Equilibrium In Chemical Reaction.

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1 Equilibrium In Chemical Reaction

2 The double arrow tells us that
N2 + 3H NH3 The double arrow tells us that this reaction can go in both directions:

3 1) Reactants react to become products,
N2 + 3H NH3 1) Reactants react to become products, N2 + 3H NH3 (‘forward’ reaction)

4 1) Reactants react to become products, N2 + 3H2 2NH3
(‘forward’ reaction) while simultaneously, 2) Products react to become reactants N2 + 3H NH3 (‘reverse’ reaction)

5 N2 + 3H NH3 In a closed system, where no reactants, products, or energy can be added to or removed from the reaction, a reversible reaction will reach equilibrium.

6 N2 + 3H NH3 At equilibrium, the rate of the forward reaction becomes equal to the rate of the reverse reaction, and so, like our escalator metaphor, the two sides, reactants and products, will have constant amounts, even though the reactions continue to occur.

7 N2 + 3H NH3 However (like the metaphor), the equilibrium amounts of reactants and products are usually not equal, they just remain unchanged.

8 N2 + 3H NH3

9 N2 + 3H NH3

10 N2 + 3H NH3

11 N2 + 3H NH3

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13

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17 reverse forward

18 reverse forward

19 reverse forward

20 reverse forward

21 reverse forward

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24 reverse forward

25 reverse forward

26 reverse forward

27 reverse forward

28 reverse forward

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30 reverse forward etc! the reactions go on continuously in both directions.

31 Changes in the concentrations of the reactants and products can be graphed; the graph indicates when equilibrium has been reached. concentration time

32 suppose you begin with the following:
For N2 + 3H NH3, suppose you begin with the following: N2 = 1 M, H2 = 1 M, and NH3 = 0 M concentration time

33 suppose you begin with the following:
For N2 + 3H NH3, suppose you begin with the following: N2 = 1 M, H2 = 1 M, and NH3 = 0 M concentration time

34 suppose you begin with the following:
For N2 + 3H NH3, suppose you begin with the following: N2 = 1 M, H2 = 1 M, and NH3 = 0 M concentration time

35 suppose you begin with the following:
For N2 + 3H NH3, suppose you begin with the following: N2 = 1 M, H2 = 1 M, and NH3 = 0 M concentration time

36 suppose you begin with the following:
For N2 + 3H NH3, suppose you begin with the following: N2 = 1 M, H2 = 1 M, and NH3 = 0 M concentration time

37 suppose you begin with the following:
For N2 + 3H NH3, suppose you begin with the following: N2 = 1 M, H2 = 1 M, and NH3 = 0 M concentration time

38 suppose you begin with the following:
For N2 + 3H NH3, suppose you begin with the following: N2 = 1 M, H2 = 1 M, and NH3 = 0 M concentration time

39 suppose you begin with the following:
For N2 + 3H NH3, suppose you begin with the following: N2 = 1 M, H2 = 1 M, and NH3 = 0 M concentration time

40 suppose you begin with the following:
For N2 + 3H NH3, suppose you begin with the following: N2 = 1 M, H2 = 1 M, and NH3 = 0 M concentration time

41 suppose you begin with the following:
For N2 + 3H NH3, suppose you begin with the following: N2 = 1 M, H2 = 1 M, and NH3 = 0 M concentration time

42 suppose you begin with the following:
For N2 + 3H NH3, suppose you begin with the following: N2 = 1 M, H2 = 1 M, and NH3 = 0 M concentration time

43 suppose you begin with the following:
For N2 + 3H NH3, suppose you begin with the following: N2 = 1 M, H2 = 1 M, and NH3 = 0 M concentration time

44 suppose you begin with the following:
For N2 + 3H NH3, suppose you begin with the following: N2 = 1 M, H2 = 1 M, and NH3 = 0 M concentration time

45 suppose you begin with the following:
For N2 + 3H NH3, suppose you begin with the following: N2 = 1 M, H2 = 1 M, and NH3 = 0 M concentration time

46 suppose you begin with the following:
For N2 + 3H NH3, suppose you begin with the following: N2 = 1 M, H2 = 1 M, and NH3 = 0 M concentration time

