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7.1 CHEMICAL SYSTEMS IN EQUILIBRIUM:
Dynamic Equilibrium in Chemical Systems
Introduction to Equilibrium
•
•
•
•
A closed system is a reaction system in which no matter can enter or leave.
Can reach an equilibrium if the reaction is reversible
Appears to be doing nothing.
But… at molecular level, equilibrium is dynamic state, where the molecules continue to react at
equal but opposite rates
(If the system is not closed, materials can enter or leave. The reaction will not reach an unchanging
state.)
However, every reaction that does not appear to change is not necessarily at equilibrium:
•
the reaction may react so much towards the products that it appears to have gone to
completion.
•
the reaction is so slow that it has not reached equilibrium yet, though it will, given
enough time.
•
if the system is open, and materials are being removed or added in such a way that there is
no net change, a steady state has been reached. This is not an equilibrium, but it is a fairly
common state for a reaction to reach.
There are three types of equilibriums:
Summary of Conditions for Systems in Equilibrium
1.
2.
3.
4.
The process is reversible, involving dynamic change at the molecular level.
The observable (macroscopic) properties of system are constant.
The system is closed, preventing the input or escape of matter and energy.
Equilibrium can be approached from either direction.
1
Omit … Percent Reaction at Chemical Equilibrium
PROBLEMS and SOLUTIONS
Problem 1
Consider the following equation for the formation of nitrogen and hydrogen from ammonia (the reverse
of the Haber process):
2NH3(g)
N2(g) + 3H2(g)
If the reaction begins with 1.000 mol/L concentration of ammonia and no nitrogen or hydrogen,
calculate the concentrations of ammonia and hydrogen at equilibrium if the equilibrium concentration
of nitrogen is measured to be 0.399 mol/L.
Set up an ICE table to organize the data.
Table. ICE table for the reaction
2NH3(g)
N2(g)
+
3H2(g)
Initial concentration (mol/L)
Change in concentration (mol/L)
Equilibrium concentration (mol/L)
[N2(g)] = x = 0.399 mol/L
[NH3(g)] = 1.000 – 2x = 1.000 -2(0.399) = 1.000 – 0.798 = 0.202 mol/L
[H2(g)] = 3x = 3(0.399) = 1.197 mol/L
The equilibrium concentrations of ammonia and nitrogen are 0.202 mol/L and 1.197 mol/L respectively.
Page 437 PP #’s 6 & 7
Page 437 SR #’s 1,3,5,7(a,b),9a omit % concentration
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7.1 CHEMICAL SYSTEMS IN EQUILIBRIUM:
Equilibrium Law in Chemical Reactions
Dynamic Equilibrium
some chemical systems in a closed system will appear as though no change is occurring, i.e. an
equilibrium has been achieved
•
•
Note:
…but in fact chemical systems in equilibrium will have a balance of “reactants moving
forward towards “products” and the “products” moving in the reverse direction towards
‘reactants”
this is considered _________________________
Equilibrium is at the same concentration on both graphs
3
1. Solubility Equilibrium
2. Phase Equilibrium
3. Chemical Reaction Equilibrium
Look at the Sample Problems on pages 433-436.
Page 437 PP #’s 6 & 7
Page 437 SR #’s 1,3,5,7(a,b),9a omit % concentration
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7.2 CHEMICAL SYSTEMS IN EQUILIBRIUM:
Equilibrium Law in Chemical Reactions
repetitive testing of various chemical reactions that reach equilibrium show that while beginning
with different initial concentrations they always reach the same equilibrium concentrations no
matter what combination of initial concentrations.
Consider the reaction: aA + bB → cC + dD
The Equilibrium Law can be expressed as;
The equilibrium law describes the behaviour of almost all gaseous and aqueous chemical systems
Homogeneous equilibrium involves reactants and products in a single phase, eg. all aqueous or all
gas
Example: The Haber Process equilibrium
N2(g) + 3H2(g)
2NH3(g)
. . . and the Kc expression is:
Heterogeneous equilibrium involves reactants and products in more than one phase, eg. liquid and
gas
•
Consider the following reaction with solid and gas phases:
3Fe(s) + 4H2O(g) ↔ Fe304 (s) + 4H2 (g)
the equilibrium law expression is:
K = [Fe304] [H2]4
[Fe]3 [H2O]4
WRONG!!!
5
Therefore….
