2nd Set of Notes
Here we will cover what is in Chapter 15 in perhaps a more general treatment than what
is present in the chapter. This more general treatment will prove useful for you in later
chapters.
In addition to the Power of Stoichiometry, the Acid-Base concept and the oxidation
reduction relationships, and kinetics; equilibrium and thermodynamics stand as
cornerstones of chemistry.
Previously in our descriptions of chemical reactions in terms of chemical eqns. we
considered the amount of product formed to be based on 100% conversion of the limiting
reagent. This is suggestive of the case:
aA + bB -> cC + dD
a, b, c, d are coefficients of the balanced eqn. where the arrow indicates that the reaction
occurs in only one direction. But as we said last before
aA + bB
->
<-
cC + dD is more indicative of the general case.
This second eqn. means that if we mix A and excess of B together we will not completely
use up the limiting reagent A to form C and D. At the end of the reaction A, B, C and D
will be present in certain amounts as determined by the
. Essentially the
reaction can go in both the forward and reverse directions, and an equilibrium is reached
when the rate in the forward direction equals the rate in the reverse direction. These rates
change as the reaction progresses since in general they are a function of the
concentrations.
Again, the limiting reagent is in fact not completely used up!. That does not mean that
the stoichiometry is not holding. No, No. In fact Stoichiometry holds if we know how
much A is used up then by the stoichiometry of the balanced eqn., one can calculate the
amount of C or D made knowing a certain amount of A reacted.
That is, the coeff. of the balanced eqn. a,b,c, and d can still calculate how much C and D
are made. Because rxns reach equilibrium they do not react completely in the forward or
reverse direction.
The theoretical yield is still however based upon ______
limiting reagent!
Consider the reaction:
CO(g) + 3H2(g)
->
<-
CH4(g) + H2O(g)
This is an equilibrium type reaction.
of the
Lets Examine what happens to this reaction if we start out with three different
conditions (Three different cases)
Case 1
If we start with 28.0g of CO and 6.06g of H2 in a 10.0L vessel, theoretically if 100%
conversion occurred, how many moles of CH4 and H2O can we generate at a particular
temp say 927o C?
___ mol CH4 and
____ mol H2
so we could form _____ mol CH4 and _____ mol H2O.
Case 2
Suppose instead that we started with only CH4 and H2O in the reaction vessel at the same
temp. as in case 1 (the reactants of the reverse reaction)! Starting with one mole of CH4
and one mole of H2O, then theoretically if 100% converted, we could make
_____ mole of CO and _____ moles of H2.
But in fact there is not complete reaction in either the forward or the reverse direction.
Case 3
Suppose we start off with just CO and H2 but there is initially more CO here than was
present in case 1, 2.000 mol CO and 3.000 mol H2 in a 10.00 L vessel at the same temp.
How much product is formed? Is it different than in case 1? Is there some quantity that
is the same as in case 1?
Case 1
It is found that at 927o C there are only 0.387 moles of H2O produced rather than 1mole.
% yield = actual/theoretical x 100% = 0.387moles H2O / 1mol H2O x 100% = ______
yield.
How much CH4 would be present at equilibrium if one started only with CO and H2?
Case 2
Here one finds that only 1.839 moles of H2 is formed rather than 3. So you see, the
equilibrium mixture contains less than 100% conversion.
So can you calculate how much CO would be present at equilibrium if one started only
with H2O and CH4?
In some cases you might start with both reactants and products present initially. In those
cases it is better to set it up as follows: CALLED the ICE treatment
CO
I nitial (starting)
1
3.00
0
0
C hange
-x
-3x
+x
+x
1.00-x
3.00-3x
x
x
E quilibrium
+
3H2
->
<-
Amount (moles)
CH4
+
H2 O
In the present cases, CASE 1, it turns out that at 927o C, x = __________ and plugging
x into the equilibrium line gives you the moles of CO, H2, CH4, and H2O.
And for Case 2: use of the ice table gives
Now a few questions may pop into your head!
1) Will the amount of product change if I start off with a different amount of reagent or
reagents – the % yield
2) Besides (amount) – what else affects the quantity of product formed? Temp.?
Pressure?
3) Can we predict the amounts of product and reactants that will be formed,
quantitatively under any condition. This represents significant predictive power.
If you were in industry it might be in your best interests to get the rxn to go to completion
if possible. It is a matter of economics.
1) Lets consider again Case I.
