Chapter 13 Review

  1. Section 13.1:  Equilibrium
    1. Most reactions are not one-way, they are reversible, the products can react and make reactants.  A + B D C + D    
    2. How is equilibrium reached?
      1. Pretend we start with only reactants A and B.  They are colliding and making C and D.  At the start [D] and [C] = zero.  There is no reverse reaction, only the forward reaction.  The forward reaction rate is at its highest. Note [ ] means concentration in moles / L.
      2. Later (could be one second or 50 years) there is enough C and D that they start colliding and reacting producing A and B.  The reverse reaction has "kicked in" and the reverse reaction rate is rising.  The forward reaction rate is slowing down. 
      3. Eventually this situation will be reached.  It could take seconds or years (that depends on kinetics) but this will eventually occur.  Everytime A + B make C + D, somewhere else C + D make A + B.  The foward reaction rate is the SAME as the reverse reaction rate.  That is what is equal at equilibrium.
      4. Note that the concentrations are no longer changing, they are constant - so it appears the reaction is over BUT that is not true.  The reaction is still going in both directions, just at equal rates. The reaction is never, ever over.
      5. Note that the concentrations are not equal to each other.  That has nothing to do with equilibrium.
      6. What happens if we start with only products C and D?  Well they can react and make A and B. Eventually the exact same equilibrium is reached. Wow! 
    3. Usually one side of the reaction dominates or is favored.  If at equilibrium 90% is products and 10% is reactants, the products are favored.  If at equilibrium 4% is products and 96% is reactants, the reactants are favored.
  2. Section 13.2 Equilibrium Constant K
    1. A better way to express these ratios (4:96) is to use equilibrium constants K.
    2. K = [products]coef / [reactants]coef where coef is the coefficient from the balanced chemical reaction. [ ] = concentration in mol/L
    3. Example:  Write the K expression for this reaction:  3 H2(g) + N2(g) D 2 NH3(g)     Answer:  K = [NH3]2 / [H2]3[N2]
    4. Equilibrium constants are just that - constant for a reaction at a certain temperature.
    5. What does it mean if K is a huge number like 4.55 x 109?  That means that [prod] >>[react], the reaction goes forward almost completely until equilibrium is reached, and products are favored.  What does it mean if K is a tiny number like 8 x 10-7?  It means that [prod]<<[react], the reaction hardly went forward at all before equilibrium set up, mostly reactants exist and are favored.
    6. Try examples 13.1 to 13.4 and Prob 13.1 to 13.4
  3. Section 13.3: Writing K expressions (Kc versus Kp)
    1. Kc is the equilibrium constant using molarities:  the expression is Kc = [prod]coef / [react]coef (don't include solids or liquids)
    2. Kp is the equilibrium constant using pressures:  the expression is Kp = Pprodcoef / Preactcoef (only include gases)
    3. Kp = Kc(RT)Dn  where Dn = moles of product gas - moles of reactant gas and R = 0.08206Latm/molK
    4. K has no units Note the expression refers to the mathematical expression above in A and B.  If I ask you to calculate K then I want a number.
  4. Section 13.4 Heterogeneous Equilibria
    1. When a reaction contains solids or liquids they are NOT included in the equilibrium expression.  Solids and liquids don't have [ ] because they are not dissolved in a solvent.  Now let's put it all together:
    2. Example:  Write Kc and Kp expressions for     2 Cu2S(s) +   3 O2(g) D   2 Cu2O(s) +   2 SO2(g)
      1. Kc = [SO2]2 / [O2]3    (remember solids don't appear in Kc expressions)
      2. Kp = PSO22 / PO23    (remember only gases appear in Kp expressions)
      3. Calculate Kc at 27.8oC if the pressure of SO2 gas is 1.25 atm and the pressure of oxygen gas is 0.784 atm.  Solution:  First we must find Kp because we are given pressures, then we can find Kc.  Kp = (1.25)2 / (0.784)3 = 3.24.  Now 3.24 = Kc [ (0.0821)(301)]-1  and thus Kc = 80.1
    3. Example:  Write Kc and Kp expressions for    CaCO3(s) + 2 HCl(aq) D CaCl2(aq) + H2O(l) + CO2(g)
      1. Kc = [CaCl2][CO2] / [HCl]2
      2. Kp = PCO2
