CHEM Thermodynamics. Entropy, S

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1 hermodynamics Change in Change in Entropy, S

2 Entropy, S Entropy is the measure of dispersal. he natural spontaneous direction of any process is toward greater dispersal of matter and of energy. Dispersal of matter: We analyze the constraints on a system. he more the system is constrained, the less dispersed it is. Dispersal of energy: We analyze the flow of heat to measure how much energy is spread out in a particular process at a specific temperature. 2

3 Entropy, S Entropy measures the dispersal of matter S = k ln W where W is the number of ways of describing the system, and k is the Boltzmann constant (1.38 x J/K). he more ways that the system can reside, the greater the Entropy, S. Note: If there is only one way to describe the system when W=1, the system is fully constrained. S=0 because ln (1)=0 he change in entropy is S. S = S 2 -S 1 where S 1 is the entropy of the initial state, and S 2 is the entropy of the final state. Like E and H, S is a state function. 3

Entropy, S Entropy measures the dispersal of matter Example 1: Melting

the molecules are constrained to fixed positions.

his is state 1 corresponding to an his is state 2 corresponding to an

4 Entropy, S Entropy measures the dispersal of matter Example 1: Melting of ice H 2 O (s) H 2 O (l) S = S 2 -S 1 S >0 In the ice crystal, all the molecules are constrained to fixed positions. In liquid water, molecules are free to move around. his is state 1 corresponding to an his is state 2 corresponding to an entropy state, S 1. entropy state, S 2. Image credit: 4

5 Entropy, S Entropy measures the dispersal of matter Example 2: Rusting of iron 4 Fe (s) + 3 O 2 (g) + 3 H 2 O (l) 2 Fe 2 O 3 3H 2 O(s) S < 0 Image credit: 5

6 Entropy, S Entropy measures the dispersal of matter Example 3: Combustion of methane CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (l) S < 0 6

Entropy, S Entropy measures the dispersal of matter Example 1: Melting of water S > 0 Example 2: Iron rusting S < 0 Example 3: Combustion of methane S < 0 All of the above processes are spontaneous

7 Entropy, S Entropy measures the dispersal of matter Example 1: Melting of water S > 0 Example 2: Iron rusting S < 0 Example 3: Combustion of methane S < 0 All of the above processes are spontaneous at room temperature and atmospheric pressure. Conclusion: When a system changes spontaneously, there may be in increase of entropy (S > 0), or a decrease in entropy (S > 0). he above entropy changes are looking at the S of the system. But in the real world, there is always heat exchanges with the surroundings. herefore, we must consider the entropy change of the universe. 7

8 Entropy, S he Second Law of hermodynamics: In any spontaneous process, there is always an increase in the entropy of the universe. S universe = S system + S surroundings > 0 For a spontaneous process, if there is a decrease in the S system, then the heat is absorbed by the surroundings must have caused an increase in the S surroundings such that overall there is an increase in the S universe (i.e. S universe > 0). 8

9 Entropy, S Entropy change measures the dispersal of energy How much energy is spread out in a particular process, or how widely spread out it becomes at a specific temperature. Every process has a preferred direction natural spontaneity. Spontaneous processes: - Contraction of a stretched elastic band - Heat flows from a warm body to a cold body hese processes increases random kinetic energy of the system. As a result, an increase of energy dispersion results. 9

10 Entropy, S Entropy change measures the dispersal of energy We analyze the flow of heat to measure how much energy is spread out in a particular process at a specific temperature. Every process has a preferred direction natural spontaneity. Let s look at what energy dispersal means on a microscopic scale with micro-states. 10

Entropy, S Entropy is a measure of available micro-states. Consider two chambers containing a few oxygen atoms. he energy of small particles are quantized.

