Complete KEY to Equilibrium Workbook

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					Chem 125                                                                                        Dr. Neff
                      Complete KEY to Equilibrium Workbook
1.What is a reversible reaction, and under what condition does equilibrium occur in such a reaction?
A reaction that can occur in both the forward and the reverse direction is said to be reversible.
When the rate of the forward reaction equals the rate of the reverse reaction, we say the
reaction is at equilibrium.

2. What is the difference between Kc and Kp for a chemical reaction?
Kc is the equilibrium constant when reactants and products are expressed in concentration
Kp is the equilibrium constant when reactants and products are expressed in pressure units.
You should be able to write expressions for both, and we can convert between them just yet.

Mathematically, Kp = Kc(RT)

3. What is the reaction quotient of a reaction, and what is the relationship between Qc and Kc?
The reaction quotient has the same form as Kc but the concentrations are not necessarily at
equilibrium. The reaction quotient can be used to predict the direction that a reaction will
move to attain equilibrium. At equilibrium Qc = Kc.

4. Is the following statement true or false? "Reactions with large equilibrium constants are very fast."
Explain your answer.
False: Kc only indicates the balance between reactants and products. In this way a large Kc
indicates that products will be favored at equilibrium but says nothing about how fast
equilibrium is achieved.

5. Suppose that the reactions A --> B and B --> A are both elementary processes with rate constants
of 9.6x102 s-1 and 3.8x104 s-1, respectively.
        (a) What is the value of the equilibrium constant for the equilibrium A  B?
       You can derive the equilibrium constant expression K using rate forward equals rate
       reverse (see text), this shows that K is equal to the ratio of rate constants for forward
       and reverse reactions:
               k fwd   9.6 x 102
        Kc =         =           = 0.0253
               krev    3.8 x 104
        (b) Which is greater at equilibrium, the concentration of A or the concentration of B? Explain.
        Since Kc < 1 the reactants (A) will be favored.

6. Write the expression for Qc for the following unbalanced reactions (balance them first!):

   a)      C3H8 (g)   +   5 O2 (g)         3 CO2 (g)   +   4 H2O (g)

             [CO2 ]3 [H 2O]4
   Qc 
               [C 3H8 ][O 2 ]5
   K would have same form, but concentrations would be equilibrium concentrations
Chem 125                                                                                          Dr. Neff
                 [CO ] [H2 O]  3                 4

        Kc 
                             2 eq                eq

                 [C 3H8 ]eq [O 2 ]5
   b)    4 KO2 (s)       +     2 H2O (g)                     4 KOH (s)    +   3 O2 (g)

                              [O 2 ]3
                 Qc 
                             [H2 O]

   c) Ti(s) + 2Cl2(g)                 TiCl4(l)        Already balanced!

                 Qc 
                             [Cl2 ]2
7. For the reaction: N2 (g) + 3 H2 (g)          2 NH3 (g)
   Kc = 2.4 x 10 at 1000 K. What are the values of Kc for the following balanced reactions?

   a) 1/3 N2 (g) + H2 (g)                  2/3 NH3 (g)
      This reaction is 1/3 x the original reaction, thus

        K'c  (Kc )1/ 3  (2.4x103 )1/ 3  1.3x101
   b) NH3 (g)               1/2 N2 (g) + 3/2 H2 (g)
      This reaction is 1/2 x the reverse of the original so

           '   1             1
        K c     1/ 2         3 1/ 2  2.0x101
            (Kc )       (2.4x10 )
8. At a typical operating temperature of an automobile's engine (~ 100 °C), N2 and O2 form NO. NO
   then combines with more O2 to form NO2, a toxic pollutant:
   1. N2 (g) + O2 (g)               2 NO (g)          K1 = 4.3 x 10-25
   2. 2 NO (g)       +       O2 (g)              2 NO2 (g)         K2 = 6.4 x 109
   a) Write the balanced equation for the net (overall) reaction, and determine K for this reaction.
      N2 (g) + O2 (g)                         2 NO (g)             K1 = 4.3 x 10-25
   + 2 NO (g) + O2 (g)                      2 NO2 (g)              K2 = 6.4 x 109
        N2 (g)   +   2 O2 (g)               2 NO2 (g)              Knet = K1 x K2 = 2.8 x 10-15
Chem 125                                                                                          Dr. Neff
b)    Prove to yourself that the expression for the overall K is the same as the product of the K
expressions for each individual step
                          2                         2
             [NO]                            [NO2 ]
       K1                       and K 2 
            [N 2 ][O2 ]                    [NO]2 [O2 ]
                   [NO]2       [NO2 ]2     [NO2 ]2
       K1  K 2                        
                  [N 2 ][O2 ] [NO]2 [O2 ] [N 2 ][O2 ]2
                [NO2 ]2
       K net             2
                             K1  K2 as shown above
               [N 2 ][O2 ]
   c) Based on the magnitude of this value for the overall K, would you say that this reaction is
   product favored or reactant favored? Explain your answer. Since Knet is very small, the reaction
   is reactant favored, as a small amount of product but large amount of reactant would result
   in a very small K value.