Determining Equilibrium Concentrations
Simple calculations involving the equilibrium constant requires knowledge of a combination of the
concentrations of the reactants and products and the equilibrium constant
•
An example involves the following reaction:
H2 (g) + CO2 (g) ↔ H2O(g) +CO(g)
Where the concentrations are given as;
[H2] = 0.24 mol/L
[H20] = 0.88 mol/L
[C02] = 1.80 mol/L
[CO] = 0.88 mol/L
the equilibrium law expression is:…
Example 2: Consider the equilibrium below:
If 1.2 mol of H2 and 1.2 mol of O2 was placed in a 1.0 L container and allowed to reach equilibrium,
what would the value of Ke be if at equilibrium [HI] = 0.40 mol/L?
H2(g) + I2(g) <=====> 2HI(g)
Solution:
SOLVE QUESTION 6, 7 p.444 in class
6
7.2 Practice (p.442-7) 1, 2,3,4,6,7
7.2 Questions (p.448) 1acd, 3 5 abc, 6abcd, 7ab 8a, 9
7.3CHEMICAL SYSTEMS IN EQUILIBRIUM:
Qualitative Changes in Equilibrium Systems
Dynamic Equilibrium:
An equilibrium classification which occurs when the number of particles entering one phase is
equal to the number of particles leaving that phase (there is no observable change).
Chemical Equilibrium:
Le Chatelier’s Principle:
Le Chatelier's Principle with a change of temperature
For this, you need to know whether heat is given out or absorbed during the reaction. Assume that our
forward reaction is exothermic (heat is evolved):…
The back reaction (the conversion of C and D into A and B) would be endothermic by exactly the same
amount.
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What would happen if you changed the conditions by increasing the temperature?
Example:
•
You increase the temperature from 300 to 500°C.
What would happen if you changed the conditions by decreasing the temperature?
Example:
•
You reduce the temperature from 500 to 400°C.
1. Change in Temperature – An increase in temperature will shift the equilibrium to consume some of
the added heat. (and vise versa)
Summary:
· Increasing the temperature of a system in dynamic equilibrium favours the_________________.
The system counteracts the change you have made by absorbing the extra heat.
·
Decreasing the temperature of a system in dynamic equilibrium favours the _________________.
The system counteracts the change you have made by producing more heat.
Using Le Chatelier's Principle with a change of concentration
Suppose you have an equilibrium established between four substances A, B, C and D.
8
What would happen if you changed the conditions by increasing the concentration of A?
What would happen if you changed the conditions by decreasing the concentration of A?
2. Change in Concentration – The equilibrium will shift to use up the excess substance being added/
produce more of the substance being removed.
Summary:
•
increasing the concentration of reactants shifts the equilibrium (to the right) in favour of the
products.
•
increasing the amount of one of the products of the reaction the position of equilibrium
would move (to the left to use up the product ) in the direction of the reactants
Using Le Chatelier's Principle with a change of pressure (or volume)
This only applies to reactions involving gases:
What would happen if you changed the conditions by increasing the pressure?
9
What happens if there are the same number of molecules on both sides of the equilibrium reaction?
3. Change in Volume (Pressure): Reducing the volume of a gaseous system causes the reaction to
decrease the number of molecules of gas, if possible.
Summary:
•
Increasing the pressure on a gas reaction shifts the position of equilibrium towards the side
with fewer molecules.
•
Decreasing the pressure on a gas reaction shifts the position of equilibrium towards the
side with more molecules.
Same numbers of molecules on both sides, no effect
•
4. Le Chatelier's Principle and adding catalysts
Le Chatelier's Principle and adding inert gases
•
Adding an inert gas (ex. a noble gas, N2) makes absolutely no difference to the position of
equilibrium
•
changes the probability of successful collisions for both the reactants and products equally,
resulting in no equilibrium shift.
7.3 Practice (p.457, 458, 459): 1, 2, 3, 4, 5, 7,
7.3 Questions (p.459, 460): 1, 2, 3, 4, 5, 6
10
Example #1 - the container holding the following reaction (already at equilibrium) has its volume
suddenly increased. Which way will the equilibrium shift to compensate?
H2 + Cl2 <===> 2 HCl
Example #2 - which way will the equilibrium shift if the system temperature goes up (heat is added):
2 SO2 + O2 <==> 2 SO3 + heat
Example #3 - using the same reaction, which way will the equilibrium shift if heat is removed (that
is, the temperature goes down).
Example #4 - which way will the equilibrium shift if more H2 is added to this reaction at equilibrium:
N2 + 3 H2 <==> 2 NH3
Example #5 - using the same reaction, which way will the equilibrium shift if some NH3 is removed
from the reaction when it is at equilibrium.
Example #6 - the container holding the following reaction (already at equilibrium) has its volume
suddenly reduced by half. Which way will the equilibrium shift to compensate?
PCl3 + Cl2 <===> PCl5
Example #7 - the system below is already at equilibrium when a catalyst is added to the system.
What happens to the position of the equilibrium? Does it shift right, left, or no change?