47 suppose you begin with the following:
For N2 + 3H NH3, suppose you begin with the following: N2 = 1 M, H2 = 1 M, and NH3 = 0 M concentration time

48 suppose you begin with the following:
For N2 + 3H NH3, suppose you begin with the following: N2 = 1 M, H2 = 1 M, and NH3 = 0 M concentration time

49 For N2 + 3H NH3, suppose you begin with the following: N2 = 1 M, H2 = 1 M, and NH3 = 0 M [N2] [H2] concentration [NH3] time

50 Question 3: at what point has equilibrium been established?
For N2 + 3H NH3, suppose you begin with the following: N2 = 1 M, H2 = 1 M, and NH3 = 0 M [N2] [H2] concentration [NH3] time Question 3: at what point has equilibrium been established?

51 For N2 + 3H NH3, suppose you begin with the following: N2 = 1 M, H2 = 1 M, and NH3 = 0 M [N2] [H2] concentration [NH3] time Question 4: what does the graph tell you about the concentration of each species once equilibrium is established?

52 For N2 + 3H NH3, suppose you begin with the following: N2 = 1 M, H2 = 1 M, and NH3 = 0 M [N2] [H2] concentration [NH3] time Question 5: what might a rate vs time graph look like for the above reaction?

53 For and still beginning with N2 + 3H2 2NH3
N2 = 1 M, H2 = 1 M, and NH3 = 0 M rate time Question 5: what might a rate vs time graph look like for the above reaction?

54 For and still beginning with
N2 + 3H NH3 N2 = 1 M, H2 = 1 M, and NH3 = 0 M rate time

55 Question 6: at what point has equilibrium been established?
For and still beginning with N2 + 3H NH3 N2 = 1 M, H2 = 1 M, and NH3 = 0 M forward rate reverse time Question 6: at what point has equilibrium been established?

56 Question 7: describe how the two graphs are related.
forward H2 concentration rate reverse NH3 time time Question 7: describe how the two graphs are related.

57 N2 H2 NH3 concentration rate time time
forward H2 concentration reverse NH3 time Question 8: do either of the two graphs indicate if Keq >1 or Keq <1?

58 Equilibrium: the extent of a reaction
In stoichiometry we talk about theoretical yields, and the many reasons actual yields may be lower. Another critical reason actual yields may be lower is the reversibility of chemical reactions: some reactions may produce only 70% of the product you may calculate they ought to produce. Equilibrium looks at the extent of a chemical reaction.

59 The Concept of Equilibrium
Consider colorless frozen N2O4. At room temperature, it decomposes to brown NO2: N2O4(g)  2NO2(g). At some time, the color stops changing and we have a mixture of N2O4 and NO2. Chemical equilibrium is the point at which the rate of the forward reaction is equal to the rate of the reverse reaction. At that point, the concentrations of all species are constant. Using the collision model: as the amount of NO2 builds up, there is a chance that two NO2 molecules will collide to form N2O4. At the beginning of the reaction, there is no NO2 so the reverse reaction (2NO2(g)  N2O4(g)) does not occur.

60 N2O4(g) 2NO2(g) As the substance warms it begins to decompose:
When enough NO2 is formed, it can react to form N2O4: 2NO2(g)  N2O4(g). At equilibrium, as much N2O4 reacts to form NO2 as NO2 reacts to re-form N2O4 The double arrow implies the process is dynamic. N2O4(g) NO2(g)

61 B A As the reaction progresses [A] decreases to a constant,
[B] increases from zero to a constant. When [A] and [B] are constant, equilibrium is achieved. A B

62 No matter the starting composition of reactants and products, the same ratio of concentrations is achieved at equilibrium. For a general reaction the equilibrium constant expression is where Kc is the equilibrium constant.

63 Kc is based on the molarities of reactants and products at equilibrium.
We generally omit the units of the equilibrium constant. Note that the equilibrium constant expression has products over reactants.

64 Write the equilibrium expression for the following reaction:

65 The Equilibrium Constant
in Terms of Pressure If KP is the equilibrium constant for reactions involving gases, we can write: KP is based on partial pressures measured in atmospheres.

