Transcript ENG1071 S1 2005 - BSAC (Bachelor of Science in Applied

General Chemistry II
2302102
Chemical Equilibrium for Gases
and for Sparingly-Soluble Ionic
Solids
Part 1
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Chemical Equilibrium
- 2 Lectures
Outline - 4 Subtopics
• Equilibrium and Le Chatelier’s Principle
• The Reaction Quotient and Equilibrium Constant
• Temperature and Pressure Effects
• Sparingly-Soluble Ionic Compounds in Aqueous
Solution
Chemical Equilibrium
Objectives - Lecture 1
By the end of this lecture AND completion of the set
problems, you should be able to:
• Understand the concepts of: the chemical equilibrium condition,
dynamic equilibrium as the balance of forward and reverse
reaction rates.
• Know the definition of Le Chatelier’s Principle, and understand its
application to the prediction of the direction of change in a
chemical reaction at equilibrium, following changes in pressure,
volume, temperature and amount of reactants and products.
• Understand the definition of the reaction quotient Q and the
equilibrium constant K (or Kp & Kc) for a chemical reaction.
• Calculate the final equilibrium conditions for gas-phase reactions,
and for heterogeneous reactions involving gases, from given nonequilibrium initial conditions.
Equilibrium in Chemical Processes
Equilibrium constants
• Take the simple process (at a given temperature)
H2O (l)
H2O (g)
– An equilibrium partial pressure of H2O known as the “vapour pressure”
po(H2O) is reached; at this point the rates of evaporation and
condensation are equal.
• The equilibrium condition is very simply stated as
pH2O = po(H2O) = Kp
where Kp is a constant whose value characterises the equilibrium.
• A more complex equilibrium
process is the chemical reaction
2NO2 (g)
N2O4 (g)
– Here, colour changes
• (NO2 is brown, N2O4 is colorless)
– and total pressure changes
(P = pN2O4 + pNO2 = n RT/V )
• (the number of molecules changes)
P can be monitored to measure
progress and find the position of
equilibrium.
• For equilibrium in the reaction
2 NO2 (g)
N2O4 (g) + heat
• At any given temperature, one finds that
pN2O4
------- has ~the same equilibrium value, Kp, whatever the pressures
pNO22
– And, as the temperature rises, the value of Kp falls
(this finding is what tells us that the reaction is exothermic).
Le Chatelier’s Principle
The Tendency Towards Equilibrium
• Any system which is not at equilibrium will tend to change
spontaneously toward a state of equilibrium (i.e. without a need
for us to adjust external variables such as temperature and
pressure).
• It follows that if a chemical system at equilibrium is disturbed or
“stressed” away from equilibrium then the system will tend to
react so as to remove the “stress” and return to equilibrium.
• This response is known as Le Chatelier’s Principle
(The approach to equilibrium may be fast, gradual or even
undetectable).
Ideal Gas Reaction Equilibrium Law
“Law of Mass Action”
aA + bB + …
at equilibrium
cC + dD + …(all gaseous)
pCcpDd…
p A ap B b…
= constant, Kp
This “ideal” equilibrium relationship between
partial pressures of products and reactants
in any gas reaction
is closely followed at normal temperatures
and total gas pressures up to about 10 atmospheres .
Ideal Gas Reaction - Concentrations & Kc
“Law of Mass Action”
Remember pBV ~ nBRT => pB ~ (nB/V) RT = [B]RT
aA + bB + …
cC + dD + …(all gaseous)
[C]c [D]d…
at equilibrium
[A]a [B]b…
= constant, Kc
This “ideal” equilibrium relationship between
concentrations of products and reactants
is closely followed in any gas reaction at normal
temperatures and total gas pressures up to about
10 atmospheres
- deviations at higher pressures arise from
intermolecular forces
Relationship between Kc and Kp
for gas phase reactions
Kp =
 = 

Kc(RT)
gaseous products
–
gaseous reactants
(= c + d + …) - (a + b + ….. ) gases only!
Pressures, Concentrations &
Activities
• For any gas in a reaction, the numerical value of partial
pressure in atmospheres is a good approximation to its
“activity” (or tendency to react, taken by convention,
relative to ideal gas behaviour at 1 atm)
Any gas:
agas , = pgas / atm
• For any solute in a reaction, the numerical value of
concentration in moles per litre is a good approximation to
its “activity” (or tendency to react, taken by convention,
relative to ideal solute behaviour at 1 mol/L)
Any solute,
asolute , = [solute] / M
For any pure solid or pure liquid in a reaction, its
“activity” (or tendency to react) is taken (by convention)
to be 1.000 [areas affect rates but not “activities”]
Any pure solid or liquid:
apure solidor liquid = 1.000
•For any solid or liquid solvent (in dilute solution) the
“activity” is taken as 1.00 (by convention, relative to pure
solvent behaviour, assuming solvent mole fraction > 98.5%,
[otherwise, Raoult’s law => ideal asolvent ~ xsolvent/1.000]
Any solvent (dilute solution):
asolvent , = 1.00
[The above “ideal” approximations are easy to use.
