NAME _________________________________ AP NOTES: UNIT 3 (2): EQUILIBRIUM:
WATCH FOR YOUR “Q” ….YOU’RE ON!!!
I-IV) Basics and the various expressions of K, the equilibrium constant to ICE Tables
V) Q vs. K
A) Q is called the reaction quotient … It is used to gauge the progress of reaction relative to
equilibrium
1) It is used to predict whether the reaction will proceed to the right (forward rxn) or to the
left (reverse rxn) … when BOTH reactant and product species are present in the reaction
mixture, often, but not always, PRIOR to equilibrium.
2) Qc is a ratio … it takes the same form of the equilibrium constant … but again its primary use
is to predict the progression of the reaction often, PRIOR to equilibrium having been
reached.
3) It follows the Law of Mass Action … HOWEVER…
a) for gases with amounts measured in atmospheres, the reaction quotient uses the partial
pressures in place of concentrations and it is then called Qp
B) Comparisons of Q and K … Well they both follow the form of the Law of Mass Action … But
1) At a given temperature, K has ONLY ONE value
of reactants and products at equilibrium
…. and
K specifies the relative amounts
2) Q is dependent upon the current state of the reaction ….. and may have many different values
as the reaction proceeds (for it is calculated with non-constant concentrations … per se)
a) e.g. In a reaction mixture with ONLY reactants, Qc = 0
Qc = [0]c[0]d
[A]a[B]b AND when the reaction mixture with ONLY products, Qc ≈ ∞
b) When all of the concentrations in the reaction mixture (both reactants and products),
equal 1 M, then Qc = 1 ….
c) Question: Under what circumstances would Qc = K? * at equilibrium
3) Q is useful, because the value of Q relative to K is a measure of the progress of the reaction
toward equilibrium.
234
VISUALIZE IT:
(Tro p.666)
VISUALIZE IT and INFER … (from B & L)
Inferences about the magnitude of Q:
When Q is a lesser value than K, the reaction has
a large amount of * reactants
and will proceed to
make * product
When Q is much larger than K the reaction already has a
large concentration of * product,
and the reaction is
predicted to shift to make more * reactant
235
TRY THIS: DETERMINING Q:
1) Given: I2(g) + Cl2(g)  2 ICl(g) where Kp = 81.9
A reaction mixture contains PI2 = 0.114 atm and PCl2 = 0.102 atm and PICl = 0.355 atm. Is the reaction
mixture at equilibrium? …. If not, in which direction will the reaction proceed, the forward or reverse?
Hint 1: * Run this problem using the
Law of Mass Action. You can substitute
in the partial pressures, in lieu of
concentrations, for the pressures are in
atmospheres
ans: * Qp = 10.8 while Kp = 81.9
Since Qp < Kp then the reaction will proceed
to the right (or will continue to make
products, because the reaction is not yet at
equilibrium)
2) Given: N2(g) + 3H2(g)  2NH3(g) For the synthesis of ammonia at 500°C, the equilibrium
constant is 6.0 x 10-2 Predict the direction in which the system will proceed (shift) to reach
equilibrium, when the experimentally determined concentrations were: [NH3] = 1.0 x 10-3M,
[N2] = 1.0 x 10-5M, [H2] = 2.0 x 10-3 M
2
-3 2
* Q = [NH3]
= (1.0 x 10 )
= 1.3 x 10
[N2][H2]3 (1.0 x 10-5)(2.0 x 10-3)3
7
Hint 1:* You are asked to predict the
direction of the reaction … this is a clue to
calculate Q … and compare its size to K
Hint 2:* Since you are given molarity, (as
opposed to partial pressures), you can solve
for Q by using the same form as the Law of
Mass Action.
ans: Q is significantly larger than K … Hence
there is way too much product, thus the
reaction will shift to the left and make more
reactant, by decomposing NH3
Assignment: Get to my man …Trivedi and check out sections 14.13-14.15 … it’s nicely done…
236
3) What if you are given moles … and asked to find Q?
A 1.0 L reaction vessel was charged with 2.20 mol N2O4(g) and 3.05 mol of NO2(g) at 400°C
Predict the direction of the reaction, given:
N2O4(g)  2 NO2(g)
Kp = 79.5
You are given moles …. You can easily determine molar concentration …but you need Kc …
and you have only Kp … and Kc = Kp only when ∆n =0.