66 of Equilibrium Constants
The Magnitude of Equilibrium Constants Therefore, the larger K the more products are present at equilibrium. Conversely, the smaller K the more reactants are present at equilibrium. If K >> 1, then products dominate at equilibrium and equilibrium lies to the right. If K << 1, then reactants dominate at equilibrium and the equilibrium lies to the left.

67 An equilibrium can be approached from any direction.
Example:

68 However, The equilibrium constant for a reaction in one direction is the reciprocal of the equilibrium constant of the reaction in the opposite direction.

69 Heterogeneous Equilibria
When all reactants and products are in one phase, the equilibrium is homogeneous. If one or more reactants or products are in a different phase, the equilibrium is heterogeneous. Consider: experimentally, the amount of CO2 does not seem to depend on the amounts of CaO and CaCO3. Why?

70 Heterogeneous Equilibria

71 Heterogeneous Equilibria
Neither density nor molar mass is a variable, the concentrations of solids and pure liquids are constant. (You can’t find the concentration of something that isn’t a solution!) We ignore the concentrations of pure liquids and pure solids in equilibrium constant expressions. The amount of CO2 formed will not depend greatly on the amounts of CaO and CaCO3 present. Kc = [CO2]  Kp = [pCO2]

72 Applications of Equilibrium Constants
Predicting the Direction of Reaction We define Q, the reaction quotient, for a reaction at conditions NOT at equilibrium as where [A], [B], [P], and [Q] are molarities at any time. Q = K only at equilibrium.

73 Predicting the Direction of Reaction
If Q > K then the reverse reaction must occur to reach equilibrium (go left) If Q < K then the forward reaction must occur to reach equilibrium (go right)

74 What we are asked for here is the equilibrium concentration of H2 ...
Note the moles into a L vessel stuff ... calculate molarity. Starting concentration of HI: 2.5 mol/10.32 L = M 2 HI H I2 Initial: Change: Equil: 0.242 M 0 0 -2x +x +x x x x What we are asked for here is the equilibrium concentration of H2 ... ... otherwise known as x. So, we need to solve this beast for x.

75 And yes, it’s a quadratic equation. Doing a bit of rearranging:
x = or – Since we are using this to model a real, physical system, we reject the negative root. The [H2] at equil. is M.

76 This type of problem is typically tackled using the “three line” approach:
2 NO + O NO2 Initial: Change: Equil.:

77 Approximating If Keq is really small the reaction will not proceed to the right very far, meaning the equilibrium concentrations will be nearly the same as the initial concentrations of your reactants – x is just about 0.20 is x is really dinky. If the difference between Keq and initial concentrations is around 3 orders of magnitude or more, go for it. Otherwise, you have to use the quadratic.

78 x = 3.83 x 10-6 M Initial Concentration of I2: 0.50 mol/2.5L = 0.20 M
I I More than 3 orders of mag. between these numbers. The simplification will work here. Initial change equil: -x x 0.20-x x With an equilibrium constant that small, whatever x is, it’s near dink, and 0.20 minus dink is 0.20 (like a million dollars minus a nickel is still a million dollars). 0.20 – x is the same as 0.20 x = 3.83 x 10-6 M

79 I2 2 I 0.20 0 Initial: -x +2x Change: 0.20-x 2x equil:
Initial Concentration of I2: 0.50 mol/2.5L = 0.20 M I I These are too close to each other ... 0.20-x will not be trivially close to 0.20 here. Initial: Change: equil: -x x 0.20-x x Looks like this one has to proceed through the quadratic ...

80 Le Châtelier’s Principle
Le Chatelier’s Principle: if you disturb an equilibrium, it will shift to undo the disturbance. Remember, in a system at equilibrium, come what may, the concentrations will always arrange themselves to multiply and divide in the Keq equation to give the same number (at constant temperature).

81 Change in Reactant or Product Concentrations
Adding a reactant or product shifts the equilibrium away from the increase. Removing a reactant or product shifts the equilibrium towards the decrease. To optimize the amount of product at equilibrium, we need to flood the reaction vessel with reactant and continuously remove product (Le Châtelier). We illustrate the concept with the industrial preparation of ammonia

82 Consider the Haber process
If H2 is added while the system is at equilibrium, the system must respond to counteract the added H2 (by Le Châtelier). That is, the system must consume the H2 and produce products until a new equilibrium is established. Therefore, [H2] and [N2] will decrease and [NH3] increases.