It is experimentally found that if you react 1.000 mole CO and 3.000 mol H2 in a 10L
vessel at 927o C you get 0.387 mol CH4, 0.387 mol H2O, with 0.613 mol CO and 1.839
mol H2 left.
Case 3
But if you take 2.000 mol CO and 3.000 mol H2 in a 10.00 L vessel at 927o C you get
0.478 mol CH4 and H2O. So, you get more product.
How much CO and H2 are left in the reaction mixture?
I
C
E
So you effectively forced the rxn to go further to completion but used more _________
.
So if we put in more of one of the reactants, then more _____________ will be formed!
Cost would prohibit one from putting in more reagent than necessary, so there is a
tradeoff here.
Le Chateliers principle is at work here – we previously talked about it
When a system in chemical equilibrium is disturbed by a change of temp., pressure, or a
concentration, the system shifts in equilibrium composition in such a way that tends to
counteract this change.
So what does this mean for us. Well qualitatively speaking, then by adding more reacant
(even just one of the reactants)
We should ____________ _____________ _________________to be made since this
lessens the amount of excess reactant present.
2) What about temp change – well remember in order to predict if the shift will be
toward more reactant or more product we must know if heat is given off or needed.
Is the reaction endothermic or exothermic Ho rxn =  niHfo (prod) -  miHfo (reactants)
Not exactly correct at 927o C From book






o rxn = -241.826 -74.87 – (-110.5 – 3(0)) =
so exothermic, as heat is given off
Since heat is given off:
CO(g) + 3H2 <-- > CH4 + H2O and H = negative value
OR
CO(g) + 3H2 <-- > CH4 + H2O + heat
exothermic H = negative value
then increasing the temp. essentially ___________ heat energy transfer, as heat, so that
you expect the reaction to counteract this by releasing less heat or going back toward
reactants in this case
DEMO
Heat + Co(H2O)62 + + 4Cl(pink)
->
<-
CoCl42 - + 6H2O
(blue)
Is this used anywhere;
2) Pressure
What if we change the pressure on the reactants and products by pushing on them with a
piston?
CO(g) + 3H2(g)
4 moles gas
->
<-
CH4(g) + H2O(g)
2 moles gas
By LeChateliers principle, if we increase the pressure, the reaction will shift to _______
the pressure. Since there are _____ _________ of gas on the left side and only _____
moles of gas on the right side, then
An increase in pressure would shift the reaction toward the side with __________
gaseous moles
So using Le Chateliers principle we can qualitatively determine whether more reactant or
product will result from a change in the concentration of the moles of reactant or product.
But can we calculate (quantitate) the changes that occur!
3) Calculating the Equilibrium Amounts
Look again at the General Expression for Equilibrium Reaction:
aA
+ bB
->
<-
cC
+ dD
A, B are reactants and C, D are products in the balanced equation.
Through experimentation it was found that an equilibrium constant can be defined as
Keq = aCc aDd / ( aAa aBb )
where
a is the activity ~ [A]/[std state] and [std state] = 1M or 1atm
note that Keq is ________________!
so define Keq  Kc = [C]c [D]d / ( [A]a [B]b )
The Law of Mass Action states that Keq and Kc are constant for a particular reaction at a
given temperature.
Note that Kc is specific to the coefficients in the balanced equation.
One can write Kc for the reaction we have been discussing:
CO(g) + 3H2(g)
->
<-
CH4(g)
+
H2O(g)
Kc = [CH4] [H2O] / ( [CO] [H2]3 )
NOTE that if we used 2CO(g) + 6H2(g)
then Keq would be:
<-
->
CH4(g)
+
H2O(g)
SINCE Keq is constant for a given reaction, we should be able to calculate it for this
reaction at 927o C.
We found from
CASE 1 that 1.00 mole of CO and 3.00 moles of H2 in a 10.0 L reaction, when given
sufficient time to reach equilibrium, gave 0.387 mol CH4 and 0.387 mol H2O.
That leaves 1- 0.387 = 0.613 moles CH4 and 3 – 3(0.387) = 1.839 mol H2
Calculating the concentrations [CH4] = 0.387 mol / 10.0 L = 0.0387 mol/L CH4
Similarly for the others
[H2O] =
[H2] =
[CO] =
And so putting these into the expression for Kc:
Kc = [CH4] [H2O] / ( [CO] [H2]3 ) =
=
Let compare this with the other information from
CASE 2
Remember here we started with only product: 1.0 mol CH4 and 1.0 mol H2O,
We found that at equilibrium 0.387 moles of CH4 and 0.387 moles of H2O were left and
thus 1- 0.387 or 0.613 mols of both CH4 and H2O reacted.
