(a) advanced calculations using the equilibrium constant; (b) le chatelier’s principle; (c)...
TRANSCRIPT
(a) Advanced Calculations using the Equilibrium Constant; (b) Le
Chatelier’s Principle; (c) Equilibria of Real Gases
Chemistry 142 B
Autumn Quarter 2004
J. B. Callis, Instructor
Lecture #19
Solving Equilibrium Problems
• Write the balanced equation for the reaction.• Write the equilibrium expression.• List the initial concentrations.• Calculate Q and determine the direction of shift to
equilibrium.• Define the change needed to reach equilibrium and define
the equilibrium concentrations. • Substitute the equilibrium concentrations into the
equilibrium expression and solve for the unknown. • Check the solution by calculating K and making sure it is
identical to the original K.
Problem 19-1 – Graphical Solution to the Quadratic Equation
Let us begin by considering the equilibrium between NO2, a red-brown gaseous pollutant formed by automobiles, and its dimer, N2O4. This equilibrium can be expressed by the chemical equation:
2 NO2(g) = N2O4(g)
The equilibrium constant for this reaction is
where PN2O4 and PNO2 are the equilibrium partial pressures of the two gases. KP has the numerical value of 8.8 at T = 25 oC when the partial pressures are expressed in atmospheres.
Problem 19-1 – Graphical Solution cont.Let us now consider the problem of finding the equilibrium partial pressures of NO2 and N2O4, given the value of KP, and the initial pressures of NO2 and N2O4 ( (PNO2)0 and (PN2O4)0 ). Then if x atm is the additional amount of N2O4 formed by the equilibrium, the partial pressure of NO2 must decrease by 2x atm.
Inserting the equilibrium partial pressures into the equilibrium expression results in the equation:
:yield toedrationaliz becan which
K
Finding the Solution to an Equilibrium Problem Using a Graphical Method
From Graph:
x = and
x =
The later is chosen because
it gives all positive concentrations.
(PNO2)eq = 0.726 atm
(PN2O4)eq = 4.637 atm
x, atm f(x)4.0 31.204.1 24.414.2 18.334.3 12.954.4 8.274.5 4.304.6 1.034.7 -1.534.8 -3.394.9 -4.555.0 -5.005.1 -4.755.2 -3.795.3 -2.135.4 0.235.5 3.305.6 7.075.7 11.555.8 16.735.9 22.616.0 29.20
=4*K*x 2̂-(4*K*PNO2o+1)*x+K*PNO2o 2̂-PN2O4o
Graphical Solution to the Equilibrium Problem
-10.00
-5.00
0.00
5.00
10.00
15.00
20.00
25.00
30.00
35.00
4.0 4.5 5.0 5.5 6.0 6.5
x, atm
f(x)
Problem 19-2: Using Exact Solution of Quadratic Equation
K.179at mequilibriu
at species theall of pressure partial thecalculate
initially, L10 in alcohol isopropyl of g 10For
K)179at 0.444(K
)()()(
:follows as writtenis gas hydorgen and acetone
toalcohol isopropyl of ondissociati The
22323
gHgCOCHgCHOHCH
Problem 19-2: Solution(a)
: tablereaction theConstruct
right the toproceeds reaction that theobvious isit - calculatednot is Q
atm 0.00 P Pproducts theof pressure partial Initial
atm, 0.617 = Plisopropano theof pressure partial Initial0H2
0Ac
0Iso
Pressure (atm)
Isopropyl alcohol, IPA Acetone, Ac Hydrogen
H2(g)
Init.
Change
Equil.
CHOH(g)CH 23 CO(g)CH 23
Problem 19-2: Continued (b)
:formula quadratic thefrom obtained be can roots (two) thewhere
form general theof quadratica is expression This
0
:right theon zero withsideleft theonx
of powers descending in rearrange and
tion,multiplica indicated theperform wex,for solve To
:expression mequilibriu theinto
table thefrom ionsconcentrat mequilibriu thengSubstituti
K
Problem 19-2: Continued (c)
444.0
:Check
atm
atm
:ionsconcentrat mequilibriu gCalculatin
atm :root negative theignore we
ions,concentrat positive all toleadsroot positive only the Since
x and x
solutions candidate twoobtain wequadratic, theFrom
2
2
IPA
HAC
IPA
HAc
P
PPK
P
xPP
x
Problem 19-3: Problems Involving Higher Order Polynomials
m.equilibriuat result that willgH
of pressure partial theDetermine ly.respective atm, 1.60 and
atm 2.30 atm, .40 1 are reaction) (before pressures partial
initial their andK, 600at chamber
evacuatedan into introduced are CO and OH ,CH Gaseous
K) 600at 108.1(K
)(3)()(
:reaction following heConsider t
2
24
7P
224
gHgCOOHgCH
Problem 19-3: Solution(a)
: tablereaction theConstruct
right the toproceeds reaction that theobvious isit - calculatednot is Q
Pressure (atm)
CH4(g) H2O(g) CO(g) 3 H2(g)
Init.
Change
Equil.