PCl3 + Cl2 <===> PCl5
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Name: _____________________________________
Date: ________________
EQUILIBRIUM PROBLEM SET #1
1. Consider the equilibrium below:
If 1.2 mol of H2 and 1.2 mol of O2 was placed in a 1.0 L container and allowed to reach equilibrium, what would
the value of Ke be if at equilibrium [HI] = 0.40 mol/L?
H2(g) + I2(g) <=====> 2HI(g)
Solution:
mol/L
H2(g)
initial
1.2
1.2
–x
–x
1.2-x=1.0
1.2-x=1.0
∆Concentration
Equilibrium
+
I2(g) <==>
2HI(g)
2x
0.40
[HI]:
2x
x
Ke =
Ke =
2.
= 0.40 mol/L
= 0.40 mol/L
2
= 0.20 mol/L
. [HI]2
[H2][ I2]
(0.40)2
(1.0)(1.0)
= 0.16
note: units are not expressed
8.0 moles of pure ammonia gas were injected into a 2.0-L flask and allowed to reach equilibrium according
to the equation shown below. If the equilibrium mixture was analyzed and found to contain 2.0 moles of
nitrogen gas, calculate the value of the equilibrium constant.
2NH3(g) <====> 2N2(g) + 3H2(g)
Solution:
3.
Consider the equilibrium below
If 1.6 mol of HI was placed in a 1.0 L container and allowed to reach equilibrium, what would the equilibrium
concentrations be for H2(g), I2(g) and HI(g) if the Ke = 36?
H2(g) + I2(g) <=====> 2HI(g)
Solution:
12
Le Chatelier's Principle
In 1884 the French chemist and engineer Henry-Louis Le Chatelier proposed one of the central concepts of
chemical equilibrium. Le Chatelier's principle can be stated as follows: A change in one of the variables that
describe a system at equilibrium produces a shift in the position of the equilibrium that counteracts the
effect of this change.
Le Chatelier's principle describes what happens to a system when a stress momentarily takes it away from
equilibrium. There are three kinds of stress which we can apply to a chemical reaction at equilibrium:
(1) changing the concentration of one of the components of the reaction
(2) changing the gas pressure (or volume)on the system
(3) changing the temperature at which the reaction is run
Problem 1
Assume that the reaction below is at equilibrium. The stress is the use of a catalyst. Indicate how the system
will respond to the stress.
2 CO (g) + 2 NO (g)
2 CO2 (g) + N2 (g)
?Ho = -10 kJ
Solution:
Problem 2
Assume that the reaction below is at equilibrium. The stress is an increase in volume. Indicate how the system
will respond to the stress.
Cl2 (g) + 3 F2 (g)
2 ClF3 (g)
?Ho = 10 kJ
Solution:
Problem 3
Assume that the reaction below is at equilibrium. The stress is the addition of CH4. Indicate how the system will
respond to the stress.
CH4 (g) + O2 (g)
H2CO (g) + H2O (g)
?Ho = -10 kJ
Solution:
13
Problem 4
Assume that the reaction below is at equilibrium. The stress is the removal of CF4. Indicate how the system will
respond to the stress.
CO2 (g) + CF4 (g)
2 COF2 (g)
?Ho = -10 kJ
Solution:
Problem 5
Assume that the reaction below is at equilibrium. The stress is the addition of CO. Indicate how the system will
respond to the stress.
CO2 (g) + H2 (g)
CO (g) + H2O (g)
?Ho = -10 kJ
Solution:
Problem 6
Assume that the reaction below is at equilibrium. The stress is the removal of CO. Indicate how the system will
respond to the stress.
CO2 (g) + H2 (g)
CO (g) + H2O (g)
?Ho = -10 kJ
Solution:
Problem 7
Assume that the reaction below is at equilibrium. The stress is the increase in temperature. Indicate how the
system will respond to the stress.
NO3 (g) + NO (g)
2 NO2 (g)
?Ho = -10 kJ
Solution:
14
Problem 8
Assume that the reaction below is at equilibrium. The stress is the decrease in temperature. Indicate how the
system will respond to the stress.
Cl5 (g)
PCl3 (g) + Cl2 (g)
?Ho = 10 kJ
Solution:
Problem 10
Sketch a proper graph of the effect of the following changes on the reaction in which SO3 decomposes.
2 SO3 (g)
2 SO2 (g) + O2 (g)
?Ho = 197.78 kJ
(a) Increasing the temperature of the reaction. Solution:
(b) Increasing the pressure on the reaction. Solution:
(c) Adding more O2 when the reaction is at equilibrium. Solution:
15
7.4 CHEMICAL SYSTEMS IN EQUILIBRIUM:
Case Study: The Haber Process: Producing Ammonia for Food and Bombs
A brief summary of the Haber Process
The Haber Process combines nitrogen from the air with hydrogen derived mainly from natural gas
(methane) into ammonia. The reaction is reversible and the production of ammonia is exothermic.
