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13.1 Equilibrium Conditions

13.1 Equilibrium Conditions. When a system is at equilibrium it may appear that everything has stopped; however, this is NOT the case. Think of chemical equilibrium like the cars on the Golden Gate Bridge and that the traffic flow in both directions is the same. No net change.

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13.1 Equilibrium Conditions

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  1. 13.1 Equilibrium Conditions • When a system is at equilibrium it may appear that everything has stopped; however, this is NOT the case. • Think of chemical equilibrium like the cars on the Golden Gate Bridge and that the traffic flow in both directions is the same. No net change

  2. Chemical Equilibrium • The state where the concentrations of all reactants and products remain constant with time. • On the molecular level, there is frantic activity. Equilibrium is not static, but is a highly dynamic situation.

  3. What is equilibrium? • Equal rates • Concentrations are not • Rates are determined by concentrations and activation energy • Concentrations do not change at equilibrium or if the reaction is verrrrrry slooooooow.

  4. 13.2 The Equilibrium Constant • For any reaction • Where K is the equilibrium constant • A,B,C & D are the concentrations of the “chemical species”a@ equilibrium • j,k,l & m are the coefficients in the balanced equation

  5. Try this: • Write the equilibrium expression for 4NH3 + 7O24NO2 + 6H2O

  6. And if I knew… • …the equilibrium concentrations for each of the reaction components, I could calculate K. Try this reaction: N2 + 3H2 ↔ 2NH3 given: [NH3]=3.1 x 10-2 mol/L [N2]=8.5 x 10-1 mol/L [H2]=3.1 x 10-3 mol/L

  7. What would K be for the reverse reaction? 2NH3 ↔ N2 + 3H2 K’ = 2.6x10-5 • How about if only 1mol of NH3 was produced? N2 + H2 ↔ 1NH3 K’’=1.9 x 102

  8. Big conclusions!! • By comparing K for the forward reaction to K for the reverse reaction, notice that • And if I multiply (or divide) my reaction by a factor then where n is the factor

  9. K is constant…??? • The equilibrium constant K always has the same value at a given temperature. • A change in the temperature will change the rate and thus a new K. • A set of concentrations at equilibrium is called an equilibrium position. • There are an unlimited (infinite) number of equilibrium positions.

  10. 13.3 Equilibrium Expressions Involving Pressures • So far we’ve been talking about equilibrium in terms of concentration… • But gases can also be described by pressures as well. PV = nRT Or Where C is molar concentration of the gas

  11. For: N2 + 3H2 ↔ 2NH3

  12. The relationship • The relationship between K and Kp comes from the fact that for an ideal gas C = P/RT. • But for the general reaction the relationship between K and Kp is Where ∆n is the sum of the coefficients of the gaseous products MINUS the sum of the coefficients of the gaseous reactants. ∆n = change in moles of gas

  13. 13.4 Heterogeneous Equilibria • Remember that homogeneous equilibria is when all the products and reactants are gases 2SO2(g) + O2(g) ↔ 2SO3(g)

  14. = 0.14 0.012 x 0.054 [COCl2] [CO][Cl2] Let’s look at a problem with homogeneous equilibria • The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl2 (g) at 740C are [CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and Kp. = 220 Kc= CO (g) + Cl2 (g) ↔ COCl2 (g) Kp = Kc(RT)Dn Kp= 220 x (0.0821 x 347)-1 = 7.7

  15. Heterogeneous equilibria • Heterogeneous equilibria is when you have reaction components that may not be in the gaseous state. CaCO3(s) ↔CaO(s) + CO2(g)

  16. CaCO3(s)CaO(s) + CO2(g) PCO PCO 2 2 does not depend on the amount of CaCO3 or CaO = Kp

  17. CaCO3 (s) CaO (s) + CO2 (g) Kc = Kc = [CO2] [CaO][CO2] [CaCO3] Kp= PCO 2 More importantly remember… • …that the concentrations of pure solids and liquid can not change therefore they are not included in the expression for the equilibrium constant [CaCO3] = constant [CaO] = constant