So by the ICE table 0.613 moles of CO must be present and 3(0.613) = 1.839 mol of H2
One can put this into the equilibrium expression to give:
Kc = [CH4] [H2O] / ( [CO] [H2]3 ) =
So we get the same value for Kc for CASE 1 and CASE 2.
What about the other experiment?
CASE 3
Here started with 2.000 mol CO and 3.000 mol H2
Found using ICE that at equilibrium the concentrations were:
[CH4] = 0.478 mol / 10.0L = 0.0478 M
[H2O] = 0.0478 M
[H2] = 1.566 mol / 10.0L =
[CO] = 1.522 mol / 10.0L =
So Kc = [CH4] [H2O] / ( [CO] [H2]3 ) = 0.0478 * 0.0478 / ( 0.1522 * (0.1566)3 )
=
INDEED Kc is a constant, independent of the starting concentration! When there are
more reactants, more products are formed keeping Kc constant.
This is what Le Chatelier’s Principle said.
The concentration of a species AT EQUILIBRIUM may change depending on what is in
the rxn. vessel initially but as long as the Temperature is constant,
the value of Kc is the SAME!
This is a very important point that we’ll see later is related to the ENTHALPY or
ENTROPY of the reaction LATER CHAPTER!
Kc determines the ratio of the product concentrations and the reactant concentrations AT
EQUILIBRIUM for any set of initial concentrations. Kc is however a function of
temperature and will have a different value at a different temp.
Again
Kc, at a certain temperature, can be calculated for the reaction at a certain temperature if
we know the value of the concentrations at equilibrium.
Example:
H2S dissociates on heating in the following way:
2 H2S(g)
->
<-
2H2(g) + S2(g)
0.100 mol H2S put into a 10.0 L vessel and heated to 1132o C it gave an equilibrium
mixture containing 0.0285 mole mol H2. What is the value of the equilibrium constant at
this temp.?
H2 S
H2
S2
I
C
E
The expression for Kc is:
Kc =
Now for gases it is usually more convenient to work in terms of pressures or partial
pressures rather than concentration in moles/liter.
Using the ideal gas law, we see that the concentration is related to the pressure:
PV = nRT
PA/ (RT) =
Ptot = PA + PB
+ PC
PA is the partial pressure of A, PB is ....
= concentration of A for example
For instance:
In the reaction:
CO(g) + 3H2(g)
->
<-
CH4(g) + H2O(g)
FIG
Kc=(n/V)CH4(n/V)H2O/{(n/V)CO(n/V)3 H2 }= (P/(RT))CH4(P/(RT))H2O/{(P/(RT)CO (P/(RT)3 h2
= PCH4 PH2O / {PCO (PH2)3 } x (RT)3 H2 (RT)CO / {(RT)CH4 (RTH2O)}
= PCH4 PH2O / {PCO (PH2)3 } x (RT)2
= Kp (RT)2
More generally then:
Kc = Kp (RT){ s u m
of n(gas react)-sum of n(gas product)}
OR Kp = Kc RT { s u m
Kp = Kc (RT)(  n )
n(gas prod) -sum of n(gas react)}
(where n is sum of stoichiometric coefficient of gaseous products –
coefficient of gaseous reactants)
In the above case
Kp = 3.92 x (0.0821 Latm/(Kmol) x (927+273K)) ( 1 + 1
- 3-1)
= if Kp is expressed in atm!
Often for gaseous rxns., the equil constant Kp is used rather than Kc.
BUT the two are related as we see here!
Example
Diphosphorus octachloride gas decomposes on heating to give phosphorus trichloride gas
and chlorine gas. If Kc equals 5.26 x 10- 2 at 191o C, what is the value of Kp at 191o C?
Also give the expression for Kc and Kp?
NOW what if there is a mixture of phases in the reactants and products, there is a
heterogeneous equilibrium!
like in:
3Fe(s) + 4H2O(g)
<-
->
Fe3O4(s) + 4H2(g)
most generally write
Keq = aFeO4(s) a4 H2(g) / { a3 Fe(s) a4 H2O(g) }
THE activity of a solid or pure liquid is generally  ___________________.
.
Because the vapor pressure of a solid or pure liquid is constant at fixed temp., so in
constant volume container, it is fixed.
Only changes of material completely reacts, that no solid or pure liquid remains.