11 Entropy, S Entropy is a measure of available micro-states. Consider two chambers containing a few oxygen atoms. he energy of small particles are quantized. Chamber 1 contains 4 oxygen atoms x w Chamber 2 b a contains y z c 3 oxygen atoms Define a system such that each atom can reside in 5 quantized states. 11

12 Entropy, S Micro-states and Entropy Let s heat chamber 2. Heating causes the atoms move faster. q b c a Heat chamber 2 After heating, there are 10 microstates with energy = 12 After heating, the oxygen atoms have more entropy. he number of available micro-states increases.(10 micro-states from 1 micro-state) 12

13 Entropy, S Recall the Second Law of hermodynamics: here is an inherent direction in which any system which is not in equilibrium moves. In any spontaneous process, the entropy of the universe will increase. In terms of micro-states, it means that all spontaneous process must increase the number of available micro-states. (i.e. he number of available micro-states after the event/reaction will always be greater than the number of available micro-states before the event/reaction). What happens if we bring the heated Chamber 2 in contact with Chamber 1? 13

14 Entropy, S Let s look at what an increase of energy dispersion mean for the process of heat spontaneous flow from a warm body to a cold body. q Chamber 1 contains Chamber 2 contains 4 oxygen atoms w b 3 oxygen atoms Energy = 8 x a y Energy =12 z c 1 micro-state 10 micro-states q he total number of micro-states for this system in the initial state is 10 micro-states (1x10). Micro-state 1: Micro-state 2: Micro-state 3 etc: 14

total number of micro-states for this system in the final state is 24 micro-states (4x6).

15 Entropy, S Let s look at what an increase of energy dispersion mean for the process of heat spontaneous flow from a warm body to a cold body. Chamber 1 contains 4 oxygen atoms Energy = 10 x z w y b c a Chamber 2 contains 3 oxygen atoms Energy =10 4 micro-states 6 micro-states he total number of micro-states for this system in the final state is 24 micro-states (4x6). Micro-state 1: Micro-state 2: Micro-state 4: Micro-state 5: Nature spontaneously proceeds toward the states that have the highest probabilities of existing. Micro-state 3: Micro-state 6 etc: 15

16 Entropy, S On a molecular l scale he number of microstates and, therefore, the entropy tends to increase with increases in: emperature Volume (gases) he number of independently moving molecules 16

17 Entropy, S Entropy is related to the various modes of motion in molecules. l ranslational: Movement of the entire molecule from one place to another. Vibrational: Periodic motion of atoms within a molecule. Rotational: Rotation of the molecule on about an axis or rotation about bonds. Image credit: Chemistry, he Central Science,, 10th edition heodore L. Brown, H. Eugene LeMay, Jr., and Bruce E. Bursten 17

18 Entropy, S More particles lead to more energy states, and therefore, more entropy Higher temperature lead to more energy states, and therefore, more entropy Less structure (gas is less structured than liquid, which is less structured t than solid) lead to more states, t and therefore, more entropy Entropy increases with the freedom of motion of molecules. herefore, S(g) > S(l) > S(s) 18

Entropy, S More spontaneous processes: S > 0 - Free expansion of a gas

- Combustion of methane S < 0 CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(l)

- Melting of ice cubes at room temperature S > 0 H 2 O (s) H 2 O (l) his

19 Entropy, S More spontaneous processes: S > 0 - Free expansion of a gas his process neither absorbs b nor emits heat. - Combustion of methane S < 0 CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(l) he process produces heat (exothermic). - Melting of ice cubes at room temperature S > 0 H 2 O (s) H 2 O (l) his process requires heat (endothermic). Conclusion: Naturally spontaneous processes could be exothermic, endothermic, or not involving heat. 19

20 Entropy, S More spontaneous processes: Photosynthesis 6622light CO H O C H O O Complex and highly-energetic compounds (compared to the starting materials, CO 2 and H 2 O) are formed, it appears that there a decrease in entropy in the process. 20

org/article/photosynthesis When the system and surroundings are considered, there is a net increase in entropy as a result of photosynthesis!

21 Entropy, S A good example is photosynthesis. 6622light 61262S<0; It appears that there a decrease in entropy in this natural process. 6CO H O C H O O Photosynthesis in a plant does not consist of a isolated system of the plant alone. Image credit: When the system and surroundings are considered, there is a net increase in entropy as a result of photosynthesis! Experiments and calculations indicate that the maximum efficiency of photosynthesis in most plants is in the 30% range. his means: 30% of the sunlight that strikes the plant is absorbed by the plant in the photosynthesis of substances that decreases entropy in the plant. 70% of the sunlight that strikes the plant is dispersed to the environment (an entropy increase in slightly heating the leaf and the atmosphere. 21

