J/mol-K

Smokescreen

or flesh-fryer?

Calculate the standard entropy change for…

P4(s,white) + 10 Cl2(g) 4 PCl5(g)

177

223

353

DSo = SnSoP – SmSoR

DSo = 4(353) – [177 + 10(223)]

= –995

J/K

White phosphorus (or “WP”) is used in bombs,

artillery shells, and mortar shells that burst

into burning flakes of phosphorus upon impact.