A flow scheme for the Haber Process looks like this:
16
Some notes on the conditions
The catalyst
The catalyst is actually slightly more complicated than pure iron. It has potassium hydroxide added
to it as a promoter - a substance that increases its efficiency.
The pressure
The pressure varies from one manufacturing plant to another, but is always high (≅ 200 atm).
Recycling
At each pass of the gases through the reactor, only about 15% of the nitrogen and hydrogen converts
to ammonia. (This figure also varies from plant to plant.) By continual recycling of the unreacted
nitrogen and hydrogen, the overall conversion is about 98%.
Explaining the conditions
The proportions of nitrogen and hydrogen
The mixture of nitrogen and hydrogen going into the reactor is in the ratio of 1 volume of nitrogen to
3 volumes of hydrogen.
Avogadro's Law says that equal volumes of gases at the same temperature and pressure contain equal
numbers of molecules. That means that the gases are going into the reactor in the ratio of 1
molecule of nitrogen to 3 of hydrogen.
That is the proportion demanded by the equation.
In some reactions you might choose to use an excess of one of the reactants. You would do this if it is
particularly important to use up as much as possible of the other reactant - if, for example, it was
much more expensive. That doesn't apply in this case.
There is always a down-side to using anything other than the equation proportions. If you have an
excess of one reactant there will be molecules passing through the reactor which can't possibly react
because there isn't anything for them to react with. This wastes reactor space - particularly space on
the surface of the catalyst.
The temperature
Equilibrium considerations
You need to shift the position of the equilibrium as far as possible to the right in order to produce
the maximum possible amount of ammonia in the equilibrium mixture.
17
The forward reaction (the production of ammonia) is exothermic.
According to Le Chatelier's Principle, this will be favoured if you lower the temperature. The system
will respond by moving the position of equilibrium to counteract this - in other words by producing
more heat.
In order to get as much ammonia as possible in the equilibrium mixture, you need as low a
temperature as possible. However, 400 - 450°C isn't a low temperature!
Rate considerations
The lower the temperature you use, the slower the reaction becomes. A manufacturer is trying to
produce as much ammonia as possible per day. It makes no sense to try to achieve an equilibrium
mixture which contains a very high proportion of ammonia if it takes several years for the reaction to
reach that equilibrium.
You need the gases to reach equilibrium within the very short time that they will be in contact with
the catalyst in the reactor.
The compromise
400 - 450°C is a compromise temperature producing a reasonably high proportion of ammonia in the
equilibrium mixture (even if it is only 15%), but in a very short time.
The pressure
Equilibrium considerations
Notice that there are 4 molecules on the left-hand side of the equation, but only 2 on the right.
According to Le Chatelier's Principle, if you increase the pressure the system will respond by
favouring the reaction which produces fewer molecules. That will cause the pressure to fall again.
In order to get as much ammonia as possible in the equilibrium mixture, you need as high a pressure
as possible. 200 atmospheres is a high pressure, but not amazingly high.
Rate considerations
Increasing the pressure brings the molecules closer together. In this particular instance, it will
increase their chances of hitting and sticking to the surface of the catalyst where they can react. The
higher the pressure the better in terms of the rate of a gas reaction.
18
Economic considerations
Very high pressures are very expensive to produce on two counts.
You have to build extremely strong pipes and containment vessels to withstand the very high
pressure. That increases your capital costs when the plant is built.
High pressures cost a lot to produce and maintain. That means that the running costs of your plant
are very high.
The compromise
200 atmospheres is a compromise pressure chosen on economic grounds. If the pressure used is too
high, the cost of generating it exceeds the price you can get for the extra ammonia produced.
The catalyst
Equilibrium considerations
The catalyst has no effect whatsoever on the position of the equilibrium. Adding a catalyst doesn't
produce any greater percentage of ammonia in the equilibrium mixture. Its only function is to speed
up the reaction.
Rate considerations
In the absence of a catalyst the reaction is so slow that virtually no reaction happens in any sensible
time. The catalyst ensures that the reaction is fast enough for a dynamic equilibrium to be set up
within the very short time that the gases are actually in the reactor.
Separating the ammonia
When the gases leave the reactor they are hot and at a very high pressure. Ammonia is easily
liquefied under pressure as long as it isn't too hot, and so the temperature of the mixture is lowered
enough for the ammonia to turn to a liquid. The nitrogen and hydrogen remain as gases even under
these high pressures, and can be recycled.
