equilibrium just the beginning. reactions are reversible a + b c + d ( forward) c + d a + b...
TRANSCRIPT
![Page 1: Equilibrium Just the Beginning. Reactions are reversible A + B C + D ( forward) C + D A + B (reverse) Initially there is only A and B so only the](https://reader030.vdocuments.us/reader030/viewer/2022032606/56649e925503460f94b98837/html5/thumbnails/1.jpg)
Equilibrium
Just the Beginning
![Page 2: Equilibrium Just the Beginning. Reactions are reversible A + B C + D ( forward) C + D A + B (reverse) Initially there is only A and B so only the](https://reader030.vdocuments.us/reader030/viewer/2022032606/56649e925503460f94b98837/html5/thumbnails/2.jpg)
Reactions are reversible A + B C + D ( forward) C + D A + B (reverse) Initially there is only A and B so only
the forward reaction is possible As C and D build up, the reverse
reaction speeds up while the forward reaction slows down.
Eventually the rates are equal
![Page 3: Equilibrium Just the Beginning. Reactions are reversible A + B C + D ( forward) C + D A + B (reverse) Initially there is only A and B so only the](https://reader030.vdocuments.us/reader030/viewer/2022032606/56649e925503460f94b98837/html5/thumbnails/3.jpg)
Rea
ctio
n R
ate
Time
Forward Reaction
Reverse reaction
Equilibrium
![Page 4: Equilibrium Just the Beginning. Reactions are reversible A + B C + D ( forward) C + D A + B (reverse) Initially there is only A and B so only the](https://reader030.vdocuments.us/reader030/viewer/2022032606/56649e925503460f94b98837/html5/thumbnails/4.jpg)
What is equal at Equilibrium? Rates are equal Concentrations are not. Rates are determined by
concentrations and activation energy. The concentrations do not change at
equilibrium. or if the reaction is verrrry slooooow.
![Page 5: Equilibrium Just the Beginning. Reactions are reversible A + B C + D ( forward) C + D A + B (reverse) Initially there is only A and B so only the](https://reader030.vdocuments.us/reader030/viewer/2022032606/56649e925503460f94b98837/html5/thumbnails/5.jpg)
Law of Mass Action For any reaction jA + kB lC + mD
K = [C]l[D]m PRODUCTSpower
[A]j[B]k REACTANTSpower
K is called the equilibrium constant. It has no units.
is how we indicate a reversible reaction
![Page 6: Equilibrium Just the Beginning. Reactions are reversible A + B C + D ( forward) C + D A + B (reverse) Initially there is only A and B so only the](https://reader030.vdocuments.us/reader030/viewer/2022032606/56649e925503460f94b98837/html5/thumbnails/6.jpg)
Try one together 2SO2(g) + O2(g) 2SO3(g)
where the equilibrium concentrations are
[SO2] = 1.50M; [O2] = 1.25M and [SO3] = 3.5M
K = [SO3]2 / [SO2]2[O2]
K = (3.50)2/(1.50)2(1.25)
K=4.36
![Page 7: Equilibrium Just the Beginning. Reactions are reversible A + B C + D ( forward) C + D A + B (reverse) Initially there is only A and B so only the](https://reader030.vdocuments.us/reader030/viewer/2022032606/56649e925503460f94b98837/html5/thumbnails/7.jpg)
Try another one together 2SO2(g) + O2(g) 2SO3(g)
where the equilibrium concentrations are
[SO2] = 1.50M; [O2] = 1.25M and [SO3] = 3.5M
Now find the k for the reverse reaction
K = [SO2]2 [O2]/ [SO3]2
K = (1.50)2(1.25)/(3.50)2
K=0.230
![Page 8: Equilibrium Just the Beginning. Reactions are reversible A + B C + D ( forward) C + D A + B (reverse) Initially there is only A and B so only the](https://reader030.vdocuments.us/reader030/viewer/2022032606/56649e925503460f94b98837/html5/thumbnails/8.jpg)
Playing with K If we write the reaction in reverse.
Then the new equilibrium constant is
K’ = 1/K
![Page 9: Equilibrium Just the Beginning. Reactions are reversible A + B C + D ( forward) C + D A + B (reverse) Initially there is only A and B so only the](https://reader030.vdocuments.us/reader030/viewer/2022032606/56649e925503460f94b98837/html5/thumbnails/9.jpg)
Try one together SO2(g) + 1/2 O2(g) SO3(g)
where the equilibrium concentrations are
[SO2] = 1.50M; [O2] = 1.25M and [SO3] = 3.5M
K = [SO3] / [SO2][O2]1/2
K = (3.50)/(1.50)(1.25)1/2
K= 2.09
![Page 10: Equilibrium Just the Beginning. Reactions are reversible A + B C + D ( forward) C + D A + B (reverse) Initially there is only A and B so only the](https://reader030.vdocuments.us/reader030/viewer/2022032606/56649e925503460f94b98837/html5/thumbnails/10.jpg)
Playing with K If we multiply the equation by a
constant, n
Then the equilibrium constant is
K’ = Kn
![Page 11: Equilibrium Just the Beginning. Reactions are reversible A + B C + D ( forward) C + D A + B (reverse) Initially there is only A and B so only the](https://reader030.vdocuments.us/reader030/viewer/2022032606/56649e925503460f94b98837/html5/thumbnails/11.jpg)
K is CONSTANT At any temperature. Temperature affects rate. The equilibrium concentrations don’t
have to be the same, only K. Equilibrium position is a set of
concentrations at equilibrium. There are an unlimited number.