You can do one of two things
a) solution 1: Use Kp = Kc(RT)∆n and substitute the molar concentrations into the
Law of Mass Action to find Q …. but (CRAP!) I
the dang equation ain’t on my tables … & I have
forgotten it.
OR
b) solution 2: Calculate partial pressures using PV=nRT and substitute into the Kp
expression to find Q
This is a really nice solution,
for it combines our work on
Dalton’s Law of Partial
Pressures, the Ideal Gas
Law, Q and Kp … with no
need to “memorize”
another equation… Thus
we continue to establish
that “Chemistry is a way”
PV = nRT….
PN2O4 = (2.20 mol)(0.08206 L atm mol-1 K-1)(673.15K)
1.0 L
PN2O4 = 122 atm
PNO2
PNO2 = (3.05 mol)(0.08206 L atm mol-1 K-1)(673.15K)
1.0 L
= 168 atm
Q = (NO2)2
(N2O4)
Conclusion:
= (168)2
122
= 231
Q at 231 is > Kp at 79.5 …. hence the reaction will
shift to the left and increase the reactant concentrations,
by decreasing the product, in order to reach equilibrium!
237
TRY THIS
___1) At 400 K, the equilibrium constant for the reaction Br2(g) + Cl2(g)  2BrCl(g) is
Kp = 7.0.
A closed vessel at 400 K is charged with 1.00 atm of Br2, 1.00 atm of Cl2, and 2.00 atm of BrCl. Use Q
to determine which of the statements below is true.
1) The equilibrium partial pressures of Br2, Cl2 , and BrCl will be the same as the initial values.
2) The equilibrium partial pressure of Br2 will be greater than 1.00 atm.
3) At equilibrium, the total pressure in the vessel will be less than the initial total pressure.
4) The equilibrium partial pressure of BrCl will be greater than 2.00 atm.
___2) Which of the following statements is true?
1) Q does not change with temperature.
2) Keq does not change with temperature, whereas Q is temperature dependent.
3) Q is the same as Keq when a reaction is at equilibrium
4) Q does not depend on the concentrations or partial pressures of reaction components.
___3) How is the reaction quotient used to determine whether a system is at equilibrium?
1) The reaction quotient must be satisfied for equilibrium to be achieved.
2) At equilibrium, the reaction quotient is undefined.
3) The reaction is at equilibrium when Q < Keq.
4) The reaction is at equilibrium when Q = Keq.
Answers:
1) 4 Q is of a lesser value than the Kp, hence the reaction should continue to the right, in order to achieve equilibrium … thus the
partial pressure of bromine monochloride should increase …beyond that of 2.00
2) 3
3) 4 …It’s essentially the same question as 2 … just a slightly different way of asking the issue.
238
C) How can calculating Q help me? OR …. The Clue To Check For Q
1) Carbon monoxide reacts with steam to produce carbon dioxide and hydrogen. At 700 K the
equilibrium constant is 5.10. Calculate the equilibrium concentrations of each of the 4
species when 1.000 mol of each component is mixed in a 1.000 L flask.
a) Write the balanced reaction equation * CO(g) + H2O(g)  CO2(g) + H2(g)
b) Write the equilibrium expression: K = [CO2][H2]
[CO][H2O]
NOTICE: Once you have the reaction
equation, the equilibrium expression, and
the initial concentrations, it is STILL
tough to jump into an ICE table …
because you are asked for the final
equilibrium concentrations … BUT you
are not given any one of the three… so
using the stoichiometric ratios is a bit
useless.
When this happens … this is the clue to
check for Q … You do not know in which
direction the reaction must shift, in order
to get to equilibrium … Q helps with that.
c) Calculate the initial concentrations …. Thank goodness,
we have 1.000 mol of each species, in a 1 L flask ,
the initial molar concentrations are each 1.000 M
d) Q = [CO2][H2] = (1.000 M)(1.000 M) = 1.000
[CO][H2O]
(1.000 M) (1.000 M)
Conclusion: Q is much smaller than K (at 5.10), so the
reaction must shift to the right in order to reach the point of
So reactant concentrations must decrease and product
concentrations must increase. THIS IS HOW Q HELPS
It tells you to subtract or to add to the reactant & product
concentrations in an ICE table
e) Ask: What must happen to the concentrations to achieve equilibrium?
NOW it is time for an ICE table … all the coefficients of the
balanced reaction equation are 1 ….notice that….