7.4 Practice (p.461, 462): 1, 2, 3b, 4, 5, 7abc,
8ab
7.4 Questions (p.462): 1ab, 2abcd, 3abcde, 4ab, 5, 6, 12
19
7.5 CHEMICAL SYSTEMS IN EQUILIBRIUM:
Quantitative Changes in Equilibrium Systems
The Reaction Quotient, Q
The reaction quotient (Q):
•
has the same form as the equilibrium constant, except concentrations are not necessarily equilibrium
concentrations.
•
can be used to determine the direction of the reaction.
•
When Q does not equal Kc, the system is not at equilibrium
•
will spontaneously go toward equilibrium.
If Q = Kc, reaction at equilibrium
If Q > Kc, reaction proceeds R to L
If Q < Kc, reaction proceeds L to R
If Q is greater than K,
the product
concentration must decrease for the values to match, thus the reaction direction is backward (to decrease
products).
If Q < K, the product concentration must increase for the value of Q to reach K, so the reaction must go forward
(create products).
When Q becomes equal to Keq, equilibrium will have been reached and no further change in concentration occurs.
In summary,
Problem 1
The equilibrium constant for the reaction represented below is 50 at 448oC
H2(g) + I2(g)
2 HI(g)
When 1.0 mol of H2 is mixed with 1.0 mol of I2 in a 0.50 L container and allowed to react at 448oC, how many
moles of HI are present at equilibrium?
Solution for Problem 1
Initial concentrations are:
[H2] = 1.0 mol/0.50 L = 2.0 mol/L
[I2] = 1.0 mol/0.50 L = 2.0 mol/L
[HI] = 0 mol/L
20
Table. ICE table for the reaction
H2(g)
+
I2(g)
↔
2HI(g)
Initial concentration (mol/L)
Change in concentration (mol/L)
Equilibrium concentration (mol/L)
[HI]2 __
[I2] [H2]
50 =
(2x)2
.
(2.0 - x)(2.0 - x)
KC =
(2x)2 ___
(2.0 - x)2
solve…..
50 =
Final reactant and product concentrations:
[H2] = 2.0 - x = 2.0 mol/L - 1.56 mol/L = 0.44 mol/L
[I2] = 2.0 - x = 2.0 mol/L - 1.56 mol/L = 0.44 mol/L
[HI] = 2 x = 2 x 1.56 mol/L = 3.12 mol/L
Therefore ….
Problem 2:
If the concentration of all products and reactants is 0.10 mol/L, what will be the direction of the gaseous reaction?
C2H4(g) + H2(g)
C2H6(g)
KC = 0.99
Solution for Problem 2:
[C2H4(g)] = [H2(g)] = [C2H6(g)] = 0.10 mol/L
21
Since Q is "too high," it can be reduced by using up products and creating reactants. Therefore the reaction goes in
reverse (i.e. left in the line diagram).
Problem 3
The equilibrium constant for the reaction represented below is 4.36 at 600oC
2SO2(g) + O2(g)
2SO3(g)
Calculate [SO3(g)] at equilibrium if the equilibrium concentrations of the other two species are [SO2(g)] = 1.50
mol/L and [O2(g)] = 1.25 mol/L.
Solution for Problem 3
First write the equilibrium constant (or the mass action) expression (or the KC equation) based on the balanced
equation for the reaction.
KC =
[SO3(g)]2 __
[SO2(g)]2 [O2(g)]2
Substitute into the equation and do the math.
[SO3(g)]2 .
4.36 =
(1.50)2 (1.25)
Solve…
22
Problem 4
What is the concentration for each substance at equilibrium for the following gaseous reaction
C2H4 + H2
C2H6
KC = 0.99
if the initial concentration of ethene is 0.335 M and that of hydrogen is 0.526 M?
Solution for Problem 4
ICE table:
Initial
Change
Equilibrium
C2H4
0.335
–x
0.335 – x
H2
0.526
–x
0.526 – x
C2H6
0
+x
x
Equilibrium constant expression:
23
Problem 5
What is the equilibrium concentration of silver ions in 1.00 L of solution when 0.010 mol of AgCl are dissolved in
a solution containing 0.010 mol Cl– .
AgCl(s)
Ag+ (aq) + Cl– (aq)
The equilibrium constant expression for the reaction is
KC = [Ag+][Cl–] = 1.8 x 10–10
Solution for Problem 5
Recall a solid, AgCl, is not included.
Initial
Change
Equilibrium
Using the equilibrium concentrations in the equilibrium constant expression
24
Problem 6
What are the equilibrium concentrations of all products and reactants for the following aqueous reaction:
HSO4–(aq) + H2O(l)
H3O+(aq) + SO42–(aq)
K = 0.012
where the initial concentrations are [HSO4–] = 0.50 M, [H3O+] = 0.020 M, [SO42–] = 0.060 M?
