1.3.1 Entropy
Entrophy
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In addition to enthalpy (heat content), there is another important thermodynamic aspect of all chemical reactions – entropy (S).
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A measure of the amount of randomness or disorder in a system.
The symbol for entropy: S
The unit for entropy: J/ K·mole
All substances, be they individual atoms of a single element or a molecule of a compound, possess some degree of disorder because
particles are always in constant motion.
Thus, S is always a positive number.
Entrophy
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when S is theoretically equal to zero? (S = 0)
only for pure crystals at absolute zero ( 0 K or -273°C)
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This is known as the Third Law of Thermodynamics
Entrophy
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Can S ever be a negative number?
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Answer - no. A substance can not be less random than not random at all.
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We are typically concerned with how entropy changes during a chemical reaction, or with ΔS rather than S:
ΔS = Sfinal – Sinitial
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Units become J/K, since now we are concerned with the entire system instead of just one mole
Entrophy
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What does the value of ΔS tell us about how entropy changes?
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Let's make up some numbers and see.
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Gas particles move much more than do particles in the liquid phase, making them more random or disordered. Let's give a gas particle
an entropy value of 10 and a liquid particle an entropy value of 5 (because it is less random, it should have a smaller number).
Entrophy
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If a system changes from a gas state to a liquid state, it becomes less random, or more ordered, because liquid particles move about
less randomly than do gas particles. Now calculate ΔS :
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ΔS = Sliquid – Sgas
ΔS = 5 – 10 = –5
Entrophy
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A negative value of ΔS indicates a decrease in entropy— the system becomes less random
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A positive value of ΔS indicates an increase in entropy— the system becomes more random
The Law of Disorder
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a.k.a Second Law of Thermodynamics
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states that systems tend to become more random over time, not more ordered.
Consider these examples:
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Think of your bedroom at home. If you're like most people, over time the room becomes messier, or more random (less ordered) over
time. It takes some effort (energy) to get things tidied up again.
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How about a brand new deck of cards, nicely arranged in suits? As you play a game of cards the deck becomes much more random.
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A sand castle on a beach? Decays into a pile of sand, a more random and disordered state.
Predicting Entropy Changes
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You can predict entropy changes by looking at several factors in an equation.
The following changes suggest an increase in entropy:
•solid → liquidChanges in state:
liquid → gas
solid → gas
solid or liquid→ aqueous state (dissolving)
Entrophy
•
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An increase in the number of moles. If the product side has more moles than the reactant side, the system has become more
random;
Increasing the temperature.
An increase in temperature increases the degree of randomness
Calculating Entropy Changes
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It is possible to calculate a value for ΔS. It is the same formula we used for Hess's Law, only now we are working with values for entropy
instead of enthalpy:
ΔS = ΣSproducts – ΣSreactants
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You'll find values for ΔS in the same Table of Thermochemical Data that you used for calculating ΔH.
Example:
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Calculate ΔS and state whether entropy increases (becomes more random) or decreases (becomes less random)?
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Do you predict a spontaneous reaction?
2 NO(g) + O2(g) →N2O4(g)
Solution:
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Look up ΔS values. Multiply values by coefficients from balanced equation.
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ΔS:
NO2
10.8× 2= 421.6
O2
205.1×1= 205.1
N2O4
304.3×1=
304.3
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Solve for ΔS
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ΔS =ΣSproducts – ΣSreactants
[N204] - [2(NO) + O2]
304.3 - [421.6 + 205.1]
304.3 - 626.7
ΔS =-322.4 J/K
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ΔS is negative, we know that entropy decreases; the system becomes less random. The reaction will not be spontaneous.
Practice Problems
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1. Predict whether entropy is increasing
(ΔS > 0) or decreasing (ΔS < 0)? Give a reason for your answer.
a)
steam condenses to water
b)
solid CO2 sublimes
c)
N2O4 (g) → 2 NO2 (g)
d)
water is heated from 25°C to 50°C
e)
C6H6 (l) + O2 (g) → 6 CO2 (g) + 3 H2O (l)
Answer
a) steam condenses to water
ΔS < 0; entropy decreases because liquid water is less random than gaseous water (steam)
b) solid CO2 sublimes
ΔS > 0; entropy increases because gaseous CO2 is more random than the solid state.
c)
N2O4 (g) → 2 NO2 (g)
ΔS > 0; entropy increases because two moles of a gas are more random than 1 mole of a gas.
d) water is heated from 25°C to 50°C
ΔS > 0; entropy increases as particles move about more when heated.
e) C6H6 (l) + O2 (g) → 6 CO2 (g) + 3 H2O (l)
ΔS > 0; entropy increases because there are more moles on the product side of the equation (9 vs 2)
Practice problem #2
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Calculate ΔS for the following reaction. Is entropy increasing or decreasing? Is the system becoming more random or less random?
Based on entropy changes only, would you predict the reaction to be spontaneous or not?
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Na (s) + ½Cl2 (g) → NaCl (s)
Solution:
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Look up S° values for all reaction participants, taking care to look in the correct column of the table of thermochemical data. Multiply
these values by coefficients in the balanced equation, then simplify:
Na
+½ Cl2
NaCl(s)
51.2
+½  223.1
72.1
=
162.8
72.1
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Use the following formula to find ΔS°:
ΔS = ΣSproducts - ΣSreactants
= 72.1 - 162.8
= -90.7 J/K
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A negative sign for ΔS° tells us that entropy is decreasing. The system becomes less random. We would NOT expect this reaction to
occur spontaneously.
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Assignment 1.3.1- entrophy