9. The following equilibrium constants were obtained for the reactions listed below at 500 K:
      H2    +    Br2             2 HBr            K1 = 7.9 x 1011
      H2             2H                           K2 = 4.81 x 10-41
      Br2                 2 Br                    K3 = 2.2 x 10-15

   What is the value of K for the reaction: H + Br      HBr?
   (NOTE this is sort of like a Hess’s Law type problem where you first have to figure out how to sum
   the three reactions together)

       We need to take 1/2 of the reverse of the second and third reactions and add those to
1/2 of the first reaction:
               H         1/2 H2                     K’ = (1/K2)1/2 = 1.44 x 1020
                Br            1/2 Br2                    K’ = (1/K3)1/2 = 2.1 x 107
+ 1/2 H2 + 1/2 Br2               HBr                     K’ = (K1)1/2 = 8.9 x 105
      H + Br                     HBr       K = (1.44 x 1020) x (2.1 x 107) x (8.9 x 105) = 2.7 x 1033
Chem 125                                                                                         Dr. Neff
10. At 127˚C, Kc = 2.6x10-5 mol2/L2 for the reaction

                                     2NH3(g)             N2(g) + 3H2(g)
Calculate Kp at this temperature.

                   n gas
K p  K c (RT )
Here ngas = 4- 2 = 2
K p  (2.6x105 )[(0.821)(400)]2
K p  2.8x102
11. When wine spoils, the ethanol is oxidized to acetic acid as O2 from the air reacts with the wine:
                            C2H5OH(aq) + O2(aq)           HC2H3O2 (aq) + H2O(l)

The value of Kc for this reaction at 25˚C is 1.2x1082.

       (a) Will much ethanol remain when this reaction has reached equilibrium? Explain.
              With a Kc that large, at equilibrium there must be a large amount of product and
       very little reactant, as K equals the concentrations of products divided by
       concentrations of reactants. So NO not much ethanol will remain at equilibrium.
        (b) With this large of a value for Kc, why doesn't a bottle of wine spoil rapidly when left open?
       K tells us nothing about the rate of a particular reaction! A bottle of wine
       doesn’t spoil rapidly because the rate of spoilage is relatively slow, but we cannot tell
       this from K.

12. In a study on the conversion of methane to other fuels, a chemical engineer mixes gaseous CH 4
and H2O in a 0.32 L flask at 1200 K:
                    CH4 (g) +        H2O (g)      CO (g) +       H2 (g)
   At equilibrium, the flask contains 0.26 mol CO, 0.091 mol H2 and 0.041 mol CH4. If K is 0.26, what
   is the [H2O] at equilibrium? Start by balancing the above equation and writing the expression for

    [CO][H 2 ]3
K               0.26
   [CH4 ][H2 O]
       [0.8125][0.2844]3    1.869x102
0.26                    
         [0.1281][H 2O]    [0.1281][H 2O]
0.26  (0.1281)  [H2 O]  1.869x102
           1.869x102     1.869x102
[H2 O]                              5.6x101
         0.26  (0.1281) 3.33x102
Chem 125                                                                                          Dr. Neff
13. The R&D department of a chemical company is studying the reaction of CH4 and H2S, two
    components of natural gas:
                     CH4 (g)     +     2 H2S (g)            CS2 (g)     +    4 H2 (g)
   In one experiment, 1.00 mol CH4, 1.00 mol CS2, 2.00 mol H2S and 2.00 mol H2 are mixed in a 250
   mL vessel at 960 °C. At this temperature, Kc = 0.036.
   a) In which direction will this reaction proceed to equilibrium? Explain.