Solution for Problem 6
ICE table,
Initial
Change
Equilibrium
HSO4–
0.50
H3O+
0.020
SO42–
0.060
25
Problem 7
What is the concentration of H+ if the initial concentration of H2CO3 is 0.14 M in the following reaction?
H2CO3 (aq)
2 H+ (aq) + CO32– (aq)
K = 2.4 x 10–17
Solution for Problem 7
The ICE table is
H2CO3
H+
CO32–
Initial
Change
Equilibrium
7.5 Practice (p.465, 466, 472, 476, 480): 1-6, 7, 10
7.5 Questions (p.481): 1, 2, 3, 4, 6
26
Name: _____________________________________
EQUILIBRIUM PROBLEM SET #2
Date: ________________
1. Calculate the reaction quotient for each reaction system. In which direction will the reaction
move? Assume Kc = 96.2 for the reaction A + B
C.
Case a: [A]0 = 0.0933 M, [B]0 = 0.123 M, [C]0 = 0.389 M
Case b: [A]0 = 0.0233 M, [B]0 = 0.0123 M, [C]0 = 0.999 M
Case c: [A]0 = 0.0333 M, [B]0 = 0.0333 M, [C]0 = 0.1067 M
2. For the following reaction system, the equilibrium concentrations of the reactants are measured
at 400K : [PCl3] = 0.5 M and [Cl2] = 0.07 M. Also, Kc = 96.2 . What is the equilibrium concentration
of the product, PCl5?
PCl3 (g) + Cl2 (g)
PCl5 (g)
3. Nitrogen monoxide, NO (g), is formed in automobile exhaust by the reaction of nitrogen and
oxygen in air. Initially, a mixture contains 0.850 mol of both N2 and O2 in a 15 L reactor vessel. At
21270C, KC = 0.0125. What are the equilibrium concentrations of all species (participants) in the
reaction?
2 NO (g)
N2 (g) + O2 (g)
4. For the same reaction system as in #2, the initial concentrations of PCl3 and Cl2, are measured at
400K : [PCl3]0 = [Cl2] 0 = 0.14 M. Also, Kc = 96.2. What is the new equilibrium concentration of
the product, PCl5 ? (Hint: Use the quadratic equation to solve for an imperfect square.)
5. For the same reaction system as in #3, the equilibrium concentrations of N2 and O2, are
measured at 400K : [N2 (g)]eq = [O2 (g)] eq = 0.0537 M and [NO(g)] eq = 0.00600 M . Also, Kc =
0.0125 as the temperature remains unchanged. The equilibrium is disturbed by adding 0.0500
mol of N2. What are the new equilibrium concentrations of all species (participants) in the
reaction? .
6. Hydrogen fluoride decomposes according to the reaction system:
2 HF (g)
H2 (g)
+
F2 (g)
The value of K at room temperature is 1.0 x 10-95. Does the decomposition occur to any great
extent at room temperature? Explain briefly. If a 1.0 L reactor vessel contains 1.0 mol of HF,
what are the equilibrium concentrations of H2 and F2? Does the result agree with your
prediction?
7. Shown below is the Haber process for the commercial production of ammonia.
N2(g) + 3H2(g) <=====> 2NH3(g) + 92 kJ
Discuss how Haber maximized the production of ammonia by manipulating this equilibrium. Also point
out the importance of this industrial process to a growing world population. Make sure to establish a
link between ammonia and protein.
1.) a. 33.9,forward b. 3490, back c.96.2, stay 2.) 3 M 3.) [N2 (g)]eq = [O2 (g)]eq = 0.054 M,
[NO (g)]eq= 0.006 0 M 4.) 0.11 M 5.) [N2 (g)]eq = 5.69 x 10-2 M, [O2 (g)]eq = 5.36 x 10-2 M, [NO
(g)]eq = 6.18 x 10-3 M 6.) No, [H2 (g)]eq = [F2 (g)]eq 3.2 x 10-48 M, Yes
27
7.6 CHEMICAL SYSTEMS IN EQUILIBRIUM:
The Solubility Product Constant
Sparingly Soluble Solutes in Animals:
1. ammonia – water-dwellers – very soluble in water so diffuses into water, very toxic but [ ammonia ] is
low
2. urea NH2 CONH2 - terrestrial animals with no shells – soluble, not
very toxic, removed from blood by kidneys – active transport!
