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Introduction to Chemical Equilibrium

Introduction to Chemical Equilibrium. CHEM 160. Chapter 15. 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 .

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Introduction to Chemical Equilibrium

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  1. Introduction to Chemical Equilibrium CHEM 160 Chapter 15

  2. 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 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.

  3. The Concept of Equilibrium

  4. The Concept of Equilibrium • The point at which the rate of decomposition: N2O4(g)  2NO2(g) equals the rate of dimerization: 2NO2(g)  N2O4(g). is dynamic equilibrium. • The equilibrium is dynamic because the reaction has not stopped: the opposing rates are equal. • Consider frozen N2O4: only white solid is present. On the microscopic level, only N2O4 molecules are present.

  5. The Concept of Equilibrium

  6. The Concept of Equilibrium • As the substance warms it begins to decompose: N2O4(g)  2NO2(g) • A mixture of N2O4 (initially present) and NO2 (initially formed) appears light brown. • 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.

  7. The Concept of Equilibrium • Consider Forward reaction: A  B Rate = kf[A] Reverse reaction: B  A Rate = kr[B] • At equilibrium kf[A] = kr[B]. • For an equilibrium we write: • 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.

  8. The Concept of Equilibrium • Alternatively: • kf[A] decreases to a constant, • kr[B] increases from zero to a constant. • When kf[A] = kr[B] equilibrium is achieved.

  9. The Equilibrium Constant • Consider • If we start with a mixture of nitrogen and hydrogen (in any proportions), the reaction will reach equilibrium with a constant concentration of nitrogen, hydrogen and ammonia. • However, if we start with just ammonia and no nitrogen or hydrogen, the reaction will proceed and N2 and H2 will be produced until equilibrium is achieved.

  10. The Equilibrium Constant

  11. The Equilibrium Constant • 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.

  12. The Equilibrium Constant • 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.

  13. The Equilibrium Constant The Equilibrium Constant in Terms of Pressure • If KP is the equilibrium constant for reactions involving gases, we can write: • KPis based on partial pressures measured in atmospheres. • We can show that PA = [A](RT)

  14. The Equilibrium Constant The Equilibrium Constant in Terms of Pressure PA = [A](RT) • This means that we can relate Kc and KP: where Dn is the change in number of moles of gas. • It is important to use: Dn = ngas(products) - ngas(reactants)

  15. The Equilibrium Constant Problem 1(a) Phosphorous Phosphorous pentachloride dissociates on heating: PCl5(g)  PCl3(g) + Cl2(g) If Kc = 3.26 x 10-2 at 191C, what is Kp at this temperature? Ans.: Kp = 1.24

  16. The Equilibrium Constant Problem 1(b) Phosphorous The value of Kc for the following reaction at 900C is 0.28. S2(g) + 4H2(g)  CH4(g) + 2H2S(g) What is Kp at this temperature? Ans.: Kp = 3.0 x 10-5

  17. The Equilibrium Constant Problem 2(a) Phosphorous Consider the following reaction at 1000C: CO(g) + 3H2(g)  CH4(g) + H2O(g) At equilibrium, the following concentrations are measured: [CO] = 0.0613 M, [H2] = 0.1839 M, [CH4] = 0.0387, [H2O] = 0.0387 M. Calculate the value of Kc for this reaction. Calculate the value of Kp. ? Ans.: Kc = 3.93, Kp = 3.60 x 10-4

  18. The Equilibrium Constant Problem 2(b) Phosphorous A 5.00 L vessel contained 0.0185 mol of phosphorus trichloride, 0.0158 mol of phosphorus pentachloride, and 0.0870 mol chlorine at 503 K in an equilibrium mixture. Calculate the value of Kc at this temperature. PCl3(g) + Cl2(g)  PCl5(g) Ans.: Kc = 49.1

  19. The Equilibrium Constant The Magnitude of Equilibrium Constants • The equilibrium constant, K, is the ratio of products to reactants. • 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.

  20. The Equilibrium Constant The Magnitude of Equilibrium Constants

  21. The Equilibrium Constant The Magnitude of Equilibrium Constants • An equilibrium can be approached from any direction. • Example: • has

  22. The Equilibrium Constant The Magnitude of Equilibrium Constants • However, • has • The equilibrium constant for a reaction in one direction is the reciprocal of the equilibrium constant of the reaction in the opposite direction.

