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Equilibrium

Equilibrium. Reversible Reactions. Many chemical reactions exist in which the reactants can be converted to products and the products converted to reactants at the same time. There are two opposing processes happening. Many biologically important reactions are reversible reactions.

robert-butler
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Equilibrium

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  1. Equilibrium

  2. Reversible Reactions • Many chemical reactions exist in which the reactants can be converted to products and the products converted to reactants at the same time. There are two opposing processes happening. • Many biologically important reactions are reversible reactions.

  3. CO + 3H2 CH4 + H2O • The double arrow shows that a reaction is reversible • If you put carbon monoxide and hydrogen into a container and let them react, they’ll form methane and water. There will be a mix of some products and some reactants. • If you put methane and water into a container and let them react, they’ll form methane and water. There will be a mix of some products and some reactants.

  4. CO + 3H2 CH4 + H2O • In fact- you can put in the reactants only, the products only or some of each and within a few limits from stoichiometry you’ll end up with a mix of products and reactants at the end. The same mix no matter where you start.

  5. Once equilibrium is reached, There are no changes in the Concentrations of products or reactants M O L E S Time

  6. R X N R A T E Time

  7. Equilibrium • A state reached by a chemical system in which the forward and reverse reaction rates are equal • This means that there is no net change in the concentration of any chemical • Equilibrium may take seconds to be reached or billions of years (we can say that catalysts get chemical systems to equilibrium faster).

  8. CO + 3H2 CH4 + H2O • Suppose you put 2.00 moles of CO and 4.50 moles of hydrogen into a 1.0 L flask. They react and after a while the concentration of methane is found to be constant. Equilibrium has been reached and from now on, no matter how long you wait, the concentration of each chemical will not change. You find that the concentration of methane is 0.65 M ([CH4] = 0.65M). Find the concentration of all the chemicals at equilibrium based on this data.

  9. Make an ICE chart • ICE stands for initial, change, equilibrium CO + 3H2  CH4 + H2O Initial 2.0 M 4.5 M 0 0 Change Equil. 0.65M

  10. ICE stands for initial, change, equilibrium CO + 3H2  CH4 + H2O Initial 2.0 M 4.5 M 0 0 Change -x -3x +x +x Equil. 0.65M The change in methane must be +0.65M (= x) Use the stoichiometric ratios (in green)

  11. ICE stands for initial, change, equilibrium CO + 3H2  CH4 + H2O Initial 2.0 M4.5 M00 Change -0.65-3(0.65)+0.65+0.65 Equil. 1.35M2.55M0.65M 0.65M ICE tables are very useful ways to organize information I n this (and other) chapters.

  12. Example 2 • One More Example 4 NH3 + 5 O2 4 NO + 6 H2O Two moles of ammonia and oxygen and three moles of water are placed in a 1L flask at room temperature. Pressure is monitored and observed to reach a constant P. At this point, there are 1.6 moles of ammonia. Calculate the concentrations of all substances.

  13. Another Example 4 NH3 + 5 O2 4 NO + 6 H2O Initial 2.00 2.00 0.00 3.00 Change Equil 1.6 This means that [NH3] decreased by 0.4 moles. So by coefficients, what must have happened to the other quantities?

  14. 4 NH3 + 5 O2 4 NO + 6 H2O Initial 2.00 2.00 0.00 3.00 Change –0.40 -0.50 + 0.40 +0.60 Equil 1.60 1.50 0.40 3.60 Things reacted in ratios according to the coefficients of the balanced rxn. Make sure you understand + and - and why each substance has a different #

  15. Equilibrium Constant • A mathematical ratio that relates amounts of products and reactants to a number (the size of which tells about relative amounts of products and reactants (and spontaneity). • May be expressed in terms of concentrations (Kc or sometimes Keq) or Pressures (Kp)

  16. Generic Reaction • aA = bB  cC + dD • Keq = [C]c[D]d [A]a[B]b Products on top, coefficients become exponents Solids and liquids are left out

