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Equilibrium In Chemical Reaction. The double arrow tells us that this reaction can go in both directions: N 2 + 3H 2 2NH 3.

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Presentasi berjudul: "Equilibrium In Chemical Reaction. The double arrow tells us that this reaction can go in both directions: N 2 + 3H 2 2NH 3."— Transcript presentasi:

1 Equilibrium In Chemical Reaction

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

3 1) Reactants react to become products, N 2 + 3H 2 2NH 3 ( forward reaction)

4 1) Reactants react to become products, N 2 + 3H 2 2NH 3 while simultaneously, N 2 + 3H 2 2NH 3 ( forward reaction) 2) Products react to become reactants N 2 + 3H 2 2NH 3 ( reverse reaction)

5 N 2 + 3H 2 2NH 3 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 N 2 + 3H 2 2NH 3 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 N 2 + 3H 2 2NH 3 However (like the metaphor), the equilibrium amounts of reactants and products are usually not equal, they just remain unchanged.

8 N 2 + 3H 2 2NH 3

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17 reverse forward

18 reverse forward

19 reverse forward

20 reverse forward

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24 reverse forward

25 reverse forward

26 reverse forward

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

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

32 For N 2 + 3H 2 2NH 3, suppose you begin with the following: N 2 = 1 M, H 2 = 1 M, and NH 3 = 0 M concentration time

33 For N 2 + 3H 2 2NH 3, suppose you begin with the following: N 2 = 1 M, H 2 = 1 M, and NH 3 = 0 M concentration time

34 For N 2 + 3H 2 2NH 3, suppose you begin with the following: N 2 = 1 M, H 2 = 1 M, and NH 3 = 0 M concentration time

35 For N 2 + 3H 2 2NH 3, suppose you begin with the following: N 2 = 1 M, H 2 = 1 M, and NH 3 = 0 M concentration time

36 For N 2 + 3H 2 2NH 3, suppose you begin with the following: N 2 = 1 M, H 2 = 1 M, and NH 3 = 0 M concentration time

37 For N 2 + 3H 2 2NH 3, suppose you begin with the following: N 2 = 1 M, H 2 = 1 M, and NH 3 = 0 M concentration time

38 For N 2 + 3H 2 2NH 3, suppose you begin with the following: N 2 = 1 M, H 2 = 1 M, and NH 3 = 0 M concentration time

39 For N 2 + 3H 2 2NH 3, suppose you begin with the following: N 2 = 1 M, H 2 = 1 M, and NH 3 = 0 M concentration time

40 For N 2 + 3H 2 2NH 3, suppose you begin with the following: N 2 = 1 M, H 2 = 1 M, and NH 3 = 0 M concentration time

41 For N 2 + 3H 2 2NH 3, suppose you begin with the following: N 2 = 1 M, H 2 = 1 M, and NH 3 = 0 M concentration time

42 For N 2 + 3H 2 2NH 3, suppose you begin with the following: N 2 = 1 M, H 2 = 1 M, and NH 3 = 0 M concentration time

43 For N 2 + 3H 2 2NH 3, suppose you begin with the following: N 2 = 1 M, H 2 = 1 M, and NH 3 = 0 M concentration time

44 For N 2 + 3H 2 2NH 3, suppose you begin with the following: N 2 = 1 M, H 2 = 1 M, and NH 3 = 0 M concentration time

45 For N 2 + 3H 2 2NH 3, suppose you begin with the following: N 2 = 1 M, H 2 = 1 M, and NH 3 = 0 M concentration time

46 For N 2 + 3H 2 2NH 3, suppose you begin with the following: N 2 = 1 M, H 2 = 1 M, and NH 3 = 0 M concentration time

47 For N 2 + 3H 2 2NH 3, suppose you begin with the following: N 2 = 1 M, H 2 = 1 M, and NH 3 = 0 M concentration time

48 For N 2 + 3H 2 2NH 3, suppose you begin with the following: N 2 = 1 M, H 2 = 1 M, and NH 3 = 0 M concentration time

49 For N 2 + 3H 2 2NH 3, suppose you begin with the following: N 2 = 1 M, H 2 = 1 M, and NH 3 = 0 M concentration time [N2][N2] [H2][H2] [ NH 3 ]

50 For N 2 + 3H 2 2NH 3, suppose you begin with the following: N 2 = 1 M, H 2 = 1 M, and NH 3 = 0 M concentration time Question 3: at what point has equilibrium been established? [N2][N2] [H2][H2] [ NH 3 ]

51 For N 2 + 3H 2 2NH 3, suppose you begin with the following: N 2 = 1 M, H 2 = 1 M, and NH 3 = 0 M concentration time Question 4: what does the graph tell you about the concentration of each species once equilibrium is established? [N2][N2] [H2][H2] [ NH 3 ]