  23. The Equilibrium Constant 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? • The concentration of a solid or pure liquid is its density divided by molar mass.

  24. The Equilibrium Constant Heterogeneous Equilibria

  25. The Equilibrium Constant Heterogeneous Equilibria • Neither density nor molar mass is a variable, the concentrations of solids and pure liquids are constant. • For the decomposition of CaCO3: • 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.

  26. Calculating Equilibrium Constants • Proceed as follows: • Tabulate initial and equilibrium concentrations (or partial pressures) given. • If an initial and equilibrium concentration is given for a species, calculate the change in concentration. • Use stoichiometry on the change in concentration line only to calculate the changes in concentration of all species. • Deduce the equilibrium concentrations of all species. • Usually, the initial concentration of products is zero. (This is not always the case.) • When in doubt, assign the change in concentration a variable.

  27. Calculating Equilibrium Constants Problem 3(a) Phosphorous Carbon dioxide decomposes at elevated temperatures to carbon monoxide and oxygen: 2CO2(g)  2CO(g) + O2(g) At 3000 K, 2.00 mol of CO2 is placed into a 1.00 L container and allowed to come to equilibrium. At equilibrium, 0.90 mol CO2 remains. What is the value for Kc at this temperature? Ans.: Kc = 0.82

  28. Calculating Equilibrium Constants Problem 3(b) Phosphorous Consider the following reaction at 1000C: CO(g) + 3H2(g)  CH4(g) + H2O(g) The original concentrations of CO and H2 were 0.2000 M and 0.3000 M, respectively. At equilibrium, the concentration of CH4 was measured to be 0.0478 M. Calculate the values of Kc and Kp. Ans.: Kc = 3.91, Kp = 3.60 x 10-4

  29. Calculating Equilibrium Constants Problem 3(c) Phosphorous Consider the following reaction at 1000C: 2HI(g)  H2(g) + I2(g) When 4.00 mol of HI was placed into the 5.0 L reaction vessel at 458C, the equilibrium mixture was found to contain 0.422 mol I2. Calculate the value of Kc for the decomposition of HI. Ans.: Kc = 1.79 x 10-2

  30. Calculating Equilibrium Constants Problem 3(d) Phosphorous Hydrogen sulfide, a colorless gas with a foul odor, dissociates on heating: 2H2S(g)  2H2(g) + S2(g) When 0.100 mol H2S was put into a 10.0 L vessel and heated to 1132C, it gave an equilibrium mixture containing 0.0285 mol H2. Calculate the value of Kc at this temperature. Ans.: Kc = 1.35 x 10-6 M

  31. Applications of Equilibrium Constants Calculation of Equilibrium Concentrations • The same steps used to calculate equilibrium constants are used. • Generally, we do not have a number for the change in concentration line. • Therefore, we need to assume that x mol/L of a species is produced (or used). • The equilibrium concentrations are given as algebraic expressions.

  32. Applications of Equilibrium Constants Problem 5(a) Nitrogen and oxygen form nitric oxide: N2(g) + O2(g)  2NO(g) If an equilibrium mixture at 25C contains 0.040 mol/L N2 and 0.010 mol/L O2, what is the concentration of NO in this mixture? Kc = 1 x 10-30 at this temperature. Ans.: 2 x 10-17 M Phosphorous

  33. Applications of Equilibrium Constants Problem 5(b) An equilibrium mixture at 1200 K contains 0.30 mol CO, 0.10 mol H2, and 0.020 mol H2O, plus an unknown amount of CH4 in each liter. What is the concentration of CH4 in this mixture if Kc = 3.92? The reaction is: CO(g) + 3H2(g)  CH4(g) + H2O(g) Ans.: 0.059 M Phosphorous

  34. Applications of Equilibrium Constants Problem 6(a) The reaction: CO(g) + H2O(g)  CO2(g) + H2(g) is used to increase the ratio of hydrogen in synthesis gas (mixtures of CO and H2). Suppose you start with 1.00 mol each of carbon monoxide and water in a 50.0 L vessel. What is the concentration of each substance in the equilibrium mixture at 1000C given that Kc = 0.58 at this temperature?   Ans.: [CO] = [H2O] = 0.0114 M, [CO2] = [H2] = 0.0086 M Phosphorous