[CO]
[H2O]
[CO2]
[H2]
Initial
1.000
1.000
1.000
1.000
Change
-x
*-x
*+x
*+x
Equilibrium 1.000-x *1.000-x *1.000+x *1.000+x
5.10 = [1.000 + x][ 1.000 + x]
[1.000 - x][1.000 - x]
So: √5.10 = √(1.000 + 𝑥)2 =
√(1.000 − 𝑥)2
2.26 – 2.26x = 1.000 + x
= (1.000 + x)2
(1.000 – x)2
2.26 = 1.000 +x
1.000-x
so 1.26 = 3.26x and x = 0.387 M
Thus the change in concentrations at equilibrium is = to 0.387 M … use
this to complete a new ICE table and to answer the question ….
[CO] [H2O] [CO2]
[H2]
Initial
1.000 1.000
1.000 1.000
Change
-0.387 *-0.387 +0.387 +0.387
Equilibrium 0.613 0.613
1.387 1.387
You can check your work by plugging into the equilibrium expression and solving for
K … You should get 5.10
239
Summarize your learning for the CLUE TO CHECK FOR Q:
Assuming you are NOT asked to find “Q” directly, as on page 212 … When should a chemist
consider finding the reaction quotient of a reaction?
*Finding the reaction quotient would be valuable when needing to know in which direction
the reaction will proceed (or shift), aiding in the attempt to determine the final equilibrium
concentrations.
C) 2) The Clue To Check For Q
Given: H2(g) + F2(g)  2HF(g) where Kp = 1.15 x 102
Assume that the reaction for the formation of gaseous hydrogen fluoride from hydrogen and
fluorine has an equilibrium constant of 1.15 x 102 at a specific temperature. In an experiment,
3.000 mol of each of the 3 substances was added to a 1.500 L flask. Calculate the equilibrium
concentration of each species.
a) You have the balanced chemical reaction equation … use it wisely … notice not everything is
in a 1:1 ratio …..
b) Kp = [HF]2
[H2][HF]
c) find molarity… 3.000 mol = 2.000 M for each of the 3 substances
1.500 L
d) NOTICE you don’t have any equilibrium concentrations from which to work … This is the
Clue to check for Q and to ask What must happen to the concentrations so as to get to
equilibrium…..
Q = [2.000]2
= 1.000 ….Conclusion: The reaction should
[2.000] [2.000]
shift to the *right
, thus *lessening the
concentration of reactants and *increasing
*remove
the concentration of the products.
Initial
Change
Equilibrium
Kc = [HF]2
[H2][F2]
[H2]
[F2]
[HF]
2.000
2.000
2.000
-x
-x
+2x
2.000 – x
2.000-x
2.000 + 2x
Did you use the
coefficient of the
reaction equation?
so: 1.15 x 102 = [2.000 + 2x]2 = √1.15 𝑥 102 = 2.000 + 2x
[2.000 – x]2
2.000 – x
x = *1.528 and thus: equilibrium concentrations of H2 & F2 = 0.472 M and the
equilibrium concentration of HF
= 5.056 M
240
C) 3) The Clue To Check For Q & using the quadratic equation
Given: H2(g) + F2(g)  2HF(g) where Kp = 1.15 x 102
Assume that the reaction for the formation of gaseous hydrogen fluoride from hydrogen and
fluorine has an equilibrium constant of 1.15 x 102 at a specific temperature. In an experiment,
3.000 mol of H2 and 6.000 mol of F2 are allowed to react in a 3.000 L flask. Calculate the
equilibrium concentration of each species.
(Zumdahl p.629)
a) You have the balanced chemical reaction equation … use it wisely … notice not everything is
in a 1:1 ratio …..
b) KP = [HF]2
[H2][ F2]
c) find molarity… 1.000 M for H2 and 2.000 M for F2, and 0 M for HF initially ….
d) Since the [HF] = 0 initially, it is silly to find Q. We can assume that the reaction, in order
to achieve equilibrium must shift to the right …
So, ask: What must happen to achieve equilibrium?
Well, the reactants must decrease by “x” while the product (HF) must increase by *2x2x
e)
Initial
Change
Equilibrium
[H2]
[F2]
[HF]
1.000
2.000
0.000
-x
-x
+2x
1.000 – x
2.000-x
+ 2x
1.15 x 102 =
(2x)2 .