52 For N 2 + 3H 2 2NH 3, suppose you begin with the following: N 2 = 1 M, H 2 = 1 M, and NH 3 = 0 M concentration time Question 5: what might a rate vs time graph look like for the above reaction? [N2][N2] [H2][H2] [ NH 3 ]

53 rate time Question 5: what might a rate vs time graph look like for the above reaction? For and still beginning withN 2 + 3H 2 2NH 3 N 2 = 1 M, H 2 = 1 M, and NH 3 = 0 M

54 For and still beginning with rate time N 2 + 3H 2 2NH 3 N 2 = 1 M, H 2 = 1 M, and NH 3 = 0 M

55 For and still beginning with rate time N 2 + 3H 2 2NH 3 N 2 = 1 M, H 2 = 1 M, and NH 3 = 0 M Question 6: at what point has equilibrium been established? forward reverse

56 rate time Question 7: describe how the two graphs are related. concentration time N2N2 H2H2 NH 3 forward reverse

57 rate time Question 8: do either of the two graphs indicate if K eq >1 or K eq <1? concentration time N2N2 H2H2 NH 3 forward reverse

58 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 59 The Concept of Equilibrium Consider colorless frozen N 2 O 4. At room temperature, it decomposes to brown NO 2 : N 2 O 4 (g) 2NO 2 (g). At some time, the color stops changing and we have a mixture of N 2 O 4 and NO 2. 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 NO 2 builds up, there is a chance that two NO 2 molecules will collide to form N 2 O 4. At the beginning of the reaction, there is no NO 2 so the reverse reaction (2NO 2 (g) N 2 O 4 (g)) does not occur.

60 60 As the substance warms it begins to decompose: N 2 O 4 (g) 2NO 2 (g) When enough NO 2 is formed, it can react to form N 2 O 4 : 2NO 2 (g) N 2 O 4 (g). At equilibrium, as much N 2 O 4 reacts to form NO 2 as NO 2 reacts to re-form N 2 O 4 The double arrow implies the process is dynamic. N 2 O 4 (g) 2NO 2 (g)

61 61 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 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 K c is the equilibrium constant.

63 63 K c 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 64 Write the equilibrium expression for the following reaction:

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

66 66 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 CO 2 does not seem to depend on the amounts of CaO and CaCO 3. Why?

70 Heterogeneous Equilibria

71 Neither density nor molar mass is a variable, the concentrations of solids and pure liquids are constant. (You cant find the concentration of something that isnt a solution!) We ignore the concentrations of pure liquids and pure solids in equilibrium constant expressions. The amount of CO 2 formed will not depend greatly on the amounts of CaO and CaCO 3 present. K c = [CO 2 ] K p = [pCO 2 ]

72 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 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 Note the moles into a L vessel stuff... calculate molarity. Starting concentration of HI: 2.5 mol/10.32 L = M 2 HI H 2 + I 2 Initial: Change: Equil: M00 -2x+x+x xxx What we are asked for here is the equilibrium concentration of H otherwise known as x. So, we need to solve this beast for x.

75 And yes, its 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 [H 2 ] at equil. is M.

76 This type of problem is typically tackled using the three line approach: 2 NO + O 2 2 NO 2 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. 77

78 Initial Concentration of I 2 : 0.50 mol/2.5L = 0.20 M I 2 2 I Initial change equil: x +2x 0.20-x 2x With an equilibrium constant that small, whatever x is, its near dink, and 0.20 minus dink is 0.20 (like a million dollars minus a nickel is still a million dollars) – x is the same as 0.20 x = 3.83 x M More than 3 orders of mag. between these numbers. The simplification will work here.

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

80 Le Chateliers 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). Le Châteliers Principle

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 82 Consider the Haber process If H 2 is added while the system is at equilibrium, the system must respond to counteract the added H 2 (by Le Châtelier). That is, the system must consume the H 2 and produce products until a new equilibrium is established. Therefore, [H 2 ] and [N 2 ] will decrease and [NH 3 ] increases.

83 83 Change in Reactant or Product Concentrations The unreacted nitrogen and hydrogen are recycled with the new N 2 and H 2 feed gas. The equilibrium amount of ammonia is optimized because the product (NH 3 ) is continually removed and the reactants (N 2 and H 2 ) are continually being added. Effects of Volume and Pressure As volume is decreased pressure increases. Le Châteliers 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 Châteliers Principle: if a system at equilibrium is disturbed, the system will move in such a way as to counteract the disturbance.

85 85 Change in Reactant or Product Concentrations

86 86

87 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 88 Effects of Volume and Pressure An increase in pressure (by decreasing the volume) favors the formation of colorless N 2 O 4. 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 N 2 O 4 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 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(H 2 O) 6 2+ is pale pink and CoCl 4 2- is blue.