    4. Example:  3 H2 (g)+ N2(g) D 2 NH3 (g)  Calculate Kc, Kp and the pressure of ammonia given [H2] = 0.104M, [N2] = 0.554M, [NH3] = 0.418M, 27.0oC, hydrogen gas pressure is 1.24 atm and nitrogen gas pressure is 2.17 atm.
      1. First we are given all molarities so we find Kc. Kc =  [NH3]2 / [H2]3[N2] = (0.418)2 / (0.104)3(0.554) = 2.80 x 102
      2. Now find Kp.  Kp = (2.80 x 102)[(0.0821)(300)]-2 = 0.462
      3. Now sub into Kp expression:  0.462 = PNH32 / (1.24)3(2.17).    solving PNH32 = (1.24)3(2.17)(0.462)   Take the square root of both sides and PNH3 = 1.38 atm
    5. Try examples 13.5 to 13.7 and prob 13.5 to 13.7    At this point given all the equilibrium concentrations you should be able to calculate K.  Also given K and all but one equilibrium concentration you should be able to calculate that unknown concentration. 
  5. Section 13.5:  Calculations with K  and Introduction to Q
    1. As mentioned earlier K can range from huge to tiny numbers.  If huge the reaction proceeds almost to completion (almost all products). If tiny the reaction really doesn't proceed at all (almost all reactants).  If K is between 10-3 and 103 then both reactants and products are present in an equilibrium mixture in appreciable amounts. Try problem 13.8.
    2. The reaction quotient Q is calculated exactly like K except we are not sure if the concentrations are at equilibrium or not.  If Q = K then we are at equilibrium.  Consider A D B with K = 500.  If we calculate Q = [B] / [A] and it is 10, then we are NOT at equilibrium, we need more products to reach the K of 500 so the reaction goes forwards to the right.  If we calculate Q = [B] / [A] and it is 1000, then we are NOT at equilibrium, we need more reactants to reach the K of 500 so the reaction goes backwards to the left. 
    3. So in general:  if Q<K reaction goes to the right until it reaches equilibrium.  If Q=K we ARE at equilibrium.  If Q>K the reaction goes to the left until it reaches equilibrium.  See Figure 13.5
    4. Example:  Which way will the reaction proceed to reach equilibrium if Kc = 5.5 x 10-2, and [SO2] = 4.58 x 10-5M,  [O2] = 4.2 x 10-4M and  [SO3] = 9.9 x 10-5M for this reaction:   2SO2 (g)+ O2 (g) D 2SO3 (g)  Answer:  Q = [SO3]2 / [SO2]2[O2] = (9.9 x 10-5)2 / (4.58 x 10-5)2(4.2 x 10-4) = 1.1 x 104 which is way bigger than K so the reaction proceeds to the left towards the reactants (goes in the reverse direction) until equilibrium is reached.
    5. Try example 13.8 and prob 13.9 and 13.10.
    6. If given initial concentrations and K, you should be able to calculate equilibrium concentrations.  THIS IS IMPORTANT.
      1. Consider the reaction  2A + B D  2C with K = 455.  If given the starting (initial) concentrations of A and B to be 1.00M, you can figure out all the equilibrium concentrations by setting up a table and using algebra skills.  Here is an example of the table which has initial concentrations, the change in concentration to reach equilibrium, and the equilibrium concentrations.
      2. K = [C]2 / [A]2[B] = (2x)2 / (1.00-2x)2(1.00-x)
      3. And so you solve by algebra skills for x.
      4. Once you get the value for x remember that [A] = 1.00-2x, [B] = 1.00-x, and [C] = 2x.  Solve for what the question asks.  You may have to use the quadratic formula. 
      5. Example:  Cl2 (g)  D  2Cl(g)  with Kc = 7.9 x 10-6.  Find all equilibrium concentrations if you start with 0.0214M chlorine gas.  If you set up your table correctly and solve for x using the quadratic formula, you will get x = 2.04 x 10-4.  Which makes the equilibrium concentrations of chlorine gas 0.0212M and chlorine atoms 4.08 x 10-4M.
      6. Example:  PCl5(g)  D PCl3(g)  +  Cl2 (g)   with Kc = 0.030.  Find all equilibrium concentrations if you start with 0.100M PCl5.  Answer:  Solving for x using the quadratic formula gets us 0.042 or -0.072 but the negative number makes no sense so x is 0.042.  So we get [PCl5] = 0.100 - 0.042 = 0.058 M and [PCl3] = [Cl2] = 0 + 0.042 = 0.042M.  
      7. Try example 13.9 to 11 and prob 13.11 to 15. 
  6. Section 13.6  Factors that affect equilibrium
    1. Once a "system" (ie chemical reaction) is at equilibrium, stressing the system will cause non equilibrium conditions, so the reaction must shift either towards reactants or products to reach equilibrium again.  This is called Le Chalelier's Principle.  What causes the stress?  Changes in concentration, pressure, volume and temperature.