(1.000-x)(2.000 –x)
= (1.15 x 102)(1.000-x)(2.000-x) = 4x2
(1.15 x102)x2 - 3(1.15 x 102)x + 2(1.15 x 102) = 4x2
= (1.11 x 102)x2 - 3.45 x102x + 2.30 x102
2
ax
+ bx
+c =
= 0
0
−𝑏 ± √𝑏 2 − 4𝑎𝑐
2𝑎
𝒙 = 3.45 𝑥 102 ± √−(3.45 𝑥 102 )^2 − 4(1.11 𝑥102 )(2.30 𝑥 102 )
2(1.11 𝑥 102 )
x = 345 ± 130. = 0.968 M
222
or 2.14 M
2.14 M makes no sense since 1 – 2.14 gives a negative concentration … thus we shall
use 0.968 …. Thus at equilibrium: [H2] = 0.032 M, [F2] = 1.032 M & [HF] = 1.936 M
241
C) 4) another example …(Atkins Chemical Principles p. 404)
3.12 grams of phosphorus pentachloride are placed in a reaction vessel of volume 500 mL and
allowed to reach equilibrium with its decomposition to phosphorus trichloride and chlorine gas
(dichlorine), at 250°C, when Kp = 78.3 for the reaction. All three substances are gases at 250°C.
a) Write the balanced chemical reaction for the decomposition:
* PCl5(g)  PCl3(g) + Cl2(g)
b) Calculate the partial pressures at the point of equilibrium.
i) Write the expression for Kp:
Kp = (PCl3)(Cl2)
(PCl5)
ii) Find the moles of PCl5 (mole mass = 208.24 g/mol) … and use this value to find the
partial pressure of the gas, with the ideal gas law
mol PCl5 = 3.12 grams| 1 mol | = 0.0150 mol
208.24
thus:
P = nRT
V
P = (0.0150)(0.08206)(523.15)/0.500L = 1.29 atm
* note: Kelvin: 523.15 and 523 yield the same pressure w/ 3 sig figs
iii) set up an ICE table
Initial
Change
Equilibrium
PCl5
atm
1.29
PCl3
atm
0
Cl2
atm
0
-x
+x
+x
1.29 – x
+x
x
iv) substitute into the equilibrium expression of Kp:
or
78.3 =
x2
1.29 - x
x2 = 101 – 78.3x
x2 + 78.3x – 101 = 0
−𝑏 ± √𝑏 2 − 4𝑎𝑐
2𝑎
𝒙 = −78.3 ± √(78.3)2 − 4(1)(−101)
2(1)
= -78.3 ± 80.8
2
x = -79.56 or x = 1.25 … Select 1.25 atm …
since “x” represents the partial pressures of the products and partial pressures must be positive
Hence the equilibrium : PPCl5 = 0.04 atm
PPCl3 = 1.25 atm and PCl2 = 1.25 atm
242
D) When That Equilibrium Constant is Really Small We Catch a Mathematical Break! OR
Sometimes even a blind squirrel gets an acorn …
1) Under certain conditions, simplifications are possible … that will help reduce, avoid, rid us
of the complicated math….
a) We can ignore “x” when the equilibrium constant is really, really small.
Here’s why
 When the equilibrium constant is small that means that the reaction reaches
equilibrium very quickly, with little change in the reactant concentration.

This change in reactant concentration is so small as to be considered to be
negligible, and thus can be ignored in the calculation.
e.g.) Gaseous nitrosyl chloride, NOCl, decomposes to form the gases NO and Cl2. At 35.0°C
the equilibrium constant is 1.6 x 10-5 . In an experiment in which 1.0 mol of NOCl is placed
in a 2.0 L reaction vessel, what are the equilibrium concentrations?
Question: Is it important to find Q … Why? * No, you are given an initial concentration …
We can assume the reaction proceeds (shifts) to the right to produce NO and Cl2
Okay, now….
a) Write the balanced chemical reaction equation: *2 NOCl(g) 2 NO(g) + Cl2(g)
b) Write the equilibrium expression:
*Kc = [NO]2[Cl2]
[NOCl]2
c) Run an ICE Table & Determine the initial concentrations…remembering the stoich!!
[NOCl] [NO] [Cl2]
Initial
0.50
0
0
Change
-2x
+2x
+x
Equilibrium 0.50-2x 2x
x
d) 1.6 x 10-5 = (2x)2(x)
(0.50-2x)2
We can avoid this hapless harridan of hairy math involving
x3, x2 and x, by recognizing that the equilibrium constant is
very small…and given that it is soooooo small, the change
of x (and even 2x) is negligible.