![Page 12: Equilibrium Just the Beginning. Reactions are reversible A + B C + D ( forward) C + D A + B (reverse) Initially there is only A and B so only the](https://reader030.vdocuments.us/reader030/viewer/2022032606/56649e925503460f94b98837/html5/thumbnails/12.jpg)
Equilibrium Constant
One for each Temperature
![Page 13: Equilibrium Just the Beginning. Reactions are reversible A + B C + D ( forward) C + D A + B (reverse) Initially there is only A and B so only the](https://reader030.vdocuments.us/reader030/viewer/2022032606/56649e925503460f94b98837/html5/thumbnails/13.jpg)
Calculate K N2 + 3H2 2NH3
Initial At Equilibrium [N2]0 =1.000 M [N2] = 0.921M
[H2]0 =1.000 M [H2] = 0.763M
[NH3]0 =0 M [NH3] = 0.157M
K = 6.03 x 10-2
![Page 14: Equilibrium Just the Beginning. Reactions are reversible A + B C + D ( forward) C + D A + B (reverse) Initially there is only A and B so only the](https://reader030.vdocuments.us/reader030/viewer/2022032606/56649e925503460f94b98837/html5/thumbnails/14.jpg)
Calculate K N2 + 3H2 2NH3
Initial At Equilibrium [N2]0 = 0 M [N2] = 0.399 M
[H2]0 = 0 M [H2] = 1.197 M
[NH3]0 = 1.000 M [NH3] = 0.203M
K = 6.02 x 10-2
![Page 15: Equilibrium Just the Beginning. Reactions are reversible A + B C + D ( forward) C + D A + B (reverse) Initially there is only A and B so only the](https://reader030.vdocuments.us/reader030/viewer/2022032606/56649e925503460f94b98837/html5/thumbnails/15.jpg)
Equilibrium and Pressure Some reactions are gaseous PV = nRT P = (n/V)RT P = CRT C is a concentration in moles/Liter C = P/RT
![Page 16: Equilibrium Just the Beginning. Reactions are reversible A + B C + D ( forward) C + D A + B (reverse) Initially there is only A and B so only the](https://reader030.vdocuments.us/reader030/viewer/2022032606/56649e925503460f94b98837/html5/thumbnails/16.jpg)
Equilibrium and Pressure2SO2(g) + O2(g) 2SO3(g)
Kp = (PSO3)2
(PSO2)2 (PO2)
K = [SO3]2
[SO2]2 [O2]
![Page 17: Equilibrium Just the Beginning. Reactions are reversible A + B C + D ( forward) C + D A + B (reverse) Initially there is only A and B so only the](https://reader030.vdocuments.us/reader030/viewer/2022032606/56649e925503460f94b98837/html5/thumbnails/17.jpg)
Equilibrium and Pressure K = (PSO3/RT)2
(PSO2/RT)2(PO2/RT)
K = (PSO3)2 (1/RT)2
(PSO2)2(PO2) (1/RT)3
K = Kp (1/RT)2 = Kp RT
(1/RT)3
![Page 18: Equilibrium Just the Beginning. Reactions are reversible A + B C + D ( forward) C + D A + B (reverse) Initially there is only A and B so only the](https://reader030.vdocuments.us/reader030/viewer/2022032606/56649e925503460f94b98837/html5/thumbnails/18.jpg)
General Equation jA + kB lC + mD
Kp= (PC)l (PD)m= (CCxRT)l (CDxRT)m
(PA)j (PB)k (CAxRT)j(CBxRT)k
Kp= (CC)l (CD)mx(RT)l+m
(CA)j(CB)kx(RT)j+k
Kp = K (RT)(l+m)-(j+k) = K (RT)n
n=(l+m)-(j+k)=Change in moles of gas
![Page 19: Equilibrium Just the Beginning. Reactions are reversible A + B C + D ( forward) C + D A + B (reverse) Initially there is only A and B so only the](https://reader030.vdocuments.us/reader030/viewer/2022032606/56649e925503460f94b98837/html5/thumbnails/19.jpg)
Homogeneous Equilibria So far every example dealt with
reactants and products where all were in the same phase.
We can use K in terms of either concentration or pressure.
Units depend on reaction.
![Page 20: Equilibrium Just the Beginning. Reactions are reversible A + B C + D ( forward) C + D A + B (reverse) Initially there is only A and B so only the](https://reader030.vdocuments.us/reader030/viewer/2022032606/56649e925503460f94b98837/html5/thumbnails/20.jpg)
Heterogeneous Equilibria If the reaction involves pure solids or
pure liquids the concentration of the solid or the liquid doesn’t change.
As long as they are not used up we can leave them out of the equilibrium expression.