  17. Examples 3 H2(g) + N2(g) 2NH3(g) Keq = [NH3]2 [H2]3[N2]

  18. CO(g) + 3 H2(g) CH4(g) + H2O(g) • CaCl2(s) Ca+2(aq) + 2Cl-(aq) • CaCO3(s)  CaO(s) + CO2(g) Keq = [CH4][H2O] [H2]3[CO] Keq = [Ca+2][Cl-1]2 Keq = [CO2]

  19. Quantitative Approaches to Equilibrium Constants For this example we will deal with the chemical system N2O4(g) 2NO2(g) (A different example is used in your book to make the same point). The equilibrium constant expression is: Keq = [NO2]2/[N2O]4

  20. N2O4(g) 2NO2(g) • Suppose we do a series of experiments in which we sometimes start with all NO2, sometimes start with all N2O4 and sometimes have some of each. If we let the system reach equilibrium (what is true at that time?), we would see something like this (based on real data)

  21. N2O4(g) 2NO2(g)

  22. Notice • At equilibrium, the ratio described by the Keq (or Kc) expression is the same. • Equilibrium will be reached whether the system starts will all products, all reactants or some mix

  23. Finding Keq Using Data • 2HI(g) H2(g) + I2(g) • 3.50 moles of HI are placed in a 5.00 L flask. At equilibrium, the flask contained 0.360 moles of iodine. Find the equilibrium concentration of each substance and the value of the equilibrium constant. • First get amounts into Molarity, then use an ICE chart

  24. [HI]0 = 3.50moles/5.00L = 0.70 M • [I2]eq = 0.36moles/5.00L = 0.072 M 2HI(g) H2(g) + I2(g) Initial 0.70 0 0 Change -0.144+0.072+0.072 Equilibrium 0.556 0.072 0.072 Keq = (0.072)(0.072) = 0.0168 (0.556)2

  25. Another Example • In a study of possible synthetic fuels the following reaction is studied: • CH4(g) + H2O(g) CO(g) + 3H2(g) • The value of Keq = 0.26. At equilibrium, the following data are obtained: • [CH4] = 0.145M, [H2] = 0.291M, [CO] = 0.88M • Find the equilibrium concentration of water vapor.

  26. Plan and Solution • Write out Keq expression • Plug in and solve • Keq = [CO][H2]3 [CH4][H2O] 0.26 = (0.88)(0.291)3 (0.145)(X) Cross-multiply and solve to get x = 0.575M

  27. Still One More Example CO +H2O  CO2 + H2 • 0.250 moles of carbon monoxide and water are placed in a 125 ml flask. The equilibrium constant for the reaction is 1.56. What will be the composition of the mixture at equilibrium? • Plan- write out a Keq expression, make an ICE chart and solve (remember to use M and not moles)

  28. Keq (or Kc) = [CO2][H2] [CO][H2O] Initial [CO]=[H2O] = 0.250 mole/.125L = 2.00M CO + H2O  CO2 + H2 Initial 2.00 2.00 0 0 Change -x -x +x +x Equilibrium 2.00-x 2.00-x x x

  29. 1.56 = (x)(x) (2.0-x)(2.0-x) This may look like you’ll need the quadratic but the alert student (that’s you) will notice that the right side is a perfect square So, take the square root of both sides and get √1.56 = 1.25 = (x)/2-x) and x = 1.11 so Plug back into the ICE chart and get

  30. [CO] = [H2O] = 2-x = 0.89M [CO2] = [H2] = x = 1.11 M Go back and plug into the original expression as a check: (1.11)(1.11) (0.89)(0.89) = 1.56 (!)

  31. So What Does anEquilibrium Constant Tell Us? • It is, roughly speaking a ratio of amounts of products vs reactants at equilibrium • The larger the value of Keq, the more the products predominate at equilibrium • If Keq = 1 then there would be equal amounts of products and reactants (depending on exponents in the expression).