83 Change in Reactant or Product Concentrations
The unreacted nitrogen and hydrogen are recycled with the new N2 and H2 feed gas. The equilibrium amount of ammonia is optimized because the product (NH3) is continually removed and the reactants (N2 and H2) are continually being added. Effects of Volume and Pressure As volume is decreased pressure increases. Le Châtelier’s Principle: if pressure is increased the system will shift to counteract the increase.a

84 Consider the production of ammonia
As the pressure increases, the amount of ammonia present at equilibrium increases. As the temperature decreases, the amount of ammonia at equilibrium increases. Le Châtelier’s Principle: if a system at equilibrium is disturbed, the system will move in such a way as to counteract the disturbance.

85 Change in Reactant or Product Concentrations

86

87 Effects of Volume and Pressure
The system shifts to remove gases and decrease pressure. An increase in pressure favors the direction that has fewer moles of gas. In a reaction with the same number of product and reactant moles of gas, pressure has no effect. Consider

88 Effects of Volume and Pressure
An increase in pressure (by decreasing the volume) favors the formation of colorless N2O4. The instant the pressure increases, the system is not at equilibrium and the concentration of both gases has increased. The system moves to reduce the number moles of gas (i.e. the reverse reaction is favored). A new equilibrium is established in which the mixture is lighter because colorless N2O4 is favored.

89 Effect of Temperature Changes
The equilibrium constant is temperature dependent. For an endothermic reaction, H > 0 and heat can be considered as a reactant. For an exothermic reaction, H < 0 and heat can be considered as a product. Adding heat (i.e. heating the vessel) favors away from the increase: if H > 0, adding heat favors the forward reaction, if H < 0, adding heat favors the reverse reaction.

90 Effect of Temperature Changes
Removing heat (i.e. cooling the vessel), favors towards the decrease: if H > 0, cooling favors the reverse reaction, if H < 0, cooling favors the forward reaction. Consider for which H > 0. Co(H2O)62+ is pale pink and CoCl42- is blue.

91 Penggabungan Rumus Tetapan Kesetimbangan
Jika diketahui: N2(g) + O2(g)  2NO(g) Kc = 4,1 x N2(g) + ½ O2(g)  N2O(g) Kc = 2,4 x Bagaimana Kc reaksi: N2O(g) + ½ O2(g)  2NO(g) Kc = ? Kita dapat menggabungkan persamaan diatas N2(g) + O2(g)  2NO(g) Kc = 4,1 x N2O(g)  N2(g) + ½ O2(g) Kc = 1/(2,4 x 10-18) = 4,2 x 1017

92 Tetapan kesetimbangan untuk reaksi bersih adalah hasil kali tetapan kesetimbangan untuk reaksi-reaksi terpisah yang digabungkan

93 Soal Latihan 1.Untuk reaksi NH3 ↔ ½ N2 + 3/2 H Kc = 5,2 x 10-5 pada 298 K. Berapakah nilai Kc pada 298 K untuk reaksi: N2 + 3H2 ↔ 2NH3 2. Senyawa ClF3 disiapkan melalui 2 tahap reaksi fluorinasi gas klor sebagai berikut (i) Cl2(g) + F2(g)  ClF(g) (ii) ClF(g) + F2(g)  ClF3(g) Seimbangkan masing-masing reaksi diatas dan tuliskan reaksi overallnya! Buktikan bahwa Kc overall sama dengan hasil kali Kc masing-masing tahap reaksi ?