91 Penggabungan Rumus Tetapan Kesetimbangan Jika diketahui: N 2(g) + O 2(g) 2NO (g) K c = 4,1 x N 2(g) + ½ O 2(g) N 2 O (g) K c = 2,4 x Bagaimana K c reaksi: N 2 O (g) + ½ O 2(g) 2NO (g) K c = ? Kita dapat menggabungkan persamaan diatas N 2(g) + O 2(g) 2NO (g) K c = 4,1 x N 2 O (g) N 2(g) + ½ O 2(g) K c = 1/(2,4 x ) = 4,2 x N 2 O (g) + ½ O 2(g) 2NO (g) K c = ?

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

93 Soal Latihan 1.Untuk reaksi NH 3 ½ N 2 + 3/2 H 2 Kc = 5,2 x pada 298 K. Berapakah nilai Kc pada 298 K untuk reaksi: N 2 + 3H 2 2NH 3 2. Senyawa ClF 3 disiapkan melalui 2 tahap reaksi fluorinasi gas klor sebagai berikut (i) Cl 2 (g) + F 2 (g) ClF(g) (ii) ClF(g) + F 2 (g) ClF 3 (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 K c dan K p 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 K p. Misalkan suatu reaksi 2SO 2(g) + O 2(g) 2SO 3(g) K c = 2,8 x 10 2 pd 1000 K

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

96 Jika penurunan yang sama dilakukan terhadap reaksi umum: aA (g) + bB (g) + … gG (g) + hH (g) + … Hasilnya menjadiK p = K c (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 SO 3 diatas kita lihat bahwa n = -1

97 Soal Latihan Hitunglah nilai Kp reaksi kesetimbangan berikut: PCl 3 (g) + Cl 2 (g) PCl 5 (g) Kc = 1,67 (at 500 K) N 2 O 4 (g) 2NO 2 (g); K c = 6,1 x (298 K) N 2 (g) + 3H 2 (g) 2NH 3 (g) K c = 2,4 x (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) + H 2 O (g) CO (g) + H 2(g) CaCO 3(s) CaO (s) + CO 2(g) K c = [CO 2(g) ] atau jika dituliskan dalam bentuk tekanan parsial menjadi K p = P CO2 K p = K c (RT) Latihan: Calculate K c for the following reaction CaCO 3 (s) CaO(s) + CO 2 (g) K p = 2,1 x 10-4 (at 1000 K)

100 Arti Nilai Tetapan Kesetimbangan

101 Soal Latihan 1)Reaksi: CO(g) + H 2 O(g) CO 2 (g) + H 2 (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 H 2 O, 2,00 mol CO 2 dan 2,00 mol H 2. Kearah mana reaksi akan berjalan dan bagaimana komposisi akhirnya? 2) Klorometana terbentuk melalui reaksi: CH 4 (g) + Cl 2 (g) CH 3 Cl(g) + HCl(g) pada 1500 K, konstanta kesetimbangan Kp = 1,6 x Didalam campuran reaksi terdapat P (CH 4 ) = 0,13 atm, P(Cl 2 ) = 0,035 atm, P(CH 3 Cl) = 0,24 atm dan P(HCl) = 0,47 atm. Apakah reaksi diatas menuju kearah pembentukan CH 3 Cl atau pembentukan CH 4.

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 453 o C. 2HI(g) H 2 (g) + I 2 (g) At equilibrium [HI] = M, calculate Kc In study of the conversion of methane to other fuels a chemical engineer mixes gaseous CH 4 and H 2 O in a 0,32 L flask at 1200 K. at equilibrium, the flask contains 0,26 mol CO; 0,091 mol H 2 and 0,041 mol CH 4. What is [H 2 O] at equilibrium? K c = 0,26 for the equation CH 4 (g) + H 2 O CO(g) + 3H 2 (g) The decomposition of HI at low temperature was studied by injecting 2,50 mol HI into a 10,32-L vessel at 25 o C. What is [H 2 ] at equilibrium for the reaction 2HI(g) H 2 (g) + I 2 (g); Kc = 1.26 x ?

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 K p 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 K 2 dan K 1 adalah tetapan kesetimbangan pada suhu kelvin T 2 dan T 1. H o adalah entalpi (kalor) molar standar dari reaksi. Nilai positif dan negatif untuk parameter ini dimungkinkan dan diperlukan asumsi bahwa H o tidak tergantung pada suhu Menurut prinsip Le Chatelier, jika H o > 0 (endoterm) reaksi kedepan terjadi jika suhu ditingkatkan, menyiratkan bahwa nilai K meningkat dengan suhu. Jika H o < 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 N 2 O 4(g) 2NO 2(g), H o = +61,5 kJ/mol dan K p = 0,113 pada 298K Berapa nilai K p pada 0 o C? (1,2x10 -2 ) Pada suhu berapa nilai K p = 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


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