3. uric acid – terrestrial animals with egg-laying reproduction – almost
insoluble and solution is more toxic than urea – ppt out so can’t harm
the organism, even the developing embryo
Pictures of people suffering from Gout: deposition of uric acid in cartilage in the
joints
Solubility Product:
Ksp: The Solubility-Product Constant
•
•
special case of equilibrium involves cases where excess solute is in equilibrium with its aqueous solution
Ksp values are for slightly soluble ionic compounds. For more soluble substances, solubility is in mol / L
or g / 100 mL
Examples of dissociation equations and their Ksp :
Sn(OH)2 (s)
Sn2+ (aq) + 2 OH¯
Ag2CrO4 (s)
2 Ag+ (aq) + CrO42¯
Fe(OH)3 (s)
Fe3+ (aq) + 3 OH¯
(aq)
(aq)
(aq)
Ksp = [Sn2+] [OH¯ ]2
Ksp = [Ag+]2 [CrO42¯ ]
Ksp = [Fe3+] [OH¯ ]3
In order to write Ksp expressions properly, you must know how ionic substances dissociate in water. That means,
you have to know your chemical nomenclature, polyatomic ions, and the charges associated with ion.
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Sample problem #1: Determine the Ksp of silver bromide, given that its molar solubility is 5.71 x 10-7 moles per
liter.
Solution:
AgBr (s)
Ag+ (aq) + Br¯
(aq)
The Ksp expression is: …
Sample problem #2: Determine the Ksp of calcium fluoride (CaF2), given molar solubility is 2.14 x 10-4 M.
Solution:
CaF2 (s)
Ca2+ (aq) + 2 F- (aq)
The Ksp expression is: …
There is a 1:1 ratio between CaF2 and Ca2+, BUT there is a 1:2 ratio between CaF2 and F-. This means that, when
2.14 x 10-4 M of CaF2 dissolves, it produces 2.14 x 10-4 M of Ca2+, BUT 4.28 x 10-4 M of F- in solution.
Sample problem #3
Calculate the molar solubility (in mol/L) of each ion IN a saturated solution of the substance tin(II) hydroxide,
Sn(OH)2
Ksp = 5.45 x 10-27
Sn(OH)2
Sn2+ + 2 OH-
Solution:…
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One last thing! The Ksp value does not have any units on it, but when you get to the value for x, be sure to put M
or mol/L (for molarity) on it.
Review: Calculating the Ksp from the gram per 100 mL Solubility
Convert g/100mL value to molar solubility: …
Predicting Precipitation: Trial ion product (T.I.P. or Qsp)
Ksp value is the upper limit of concentration of the soluble ions. When [ions] exceeds Ksp they will form a
precipitate in order to reduce concentrations …back to equilibrium
Predicting Precipitation:
-
Trial ion product (T.I.P. or Qsp): solved like Ksp
if T.I.P. or Qsp is greater than Ksp, …salt will ppt …
T.I.P. or Qsp is less than Ksp, unsaturated… no ppt
Problem 1:
A solution contains Ag+ (aq) and Cl¯
(aq)
ions, each at a concentration of 4.1 x 10-5 mol/L. Will a precipitate form?
Ag+(aq) + Cl-(aq)
Dissociation equation:
AgCl(s)
T.I.P. (Qsp) expression:
T.I.P. = [Ag+][Cl-]
From App. C8 of text p.802:
Ksp = 1.77 x 10-10
Therefore… the [Ag+] and [Cl¯ ] must exceed the Ksp of 1.77 x 10¯ 10, for a ptt. to form (solid).
Q
= [Ag+] [Cl¯ ] = [4.1 x 10-5 M]2 = 1.7 x 10-9
Therefore precipitate forms since Q greater then Ksp
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Problem 2:
The insoluble salt, CaCl2 has a Ksp of 1.45 x 10-7. If equal volumes (100 mL) of 0.0018 M Ca(NO3)2 and 0.003 M
NaCl are mixed, will a precipitate (solid) form?
The Common Ion Effect
Solubility of insoluble substances can be decreased by the presence of a common ion.
Problem 2:
If AgCl was ALREADY dissolved into a solution with 0.0100 M chloride ion, what is the new solubility of AgCl?
By the way, the source of the chloride is unimportant (at this level). Let us assume the chloride came from some
excess dissolved sodium chloride, sufficient to make the solution 0.0100 M. So, on to the solution . . . .
The dissociation equation for AgCl is:
Ag+ (aq) + Cl¯ (aq)
AgCl (s)
Ksp
= [Ag+] [Cl¯ ]
1.77 x 10¯ 10 = [Ag+] [Cl¯ ]
1.77 x 10¯
10
…[Cl¯ ] is already 0.0100 M plus the extra
= (x) (0.0100 + x)
1.77 x 10¯ 10 = (x) (0.0100)
8
x =[Ag+]= 1.77 x 10¯ M
quadratic equation but
assume 'x' is small number,
…so much faster.