22 Entropy, S For any spontaneous process, S universe = S system + S surroundings > 0 Spontaneous process, favoured by a decrease in H (exothermic) favoured by an increase in S Nonspontaneous process, favoured by an increase in H (endothermic) favoured by an decrease in S 22

23 Entropy, S Entropy measures the dispersal of energy At a specific temperature, the change in entropy, S, is S q rev where q rev is the heat transferred at a constant temperature in Kelvin. What is q rev? 23

24 Entropy, S What is q rev? S q rev Consider a gas in a piston and cylinder configuration. he gas is held in place by a pile of wheat grains. he gas undergoes isothermal expansion. Since E = 0 for an isothermal process, q is the same magnitude of w but opposite in sign. Let s take the PV curve of five different paths to take the gas from State I to State F. Work done by the gas (w<0) is the area under the PV curve. w rev w irrev his is the reversible path. he area under the PV curve is the greatest. his path yields the smallest w (most negative), and the largest q. We refer to the heat transfer of a reversible path reversible q, q rev. Of all the different possible paths to take the gas from State I to F, q rev is the limit of the observed heat transferred. S q rev q irrev 24

25 Entropy, S What is q rev? S q rev Likewise, consider a gas in a piston and cylinder configuration. he gas is held in place by a pile of wheat grains. he gas undergoes isothermal compression. Since E = 0 for an isothermal process, q is the same magnitude of w but opposite in sign. Let s take the PV curve of five different paths to take the gas from State I to State F. Work done on the gas (w>0) is the area under the PV curve. w rev w irrev his is the reversible path. he area under the PV curve is the smallest. his path yields the smallest w, and the largest q. We refer to the heat transfer of a reversible path reversible q, q rev. Of all the different possible paths to take the gas from State I to F, q rev is the limit of observed heat transferred. S q rev q irrev 25

26 Entropy, S What is q rev? S q rev A reversible change is one that is carried out in such a way, such that, when the change is undone, both the system and the surroundings remain unchanged. his is achieved when the process proceeds in Infinitesimal steps that would take infinitely long to occur. Reversibility is an idealization that is unachievable in real process, except when the system is in equilibrium. A process that is carried out reversibly has the smallest w rev, and the largest q rev. 26

Entropy, S Entropy measures the dispersal of energy At a specific temperature, the change in entropy, S, is q rev S S where q rev is the heat transferred at a

Consider ice melting at 0 o C, where H fusion = 79.71 cal/g = 6.003 kj/mole his is a reversible process.

27 Entropy, S Entropy measures the dispersal of energy At a specific temperature, the change in entropy, S, is q rev S S where q rev is the heat transferred at a constant temperature in Kelvin. Although most real processes are irreversible, there are real reversible processes. Consider ice melting at 0 o C, where H fusion = cal/g = kj/mole his is a reversible process. H 2O (s) H 2O (l) If the heat is absorbed in small enough increments such that it does not disturb the equilibrium, this is a reversible process. 603q H. 20S K kj rev fusion 273mole. J mole K here is an increase in the entropy of water when ice melts to form liquid water. 27

28 Entropy, S Similarly, il l consider water vapourizing i at 100 o C, where H vapourization = 40.7 kj/mole his is a reversible process. H 22O (l) H O (g) If the heat is absorbed in small enough increments such that it does not disturb the equilibrium, this is a reversible process. 07 K kj q H rev vap. 1mole S J mole K here is an increase in the entropy of water when water is converted to steam. Note: S for the vapourization of water is much higher than that of fusion. 28

Entropy, S Expansion of an Ideal Gas Recall how we define a reversible process. For the expansion of a gas in a piston and cylinder configuration under constant temperature condition.