94 Hubungan Tetapan Kesetimbangan Kc dan Kp
Tetapan kesetimbangan dalam sistem gas dapat dinyatakan berdasarkan tekanan parsial gas, bukan konsentrasi molarnya Tetapan kesetimbangan yang ditulis dengan cara ini dinamakan tetapan kesetimbangan tekanan parsial dilambangkan Kp. Misalkan suatu reaksi 2SO2(g) + O2(g)  2SO3(g) Kc = 2,8 x 102 pd 1000 K

95 Sesuai dengan hukum gas ideal, PV = nRT
Dengan mengganti suku-suku yang dilingkari dengan konsentrasi dalam Kc akan diperoleh rumus; Terlihat ada hubungan antara Kc dan Kp yaitu:

96 Jika penurunan yang sama dilakukan terhadap reaksi umum:
aA(g) + bB(g) + …  gG(g) + hH(g) + … Hasilnya menjadi Kp = Kc (RT)n Dimana n adalah selisih koefisien stoikiometri dari gas hasil reaksi dan gas pereaksi yaitu n = (g+h+…) – (a+b+…). Dalam persamaan reaksi pembentukan gas SO3 diatas kita lihat bahwa n = -1

97 Soal Latihan Hitunglah nilai Kp reaksi kesetimbangan berikut:
PCl3(g) + Cl2(g)  PCl5(g) Kc = 1,67 (at 500 K) N2O4(g)  2NO2(g); Kc = 6,1 x 10-3 (298 K) N2(g) + 3H2(g)  2NH3(g) Kc = 2,4 x 10-3 (at 1000 K).

98 Kesetimbangan yang melibatkan cairan dan padatan murni (Reaksi Heterogen)
Persamaan tetapan kesetimbangan hanya mengandung suku-suku yang konsentrasi atau tekanan parsialnya berubah selama reaksi berlangsung Atas dasar ini walaupun ikut bereaksi tapi karena tidak berubah, maka padatan murni dan cairan murni tidak diperhitungkan dalam persamaan tetapan kesetimbangan.

99 C(s) + H2O(g)  CO(g) + H2(g) 
CaCO3(s)  CaO(s) + CO2(g) Kc = [CO2(g)] atau jika dituliskan dalam bentuk tekanan parsial menjadi Kp = PCO2 Kp = Kc(RT) Latihan: Calculate Kc for the following reaction CaCO3(s)  CaO(s) + CO2(g) Kp = 2,1 x 10-4 (at 1000 K)

100 Arti Nilai Tetapan Kesetimbangan

101 CH4(g) + Cl2(g)  CH3Cl(g) + HCl(g)
Soal Latihan Reaksi: CO(g) + H2O(g) ↔ CO2(g) + H2(g) pada suhu 1100 K nilai Kc = 1,00. Sejumlah zat berikut dicampur pada suhu tersebut dan dibiarkan bereaksi: 1,00 mol CO, 1,00 mol H2O, 2,00 mol CO2 dan 2,00 mol H2. Kearah mana reaksi akan berjalan dan bagaimana komposisi akhirnya? 2) Klorometana terbentuk melalui reaksi: CH4(g) + Cl2(g)  CH3Cl(g) + HCl(g) pada 1500 K, konstanta kesetimbangan Kp = 1,6 x 104. Didalam campuran reaksi terdapat P (CH4) = 0,13 atm, P(Cl2) = 0,035 atm, P(CH3Cl) = 0,24 atm dan P(HCl) = 0,47 atm. Apakah reaksi diatas menuju kearah pembentukan CH3Cl atau pembentukan CH4.

102 Study Check In a study of hydrogen halide decomposition; a researcher fills an evacuated 2,00 L flask with 0,2 mol HI gas and allows the reaction to proceed at 453oC. 2HI(g)  H2(g) + I2(g) At equilibrium [HI] = M, calculate Kc In study of the conversion of methane to other fuels a chemical engineer mixes gaseous CH4 and H2O in a 0,32 L flask at 1200 K. at equilibrium, the flask contains 0,26 mol CO; 0,091 mol H2 and 0,041 mol CH4. What is [H2O] at equilibrium? Kc = 0,26 for the equation CH4(g) + H2O  CO(g) + 3H2(g) The decomposition of HI at low temperature was studied by injecting 2,50 mol HI into a 10,32-L vessel at 25oC. What is [H2] at equilibrium for the reaction 2HI(g)  H2(g) + I2(g); Kc = 1.26 x 10-3?