  35. Applications of Equilibrium Constants Problem 6(b) Hydrogen iodide decomposes to hydrogen gas and iodine gas: 2HI(g)  H2(g) + I2(g) At 800 K, Kc for this reaction is 0.016. If 0.50 mol HI is placed into a 5.0 L flask, what will be the composition of the mixture at equilibrium? Ans.: [HI] = 0.080 M, [H2] = [I2] = 0.010 M Phosphorous

  36. Applications of Equilibrium Constants Problem 7(a) N2O4 decomposes to NO2 according to the reaction: N2O4(g)  2NO2(g) At 100C, Kc = 0.36. If a 1.00 L flask initially contains 0.100 mol N2O4, what will be the concentrations of N2O4 and NO2 at equilibrium? Ans.: [N2O4] =0.040 M, [NO2] = 0.12 M Phosphorous

  37. Applications of Equilibrium Constants Problem 7(b) Hydrogen and iodine react according to the equation: H2(g) + I2(g)  2HI(g) Suppose 1.00 mol H2 and 2.00 mol I2 are placed into a 1.00 L vessel. How many moles of each substance are in the equilibrium mixture at 458C if Kc = 49.7 at this temperature? Ans.: 1.86 mol HI, 0.07 mol H2, 1.07 mol I2 Phosphorous

  38. Applications of Equilibrium Constants Problem 7(c) Iodine and bromine react to give iodine monobromide: I2(g) + Br2(g)  2IBr(g) What is the equilibrium composition of a mixture at 150C that initially contained 0.0015 mol each of iodine and bromine in a 5.0 L vessel if Kc = 1.2 x 102 at this temperature? Ans.: [IBr] = 5.1 x 10-4 M, [Br2] = [I2] = 4.7 x 10-5 M Phosphorous

  39. Applications of Equilibrium Constants Predicting the Direction of Reaction • We define Q, the reaction quotient, for a general reaction as where [A], [B], [P], and [Q] are molarities at any time. • Q = K only at equilibrium.

  40. Applications of Equilibrium Constants Predicting the Direction of Reaction • If Q > K then the reverse reaction must occur to reach equilibrium (i.e., products are consumed, reactants are formed, the numerator in the equilibrium constant expression decreases and Q decreases until it equals K). • If Q < K then the forward reaction must occur to reach equilibrium.

  41. Applications of Equilibrium Constants Problem 4 • The following reaction has an equilibrium constant Kc of 3.07 x 10-4 • at 24C: • 2NOBr(g)  2NO(g) + Br2(g) • For each of the following compositions, decide whether the reaction • is at equilibrium. If not, decide which direction the reaction should go. • [NOBr] = 0.0610 M,[NO] = 0.0151 M, [Br2] = 0.0108 M • [NOBr] = 0.115 M,[NO] = 0.0169 M, [Br2] = 0.0142 M • NOBr] = 0.181 M,[NO] = 0.0123 M, [Br2] = 0.0201 M • Ans.: a.) goes left; b.) equilibrium; c.) goes right Phosphorous

  42. Le Châtelier’s Principle • 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. • Can this be predicted? • Le Châtelier’s Principle: if a system at equilibrium is disturbed, the system will move in such a way as to counteract the disturbance.

  43. Le Châtelier’s Principle

  44. Le Châtelier’s Principle Change in Reactant or Product Concentrations • 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.

  45. Le Châtelier’s Principle Change in Reactant or Product Concentrations

  46. Le Châtelier’s Principle 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

  47. Le Châtelier’s Principle Change in Reactant or Product Concentrations

  48. Le Châtelier’s Principle Change in Reactant or Product Concentrations • N2 and H2 are pumped into a chamber. • The pre-heated gases are passed through a heating coil to the catalyst bed. • The catalyst bed is kept at 460 - 550 C under high pressure. • The product gas stream (containing N2, H2 and NH3) is passed over a cooler to a refrigeration unit. • In the refrigeration unit, ammonia liquefies but not N2 or H2.

  49. Le Châtelier’s Principle 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.

  50. Le Châtelier’s Principle Problem 8 Predict the effect of the following concentration changes on the reaction below. CH4(g) + 2S2(g)  CS2(g) + 2H2S(g) a) (a) Some CH4(g) is removed. (b) Some S2(g) is added. (c) Some CS2(g) is added. (d) Some H2S(g) is removed. (e) Some argon gas is added. Ans.: a.) goes left; b.) goes right; c.) goes left; d.) goes right ; e.) no effect Phosphorous

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