29 Entropy, S Expansion of an Ideal Gas Recall how we define a reversible process. For the expansion of a gas in a piston and cylinder configuration under constant temperature condition. Since E = 0 for an isothermal process, q = -w. Work done by the gas (w<0) is the area under the PV curve. w rev w rev w rev V 2 It follows that - P ext dv V V1 q rev nr ln 1V2 nr - dv V1 V V 2 dv -nrv 1 V V -nr ln 2V w rev w rev 1Entropy increases with an increase in volume or a decrease in pressure. V 2V q 2rev V S nr ln 1 V or q rev P S nr ln 2 P 1P 29

30 Change in Entropy in the surroundings, S surroundings Entropy Changes in Surroundings Heat that flows into or out of the system also changes the entropy of the surroundings. For an isothermal process: S surroundings q system At constant pressure, q sys is H for the system. S surroundings q system H 30

31 Change in Entropy in the surroundings, S surroundings A phase change is isothermal (no change in ). H 2O (s) H 2O (l) For H 2 O: H fusion = kj/mole H vapourization = 40.7 kj/mole If this is done reversibly: S surr = S sys S surroundings q system H 31

Entropy, S 1.00 mole of a monatomic ideal gas at 273 K is expanded via two paths: (a) isothermally and reversibly from 22.4 L to 44.8 L.

External pressure, P ext (w < 0) Under isothermal condition, =0, therefore E = H = 0 Since E = nc v and H = nc p q (q > 0) his is because work

q V 18314257 S V rev 2J nr ln ( mole) (.. mole K ) ln 1 64J K his is S system. What is S surroundings? What is S universe? q rev = S = 5.

32 Entropy, S 1.00 mole of a monatomic ideal gas at 273 K is expanded via two paths: (a) isothermally and reversibly from 22.4 L to 44.8 L. Find w, q, E, H, S system, S surroundings, and S universe. External pressure, P ext (w < 0) Under isothermal condition, =0, therefore E = H = 0 Since E = nc v and H = nc p q (q > 0) his is because work of expansion done by the gas equals to the heat flow into the system. Net overall change in internal energy and enthalpy is zero. q V S V rev 2J nr ln ( mole) (.. mole K ) ln 1 64J K his is S system. What is S surroundings? What is S universe? q rev = S = = kj Since E = 0, q = -w = kj Since the amount of heat that flowed out of the surroundings is q = kj, S 273 S S S surr. J K universe system surr his is true for a reversible process. he system is at equilibrium. 32

Entropy, S 1.00 mole of a monatomic ideal gas at 273 K is expanded via two paths: (b) via free expansion from 22.4 L to 44.8 L. Find w, q, E, H, and S system, S surroundings, and S universe.

33 Entropy, S 1.00 mole of a monatomic ideal gas at 273 K is expanded via two paths: (b) via free expansion from 22.4 L to 44.8 L. Find w, q, E, H, and S system, S surroundings, and S universe. his is an irreversible process with no change in temperature, =0. As a result, E = H = 0 w = 0 because P ext = 0 w = -q, q =0 Since E = nc v and H = nc p o calculate S, we need to find a reversible path that will take the gas from the same initial state to the same final state. Part (a) is such a path. q V J S nr ln 1 ( mole) (. mole K ) ln V rev. 64J K his is S system. It has the same disorder as calculated in Part (a). his is a What is S surroundings? spontaneous What is S universe? process. S universe > 0 Since the amount of heat that flowed out of the surroundings zero 5zero,0764S surr S universe S system S surr. J K 33

34 Entropy, S Entropy associated with temperature change 22S S ds S 11 dq rev Recall, q = n C where n = number of moles C = molar heat capacity = temperature difference nc S d the heat capacity is constant 1and over the temperature range: 2For a change in temperature of n moles of substance from 1 to 2 S q rev At a specific temperature Under constant volume condition Under constant pressure condition 2 nc d 2v nc pd S 1S S nc ln v 1 S nc p ln 1 34

35 Entropy, S Entropy associated with temperature change S nc d 2For a change in temperature of n moles of substance from 1 to 2 S q rev At a specific temperature and the heat capacity is expressed as a power series, a + b + c nc S d 1 in Kelvin 22 n(a b c )d S 2122 S n a 1 d b 1 d c 1 d c 1 S n a ln 1 b( ) ( 2 ) H nc d 1 22p 21H n (a b c )d 2222 H na 11d b 1 d c 1 d b c H n a( ) ( 2 11) 3( 2 ) 2S H Compare 35