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Validate:
%=(0.0100 M - 1.77 x 10¯ 8 M) x 100 = 99.9998% = 100%
(0.0100 M)
Valid since within 5%
7.6 Practice (p.486 - 492): 1, 3, 4a, 5c, 6, 7, 9abc, 10, 11, 12
7.6 Questions (p.493): 1, 2ab, 3, 8, 10a, 11
7.7 CHEMICAL SYSTEMS IN EQUILIBRIUM: Energy and Equilibrium
The Laws of Thermodynamics
1st Law of Thermodynamics – The total energy of the universe is a constant
Entropy (S) – Describes the degree of randomness of a system.
•
If entropy is positive, then randomness is increasing.
•
If entropy is negative, then t randomness is decreasing.
Entropy is really a measurement of the number of ways in which energy
can be scattered. The more dispersed or spread out the energy, the
greater the entropy. Changes in entropy, S, are the real reason behind
all change.
Ludwig Boltzmann showed that entropy is really the consequence of
probability. If you have molecules moving about at random, then it is
more likely that they will become mixed than stay separate. As entropy
increases, the distribution of energies that can exist in molecules become
more "mixed up".
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So where does the energy get dispersed or spread-out into? It
goes into the atoms and molecules of which the substance is
made– rotation, translation, and vibration. An increase in
entropy is an increase in the number of ways in which heat
energy is distributed amongst molecules.
2nd Law of Thermodynamics – All changes either directly or indirectly increase the entropy of the universe. (The
entropy of the universe is increasing)
3rd Law of Thermodynamics – The entropy of a perfect crystal at 0 Kelvin is zero.
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Spontaneous change – Any change that occurs without the constant input of energy from the surroundings.
Increases randomness of the particles .
Fig 1: When the valve is opened, gas in A expands spontaneously into evacuated bulb B. Reverse process,
compression of the gas, is nonspontaneous.
Events which will increase randomness:
1. Dissolving Solids
2. Mixing liquids (usually)
3. Changing state from
solidliquidgas
4. An increase in temperature
5. The expansion of a gas
6. Increasing surface area
Gibb’s Free Energy (G) – Energy from a chemical reaction which is “available” to do useful work.
A chemical reaction will always be spontaneous if it is able to deliver usable energy to the surroundings. As long
as ∆G is negative, then energy will be available to do useful work, thus the reaction will be spontaneous.
Condition
Result (forward direction)
∆G > 0 (positive) non-spontaneous reaction
∆G < 0 (negative) spontaneous reaction
∆G = 0
reaction at equilibrium
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•
You can't really tell by just looking at a chemical equation whether its entropy will be increasing or
decreasing. But you can calculate the change in ∆G very easily using a table of thermodynamic data
which will list values of ∆Hfo (standard heat of formation) and So (molar enthalpy) that can be used to
calculate the overall ∆G for a reaction.
The Difference Between Enthalpy of Reaction and Entropy of Reaction Calculations
Example 1:
What are the standard enthalpies of formation and standard (or absolute) entropies for H2(g), O2(g),
I2(s), C(graphite), C(diamond), H2O, and HI?
Solution
The data are usually listed in a thermodynamic data table (Table C6 on p. 799-800 of the text).
Substance
H2(g)
I2(s)
O2(g)
C(graphite)
C(diamond)
H2O(l)
HI(g)
∆Hfo
kJ/mol
∆So
J/(mol K)
0
0
0
0
1.985
-285.83
26.50
130.680
116.14
205.152
5.74
2.377
69.95
206.590
Note: units for the two quantities. The standard entropies in J/(mol K)
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Example 2:
Find the enthalpy and entropy changes for the reaction:
H2(g) + I2(s)
2 HI(g)
Solution
∆Horeaction = ∑n∆
∆Hfo(products) - ∑n∆
∆Hfo(reactants)
= [(2 mol)(26.5 kJ/mol)] – [(1 mol)(0 kJ/mol) + (1 mol)(0 kJ/mol)]
= + 53.0 kJ
The standard entropy of reaction is measured at standard conditions. ∆Soreaction can be evaluated in a
similar fashion.
H2(g) + I2(s)
2 HI(g)
∆Soreaction = ∑n∆
∆So(products) - ∑n∆
∆So(reactants)
solve:…
Now convert the calculated value of ∆Soreaction to units of kJ K-1.
Example 3:
Calculate the standard Gibb's free energy change associated with this reaction of hydrogen with
iodine.
Solution:
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Example 4:
How could you make a nonspontaneous reaction like this one happen spontaneously?
Solution
7.7 Practice (p.498, 508, 511): 1abcd, 2abc
7.7 Questions (p.512): 1, 2, 3abc, 4abc, 5ab, 6ab, 7, 8, 10abc, 11a, 15,
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