More accurate values can be obtained from measurements
of vapour pressures, osmotic pressures, m.pts, b.pts, etc.]
Ideal Solution Reaction Equilibrium
Law
Many reactions occur between solutes in solutions
aA + bB + …
solvent)
cC + dD + …(all in a liquid
here we use concentrations/(mole per litre) to represent activities
( partial pressures are NOT relevant within solutions) - Thus
“Reaction Quotient” Q =
([C]/M)c ([D]/M)d…
([A]/M)a ([B]/M)b…
--> constant, K, at equilm
This “ideal” relationship between the quotient of
numerical values of concentrations/ (mol/L) at equilibrium,
and the equilibrium constant, K,
is closely followed in dilute solutions .
10.3 The Equilibrium Constant
• Suppose we mix 0.6 atm of CO and 1.1 atm of
Cl2. Say we can measure the equilibrium
partial pressure of COCl2 and find that it is 0.1
atm. The temperature is 600oC.
CO(g) + Cl2(g)
0.6-0.1 atm 1.1-0.1 atm
COCl2(g)
0.1 atm at equilm
• The equilibrium expression in terms of partial
pressures is
Q = (pCOCl2/atm)/ (pCO/atm) x (pCl2 /atm) --> K
• Using the known stoichiometry we now insert the
activities (as pressure/atm data), as follows:
K =
0.1
= 0.2
(0.6 - 0.1)(1.1- 0.1)
(Take care if the stoichiometric coefficients are not all
equal to one!!)
[For reasons to be covered in later chapters,
K should be dimensionless. This is ensured by
using “activities” (approximated here by dividing all partial
pressures by the reference pressure of 1 atm, or the
equivalent pressures 760 torr, 101.3 kPa, etc, to obtain purely
numerical values).
A partial pressure of 1 atm is often referred to as the standard
state for a gaseous substance.]
Favoured Reactions
• If K (for a gas reaction) is:
>> 1 then equilibrium tends to favour “products”
<< 1 then equilibrium tends to favour “reactants”
~ 1 then substantial amounts of both products and
reactants will tend to be present at equilibrium
[this depends to some extent on the initial mixture
that is used]
Relationships among Equilibrium
Expressions
(i)
forward reaction
reversed equation
K1
K2
A
A


B
B
K2 = 1/K1
(ii)
Multiplying an equation by a factor n :
– reaction (1)
– reaction (x n)
K1
Kn = K1n
A
nA


B
nB
(iii) addition and subtraction of reactions
– reaction 1
– reaction 2
1+2
1-2
K1
K2
K1+2 = K1K2
K1-2 = K1 / K2
A 
C 
A+C 
A+D 
B
D
B+D
B+C
Equilibrium Calculations for GasPhase Reactions
• Sometimes equilibrium problems lead to complex
equations which can be solved via simple
assumptions. For example
CH4(g) + H2O(g)
CO(g) + 3H2(g)
– for which K at 600 K is 1.8 x 10-7.
• Suppose our initial conditions are as follows:
Gas
CH4
H2O
CO
H2
p/(atm)
1.4
2.3
1.6
0.0
• Let's assume the partial pressure of CH4 decreases by y atm
to reach equilibrium. Again we write an equilibrium
expression,
• At equilibrium the following pressures will apply:
Gas
p/(atm)
CH4
1.4-y
H2O
2.3-y
CO
1.6+y
H2
3y
• By substitution we obtain
(1. 6 + y)(3 y) 3
K =
= 1. 8 x 10 - 7
(1. 4 - y)(2. 3 - y )
This looks like real trouble ( a quartic poynomial in y) !!
DON'T TRY TO SOLVE THIS DIRECTLY.
There is a much easier way.
• We'll make the assumption that y<<1.4 and check at
the end that everything is OK.
i.e.
(1. 6 )(27 y 3 )
= 1. 8 x 10 - 7
(1. 4 )(2 .3 )
Which transposes to y3 = 1.34 x 10-8
or y = 2.38 x 10-3
(cube root)
Clearly our assumption of y<<1.4 is quite reasonable.
• At equilibrium, then, the partial pressure will
be pH2/atm = 3y = 7.1 x 10-3 .
Example
2NO2 (g)
N2O4 (g) + heat
• A sample of atmospheric NO2 is concentrated by freezing as
N2O4 and then vaporizing. Immediately upon vaporizing the
vessel would contains 200 kPa of N2O4(g) before any reaction.
– What are the pressures of N2O4(g) and NO2(g) at equilibrium
for T=298 K ??