103 Prinsip Le Chatelier Usaha untuk mengubah suhu, tekanan atau konsentrasi pereaksi dalam suatu sistem dalam keadaan setimbang merangsang terjadinya reaksi yang mengembalikan kesetimbangan pada sistem tersebut

104 Pengaruh perubahan Jumlah spesies yang bereaksi
Kesetimbangan awal Gangguan Kesetimbangan akhir

105 Pengaruh Perubahan Tekanan
Jika tekanan pada campuran kesetimbangan yang melibatkan gas ditingkatkan reaksi bersih akan berlangsung kearah yang mempunyai jumlah mol gas lebih kecil begitupun sebaliknya

106 Pengaruh Gas Lembam (inert)
Pengaruh tidaknya gas lembam tergantung pada cara melibatkan gas tersebut Jika sejumlah gas helium ditambahkan pada keadaan volume tetap, tekanan akan meningkat, sehingga tekanan gas total akan meningkat. Tetapi tekanan parsial gas-gas dalam kesetimbangan tetap Jika gas ditambahkan pada tekanan tetap, maka volume akan bertambah. Pengaruhnya akan sama dengan peningkatan volume akibat penambahan tekanan eksternal. Gas lembam mempengaruhi keadaan kesetimbangan hanya jika gas tersebut mengakibatkan perubahan konsentrasi (atau tekanan parsial) dari pereaksi-pereaksinya

107 Pengaruh Suhu Penambahan kalor akan menguntungkan reaksi serap-panas (endoterm) Pengurangan kalor akan menguntungkan reaksi lepas-panas (eksoterm) Peningkatan suhu suatu campuran kesetimbangan menyebabkan pergeseran kearah reaksi endoterm. Penurunan suhu menyebabkan pergeseran kearah reaksi eksoterm

108 Pengaruh Suhu pada Kesetimbangan
Umumnya tetapan kesetimbangan suatu reaksi tergantung pada suhu Nilai Kp untuk reaksi oksidasi belerang dioksida diperlihatkan pada tabel berikut

109 Hubungan pada tabel tersebut dapat dituliskan dengan:
Persamaan garis lurus y = m .x b Dan jika ada dua keadaan yang berbeda kita dapat menghubungkan dengan modifikasi sederhana hingga diperoleh:

110 K2 dan K1 adalah tetapan kesetimbangan pada suhu kelvin T2 dan T1
K2 dan K1 adalah tetapan kesetimbangan pada suhu kelvin T2 dan T1. ∆Ho adalah entalpi (kalor) molar standar dari reaksi. Nilai positif dan negatif untuk parameter ini dimungkinkan dan diperlukan asumsi bahwa ∆Ho tidak tergantung pada suhu Menurut prinsip Le Chatelier, jika ∆Ho > 0 (endoterm) reaksi kedepan terjadi jika suhu ditingkatkan, menyiratkan bahwa nilai K meningkat dengan suhu. Jika ∆Ho < 0 (eksoterm) reaksi kebalikan terjadi jika suhu ditingkatkan dan nilai K menurun dengan suhu Persamaan diatas menghasilkan nilai kuantitatif yang sesuai dengan pengamatan kualitatif dari prinsip Le Chatelier.

111 Soal Latihan Untuk reaksi N2O4(g)  2NO2(g), ∆Ho = +61,5 kJ/mol dan Kp = 0,113 pada 298K Berapa nilai Kp pada 0oC? (1,2x10-2) Pada suhu berapa nilai Kp = 1,00 (326 K)

112 Pengaruh Katalis pada Kesetimbangan
Katalis dalam reaksi reversibel dapat mempercepat reaksi baik kekanan atau kekiri. Keadaan kesetimbangan tercapai lebih cepat tetapi tidak mengubah konstanta kesetimbangan dari spesies-spesies yang bereaksi. Peranan katalis adalah mengubah mekanisme reaksi agar tercapai energi aktivasi yang lebih rendah. Keadaan kesetimbangan tidak bergantung pada mekanisme reaksi Sehingga tetapan kesetimbangan yang diturunkan secara kinetik tidak dipengaruhi oleh mekanisme yang dipilih.

113 Terima Kasih Selamat Belajar