quantitative changes in equilibrium systems
DESCRIPTION
Quantitative Changes in Equilibrium Systems. Chapter 7. Testing to see if an equilibrium has been established. The reaction quotient ( Q ), uses the equilibrium expression with the available concentrations to determine if the reaction has attained equilibrium or not. - PowerPoint PPT PresentationTRANSCRIPT
Quantitative Changes in Equilibrium Systems
Chapter 7
Testing to see if an equilibrium has been established.
The reaction quotient (Q), uses the equilibrium expression with the available concentrations to determine if the reaction has attained equilibrium or not.For the general chemical reaction:
aA + bB D cC + dD
BA
DC ba
dcQ
Predicting the direction of an equilibrium
Reaction Quotient Description of equilibrium
Q=Keq The system is at equilibrium
Q>Keq
The system is “product rich”. It must “shift” to the left to achieve equilibrium.
Q<Keq
The system is “reactant rich”. It must “shift” to the right to achieve equilibrium.
Quantitative analysis of equilibria
Using Le Châtelier’s Principle & the Equilibrium Law one can quantitatively assess;The equilibrium constant (Keq)The position or progress of the equilibrium
system with the reaction quotient (Q)The equilibrium concentrations
Determining the position or status of an equilibrium system. Calculate the reaction quotient using the
equilibrium concentrations. Compare the reaction quotient value to the
equilibrium constant value. Determine whether the reaction is reactant or
product rich and adjust the equilibrium to establish the equilibrium.
At 500o C, the value of Keq for reaction between nitrogen and hydrogen to produce ammonia is 25.0. The equilibrium concentrations for the chemicals are; [N2] = 0.10 mol/L, [H2] = 0.30 mol/L & [NH3]=0.20 mol/L. Is the mixture at equilibrium? If not, which direction must it be adjusted to acquire an equilibrium? Balanced equation for the reaction
N2 (g) + 3 H2 (g) 2 NH3 (g)
Generate the equilibrium expression and reaction quotient expression
At 500o C, the value of Keq for reaction between nitrogen and hydrogen to produce ammonia is 25.0. The equilibrium concentrations for the chemicals are; [N2] = 0.10 mol/L, [H2] = 0.30 mol/L & [NH3]=0.20 mol/L. Is the mixture at equilibrium? If not, which direction must it be adjusted to acquire an equilibrium? Substitute the supplied concentrations into the
expression to solve for Q.
Compare the Q value to the known Keq value.Q < Keq
14.8 < 25.0
At 500o C, the value of Keq for reaction between nitrogen and hydrogen to produce ammonia is 25.0. The equilibrium concentrations for the chemicals are; [N2] = 0.10 mol/L, [H2] = 0.30 mol/L & [NH3]=0.20 mol/L. Is the mixture at equilibrium? If not, which direction must it be adjusted to acquire an equilibrium? Use the Q and Keq values to determine the position of the
equilibrium system.
Since Q < Keq, the system has fewer products than the equilibrium state. Therefore the reaction needs to progress toward the products to attain equilibrium. (i.e.- Shift to the right.)
Calculating equilibrium concentrations
To determine the equilibrium concentrations, one must use:A balanced equation for the equilibrium
reactionAn equilibrium expression and constant, Keq
An Initial concentration, Change in concentration, and Equilibrium concentration (ICE) table.
Calculating equilibrium concentrations
In cases where the Keq value is very small the addition and subtraction of the change value (x) may be insignificant and thereby omitted. Use the ratio of the smallest initial concentration and the
Keq to assess the affect of the change.
clueionconcentrat initialsmallest
eqK
Calculating equilibrium concentrations
If clue > 500, the addition or subtraction of “x” is insignificant and may be ignored.
If 100<clue<500, the addition or subtraction of “x” is probably insignificant, but should be checked.
If clue<100, the addition or subtraction of “x” is significant and must be included in the calculations.
clueionconcentrat Initial smallest
eqK
Calculating equilibrium concentrations
In cases where the Keq expression results in the resolution of a quadratic equation, the use of the quadratic formula my prove helpful.
When the quadratic formula produces two answers select the one that is viable. Remembering that you can not have a negative concentration!
aacbbx
242
Example – At high temperatures, as with lightning, nitrogen and oxygen will react to produce nitrogen monoxide. A chemist puts 0.085 moles of N2(g) and 0.038 mol of O2(g) in a 1.0 L flask at high temperature, where the Keq= 4.2 x 10-8. What is the concentration of the NO(g) in the mixture at equilibrium? Strategy
1.Divide the smallest initial concentration by the Keq to determine if the change in concentration can be ignored.
2.Write a balanced chemical reaction for the equilibrium reaction and determine the equilibrium expression.
3.Set up an ICE table letting “x” represent the change in concentrations.
4.Substitute the equilibrium concentrations (ICE Table) into the expression and solve for “x”.
5.Calculate the required value(s).
1. Divide the smallest initial concentration by the Keq to determine if the change in concentration can be ignored.
58
100.9102.4
038.0 x x
clueionconcentrat Initial smallest
eqK
The clue is far greater than 500 so we can ignore the changes in N2(g) and O2(g).