             [CS2 ][H 2 ]4     4.00(8.00)4
       Qc                 2
                                        2
                                            64.0
            [CH 4 ][H 2 S]     4.00(8.00)

       As this shows, Qc>>Kc, so reaction will proceed in reverse direction to
       reach equilibrium. We use this knowledge to help us solve b.

b) If [CH4] = 5.56 M at equilibrium, what are the equilibrium concentrations of all the other
substances? (Start with a reaction table!!!!)

                     CH4 (g)       +     2 H2S (g)               CS2 (g)     +       4 H2 (g)

Initial              4.00                  8.00                   4.00                  8.00
Change               +x                    +2x                    -x                    -4x
Equilibrium          4.00+x = 5.56         8.00+2x               4.00-x                 8.00-4x

We’re told that [CH4] = 5.56 M at equilibrium, this tells us that reactant
concentration has increased as the reaction progressed, that along with
magnitude of Q above helps us decide sign on x for reactants and products.
Thus 5.56 = 4.00+x, or x = 1.56. Then [H2S]=8.00+2x = 11.12 M, [CS2] = 4.00-x =
2.44M, [H2] = 8.00-4x = 1.76M

       Plug these values back into expression for K and solving for K , make sure
       it matches original K value given. It doesn’t match exactly, but its close

        [CS2 ][H2 ]4     2.44(1.76)4    23.4121
Kc                  2
                                   2
                                                0.0341 0.036
       [CH4 ][H2 S] 5.56(11.12) 687.5184
Chem 125                                                                                         Dr. Neff
14. A mixture of CO and Cl2 is placed in a reaction flask: [CO] = 0.0102 mol/L and [Cl2]= 0.0069
mol/L. When the reaction has come to equilibrium at 600 K, [Cl2] = 0.00301 mol/L.
        CO (g) + Cl2 (g)       COCl2 (g)
    a) Calculate the concentrations of CO and COCl2 at equilibrium. Start with an ICE table!
                                       CO                       Cl2                    COCl2
    Initial                          0.0102                   0.0069                      0
    Change                             -x                        -x                      +x
    Equilibrium                    0.0102 – x                0.00301                      x
        Since we know the equilibrium concentration of Cl2, this allows us to calculate x. This
is in fact the only way to calculate x, as we don’t know what the value of K is.
0.0069 – x = 0.00301
x = 0.00389 M              so [CO]eq = 0.0102 –x = 0.0102-0.00389 = 0.00631 M
                           [COCl2]eq = x = 0.00389 M

b) Calculate the equilibrium constant Kc.
      Just plug the equilibrium concentrations we just calculated into the expression for K:
          [COCl 2 ]       0.00389
       K                                2.0 10 2
          [CO][Cl2 ] (0.00631 )(0.00301)
15. The air pollutant nitrogen monoxide is produced in automobile engines because of the high
temperature reaction
      N2 (g) + O2 (g)          2 NO (g)        Kc = 1.7 x 10-3 at 2300 K.
If the initial concentrations of both N2 and O2 are 1.40 M, what are the concentrations of NO, N2 and
O2 when the reaction mixture reaches equilibrium?
Make this your mantra: SET UP THE TABLE
                           N2                                    O2                        2NO
      Initial             1.40                                  1.40                         0
     Change                -x                                    -x                        +2x
   Equilibrium           1.40-x                                1.40-x                       2x
                                     2                     2                           2
                       [NO]             (2x)               (2x)
       K  1.7  10     3
                                                     =
                      [N2 ][O2 ] (1.40  x)(1.40  x)   (1.40  x)

                         (2x)2                                             (2x)
           1.7  10 
                                  or                4.123  102 
                      (1.40  x)2                                       (1.40  x)
       4.123  102 (1.40  x)  2x
       5.772 102  4.123  102 x  2x
       5.772 102  2.04123x
       2.83  10  x
Thus [N2]eq=[O2]eq=1.40-x = 1.37 M (which is essentially 1.4 to 2 sig figs…) and
[NO]eq = 2x = 5.66x10-2 M
Chem 125                                                                                     Dr. Neff
16. The decomposition of nitrogen monoxide to form nitrogen and oxygen gases has an equilibrium
constant Kc of 2.44 x 103 at 2000˚C:
                     2 NO (g)              2 (g) + O2 (g)

a) If 3.7 moles of N2, 5.5 moles of O2 and 0.10 moles of NO are initially placed in a sealed,
evacuated 1.0 L flask, which way (if any) will the reaction proceed to achieve equilibrium? Show your
work to justify your answer.
We calculate Qc and compare to Kc:
              [N 2 ][O2 ] 3.7(5.5)         3
       Qc            2        2  2.0 10  Kc
               [NO]       (0.10)
Since Q< K, reaction must proceed in forward direction (to the right) to reach equilibrium.