So
Keq  a4 H2(g) / a4 H2O(g)
and assuming the activities of the remaining reactants and
products are equal to the concentrations
Keq  Kc 
DON’T INCLUDE concentrations of pure solid or pure liquid!
Example
In the Mond Process Ni(CO)4 is formed from Ni metal and carbon monoxide gas.
Write Kc for the reaction:
Ni(s) + 4CO(g)
<-
->
Ni(CO)4(g)
Example
Write the Kc expression of the dissociation of an ionic compound in H2O. A weak
electrolyte or insoluble compound will have some small solubility and can be thought of
as an equilibrium where there is little to no product formed.
PbI2(s) <-- > Pb2 + (aq) + 2I- (aq)
Kc =
One final consideration regarding the Kc expression, what if we have a reaction C with an
unknown value of Kc, but this rxn C is the sum of two other reactions, rxn A and rxn B
where the Kcs for those reactions are known!
Can we find the Kc of rxn C from this information
Consider the rxn C):
not known!
HCN(aq) <-- > H+ (aq) + CN- (aq) and the value of Kc for this rxn is
the expression for rxn C is Kc is Kc(C) =
but two other reactions:
A) HCN(aq) + OH- (aq)
->
<-
CN- (aq) + H2O(l)
B) H2O(l) <-- > H+ (aq) + OH- (aq)
Kc(A) = 4.9 x 10+ 4
Kc(B) = 1.0 x 101 4
The sum of rxns. A) and B) give: HCN(aq) <-- > H+ (aq) + CN- (aq)
which is rxn. C)
The expressions of Kc for these reactions are:
Kc(A) =
Kc(B) =
And Note that KC(A) x Kc(B) = KC(C)
SO Generalizing this idea:
If a given chemical equation can be obtained by taking the sum of the other eqns., the
value of the equil. constant for the reaction of interest is given by the product of the equil.
constants of the other rxns.
ALSO, How is the value of the equilibrium constant of a reverse reaction related to the
value of the equilibrium constant in the forward direction?
K(reverse) = 1/ K(forward)
Summary:
So far, we’ve seen that the equilbrium constant defines the equil concentrations of the
non-solid and non-pure liquid species.
One can imagine using it in the following ways:
1) Qualitatively from the magnitude of the value of Kc - can decide if a particular rxn.
will procede in a forward direction very much!
2) By comparing Q - the reaction quotient calculated at a condition before the system has
reached equilibrium, with Kc, one can decide if the reaction is moving toward products or
reactants.
Q = [product A]a [product B]b / ([reactant C]c [reactant D]d ) not at equilibrium
3) To calculate the equilibrium concentrations of the non-solid, non-pure liquid product
and reactant species.
Save the best for last!
Consider application 1):
For N2(g) + O2(g)
->
<-
What does the value of Kc, Kp, and/or Keq qualitatively tell us?
2NO(g)
Kc =
=
4.6 x 10- 3 1
Remember - these concentrations are the equilibrium concentrations.
Since Kc is small the conc. of NO must be __________ relative to the conc. of N2 and O2!
Thus N2 & O2 don’t react together very readily since very little NO exists at equilibrium!
Example
Consider the reaction of HCl with water:
HCl (aq) + H2O(l)
<-
->
H3O+ (aq) + Cl- (aq)
Does HCl dissociate very much in water?
Kc  1 x 103
Is it a strong acid?
So summarizing:
If Kc for a particular reaction is large, then at equil. mostly products are present!
If Kc is 1 then approx equal concentrations of products and reactants at equil.
If Kc is a small fractional value less than 1, then at equil., mostly reactants present!
Application 2)
2) Has to do with a reaction mixture that has not yet reached equilibrium.
IF one knows the ratio of products to reactants defined similarly to the expression for Keq
except that the concentrations are not the equilibrium concentrations, then one can
determine in which direction the reaction will proceed to reach equilibrium.
Consider the rxn. above:
N2(g) + O2(g)
->
<-
2NO(g)
Kp = P2 NOeq / (PO2eq PN2eq )
Kc = [NO]2 eq / {[N2]eq [O2]eq}
Kc = 4.6 x 10- 3 1
Suppose the calculated value of the reaction quotient, Q, at some point during the
reaction
Qp is P2 NOeq / (PO2eq PN2eq )
Q is
more generally
= [NO]2 / {[N2] [O2]}
IF Q = 1
Then one could say
a) rxn not at equilibrium yet
b) too much product is present since Q > Kc
(numerator is too large to be at equilibrium)
On the other hand if:
Q is calculated to be = 4 x 10- 3 4
Then one could say
a) rxn not at equilibrium yet
b) too much reactant is present since Q < Kc
(denominator is too large to be at equilibrium)
If Q = Kc then the reaction is ________ _______________________!