2. Write a balanced chemical reaction for the equilibrium reaction and determine the equilibrium expression.
N2(g) + O2(g) NO(g)N2(g) + O2(g) 2 NO(g)
8-
22
22 10 x 4.2 ON
NO eqK
3. Set up an ICE table letting “x” represent the change in concentrations.
Concentration (mol/L) N2(g) + O2(g) 2 NO(g)
Initial concentration 0.085 0.038 0
Change in concentration -x -x 2x
Equilibrium concentration
0.085 – x ≈0.085
0.038 – x ≈0.038 2x
4. Substitute the equilibrium concentrations (ICE Table) into the expression and solve for “x”.
6
2
28-
22
22
10 x 82.5 4
00232.000232.04
)038.0)(085.0 ()2( 10 x 4.2
ON
NO
x
x
x
x
Keq
The negative value is impossible as one can
not have a negative concentration.
5. Calculate the required value(s).
5
6
102.1
)1082.5(2
2
x
x
[NO] xWhat is the NO(g) concentration at equilibrium?
The NO(g) concentration at equilibrium is 1.2 x 10-5 mol/L.
Example – In a 1.00 L flask, 2.00 mol of H2(g) is combined with 3.00 mol of I2(g) which produces HI(g). The Keq for the reaction is 25 at 1100 K. What is the What is the concentration of each gas in the mixture at equilibrium?
Strategy1.Divide the smallest initial concentration by the Keq to
determine if the change in concentration can be ignored.2.Write a balanced chemical reaction for the equilibrium
reaction and determine the equilibrium expression. 3.Set up an ICE table letting “x” represent the change in
concentrations.4.Substitute the equilibrium concentrations (ICE Table) into
the expression and solve for “x”.5.Calculate the required value(s).
1. Divide the smallest initial concentration by the Keq to determine if the change in concentration can be ignored.
08.02500.2
clueionconcentrat Initial smallest
eqK
The clue is much less than 500 so we can not ignore the changes in H2(g), I2(g) and HI(g).
2. Write a balanced chemical reaction for the equilibrium reaction and determine the equilibrium expression.
H2(g) + I2(g) HI(g)H2(g) + I2(g) 2 HI(g)
52
IHHI
22
2
eqK
3. Set up an ICE table letting “x” represent the change in concentrations.
Concentration (mol/L) H2(g) + I2(g) 2 HI(g)
Initial concentration 2.00 3.00 0
Change in concentration -x -x 2x
Equilibrium concentration 2.00 – x 3.00 – x 2x
4. Substitute the equilibrium concentrations (ICE Table) into the expression and solve for “x”.
600.5840.00564
003002225
IH
HI
2
2
2
2
22
2
xxxx
x
x).-x)(.(x)(
Keq
4. Substitute the equilibrium concentrations (ICE Table) into the expression and solve for “x”.
7.13.4)6(2
)6)(84.0(4)5()5(
24
600.5840.00
2
2
2
or
x
aacbbx
xx
4. Substitute the equilibrium concentrations (ICE Table) into the expression and solve for “x”.
4.3 and 1.7 are determined through the quadratic formula calculation, but only one is correct!
3.00 – 4.3 = negative value so 1.7 mol/L is the correct value. (A negative concentration is impossible!)
Therefore the final equilibrium concentrations are: [H2]eq = 2.00 – 1.7 = 0.3 mol/L[I2]eq = 3.00 – 1.7 = 1.3 mol/L[HI]eq = 2( 1.7) = 3.4 mol/L
SUMMARY OF LESSONS 1,2 AND 3
EQUILIRIUM EXPRESSIONSolids and liquids are not included in equilibrium expression ONLY gases and
aqueous. units: use M for Kc and or atm for Kp
SIZE OF KK >1 → products favored in equilibrium system
K < 1 → reactants favored in equilibrium system
ENDOTHERMICK increases with increase temperature.K decreases with decrease temperature
EXOTHERMICK decreases with increase temperature.K increases with decrease temperature
SUMMARY OF LESSONS 1,2 AND 3
HOW “K” CHANGES AS THE REACTION CHANGES1.Reverse rxn: K →1/K
2.Multiply rxn by n: K → Kn
3.Add rxs: K3 = K1 x K2
REACTION QUOTIENT (Q): a snapshot of the reaction; shows which way reaction proceeds (L, R, no shift)
Q > K → leftQ < K → right
Q = K → at equilibrium (no shift)
CALCULATIONS4.given [ ]eq → find K
5.given [ ]o + other information → find [ ]eq → K6.given K → [ ]eq using one of three methods
a) perfect squaresb) Approximation: 100 rule
c) Quadratic equation
SUMMARY OF LESSON 1,2 AND 3
Le CHATELIER’S PRINCIPLE: reaction shifts to restore equilibrium (opposes stress on the system)
FACTORS AFFECTING EQUILIBRIUMConcentration; Temperature; Pressure
Concentration: • add/remove reactants/products part of K expression; pure liquids or solids usually do
not affect equilibrium• increase concentration of A, move away from A ( A )
• Decrease concentration of A, move towards A ( A)
Temperature: • endothermic: heat written as a reactant; • exothermic: heat written as a product
• Increase temp always favours the endothermic reaction
Pressure: Must be gaseous and must have unequal number of moles.P ↑ move in a direction with less gas mol
if possible; convert V changes to P
FACTORS NOT AFFECTING EQUILIBRIUMCatalyst, Pure liquid or solid (water sometimes can shift rxn). Inert gases does not affect
equilbrium