Transcript Equilibrium - Schoolwires
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 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
Forward Reaction Equilibrium Time Reverse reaction
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.
Law of Mass Action
For any reaction
j
A +
k
B
l
C +
m
D K = [C]
l
[D]
m
PRODUCTS power [A]
j
[B]
k
REACTANTS power K is called the equilibrium constant. It has no units.
is how we indicate a reversible reaction
Try one together
2SO 2(g) + O 2(g) 2SO 3(g) where the equilibrium concentrations are [SO 2 ] = 1.50M; [O 2 ] = 1.25M and [SO 3 ] = 3.5M
K = [SO 3 ] 2 / [SO 2 ] 2 [O 2 ] K = (3.50) 2 /(1.50) 2 (1.25) K=4.36
Try another one together
2SO 2(g) + O 2(g) 2SO 3(g) where the equilibrium concentrations are [SO 2 ] = 1.50M; [O 2 ] = 1.25M and [SO 3 ] = 3.5M
Now find the k for the reverse reaction K = [SO 2 ] 2 [O 2 ]/ [SO 3 ] 2 K = (1.50) 2 (1.25)/(3.50) 2 K=0.230
Playing with K
If we write the reaction in reverse.
Then the new equilibrium constant is K ’ = 1/K
Try one together
SO 2(g) + 1/2 O 2(g) SO 3(g) where the equilibrium concentrations are [SO 2 ] = 1.50M; [O 2 ] = 1.25M and [SO 3 ] = 3.5M
K = [SO 3 ] / [SO 2 ][O 2 ] 1/2 K = (3.50)/(1.50)(1.25) 1/2 K= 2.09
Playing with K
If we multiply the equation by a constant, n Then the equilibrium constant is K’ = K n
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.
Equilibrium Constant
One for each Temperature
Calculate K
N 2 + 3H 2 Initial 2NH 3 At Equilibrium [N 2 ] 0 =1.000 M [N 2 ] = 0.921M
[H 2 ] 0 =1.000 M [H 2 ] = 0.763M
[NH 3 ] 0 =0 M [NH 3 ] = 0.157M
K = 6.03 x 10 -2
Calculate K
N 2 + 3H 2 Initial 2NH 3 At Equilibrium [N 2 ] 0 = 0 M [N 2 ] = 0.399 M [H 2 ] 0 = 0 M [NH 3 ] 0 [H 2 = 1.000 M [NH ] = 1.197 M 3 ] = 0.203M
K = 6.02 x 10 -2
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
Equilibrium and Pressure
2SO 2 (g) + O 2 (g) 2SO 3 (g) Kp = (P SO3 ) 2 (P SO2 ) 2 (P O2 ) K = [SO 3 ] 2 [SO 2 ] 2 [O 2 ]
Equilibrium and Pressure
K = (P SO3 /RT) 2 (P SO2 /RT) 2 (P O2 /RT) K = (P SO3 ) 2 (1/RT) 2 (P SO2 ) 2 (P O2 ) (1/RT) 3
K
= Kp (1/RT) 2 =
Kp RT
(1/RT) 3
General Equation
j
A +
k
B
l
C +
m
D K p = (P C )
l
(P D )
m =
(C C xRT )
l
(C D xRT )
m
(P A )
j
(P B )
k
(C A xRT )
j
(C B xRT )
k
K p = (C C )
l
(C D )
m
x(RT)
l+m
K p (C A )
j
(C B )
k
x(RT)
j+k
= K (RT) (
l+m)-(j+k) =
K (RT) D n D n= (l+m)-(j+k) = Change in moles of gas
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.
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
For Example
H 2 (g) + I 2 (s) 2HI(g) K = [HI] 2 [ H 2 ][ I 2 ] But the concentration of I 2 does not change.
K= [HI] 2 [ H 2 ]
Write the equilibrium constant for the heterogeneous reaction 2 N aH C O (s) 3 N a C O (s) + C O (g ) + H O (g ). 2 3 2 2 A. CO 2 1 H O 2 2 3 P C O 2 H O 2 D. Na CO 2 3 CO NaHCO 2 3 C . P C O 2 P E. Na CO 2 3 CO NaHCO 2 3 2 2 2
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
What Q tells us
If Q
Example
for the reaction 2NOCl(g) 2NO(g) + Cl 2 (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 Cl 2 are mixed in 2.0 L flask.
Which direction will the reaction proceed to reach equilibrium?
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 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. Consider the following reaction at 600ºC 2SO 2 (g) + O 2 (g) 2SO 3 (g) In a certain experiment 2.00 mol of SO 2 , 1.50 mol of O 2 SO 3 and 3.00 mol of were placed in a 1.00 L flask. At equilibrium 3.50 mol of SO 3 were found to be present. Calculate The equilibrium concentrations of O 2 and SO 2 , K and K P K = [SO 3 ] 2 / [SO 2 ] 2 [O 2 ] K = (3/3.5) 2 /(2/3.5) 2 (1.5/3.5) K = (.857) 2 / (.571) 2 (.429) K = 5.25 K p = k(RT) D n K p = 5.25[.0821*873] -1 K p = .073Solving Equilibrium Problems
Solution