Name: _________________________________________
Date:____________________________
Optional Homework 6 – do not turn in!!
1.) What is the first thing you must do to a reaction in order to write an accurate equilibrium (K)
expression? Give a real example of a K expression:
You must first balance the equation. You cannot write a K expression unless the equation is
balanced!!
2H2 (g) + O2 (g)  2H2O (g)
Keq =
[H 2 O]2
[H 2 ] 2 [O 2 ]
2.) Explain how K can tell you the direction the reaction is going and if the reaction is at equilibrium
(e.g. explain what a large, small, and intermediate values of K mean for determining reaction
direction).
K values are given by a ratio of product raised to their coefficients divided by reactants raised to
their coefficients.
For a generic equation: aA + bB … cC + dD
K
…
[C]c [D]d
[ A ]a [B]b
If K is large: then the numerator must be a larger quantity than the denominator. This means
that we have product being formed. If the K value is larger than 10+10 in value, we would say
that we have hardly any appreciable reformation of reactants, and the reaction is going to
completion (forming products).
If K is small: then the denominator must be a larger quantity than the numerator. This means
that we have reactants that are not reacting to form products. If the K value is smaller than 10 -10
we might say that this reaction is NOT happening. We might say “no reaction”. Which means
that we really only have measurable amounts of reactants.
If the K is intermediate: we have both reactants and products present. Reactants are turning
into products, products are turning into reactants. We have a system at equilibrium!
3.) Define Q in your own words. How is Q related to K? Is Q always the same numerical value as K?
If Q has different numerical values than K, explain why. Write your OWN chemical reaction and
write the Q expression, then write the Kc expression.
Q is the reaction quotient. It is used to monitor a reaction as it proceeds towards equilibrium.
Examining Q will tell us which way the reaction needs to proceed in order to REACH
equilibrium. The set-up for Q is the exact same set-up for K mathematically. But, Q ONLY
equals K under one condition. The condition of equilibrium. Let’s examine a chemical reaction
below:
2H2 (g) + O2 (g)  2H2O (g)
Q=
[H 2 O]2
[H 2 ]2 [O 2 ]
K=
[H 2 O]2
[H 2 ]2 [O 2 ]
Since the equations are the same, the difference becomes what numbers go into the equations?
At equilibrium, the concentrations will have stabilized (will not be changing so therefore the
ratio of products/reactants will give us the same numerical output answer over and over and
over again). This represents equilibrium. However, when the reaction first begins, we have A
LOT of reactant available. This will make the denominator larger so Q will be a small number.
As the reaction proceeds, the concentration of products will increase, so the Q value will change
from being a small number to growing in size (how big Q gets depends on the reaction!!).
Eventually, calculations for Q will STOP changing. The concentrations of species will have
stabilized and equilibrium will have been reached.
Moral of the story: Q is not always the same number as K. It can be smaller or larger.
4.) When does Q = K? This is the special circumstance when the system is at equilibrium!
5.) The equilibrium constant for the following reaction:
½ H2 (g) + ½ I2 (g)

HI(g)
at 718 K is 14.21
a.) What is the numerical value of Kc for the reverse reaction?
The reverse reaction means that we take the inverse of the given K value: K rev =
1/Koriginal
K=
1
= 0.07037
14.21
b.) What is the numerical value of Kc for the above reaction if the above reaction is doubled?
If the reaction is doubled, then we must RAISE the K value to that exponent:
Kx2 = (Koriginal)2
K = (14.21)2 = 201.9
6.) The equilibrium constant for the following reaction:
½ N2 (g) + 3/2H2O(g)

NH3 (g) + ¾ O2 (g)
at 900 K is 2.48 x 10-19. What is the value of Kc at 900 K for the following reaction?
4NH3 (g) + 3 O2 (g)

The reaction is reversed:
2 N2 (g) + 6H2O(g)
1
2.48 x 10 19
= 4.03 x 1018
The reaction is also multiplied by 4: so we will raise the K value to the 4th power
(4.03 x 1018)4 =
2 .64 x 10 74
7.) In a study of hydrogen iodide gas undergoing decomposition, a researcher fills a 2.00 L flask with
0.200 moles of HI. The reaction is allowed to reach equilibrium.
2HI (g)  H2 (g) + I2 (g)
At equilibrium : [HI] = 0.078 M Calculate Kc
Let’s go through all the steps: step 1: make sure equation is balance (it was given to us!)
step 2: we always need concentrations so turn moles → M
0.200 moles
= 0.100 M HI
2.00 L
[HI] =
step 3: make an ICE table
Equation
I
C
E
2HI
H2

0.100
- 2x
0.078
+
0
+x
x
I2
0
+x
x
Since there are no products to begin, we must form products (gain products) so we will be
adding to the product side. In order to form the products, we must lose reactants, so we subtract
from the reactant side. Remember the change line is tied to the stoichiometry of the reaction!!
Let’s find out what x is! 0.100 – 2x = 0.078
0.100 – 0.078 = 2x
0.022 = 2x
0.011 = x
So now we know how much species changed by: let’s make a new ICE table (not necessary but
it does help!)
Equation
I
C
E
2HI
H2

0.100
- 2(0.011)
0.078
+
0
+0.011
0.011
Now we have all the equilibrium concentrations, plug and chug to K
Kc =
Kc =
[H 2 ][I 2 ]
[HI]2
[0.011][0.011]
[0.078] 2
Kc = 2.0 x 10-2
I2
0
+0.011
0.011
8.) The atmospheric oxidation of nitrogen monoxide 2NO (g) + O2 (g)  2NO2 (g) was studied at 184oC
with initial pressures of 1.000 atm of NO and 1.000 atm of O2. At equilibrium, PO2 = 0.506 atm.
Determine Kp
Same idea as above! Except we use pressures instead of concentrations!
Equation
I
C
E
2NO
+
1.000
- 2x
1.000-2x
O2

1.000
-x
0.506
2NO2
0
+2x
2x
We need to find x. Using the only column that has a complete set of information: O2: let’s find x
1.000 – x = 0.506
1.000 – 0.506 = x = 0.494
x = 0.494 therefore let’s plug that info into a new ICE table
Equation
I
C
E
2NO
+
1.000
- 2(0.494)
0.012
O2

1.000
-0.494
0.506
2NO2
0
+2(0.494)
0.988
Now we have equilibrium pressures of all species, plug it into the K expression and chug out a
Kp number!
Kp =
Kp =
(PNO2 ) 2
(PNO ) 2 (PO2 )
(0.988) 2
(0.012 ) 2 (0.506 )
Kp = 1.3 x 104