Example
The reaction of methane and hydrogen disulfide yields carbon disulfide and hydrogen
gas. Kc for this reaction is 3.59 at 900o C.
During the reaction the following reaction compositions were measured:
[CH4] = 1.15M
[H2S] = 1.20M
[CS2] = 1.51M
[H2] = 1.08M
Decide in which direction the reaction is proceeding at this point in the reaction?
Application 3)
3) Lastly, but certainly not least, use Kp or Kc to determine what the reaction mixture
concentrations will be when equilibrium has been reached. The concentrations of the
reactants and products at equilibrium will depend on the initial concentrations, even
though Kc does not change unless the temp. changes.
That is Kc has to be a constant meaning the ratio of products to reactants raise to the
appropriate power does not change, but the individual concentrations at equilibrium will
be different depending on the initial concentrations.
So use the following method to solve for the equilibrium concentrations, if you have a
given set of initial concentrations!
A) Set up an ICE table with
I) Starting # of moles (or sometimes initial conc.) of reactants and products
C) the change in the moles of each that must occur to reach equilibrium. Since
this is an unknown use a variable such as x or y to represent the number of moles
reacting to get to equilibrium. (Stoichiometry is important here!)
E) Write down the equilibrium # of moles in terms of x and initial conditions.
Might want another line for the equilibrium concentrations of the reactants and
products in terms of x or y by dividing by the volume
4) Put the equilibrium concentrations in E) into the Kc expression.
5) Solve for x or y, and then plug back into E to get the equil. concentrations.
Example:
Phosphorus pentachloride, PCl5, decomposes when heated:
PCl5(g)
->
<-
PCl3(g) + Cl2(g)
Initially 1.00 mol/L of PCl5 is present in a rigid reaction vessel, what is the equilibrium
composition of the gaseous mixture at 160o C, if the equilibrium constant at 160o C is
0.211?
Pick Basis of 1L
PCl5
I
C
E
Equil Conc
PCl3
Cl2
Use Quadratic Formula
If you don’t like to solve the quadratic formula you can make an approximation that x <<
initial concentration and simplify.
Need to check if this introduces a large error. Use 5% rule to decide if the approximation
is okay
5% rule
You can make approximation that
(initial conc. - x)  initial conc.
If the concentrations obtained using the x obtained from the approximation are less than
5% of the initial concentration, then you can approximate
1.00 - x by just 1.00 for example.
Restatement of the 5% rule:
If the quantity ignored in an equilibrium calculation is equal to or less than 5% of the
quantity from which it is subtracted, (or to which it is added), then it is ok to use the
simplification
So if we had for example
4.31 x 10- 4 = 2x2 / {(0.802-x) (0.373-3x)3 }
4.31 x 10- 4 = 2x2 / {0.802 x 0.373)3
Use simplification
x = 2.20 x 10- 3
Check if this is okay
2.20 x 10- 3 / 0.802 x 100% = 0.27 %
less than 5% so a good approx. to simplify.
(3 x 2.20 x 10- 3 ) / 0.373 x 100% = 1.77% so less than 5% so a good approximation.
Example
For the reaction:
I2(g) + Br2(g) <-- > 2IBr(g)
Kc = 1.2 x 102 at 50.o C A reaction vessel initially contains 0.0015 mol each of iodine
and bromine in a 5.0 L vessel. What is the equilibrium composition?
I2
I
C
E
Equil Conc.
Example
Br2
2IBr
Ammonia is produced in the following reaction
N2(g) + 3H2(g) <-- > 2 NH3(g)
Kc = [NH3]2 / [N2] [H2]3 = 0.159 at 200.o C. If 1.0 mol of N2 and 3.0 mol H2 are put in a
1.0 L reaction vessel, what are the equilibrium concentrations of the reactants and
products
N2
H2
2NH3
I
C
E
Equil Conc.
Remember that Le Chatelier’s Principle can be used to predict how the equilibrium
constituency can change as a function of Temperature, Concentration, and Pressure
Remember the equilibrium constant Kc does not change, although Kp can change since
Kp = Kc (RT) n
What is the effect of a catalyst. Recall that a catalyst can speed up a reaction, but
___________ ______________ ________________ Kc
A catalyst may affect the product distribution, however, if there are several reactions
happening at the same time. It may speed up only one of all the possible paths.