    2. Example:  A D B with K = 4.  
      1. At equilibrium let [A] = 5 and [B] = 20.  As you can see K = [B] / [A] = 20/5 which = 4.  Yes we are at equilibrium.  
      2. What happens if I remove 5 of the B's?  Well then [B] = 15 while [A] is still 5.  Note Q = 15 / 5 = 3 which is NOT equilibrium.  The change in concentration cause us to not be at equilibrium - the system is stressed!  What to do?
      3. Note that Q<K, so the reaction must go right towards the products.  (We must replace B since we removed it).  One A will turn into one B.
      4. Now after the shift, [A] = 4 and [B] = 16 so Q = 16 / 4 = 4!  Hurray, we are at equilibrium one again.  Amazing!
  7. Section 13.7 Changes in concentration 
    1. If we remove a chemical we must shift to replace it.  
    2. If we add a chemical, there is more of it to react so it will react
  8. Section 13.8  Changes in V and P
    1. Changes in Volume (affects reactions with at least one gas).  Increase volume - there is more space so there can be more gases.  Shift towards the side with the most gas.  Decrease volume - there is less space for gases.  Shift towards the side with the least gas.
    2. Changes in Pressure (affects reactions with at least one gas).  Decrease pressure - it is less crowded, there is more space so there can be more gases.  Shift towards the side with the most gas.  Increase pressure - it is crowded, there is less space for gases.  Shift towards the side with the least gas.
    3. Note if there are the same number of moles of gas on each side, then P and V won't affect the equilibrium and no shifting will occur. Also note that adding another gas won't change anything either.  
  9. Section 13.9 Changes in T
    1. Changes in Temperature (depends on if the reaction is exothermic or endothermic).  If you increase the temperature there is more heat to react, just like increasing concentration.  If you decrease the temperature heat is removed so we must replace it.  Remember that exothermic reactions have heat exiting as a product - heat is on the product side.  Endothermic need heat to enter as a reactant - heat is on the reactant side. 
    2. Unlike changing Concentration, V and P, changing the Temperature will actually change the K value.  K is constant only at constant T so if the T changes, the value of K also changed
    3. Try example 13.12 to 13.14  and prob 13.16 to 13.22.
  10. Section 13.10 Catalyst ( a substance that speeds up a reaction)
    1. A catalyst helps a reaction REACH equilibrium faster.  Once at equilibrium a catalyst has no effect. So a catalyst will not cause any shifting to occur nor have any affect on K.. See Figure 13.14. 
    1. Examples of Le Chatelier's Principle:
    2. Consider this exothermic reaction (heat is a product):  A(s) + 2 B(g)  D C(g) + 3 D(g) + heat
      1. add A, nothing - solids aren't in K expressions  Got ya! 
      2. add C, shift left
      3. remove B, shift left
      4. remove D, shift right
      5. increase T, shift left
      6. decrease T, shift right
      7. decrease P, shift right
      8. increase V, shift right
      9. increase P, shift left
      10. add catalyst, nothing
      11. add helium gas, nothing
    3. Consider this endothermic reaction:  3 H2(g) + N2(g)    D  2 NH3(g)
      1. add nitrogen, shift right
      2. remove hydrogen, shift left
      3. add ammonia, shift left
      4. remove ammonia, shift right
      5. raise temp, shift right  (remember heat is on the reactant side)
      6. lower temp, shift left
      7. remove nitrogen, shift left
      8. add argon gas, nothing
      9. decrease V, shift right
      10. decrease P, shift left
    4. Which of the changes above changed the value of K???   Only the ones involving changing T.

Section 13.11:  How K and k relate

    1. Consider this equilibrium:  A D B
    2. For the forward reaction A g  B, the rate law is forward rate = kf[A]
    3. For the reverse reaction B g A, the rate law is reverse rate = kr[B]
    4. At equilibrium the forward and reverse rates are equal, so kf[A] = kr[B]
    5. Rearranging we get kf / kr = [B] / [A]
    6. The equilibrium constant K also = [B] / [A]
    7. Thus K = kf / kr !  Wow I know you are amazed!   Example 13.15

 

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