36 Entropy, S Calculate S for the change in state H O o (liq, 1 atm, 25 C ) H 22 S O (g, 1 atm, 25 o C ) Increase the volume of the container. he amount of gaseous H 2 O present is governed by the vapour pressure of water at 25 o C. Given C p (H 2 O,liq),q) = 18 cal mole -1 K -1 C p (H 2 O, g) = 9.0 cal mole -1 K -1 H vap = 9713 cal/mole (at 373K) Let s devise a reversible process to calculate S. H O o (liq, 1 atm, 25 C) H 22 S O (g, 1 atm, 25 o C ) Warming liquid water, S 1 Cooling gaseous water, S 3 H O o (liq, 1 atm, 100 C ) H 22 S 2 O (g, 1 atm, 100 o C ) S = S 1 + S 2 + S 3 36

37 Entropy, S Calculate S for the change in state H S > 0 2o o O (liq, 1 atm, 25 C ) S H 2O (g, 1 atm, 25 C ) Given C = -1-1 p (H 2 O,liq) 18 cal mole K C p (H 2 O, g) = 9.0 cal mole -1 K -1 H vap = 9713 cal/mole (at 373K) S < 0 H 2S O o (liq, 1 atm, 25 C ) H 2O (g, 1 atm, 25 o C ) Warming liquid water, S 1 Cooling gaseous water, S nc ln 8ln 1 H O S 3 nc p ln ln S p 298o (liq, 1 atm, 100 C ) H 22 S 2 S > 0 S 2 H vap cal mole K O (g, 1 atm, 100 o C ) S = S S 2 + S 3 = = 28.1 cal K 37

38 Entropy, S routon s Rule A useful generalization that works for many liquids at their normal boiling point (i.e. boiling point at 1 atm) is that, the standard molar entropy of vapourization has a value of about 22 cal mole -1 K -1 or 88 J mole -1 K -1 Since S o vapourization H vapourization boiling po int he idea is, if the degree of disorder associated in the phase transformation of 1 mole of liquid to 1 mole of vapour at 1 atm is the same, then the S o vapourization of different liquids should be similar. S o vapourization = 22 cal mole -1 K -1 or 88 J mole -1 K -1 routon s rule is useful method for estimating the enthalpy of vapourization of a liquid if its normal boiling point is known. Example: We can estimate the enthalpy of vapourization of Hg using routon s rule from the normal boiling point of Hg. he normal boiling point of Hg is 357 o C. S 88H vapourization H Joule 3mole K o vapourization 6boiling point vapourization H vapourization = 630 K 88 J mole -1 K -1 = 55 kj mole -1 0K Compare: Heat of vaporization of mercury: kj mol -1 Cited value: Wikipedia 38

Entropy, S routon s Rule S o vapourization = 22 cal mole -1 K -1 or 88 J mole -1 K -1 Substance H vap (kcal mole -1 ) b ( o C) S vap (cal mole -1 K -1 ) O 2 1.630-182.97 18.07 CH 4 1.955-161.49 17.

39 Entropy, S routon s Rule S o vapourization = 22 cal mole -1 K -1 or 88 J mole -1 K -1 Substance H vap (kcal mole -1 ) b ( o C) S vap (cal mole -1 K -1 ) O CH H 2 O NH C 2 H 5 OH CHCl C 6 H (C 2 H 5 ) 2 O ether (CH 3 ) 2 CO acetone In liquid state, hydrogen bonding between molecules produces a greater degree of order than expected. As a result, the degree of disorder produced in the vapourization process is generally greater than the nominal 22 cal mole

40 Entropy, S he hird Law of hermodynamics: he entropy of a pure perfect crystal is zero when the crystal is at zero Kelvin (0 K) A pure perfect crystal is one in which every molecule is identical, and the molecular alignment is perfectly even throughout the substance. For non-pure crystals, or those with less-than perfect alignment, there will be some dispersal of the imperfection, so the entropy cannot be zero. 40

41 Entropy, S he entropy of the crystal gradually increases with temperature as the average kinetic energy of the particles increases. As the crystal warms to temperatures above 0 K, the particles in the crystal start to move, generating some disorder. he entropy of the liquid gradually increases as the liquid becomes warmer because of the increase in the vibrational, rotational, and translational motion of the particles. At the melting point, m, the entropy of the system increases in entropy without a temperature change as the compound is transformed into a liquid, id which h is not as well ordered as the solid. At the boiling point, there is another increase in the entropy of the substance without a temperature change as the compound is transformed into a gas. 41