  18. Write the equilibrium constant for the heterogeneous reaction

  19. Kp = P NH3 P H2S Try this one. • Consider the following equilibrium at 295 K: • The partial pressure of each gas is 0.265 atm. Calculate Kp and Kcfor the reaction? Hint: this time start with Kpand then find Kc NH4HS (s)↔ NH3(g) + H2S (g) = 0.265 x 0.265 = 0.0702 Kc = 0.0702 x (0.0821 x 295)-2 = 1.20 x 10-4

  20. 13.5 Applications of the Equilibrium Constant • Knowing the equilibrium constant for a reaction allows us to predict several important features of the reaction: • Tendency of the reaction to occur • If concentrations given represent equilibrium • Equilibrium position from the initial concentrations

  21. Important feature 1:Tendency of a reaction to occur • The tendency of a reaction to occur can be indicated by the magnitude of K K >> 1 Lie to the right Favor products K << 1 Lie to the left Favor reactants Reaction goes to completion Reaction does not occur to any significant extent

  22. Remember • The size of K and the time it takes for the reaction to reach equilibrium are NOT directly related. • The time it takes to reach equilibrium depends on the reaction rate and therefore ultimately the activation energy

  23. Reaction Quotient - Q • Tells us the direction the reaction needs to go in order to reach equilibrium • It is calculated the same as the equilibrium concentration; however, it uses initial concentrations rather than equilibrium concentrations.

  24. Compare Q to Kc • Qc < Kcsystem proceeds from left to right to reach equilibrium Not enough products Forward reaction occurs • Qc = Kcthe system is at equilibrium No net change • Qc > Kcsystem proceeds from right to left to reach equilibrium Too many products Reverse reaction occurs

  25. Try this…N2 (g)+ 3H2 (g)↔ 2NH3 (g) • At the start of a reaction, there are 0.249 mol N2, 3.21x10-2 mol H2, and 6.42x10-4 mol NH3 in a 3.50 L reaction vessel at 375oC. If the equilibrium constant (Kc) for the reaction is 1.2 at this temperature, decide whether the system is at equilibrium. IF it is not, predict which way the net reaction will proceed.

  26. How did you do? • First: find the molarity of each gas. • Then find Q • Not at equilibrium • Shift left to right

  27. Important feature 2: given the equilibrium constant and initial concentrations, find equilibrium concentrations/pressures of products & reactants • Let’s do a few problems • Gaseous N2O4 was placed in a flask and allowed to reach equilibrium @ a temperature where Kp=0.133. At equilibrium, the pressure of N2O4 was found to be 2.71atm. Calculate the equilibrium pressure of NO2. N2O4 (g)↔NO2(g)

  28. Here’s how • First: Given Kpand PN2O4 = 2.71atm

  29. Here’s another type of problem • At a certain temperature a 1.00L flask initially contained 0.298M PCl3 and 8.7x10-3M PCl5. After the system had reached equilibrium, 2.00x10-3M Cl2 was found in the flask. Gaseous PCl5 decomposes according to the following reaction: PCl5 (g) ↔ PCl3 (g) + Cl2 (g) Calculate the equilibrium concentrations for all species and Kc

  30. The steps to calculating equilibrium concentrations – ICE diagrams • Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, which represents the change in concentration. • Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x. • Having solved for x, calculate the equilibrium concentrations of all species.

  31. Let’s set one up Since the concentration of Cl2 we know has changed by 2x10-3, the balanced reaction tells us that since it is a 1-1-1 reaction that the reactants decompose by 2x10-3 and the products concentration increases by 2x10-3

  32. Once we know the equilibrium concentrations… • …we can find the equilibrium constant.

  33. Let’s try another using an ICE diagram • There are some instances when you may not be given ANY equilibrium concentrations. Here’s how to do those. Let’s get started!!