   b) This problem concerns the same equilibrium system but is unrelated to part a!!! An
       unknown amount of NO is placed in a reaction vessel (with no products present) and is
       allowed to come to equilibrium at 2000˚C. The equilibrium concentration of N2 is then
       determined to be 0.45 M. Determine the initial concentration of NO. Show your work
       below. (Warning: Carry all sig figs through the calculation until the end!)
This first sentence tells us we can use the same value of K, but the concentrations given in a)
have no bearing here.
So we start from scratch – SET UP THE ICE TABLE!
                                    2 NO                        N2                     O2
          Initial                     y                          0                      0
        Change                       -2x                        +x                     +x
      Equilibrium             y -2x = y -0.90                x=0.45                 x=0.45

This problem is a little different, as we don’t know the initial concentration of reactant. We do
know the final concentration of one of the products, this is x, so we still have only one
variable to solve for, y. Here y is equal to the initial concentration of NO. So we still plug our
equilibrium concentrations into the expression for K and solve for our variable. (There are
other ways to do this)
                     [N2 ][O 2 ] 0.45(0.45)
Kc = 2.44  103                
                      [NO]2       (y  0.90) 2
  2.44  10 3 
                   (y  0.90)2
49.39635           so 49.39635 (y  0.90)  0.45
         (y  0.90)
49.39635y 44.45672  0.45
49.39635y  44.90672
y  0.9091 M 0.91 M = [NO]initial
Chem 125                                                                                          Dr. Neff
17. State Le Chatelier’s Principle in your own words.
Le Chatelier’s Principle states that if we stress an equilibrium system by changing the
temperature, the pressure (volume) or the concentrations of one species involved, that
equilibrium will shift in order to relieve the stress.

18. In the first step of the Ostwald process for synthesis of nitric acid, ammonia is oxidized to nitric
oxide by the reaction
               4 NH3 (g) + 5 O2 (g)        4 NO (g) + 6 H2O (g)                      -905.6 kJ
a) How does the equilibrium amount of NO vary with an increase in temperature? Explain why.
If we increase the temperature, that’s like adding heat to the reaction system. Since this is an
exothermic reaction, heat can be thought of as a product in this reaction. Adding more of that
product (heat) causes the reaction to shift in the direction that uses up heat, to the leftt, back
to reactants. Thus the equilibrium amount of NO will decrease if we increase the temperature.

b) How would you increase the amount of NO produced by a change in temperature? Explain.
To increase the amount of NO produced, we need to stress the system such that it will shift to
relieve that stress. If we decrease the temperature we’re removing heat from the system, so
the system will respond by shifting in such a way as to make more heat, will shift to the right,
producing more heat and more NO.

19. The reaction of iron(III) oxide with carbon monoxide occurs in a blast furnace when iron ore it
reduced to iron metal:
              Fe2O3 (s) + 3 CO (g)          2 Fe (l) + 3 CO2 (g)
Use Le Chatelier’s Principle to predict the direction of reaction when an equilibrium mixture is
disturbed by :
       a) Adding CO (g) If we add CO, reaction will shift to the right to use up added CO. So
       reaction will proceed to the right, to products
       b) Removing CO2 (g) If we remove CO2, reaction will shift to make more CO2, so again
       reaction will shift to the right.
       c) Adding Fe2O3 (s) This is a solid, it concentration has no effect on the equilibrium

20. Does the number of moles of products increase, decrease or remain the same when each of
the following equilibria is subjected to an increase in pressure by decreasing the volume? Explain
your answer in each case.
        a) CO (g) + H2O (g) )          CO2 (g) + H2 (g)
        If we increase the pressure, number of moles of product will be unaffected. There are
        same number of moles of gas on both sides of the equation, so shifting in one direction
        or the other will have the same effect on pressure, so there is no way to relieve a
        pressure change. Thus changing pressure doesn’t affect the equilibrium.

       b) 2 CO (g)         C (s) + CO2 (g)     Moles of product will increase if we increase the
       pressure. Here, if reaction shifts to products, where there are fewer moles of gas,
       pressure will be released.

       c) N2O4 (g)                 2 (g)       Moles of product will decrease if we increase the
       pressure. Here, if reaction shifts to reactants, where there are fewer moles of gas,
       pressure will be released.

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