42 Standard Molar Entropy, S o Standard entropies (measured at 298K) have been carefully measured for many substances. For example, the standard molar entropy of some solids: S o Joules K -1 mole -1 C(diamond) 2.38 C (graphite) 5.74 Sodium 51.3 Phosphorus (white) 41.1 Sulfur (rhombic) 31.8 Silver

43 Standard Molar Entropy, S o Standard entropies (measured at 298K) have been carefully measured for many substances. For example, the standard molar entropies (standard entropy per mole) for gases are usually higher because heat of melting and heat of vapourization must be included. he standard molar entropies for noble gases are: S o Joules K -1 mole -1 He Ne Ar Kr Xe For liquids and gases S o are usually higher because heat of melting and heat of vaporization must be included. 43

44 Standard Molar Entropy, S o Standard entropies (measured at 298K) have been carefully measured for many substances. For example, the standard molar entropies (standard entropy per mole) for diatomic gases. S o Joules K -1 mole -1 H N O Halogens: F Cl Br I For diatomic gases, they are usually higher than those of monatomic noble gases as there are more degree of freedom of motion. 44

45 Standard Molar Entropy, S o For example, the standard molar entropies (standard entropy per mole) for compounds. S o Joules K -1 mole -1 H 2 O (l) H 2 O 2 (l) CH 3 OH(l) CH 3 Cl(l) CH 3 Cl 3 (l) S o Joules K -1 mole -1 H 2 O (g) CO (g) CO 2 (g) NO (g) NO 2 (g) N 2 O 4 (g) SO 2 (g) A NO molecule has only one type of vibrational molecule A NO 2 molecule has three types of vibration. 45

46 Standard Molar Entropy, S o Standard entropies (measured at 298K) have been carefully measured for many substances. For example, the standard molar entropies (standard entropy per mole) for more gases. S o Joules K -1 mole -1 CH 4 (g) 186 C 2 H 2 (g) 201 C 2 H 4 (g) 221 C 2 H 6 (g) In general, the more complicate the molecule and the heavier the molar mass, the higher the standard entropy. S o reaction n products products S o ( products ) n reactants reactants S o ( reactants ) 46

47 Standard Molar Entropy, S o Calculate the standard entropy change accompanying the burning of ethane, C 2 H 6. C 2 H 6 (g) + 7 / 2 O 2 (g) 2 CO 2 (g) + 3 H 2 O (g) S =? S o Joules K -1 mole -1 C 2H 6 (g) In general the more complicate O 2 (g) CO 2 (g) H 2 O (g) In general, the more complicate the molecule and the heavier the molar mass, the higher the standard entropy. S o reaction n products products S o ( products) n reactants reactants t S o ( reactants) S o = 2S o (CO 2 (g)) + 3 S o (H 2 O (g)) (S o (C 2 H 6 (g)) + 7 / 2 S o (O 2 (g))) S o = 2(213.6) + 3 (188.7) ( / 2 (205.1)) S o = 46.3 Joules K -1 47

48 Summary: he Second Law of hermodynamics: In any spontaneous process, there is always an increase in the entropy of the universe. S universe = S system + S surroundings > 0 Change in entropy, S, at a specific temperature,. Change in entropy, S, measured over a temperature range from 1 to 2. 2S 1 nc v d S q rev Assuming that C v and C p are constant from 1 to 2. S 2 nc ln v S nc 2p ln 1 1 For an isothermal process expansion or compression of a gas. V S nr ln 21 V P S nr ln 12 P he hird Law says the entropy of a pure perfect crystal is zero when the crystal is at zero Kelvin (0 K). For a chemical reaction, S o can be calculated from the reactants and products standard molar entropies. S o reaction n products products S o ( products) n reactants reactants S o ( reactants) 48

S = k log W 11/8/2016 CHEM Thermodynamics. Change in Entropy, S. Entropy, S. Entropy, S S = S 2 -S 1. Entropy is the measure of dispersal.

S = k log W 11/8/2016 CHEM Thermodynamics. Change in Entropy, S. Entropy, S. Entropy, S S = S 2 -S 1. Entropy is the measure of dispersal. Entropy is the measure of dispersal. The natural spontaneous direction of any process is toward greater dispersal of matter and of energy. Dispersal of matter: Thermodynamics We analyze the constraints