Step 1: From tables of chemical data
K298K = 11.3
Step 2: Rough prediction.
– at time = 0, we have only product (N2O4 and no reactant).
– so reaction must go backwards to reach equilibrium.
Step 3: Write down equilibrium expression.
(pN2O4/atm)eq
(pNO2/atm)eq2
=
K = 11.3
Step 4: Work out expressions for the pressures.
– We start with 202 kPa of N2O4
– Say that a fraction  reacts => 1-  unreacted
pN2O4 = (1- ) 202 kPa = (1- ) x 2.00 atm.
– Next consider NO2:
pNO2 = (2 ) 202 kPa = 2  x 2.00 at = 4.00  atm
Step 5: Substitute each p/atm in the equilibrium expression
2.00 (1 - ) / (4.00 )2 = 11.3
2 (1 - )
.
=

11.3 x (16.00  2) = 2.00 - 2.00  (4 )2

or
90.4  2 + 1.00  - 1.00 = 0, a quadratic, in which
 = 0.0998 (discarding the negative root*)
(i.e. about 10% of N2O4 has reacted)
– *note: a negative  is not physically sensible => reject
• So, at equilibrium:
pNO2(eq) = 2  202 kPa
= 39.9 kPa
pN2O4(eq) = (1- ) 202 kPa = 180.1 kPa
 Ptotal
= 0.399 atm
= 1.801 atm
= 2.200 atm = 222.9 kPa I.e. 223 kPa
• Notice that the total pressure has gone up during the
reaction.
• Finally, it's a good idea to check our calculations by backsubstituting each calculated value of pressure/atm (not /kPa)
into the equilibrium constant expression:
1.801
= 11. 313 =
2
0.399
K
• All is well!!
The Reaction Quotient
The "reaction quotient”, Q, contains
“activities of products” / “activities of reactants”
-
activity of gas, a(gas)
= p(gas)/atm
activity of solute, a(solute) = [solute]/M
Pure liquid and solid substances need not appear in
reaction quotients (their activities can be taken as 1.000
and normal pressures have negligible effect and areas ).
Activity of a solvent (close to its mole fraction) is taken as 1.00.
• By definition, the value of Q at equilibrium is the "equilibrium
constant", K.
• Reactions always proceed in such a direction as to make
Q = K. eg:
– if Q < K it follows that the reaction can only go forward.
(eg. when we start with only “reactants”)
– if Q > K it follows that the reaction can only go backward.
(eg. When we start with only “products”)
– if Q = K it follows that the reaction is at equilibrium
• Therefore by finding Q at any point in time compared with K
we can predict the direction in which the reaction tends to
proceed
[The value of Q compared with K tells us nothing about the rate
of reaction, but the minimum work required to drive reaction is
G = RT ln Q/K (negative G => spontaneous reaction).]
• Example:
• The reaction:
SO2Cl2(g)
SO2(g) + Cl2(g)
has an equilibrium constant K = 2.4
• Suppose the initial values of partial pressure/atm are:
pSO2Cl2 /atm = 1.0
pSO2 /atm = 0
pCl2
/atm = 0
• We would proceed by computing the reaction quotient
p SO 2 p Cl2
(0 )(0 )
Q =
=
= 0
p SO 2 Cl2
1.0
• Therefore since Q < 2.4 the reaction tends to proceed
spontaneously (unaided, without being driven) to the right.
• Alternatively, if the conditions had been
pSO2Cl2
= 0.01 atm
pSO2
= 0.1 atm
pCl2
= 0.4 atm
pSO2/atm x pCl2/atm
Q = ----------------------pSO2Cl2/atm
0.1 x 0.4
= -------0.01
= 4
Since Q > 2.4 the reaction now tends to proceed
spontaneously to the left (this reaction approaches
equilibrium quite rapidly and without need of
catalyst).
External Effects and Le Chatelier's “Principle”
‘When a system at equilibrium is subject to
change it will respond in such a way as to reduce
the change.’
• For example, in the reaction
2NO2 (g)
N2O4 (g) initially at equilibrium
• Some possible changes we can explore are
– addition of NO2
– increase in total volume
– addition of an inert gas
pN2O4/atm
• First we need the reaction quotient
Q = ----------------
(pNO2/atm)2
• If we increase pNO2 , Q will decrease to less than K
and the reaction will adjust by forming more N2O4 (g).
• If we increase the volume of the system, then each
partial pressure drops. Now Q increases, and the
reaction must proceed in the reverse direction.
• What happens if we increase Ptotal (by adding an inert
gas)?
Nothing! The partial pressures do not change!!
• Another example:
•
•
H2(g) + I2(g)
2HI(g)
p HI2
Q =
p H 2 pI2
• Stage 1: reactants only => Q = 0 (Q < K )
 react.  right
•
Equilibrium reached -> Q = K
• Stage 2: inject iodine -> Q < K  react.  right
• System always adjusts to counteract the perturbation
which takes it out of balance.
