Idea Transcript
15.3 ENTROPY The study of energy and how it changes is called thermodynamic. In thermodynamics, the object / reaction / change / process being studied is called the system and everything else is called the surroundings. The energy of the system and the surrounding is referred to as the ‘universe’. Chemical substances have two types of energy 1. Kinetic – energy of molecular movement. There are three types – rotation, translation and vibration. Kinetic energy is useable energy because it makes particles move and can be transformed into other kinds of useful energy. 2. Potential – energy stored in chemical bonds. The amount of potential energy stored in a bond depends on the amount of electron-electron repulsion, electron-nucleus attraction and nucleus-nucleus repulsion and therefore depends on the radius, nuclear charge and number of bonding electrons. The total energy of a system is the sum of the potential and kinetic energies. Laws are generalizations, principles or patterns in nature and theories are the explanations of those generalizations. Many theories if sufficiently tested can become laws, but a theory is not needed to bestow lawlessness. The Laws of Energy/Thermodynamics These laws can be used to account for the change in energy that accompanies bond breaking and making during a chemical reaction. 1. The first law of thermodynamics or Law of conservation of energy Energy cannot be created or destroyed but can be transformed (converted) from one form to another. This means that the energy of the ‘universe’ does not change even though energy is being transformed from one form to another. If you lift a book off the table, and let it fall, the total energy of the ‘universe’ remained unchanged. 2. The second law of thermodynamics When energy is transformed from one form to another, some of its transferable ability (work) is lost as heat energy. The second law tells us that heat differs from other forms on energy in that it is not convertible. For example potential energy can be converted to kinetic energy and heat energy, however, complete conversion of heat into other forms of energy is not possible. One of the interesting things about thermodynamics is that although it deals with matter, it makes no assumptions about the nature of that matter. This is an important quality, because it means that laws and theories of thermodynamics are unlikely to change as new knowledge about the structure and bonding of matter comes to light. In thermodynamics entropy, S is a measure of the number of ways (probability) of arranging molecules so as to share their kinetic energy. As the probability increases, the entropy increases. Entropy is measured in JK-1mol-1 or JK-1.
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Generally speaking gases have higher entropy than liquids which have higher entropy than solids. The more spread-out, mixed up or disordered a system, the more ways the kinetic energy of the molecules can be distributed.
Increasing disorder, increasing entropy
Disorder therefore is more probable than order because disorder produces more ways of distributing kinetic energy. Thus coins and cards tend to assume random configurations when tossed or shuffled. Everybody knows that a gas will expand and fill the volume within which it is confined completely and uniformly. Entropy drives this expansion. When there is more space available; the individual molecules will disperse them throughout the space because the larger volume increases the number of ways of distributing the total kinetic energy of the molecules. Warm water contains more heat energy than cold water of the same volume. Since heat energy is proportional to the average kinetic energy of the molecules, warm water has greater entropy. Warm water disperses into cold water because the number of ways the total kinetic energy of the water molecules can be distributed increases when the volume is increased. Living organisms are made up of atoms organized into molecules (DNA, proteins, lipids, cell membrane).
Like all other biological structures a cell membrane is highly specialized. The phospholipids and protein molecules involved have their own unique structure and organization. The cell membrane has low entropy.
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According to the second law of thermodynamics when energy is transformed disordered heat energy is produced. Therefore according to this law all process/reactions in living and non-living things tend to favor the direction that increases the amount of disorder or entropy. During the process of photosynthesis carbon dioxide gas, water vapor and energy create more ordered liquid glucose (C6H12O6) and gaseous oxygen with lower entropy. Respiration releases heat energy into the surroundings increasing its entropy. The process of respiration (system) has a greater probability of occurring because it produces products that are more stable because they have higher entropy. photosynthesis Heat 6 CO2 (g) + 6 H2O (g) + C6H12O6 (l) + 6 O2 (g) Energy respiration Entropy rules In general in chemical and physical changes: 1.
The more disordered the state, the greater the entropy and the greater it’s probability of occurring. This produces a more positive value for S. Decrease in order (increase in disorder), Increase in Entropy
Solid (ice) S= 48 JK-1mol-1 2.
Liquid (water) S = 70 JK-1mol-1
Gas (steam) S = 189 JK-1mol-1
The entropy will increase when there is an increase in the number of moles of gas molecules produced. The greater the number of moles of gas the more positive the entropy. e.g.
N2O4(g)
2NO2(g)
→
1 mole gas
2 moles gas
Conversely the entropy will decrease if there is a decrease in the number of moles of gas produced.
CH4(g)
+
2O2(g) 3 moles gas
→
CO2(g)
+
2H2O(l)
1 mole gas
If there is no change in the number of moles of gas there will be no change in entropy.