Hence we can assume that 0.50 – 2x ≈ 0.50
Yielding:
1.6 x 10-5 = (2x)2(x) = 4.0 x 10-6 = 4x3
(0.50)2
4.0 x 10-6 = x3 thus x = 1.0 x 10-2
4
This means that the equilibrium concentrations are as follows: [NOCl] = 0.50 – 2x ≈ 0.50 M
[NO] = 2(1.0 x 10-2) = 2.0 x 10-2 M
[Cl2] = 1.0 x 10-2
Plug the values into the equilibrium expression to check..
243
SUMMARY RE: COMPLETING EQUILIBRIUM PROBLEMS DEALING WITH K OR
EQUILIBRIUM CONCENTRATONS…
1) Write the balanced reaction equation, if necessary….
2) Write the equilibrium expression using the Law of Mass Action
3) Find the Molar Concentrations and Run an ICE table …. But … there are different approaches:
a) When you know the reaction is at equilibrium, and have initial concentrations and at least one
equilibrium concentration just run an ICE Table. Fill in the table using the stoichiometric ratios.
Finding Q is not necessary.
b) When you are given concentrations but you do NOT have any equilibrium concentrations, you
need to assess for Q and then run an ICE Table … but you will need to use “x”. Some problems
will require the use of the quadratic equation. Don’t bother assessing for Q, when the reaction vessel has 0 product
… you can assume the reaction must shift to the right to reach a point of equilibrium.
c) When you are asked to find the final equilibrium concentrations, but the value for K is really
small you may simplify the calculations by ignoring “x”.
4) Define the change needed to reach equilibrium and define the equilibrium concentrations by applying
the change to the initial concentrations … This completes your ICE table
5) Plug and Chug into the equilibrium expression
6) You can check your work by using your calculated equilibrium concentrations by making sure they give
the correct value for K
****************************************************************************************************
In my ever-present attempt to show you the Chemistry is “a way” …the next two problems are dedicated
to helping you see the variations in C 3 a)-b). These are repeated from earlier in the packet, because
you know how they work. My goal is to show you how they blend with the summary.
EXEMPLAR OF C) 3a When you know the reaction is at equilibrium, and have initial concentrations
and at least one equilibrium concentration just run an ICE Table. Fill in the table
using the stoichiometric ratios. Finding Q is not necessary.
Consider the following reaction, at equilibrium: CO(g) + 2 H2(g)  CH3OH(g)
A reaction mixture at 780°C initially contains [CO] = 0.500 M and [H2] =1.00 M.
At equilibrium the CO concentration is experimentally determined to be 0.150 M.
What is the value of Kc?
Notice that you are given molar concentrations
Kc = [CH3OH]
[CO] [H2]2
Initial
Change
Equilibrium
[CO]
[H2]
[CH3OH]
0.500
1.00
0.00
*-0.350
*-0.700
*+0.350
0.150
*0.300
*0.350
(molarities) of REACTANTS … You are given the
balanced equation and at least one equilibrium
concentration. You may assume the initial
concentration of CH3OH is 0.00, thus finding Q
is not required.
Given all of this information, you have enough
to complete an ICE table and to determine the
equilibrium constant.
Again, there is no real advantage to solving for
Q, since we know the equilibrium
concentration of at least one reactant species
and 0 produce initially. Hence we may assume
we are at equilibrium
244
EXEMPLAR OF C) 3b When you are given concentrations but you do NOT have any equilibrium
concentrations, you need to assess for Q and then run an ICE Table …
Given: H2(g) + F2(g)  2HF(g) where Kp = 1.15 x 102
Assume that the reaction for the formation of gaseous hydrogen fluoride from hydrogen and fluorine has
an equilibrium constant of 1.15 x 102 at a specific temperature. In an experiment, 3.000 mol of each of
the 3 substances was added to a 1.500 L flask. Calculate the equilibrium concentration of each species.
Notice you are “plopped down” into the middle of
a reaction, in which concentrations (or partial
pressures in atm) of each species is given you.
This is the Clue To Check For Q. Find Q and
compare it to K to figure out how the reaction
must shift.
It is important to ask: What is in the beaker …
how will it react achieve equilibrium ?….
Knowing the direction of the reaction’s
procession, will allow you to either subtract from
the reactant concentration or add to it.
Given all of this information, you have enough to
complete an ICE table and to determine the
equilibrium constant.
245