Problem 19-3: Continued (b)
.polynomialorder fourth a is expression This
:right on the zero with sideleft on they
of powers descendingin rearrange and
tion,multiplica indicated theperform wefor x, solve To
108.1
:expression mequilibriu theinto
table thefrom ionsconcentrat mequilibriu thengSubstituti
7PK
Problem 3- Continued (c)
atm P
thenm,equilibriuAt
pressures. starting with thecompared small indeed is valueThis
y
:gives sidesboth ofroot cube theTaking
y
yieldsy for Solving 108.1
equation eapproximat theuse weThus
small. isy that imagine wei.e. right, themuch to
shift thatnot doesreaction thesmall, isK sincey Fortunatel
H2
3
7
Le Chatelier’s Principle
If a change in conditions (a ‘stress’) is imposed on a system at equilibrium, the equilibrium position will shift in a direction that tends to reduce that change in conditions.
Henri Le Chatelier, 1884
The Effect of a Change in Concentration
• If a gaseous reactant or product is added to a system at equilibrium, the system will shift in a direction to to reduce the concentration of the added component.
• If a gaseous reactant or product is removed from a system at equilibrium, the system will shift in a direction to to increase the concentration of the removed component.
Problem 19-4: The Effect of a Change in Concentration
Consider the following reaction:
2 H2S(g) + O2(g) = 2 S(s) + 2 H2O(g)
What happens to:
(a) [H2O] if O2 is added?
Ans:
(b) [H2S] if O2 is added?
Ans:
(c) [O2] if H2S is removed?
Ans:
(d) [H2S] if S is added?
Ans:
Effect of a Change in Pressure
1. Add or remove a gaseous product at constant volume.
2. Add an inert gas (one not involved in the reaction) at constant volume.
3. Change the volume of the container.
There are three ways to change the pressure of a reaction system involving gaseous components at a given temperature:
Problem 19-5: The Effect of a Change in Pressure
How would you change the total pressure of each of the following reactions to increase the yield of the products:
(a) CaCO3 (s) = CaO(s) + CO2 (g)
Ans:
(b) S(s) + 3 F2 (g) = SF6 (g)
Ans:
(c) Cl2(g) + I2(g) = 2 ICl (g)
Ans:
The Effect of a Change in Temperature
• Changes in concentration or pressure alter the equilibrium position.
• In contrast, changes in temperature alter the value of the equilibrium constant.
Exothermic Reactions
• Releases heat upon reaction. H is negative.
• Addition of heat to an exothermic reaction shifts the equilibrium to the left.
• The value of K decreases in consequence.
Endothermic Reactions
• Absorbs heat upon reaction. H is positive.
• Addition of heat to an endothermic reaction shifts the equilibrium to the right.
• The value of K increases in consequence.
Using Le Chatelier’s Principle to Describe the Effect of a Temperature Change on a System in Equilibrium
• Treat the energy as a reactant (in an exothermic process) or as a product (in an endothermic process).
• Predict the direction of the shift as if an actual reactant or product has been added or removed.
Problem 19-6: The Effect of a Change in Temperature on the Position of Equilibrium
How does an increase in temperature affect the equilibrium concentration of the indicated substance and K for the following reactions:
(a) CaO(s) + H2O (l) = Ca(OH)2 (aq) H0 =-82 kJ
Ans:
(b) (a) CaCO3 (s) = + CaO(s) + CO2 (g) H0 = 178 kJ
Ans:
(c) SO2 (g) = S(s) + O2(g) H0 = 297 kJ
Ans:
Equilibria Involving Real Gases
• We have thus far assumed that gas phase equilibria involve gases that behave as ideal gases.
• The effect of non-ideal behavior is to cause real equilibrium constants to become non-constant under differing conditions of total pressure.
The Activity Coefficient
The activity of the ith gaseous component of the equilibrium system is represented as:
ref
obsii
i P
Pa
Where i is called the activity coefficient for correcting Pi
obs to the ideal value.
For equilibrium pressures of 1 atm or less, the value of Kp calculated from the observed pressures is expected to be within about 1% of the true value
Answers to Problems in Lecture 19
1. From graph: x = 5.391 and x = 4.637. The later is chosen because it gives all positive concentrations.
2.
3. 4. (a) The reaction proceeds to the right so H2O increases. (b) Some H2S reacts with the
added O2 to move the reaction to the right, so [H2S] decreases. (c) The reaction proceeds to the left to re-form H2S, more O2 is formed as well, O2 increases. (d) S is a solid, so its concentration does not change. Thus, [H2S] is unchanged.
5. (a) The only gas is the product CO2. To move the reaction to the right decrease the pressure. (b)With 3 moles of gas on the left and only one on the right, we increase the pressure to form more SF6. (c) The number of moles of gas is the
same on both sides of the equation, so a change in pressure will have no effect.
6. (a) Add heat to the right side. Adding heat shifts the system to the left. [Ca(OH)2] and K will decrease. (b) Add heat to the left side. Adding heat shifts the system to the right. [CO2] and K will increase. (c) Add heat to the left side. Adding heat shifts the system to the right. [SO2] will decrease and K will increase.
atm 101.7P 3H2
atm 270.0atm 347.02 IPAHAc PPP