  34. H2 (g) + I2 (g) ↔ 2HI (g) • A mixture of 0.00623M H2, 0.00414M I2 and 0.0224M HI was placed in a 1.00L stainless-steel flask at 430oC. The equilibrium constant for the reaction is 54.3 at that temperature. Calculate the concentrations of each species at equilibrium.

  35. The first step • Calculate Q Since 19.5<<<54.3, the reaction will proceed from left to right. The hydrogen and iodine will be depleted and there will be a gain in HI. H2 (g) + I2 (g) ↔ 2HI (g)

  36. Set up an ICE diagram Now for the Algebra!!

  37. Next: the equilibrium constant is… After multiplying all of this out… Now do the quadratic formula…

  38. And the x values are… x = 0.0114M and x = 0.00156M The first value of x can NOT be correct since it is larger than the original concentration of hydrogen and iodine. Therefore x = 0.00156 This is the change in concentration. Now back to the ICE diagram!!

  39. You can check your work by plugging the values in to find Kc. Remember Kc = 54.3

  40. 13.6: Solving equilibrium problemsA more complete set of steps to solve equilibrium problems. • Write the balanced equation for the reaction. • Write the equilibrium expression using the law of mass action • List the initial concentrations • Calculate Q, and determine the direction of the shift to equilibrium • Define the change needed to reach equilibrium, and define the equilibrium concentrations by applying the change to the initial concentrations. • Substitute the equilibrium concentrations into the equilibrium expression, and solve for the unknown • Check your calculated equilibrium concentrations by making sure the give the correct value of K.

  41. Let’s do a few more ICE problems Consider the reaction represented by the equation Fe3+(aq) + SCN-(aq) FeSCN2+(aq) • Trial #1 6.00 M Fe3+(aq) and 10.0 M SCN-(aq) are mixed and at equilibrium the concentration of FeSCN2+(aq) is 4.00 M. What is the value for the equilibrium constant for this reaction? You’ll need this for the next few problems!!

  42. Fe3+(aq) + SCN-(aq) ↔ FeSCN2+(aq) • Initial 6.00 10.00 0.00 • Change -4.00 -4.00 +4.00 • Equilibrium 2.00 6.00 4.00

  43. Consider the reaction represented by the equation Fe3+(aq) + SCN-(aq) ↔ FeSCN2+(aq) • Trial #2: Initial: 10.0 M Fe3+(aq) and 8.00 M SCN−(aq) Equilibrium: ? M FeSCN2+(aq) 

  44. Consider the reaction represented by the equation Fe3+(aq) + SCN-(aq) FeSCN2+(aq) • Trial #3: Initial: 6.00 M Fe3+(aq) and 6.00 M SCN−(aq) Equilibrium: ? M FeSCN2+(aq)

  45. Consider the reaction represented by the equation Fe3+(aq) + SCN-(aq) ↔ FeSCN2+(aq) Fe3+ SCN- FeSCN2+ Trial #1 9.00 M5.00 M 1.00 M Trial #2 3.00 M2.00 M 5.00 M Trial #3 2.00 M9.00 M6.00 M Find the equilibrium concentrations for all species.

  46. Let’s see what happen here… H2(g) + I2(g) ↔ 2HI(g) K=7.1 x 102 Calculate the equilibrium concentrations if a 5.00L container initially contains 15.8g of H2 and 294g of I2 [HI]0 = zero

  47. Since Q = 0 and Q < K, more product will be formed. We know that: K is large so the rxn will almost go to completion I2 is the limiting reagent. It will be the smallest at equilibrium…let it’s equilibrium concentration = x

  48. Fill in ICE based on stoichiometry Now use K = 7.1 x 102 and algebra to find the value of x. BUT…we know x will be very small…sooo..

  49. Because x is so small, it is negligible when added to another number and will essentially have little effect on the outcome. Giving us the following for our equation… • Making the algebra MUCH, MUCH easier. • Now , find the other concentrations.

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