Heterogeneous Equilibrium
• So far we have concentrated on homogeneous
equilibria involving gases. We now consider
heterogeneous reactions in which at least two
different physical states of matter are present.
• Pure solids and liquids appear in Q as 1.000 :
H2(g) + I2(s)
2HI(g)
Q = (pHI/atm)2 / (pH2/atm) x 1.000
C(s) + H2O(l) + Cl2(g)
COCl2(g) + H2(g)
Q =(pCOCl2/atm) x (pH2/atm) / 1.000 x 1.00 x (pCl2/atm)
For a liquid solvent in a reaction equation
its mole fraction should be used in Q,
this can normally be taken as 1.00
I2(s)
Q =
=
I2(aq)
[I2(aq)] / 1.00 ---> solubility/(mol/L) at equilm
CaCO3(s)
=
CaO(s) + CO2(g)
Q = 1.000 x pCO2/atm / 1.000 ---> equilm pressure/atm
Q
<K?
=K?
>K?
The Lime Kiln
CaCO3(s)
An industrially
important example of
heterogeneous
equilibria involves the
production of calcium
oxide (CaO - builders’
lime, quicklime) from
limestone (CaCO3 garden lime,
agricultural lime).
CaO(s) + CO2(g)
The Lime Kiln
CaCO3(s)
CaO(s) + CO2(g)
At any given temperature the equilibrium pressure of CO2 (/atm) is
equal to the equilibrium constant K and is independent of the
relative amounts of CaO and CaCO3 present. K increases with T
(=> endothermic reaction) and exceeds 1 at above 1000oC (-->
equilm pCO2 > 1 atm). It is then simply necessary to allow the CO2
to escape to obtain more CaO from the CaCO3.
• In all equilibria involving pure solids with gases, the
partial pressures of gases are independent of the
amounts of solid present. (n.b. as long as we have
some solid present)
• Similar behaviour is found for solid-liquid equilibria.
For example, a sparingly soluble salt like PbSO4
dissolves in water to form ions:
PbSO4(s)
Pb2+(aq) + SO42-(aq)
Q = ([Pb2+]/M) x ([SO42-]/M) / 1.000 ---> K
(at equilm)
– The concentrations of Pb2+(aq) and SO42-(aq) are
independent of the amount of PbSO4(s) that is present.
• Some more general rules for the Law of Mass Action
follow from these observations.
1. Each gas enters a reaction quotient as a numerical activity
(partial pressure in atmospheres, p(gas)/atm)
2. Each dissolved species enters as a numerical activity
(concentration in moles per litre, [solute]/M)
3. Activity of a pure solid or pure liquid need not appear in a
reaction quotient (except as a = 1.000), neither need a
solvent taking part in the reaction, provided the solution is
dilute (take a(solvent) as 1.00).
4. The reaction quotient Q contains the numerical activity of
each “product” in the numerator and of each “reactant” in the
denominator (each raised to the power of its coefficient in
the balanced chemical equation).
• Example 1:
• Consider the equilibrium
•
H2(g) + I2(s)
2HI(g)
• Given K = 0.345 at 25°C
• If pH2 = 1.00 atm and I2(s) is present, determine pHI
• Solution
• For the reaction given we have
Q=
• so that pHI/atm = 0. 345 pH2/atm.
p HI2
= K
p H2 x 1.000
= 0.5874 x 1.00
thus pHI = 0.587 atm
Example 2:
Find the equilibrium expression for the extraction of
gold from ore with cyanide via the reaction:
4 Au(s) + 8 CN–(aq) + O2(g) + 2H2O(l)
4 Au(CN)2–(aq) + 4 OH–(aq)
4 Au(s) + 8 CN–(aq) + O2(g) + 2 H2O(l)
4 Au(CN)2–(aq) + 4 OH–(aq)
Answer:
[Au(CN) -2] 4 [OH -] 4
Q= 4
1.000 [CN -] 8.p O 1.002
=K
2
Chemical Equilibrium - End of Lecture 1
Objectives Covered in Lecture 1
After studying this lecture you should be able to:
• Understand the concepts of: the chemical equilibrium condition,
dynamic equilibrium as the balance of forward and reverse
reaction rates.
• Know the definition of Le Chatelier’s Principle, and understand
its application to the prediction of the direction of change in a
chemical reaction at equilibrium, following changes in pressure,
volume, temperature and amount of reactants and products.
• Understand the definition of the reaction quotient Q and the
equilibrium constant K (or Kp and Kc) for a chemical reaction.
• Calculate the final equilibrium conditions for gas-phase
reactions, and for heterogeneous reactions involving gases, from
given non-equilibrium initial conditions.