For example
![Page 21: Equilibrium Just the Beginning. Reactions are reversible A + B C + D ( forward) C + D A + B (reverse) Initially there is only A and B so only the](https://reader030.vdocuments.us/reader030/viewer/2022032606/56649e925503460f94b98837/html5/thumbnails/21.jpg)
For Example H2(g) + I2(s) 2HI(g)
K = [HI]2 [H2][I2]
But the concentration of I2 does not
change.
K= [HI]2 [H2]
![Page 22: Equilibrium Just the Beginning. Reactions are reversible A + B C + D ( forward) C + D A + B (reverse) Initially there is only A and B so only the](https://reader030.vdocuments.us/reader030/viewer/2022032606/56649e925503460f94b98837/html5/thumbnails/22.jpg)
Write the equilibrium constant for the heterogeneous reaction
2 2C O H OC . P P
2 2
1A.
CO H O
22 3 C O 2B. N a C O P H O
3 2 3 2 22 N aH C O (s) N a C O (s) + C O (g ) + H O (g ).
2 3 2 2
3
Na CO CO H O D.
NaHCO
2 3 2 22
3
Na CO CO H O E.
NaHCO
![Page 23: Equilibrium Just the Beginning. Reactions are reversible A + B C + D ( forward) C + D A + B (reverse) Initially there is only A and B so only the](https://reader030.vdocuments.us/reader030/viewer/2022032606/56649e925503460f94b98837/html5/thumbnails/23.jpg)
The Reaction Quotient, Q Tells you the direction the reaction
will go to reach equilibrium Calculated the same as the
equilibrium constant, but for a system not at equilibrium
Q = [Products]coefficient
[Reactants] coefficient
Compare value to equilibrium constant
![Page 24: Equilibrium Just the Beginning. Reactions are reversible A + B C + D ( forward) C + D A + B (reverse) Initially there is only A and B so only the](https://reader030.vdocuments.us/reader030/viewer/2022032606/56649e925503460f94b98837/html5/thumbnails/24.jpg)
What Q tells us If Q<K
Not enough products Shift to right
If Q>K Too many products Shift to left
If Q=K system is at equilibrium
![Page 25: Equilibrium Just the Beginning. Reactions are reversible A + B C + D ( forward) C + D A + B (reverse) Initially there is only A and B so only the](https://reader030.vdocuments.us/reader030/viewer/2022032606/56649e925503460f94b98837/html5/thumbnails/25.jpg)
Example for the reaction 2NOCl(g) 2NO(g) + Cl2(g) K = 1.55 x 10-5 M at 35ºC In an experiment 0.10 mol NOCl,
0.0010 mol NO(g) and 0.00010 mol Cl2 are mixed in 2.0 L flask.
Which direction will the reaction proceed to reach equilibrium?
![Page 26: Equilibrium Just the Beginning. Reactions are reversible A + B C + D ( forward) C + D A + B (reverse) Initially there is only A and B so only the](https://reader030.vdocuments.us/reader030/viewer/2022032606/56649e925503460f94b98837/html5/thumbnails/26.jpg)
Solution K = 1.55 x 10-5 M Q = [NO]2[Cl] / [NOCl]2
Q = (.001mol/2.0L)2(.0001mol/2.0L) / (.1mol/2.0L)2
Q = (5 x 10 -4)2(5 x 10-5) / (5 x 10-2)2
Q = 5 x 10-9
Q<K, therefore more product will need to be formed to reach equilibrium
![Page 27: Equilibrium Just the Beginning. Reactions are reversible A + B C + D ( forward) C + D A + B (reverse) Initially there is only A and B so only the](https://reader030.vdocuments.us/reader030/viewer/2022032606/56649e925503460f94b98837/html5/thumbnails/27.jpg)
Solving Equilibrium Problems Given the starting concentrations and
one equilibrium concentration. Use stoichiometry to figure out other
concentrations and K. Learn to create a table of initial and
final conditions.
![Page 28: Equilibrium Just the Beginning. Reactions are reversible A + B C + D ( forward) C + D A + B (reverse) Initially there is only A and B so only the](https://reader030.vdocuments.us/reader030/viewer/2022032606/56649e925503460f94b98837/html5/thumbnails/28.jpg)
Consider the following reaction at 600ºC
2SO2(g) + O2(g) 2SO3(g)In a certain experiment 2.00 mol of SO2, 1.50 mol of O2 and 3.00 mol of SO3 were placed in a 1.00 L flask. At equilibrium 3.50 mol of SO3 were found to be present. Calculate The equilibrium concentrations of O2 and SO2, K and KP
![Page 29: Equilibrium Just the Beginning. Reactions are reversible A + B C + D ( forward) C + D A + B (reverse) Initially there is only A and B so only the](https://reader030.vdocuments.us/reader030/viewer/2022032606/56649e925503460f94b98837/html5/thumbnails/29.jpg)
SolutionK = [SO3]2 / [SO2]2[O2]
K = (3/3.5)2/(2/3.5)2(1.5/3.5)
K = (.857)2 / (.571)2 (.429)
K = 5.25
Kp = k(RT)n
Kp = 5.25[.0821*873]-1
Kp = .073