  32. For Example • Keq = 3 x 10-6 suggests primarily reactants at equilibrium • Keq = 6 x 107 suggests primarily products are equilibrium • Keq = 1.2 suggests significant amounts of both products and reactants at equilibrium (this is usually said to be the case if Keq is between 0.1 and 10)

  33. Pressure • The equilibrium constant can also be written in terms of the partial pressures of each gas in a mixture. • The general rules still apply • The Kc and Kp expressions can be related by the equation: • Kp =Kc(RT)Δn, where Δn = change in # moles of gas as the reaction goes from reactants to products

  34. Predicting The Direction of a Reaction • You can also use the equilibrium constant idea to determine if a system is at equilibrium, and if not whether it will tend to reach equilibrium via the forward or reverse reaction. • The term Keq or Kc means we know that a system is at equilibrium (no net change in [reactants] or [products], forward and reverse rate are =

  35. Q - the reaction quotient We define a term Q, the reaction quotient to be the same ratio as the equilibrium constant, just not known to be at equilibrium We can compare Q to K and one of three things must be true Q > K (not at equilibrium goes to left) Q < K (not at equilibrium goes to right) Q = K (at equilibrium)

  36. Example • Ammonia is made by the following reaction • 3 H2(g) + N2(g) 2 NH3(g) , which as an equilibrium constant value of 0.95 at 300K • A mixture in a 5.0 L flask has the following composition (at 300K): • 2.0 moles hydrogen, 4.0 moles nitrogen and 1.0 moles ammonia. • Is the system at equilibrium and if not will the forward reaction (shift right) or reverse reaction (shift left) happen to reach equilibrium?

  37. Q vs K • Write out Q • Plug in and solve for Q • Compare Q vs K • Explain • Q = [NH3]2 = (0.2)2 = .78 [H2]3[N2] (0.4)3(0.8)

  38. Q vs K • Q is less than K so the reaction mix has too few products (or too many reactants) to be at equilibrium • The forward rxn will happen until equilibrium is reached (shift right) • This does not tell us how fast this will happen

  39. Example:  A mixture consisting initially of 3.00 moles NH3, • 2.00 moles of N2, and 5.00 moles of H2, in a 5.00 L container was • heated to 900 K, and allowed to reach equilibrium.  Determine • the equilibrium concentration for each species present in the equilibrium • mixture. • 2 NH3(g)       N2(g) + 3 H2(g)    Kc = 0.0076 @ 900 K • Convert the initial quantities to molarities as shown for NH3. • Create a chart as illustrated below and enter in the known quantities. • Calculate Qc and compare to Kc to determine the direction the reaction • will proceed.

  40. Le Chatelier’s Principle • Equilibrium systems have an ability to restore equilibrium if some change happens that makes the system not at equilibrium (a chemical is added or removed, the temperature or pressure is changed). • The formal statement of this is called LeChatelier’s principle

  41. LeChatelier • A system at equilibrium, when stressed, will reacheive equilibrium by having either the forward or reverse reaction increase in rate, until equilibrium is restored

  42. Adding A Chemical • Adding a reactant will mean that the ratio of [products]/]reactants] is too low (excess of reactants or denominator) • How does this get solved? • You need to “use” up the excess reactants and this is done by having the forward reaction take place, until equilibrium is restored. We say the system “shifts right.”

  43. Like This equilibrum excess of reactants equilibrium restored

  44. Adding a Reactant • Causes the forward reaction to increase in rate until the equilibrium ratio is restored • This uses up the excess reactants, creating more products

  45. Adding A Product

  46. Removing a Reactant • This creates a “shortage” of reactants (or an excess of products) • How to get more reactants- reverse reaction, uses up products and forms reactants

  47. Reactant/Product Summary • Add a Reactant • Shift right • Add a Product • Shift left • Remove a Reactant • Shift left • Remove a Product • Shift right

  48. 3H2 +N2 2 NH3

  49. Temperature Effects • This is not dealing with the rate of the reaction • This is the effects of changing temperature on the composition of an equilibrium mixture • You need to know if the rxn is endothermic or exothermic • Endothermic- heat goes in and is treated like a reactant • Exothermic- heat is released and is treated like a product

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