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3.
When the entropy increases, the entropy change, ∆S, the difference in entropy of the reactants and products becomes more positive. The products have more entropy than the reactants and therefore are more stable than the reactants. Solid (ice) → -1 -1 S= 48 JK mol
Liquid (water) S = 70 JK-1mol-1
Gas (water) ∆S = 189 – 48 = 141 JK-1mol-1 S = 189 JK-1mol-1 S becomes more positive
Solid (ice) → -1 -1 S= 48 JK mol 4.
∆S = 70 – 48 = 22 JK-1mol-1 S becomes more positive
When the entropy decreases, the entropy change, ∆S is negative. The reactants have more entropy than the products and are therefore more stable than the products. Solid ice has less entropy than liquid water. The liquid water is more stable than the solid ice because it has more entropy. Liquid (water) S = 70 JK-1mol-1
→
Solid (water) S= 48 JK-1mol-1
∆S = 48 – 70 = -22 JK-1mol-1
Dissolving sucrose in water increases the disorder and increases the entropy. The entropy change is positive. C12H22O11(s)
→
C12H22O11(aq)
∆S = positive
If there is no change in the entropy of the reactants and products then ∆S is zero Fe2O3(s)
+ 2 Al (s)
2 Fe (s)
+ Al2O3 (s)
Both reactants and products are solids 5.
An increase in the temperature of the system will increase the average kinetic energy of particles, increasing the entropy.
In general the greatest increase in entropy usually occurs when the number of particles in the gaseous state increases.
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Standard Entropy Change (∆ ∆Sө)
15.3.2
Entropy, S is measured in JK-1mol-1 or JK-1. Joules tend to be used over KJ because entropy values are typically quite small. Like enthalpy, entropy can be determined from experiments. For any chemical or physical change there will be an enthalpy change, ∆H and an entropy change, ∆S. The standard entropy is defined as the absolute entropy of one mole of a substance under standard conditions of temperature and pressure (P = 1.01 x 105 Pa (or 1atm), Temperature, T = 298 K (or 25˚C). The standard entropy change for a reaction has the symbol ∆Sө and can be calculated from the formula:
∆Sө
=
Σ ∆Sө products
-
Σ ∆Sө reactants
(JK-1mol-1)
Standard entropy values can be found in the Chemistry data booklet. The standard entropies of all elements and compounds are positive values. Example 1 Calculate the standard entropy change per mole of ammonia associated with the Haber process. N2 (g)
+
3 H2 (g)
↔
2 NH3 (g)
[Standard entropies: N2 (g) =192 Jmol-1K-1; H2 (g) =131 Jmol-1K-1 ; NH3 (g) =192 Jmol-1K-1] Answer ∆S° (NH3)
=
[(2 x 192)] – [ (3 x 131) + 192 ]
=
- 201 JK-1mol-1 ÷ 2
=
-101 JK-1mol-1
Example 2 Calculate the standard entropy change for the reaction: CH4 (g)
+ H2O(g)
3 H2(g) + CO2(g)
[Standard entropies: H2O (g) = 70 Jmol-1K-1; CO2 (g) = 214 Jmol-1K-1; H2 (g) =131 Jmol-1K-1 ; CH4 (g) =186 Jmol-1K-1]
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Problems 1.
Below are six pairs of pictures. Choose the picture in each pair that represents the most high entropy and the lowest entropy.
2.
Which one of the following does not generally lead to an increase in the entropy of a system? A. An increase in the total number of moles of particles. B. The formation of a solution. C. The formation of a gaseous product. D. The formation of solid products.
3.
A short length of magnesium ribbon burns in a crucible over a Bunsen burner. a) Write a balanced equation for the reaction. b) Is the reaction exothermic or endothermic? c) Do you think the product, MgO, can be classified as more or less ordered as the reactants, metallic magnesium and gaseous oxygen? d) Give the sign of the enthalpy and entropy change in this reaction.
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4.
For each of the following reactions, state with a reason whether you would expect the entropy of the products to be greater or less than that of the reactants.
A.
Br2 (l)
B.
+
Ag (aq)
C.
2 NO2 (g)
N2O4 (g)
D.
2 OH-(aq)
+ CO2(g)
E.
H2 (g)
Cl2 (g) 2 HCl (g)
Br2 (g)
+ Cl-(aq)
+
AgCl (s)
H2O (l)
+ CO32- (aq)
5.
