introduction to equilibrium and keq... have your packet out
TRANSCRIPT
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Introduction to Equilibrium and Keq...
Have your packet out.
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On Equilibrium Island lived 20 single girls, 10 single guys, and 10 couples
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Sometimes a couple split up….
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….and just as frequently, sometimes a new couple would form…
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RV PALMA
One day, a ship sank not too far from Equilibrium Island….
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The wreck of the RV Palma added 105 guys to the island.
How will this change alter the
number of “single” guys, “single” girls,
and couples over time?
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Example: 5 single girls, 100 single guys, and 25 couples
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The End
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Point 1
Reactions are reversible!!girl + guy couple
The forward and reverse reactions are happening at the same time.
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Point 2Changing amount of something in a
reaction changes amount of everything else.
Le Chatelier’s PrincipleIf a stress is applied to a reaction at
equilibrium, the system is no longer at equilibrium. The system will respond (change or shift) to partially relieve or undo the stress.
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single + single couples
girls guys
5 7 3
Silly Suzy and Bozo Joe break up over the weekend………
6 8 2
Lazy Larry and Ditzy Daisy hold hands at the movie over the weekend…..
5 7 3
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At equilibrium:
1.The reactions are still taking place
2. No change in [reactants] or [products]
3. Rate of the forward rxn = rate reverse rxn
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Assume this system is at equilibrium:
A(g) + B(aq) C(s) + D(aq) + 453 kJ (exothermic)
What happens to the equilibrium position when more of compound C is added to the reaction chamber?
To relieve stress, the equilibrium position shifts to the left (the reactants side) , causing the rate of the reverse reaction to speed up.
What happens to equilibrium, A, C, and D when the [ ] of B is increased?
Equilibrium shifts right (toward the product side). The forward reaction speeds up. More C and D are made, and more of A is consumed.
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**refer to notes written out in packet**What happens to the equilibrium position when the
temperature of the reaction is increased ?A(g) + B(aq) C(s) + D(aq) + 453 kJ (exothermic)
Consider energy to be a product in this case! The equilibrium will have to shift to the left to relieve the stress
What if pressure was increased (or volume decreased)?
Equilibrium shifts towards the side with the fewest total number of moles of gas (in THIS case, the products side).
Equal moles gas? P or V changes have no effect.
Heat is added
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Predict the reaction shift (left, right, or none) when the following equilibria are stressed:Reaction #1: CO2 (g) + C (s) + heat ↔ 2CO (g)
Reaction #2: H2 (g) + Cl2 (g) ↔ 2HCl (g) + heat
Reaction #3: N2 (g) + O2 (g) + heat ↔ 2NO (g)
Increase temperature
Decrease [O2]
Increase pressure
add a catalyst
Increase [H2]
Increase volume of the system
Increase pressure
left
right
No change
right
No change
left
No change
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The Equilibrium Constant
K, Kc, or Keq : general reactions
Ksp: solubility product constant (dissociation constant for salts that aren’t very soluble)
Ka or Kb : dissociation constant for acids or bases
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The value of Keq tells us what direction the reaction prefers to go in.
It’s calculated from the [products and reactants] at equilibrium.
If Keq >>1, the system makes lots of products. Forward reaction goes almost to completion.Ag+(aq) + Cl-(aq) AgCl(s) Ksp = 1.0 x 1010
If Keq << 1, very little forward reaction occurs. The system favors reactants.CH3COOH H+ + CH3COO- Ka = 1.8 x 10-5
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Writing Keq Expressions
Keq = [C]c[D]d aA + bB cC + d D [A]a[B]b
For the reaction: N2(g) + 3Cl2(g) 2NCl3(g)a. Write the equilibrium expression
Keq = [NCl3]2
[N2]1 [Cl2]3
Rule: Only include g and aq substances; amounts of solid and liquids do not effect the eq position
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Keq = [NCl3]2 plug in [N2]1[Cl2]3concentrations!
Keq = (1.41 x 10-1)2
(1.04 x 10-4)1 (2.01 x 10-4)3
b. Calculate Keq if [N2] = 1.04 x 10-4 M
[Cl2] = 2.01x 10-4 M [NCl3] = 1.41x 10-1 M
…first simplify all values in parenthesis…
Keq = (1.9881 x 10-2)
(1.04 x 10-4) (8.120601 x 10-12)
Keq = 2.35 x 1013
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c. The container is opened and some chlorine gas escapes. Once equilibrium is reestablished, the [Cl2] is 5.01x 10-7 M and the new [N2] is measured to be 3.24 x 10-5 M.
i. Explain why. N2(g) + 3Cl2(g) 2NCl3(g)
ii. What is the new [NCl3] ?
Keq = 2.35 x 1013 = [NCl3]2 [N2]1[Cl2]3
2.35 x 1013 = [NCl3] 2 .
(3.24 x 10-5)1 (5.01x 10-7)3
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2.35 x 1013 = [NCl3] 2 .
(3.24 x 10-5)1 (5.01x 10-7)3
2.35 x 1013 = [NCl3] 2 .
(4.074386 x 10-24)
9.5747192 x 10-11 = [NCl3] 2
[NCl3] = 9.79 x 10-6 M
If you plug in all the new concentrations once equilibrium is reestablished, you will come up with the
SAME equilibrium constant