5 cm3 of hydrochloric acid is put in a test tube and the temperature of the acid recorded. 3 g of solid ammonium carbonate was added and the mixture stirred. The temperature of the reaction decreased. a) Write a balanced equation for the reaction. b) Consider the surroundings. Would the surroundings gain or lose energy as the chemicals returned to room temperature. c) Comment on the relative change in the magnitude of the entropy and enthalpy and their sign.
6.
From an entropy point of view which one of the following reactions is the products more stable than the reactants.
7.
A.
2C(s)
+
O2(g)
→
2CO(g)
B.
CaO(s)
+
CO2(g)
→
CaCO3(s)
C.
H2(g)
+
Cl2(g)
→
2HCl(g)
D.
2SO2(g)
+
O2(g)
→
2SO3(g)
(M04) Consider the following reaction for the Haber process. The magnitude for the entropy change, ∆S at 27ºC is - 62.7 JK-1mol-1. State with a reason, the sign of ∆S. [2] N2 (g)
8.
+
3 H2 (g)
↔
2 NH3 (g)
Consider each of the following reactions and determine which would have the most positive value of ∆S. A.
CH4 (g)
+ 2 O2(g)
B.
CO2 (g)
+ 3 H2(g) CH3OH(g) + H2O(g)
C.
2 Al (s)
+ 3 S(s)
D.
CH4 (g)
+ H2O(g)
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CO2(g) + 2 H2O(g) Al2S3 (s) 3 H2(g) + CO2(g)
9.
Calculate the standard entropy change for the following reactions at 25°C. A.
CH4 (g)
+ 2 O2(g)
B.
CO2 (g)
+ 3 H2(g) CH3OH(g) + H2O(g)
CO2(g) + 2 H2O(g)
CH4(g) = 186 JK-1mol-1
S∅
O2(g) = 205 JK-1mol-1 CO2(g) = 214 JK-1mol-1 CH3COH(g) = 240 JK-1mol-1 H2O(g) = 189 JK-1mol-1 H2 (g) =131 Jmol-1K-1 10.
Dinitrogen oxide is a greenhouse gas produced from the combustion of fossil fuels. Once in the atmosphere it takes about 150 years to decompose into its elements according to the reaction. N2O(g) N2(g) + ½ O2(g) → ∆H∅ = -162.2 kJ
S∅ for
N2O(g) = 0.220 kJK-1mol-1 N2(g) = 0.192 kJK-1mol-1 O2(g) = 0.205 kJK-1mol-1
Calculate the value for ∆S∅ in kJK-1mol-1and JK-1mol-1 for the decomposition of N2O into its elements. [3] 11.
12.
M99/420/H(2) a) Write the balanced equation for the formation of butanoic acid, C3H7COOH from its elements. b) Calculate the standard entropy change, ∆S∅f for the formation of butanoic acid at 25oC. Substance
Absolute Entropy, S∅ (Jmol-1K-1)
C (s)
5.7
H2 (g)
130.6
O2 (g)
205.0
C3H7COOH (l)
226.3
(N00//H) Define standard entropy, S∅ and explain why the symbol ∆ is not included. [2]
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15.3 Entropy ANSWERS 2. D 3. a) 2 Mg (s) + O2 (g) ↔ 2 MgO (s) b) exothermic, is a combustion reaction. All combustion reactions are exothermic c) MgO is more ordered than Mg and O2, lower entropy. 4. A. B. C. D. E.
Products greater entropy Products less entropy Products less entropy Products less entropy reactants and products have the same entropy (NH4)2CO3(s) + 2 HCl(aq) 2 NH4Cl(aq) + H2O(l) + CO2(g) loses energy ∆H = + and ∆S = +
5.
a) b) c)
6. 7.
A More stable means greater entropy negative; decrease in the number of moles of gas molecules.
8.
D
9.
a)
∆Sө = - 4 JK-1mol-1
b)
∆Sө = - 176 JK-1mol-1
10.
∆Sөf
11.
a)
4 C(s) + 4 H2(g) + O2(g) C3H7COOH(l)
b)
∆Sөf
12.
= = =
∑ Sө products - ∑ Sө reactants [ ½ (0.205) + (0.192) ] – (0.220) + 74.5 JK-1mol-1 (+ 0.0745 kJK-1mol-1) (correct state symbols must be used)
=
∑ Sө products - ∑ Sө reactants
=
(226.3) – [ (4 x 5.7) + 4(130.6) + (205.0) ] Jmol-1K-1
=
226.3 – 750.2 Jmol-1K-1
=
- 523.9 JK-1mol-1
(1 dp)
Standard entropy is related to (a measure of) the disorder of particles; OR It is the quantity of energy owned by one mole of an element or a compound in its standard state at 298K. ∆ is not included because entropy has absolute values OR S can be measured;
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