Equilibrium The student will: 1. compare and contrast forward reactions with reversible reactions 2. express the equilibrium constant equation 3. calculate the equilibrium constant

Reversible reaction A chemical reaction in which the products can react to form the reactants. Examples in your life : baby make-up, match box cars, pencils, shirts, prescription sunglasses. Equilibrium: Two opposing changes occur at ‘equal rates’ Q: When is a reaction that is reversible considered to have reached equilibrium? A: When forward reaction rate = reverse reaction rate. Caution:: Equilibrium….equal… it may appear all has stopped but not so……. Very dynamic, frantic activity, not static.

Steam + CO, closed container, high temp H CO H 2 + CO 2 add more causes more collisions causes more reactions shift reaction to the right as it makes more H 2 + CO 2 causes more collisions causesmore reactions shifts back to the left Until equilibrium forward and reverse are constant

Equilibrium depends on: 1.entropy S organization (s), (l), (g) 2.potential energy in chemical 3. [concentration]……[M] molarity Rate forward = Rate reverse NOT concentration = concentration

Writing Equilibrium Equations Equilibrium constant: K A numerical ratio Of the concentration of the products To the concentration of the reactants 1864 – Law of Mass Action 3A + 2B4F + E [F] 4 [E] [A] 3 [B] 2 K =

Writing the equilibrium expression K 1.Write the concentration of the products using their coefficients as exponents on top. 2.Write the concentration of the reactants using their coefficients as exponents on bottom. 1. N 2 O 4(g) 2NO 2(g) 2.CO (g) + 3H 2(g) CH 4(g) + H 2 O (g) 3.8H 2 S (g) 8H 2(g) + S 8(g) Notice all are gases

4.N 2(g) + 3H 2(g) 2NH 3(g) 5.3A + 2B 4DC + 7DE 4.H 2(g) + I 2(g) 2HI (g) Notice all are gases

Heterogenious Equilibrium solids & liquids do not affect K Why? Pure solids and pure liquids cannot change. is only with gases (g) and aqueous solutions (aq) ions 1.CO (g) + H 2 O (l) CO 2(g) + H 2(g) 2.2SO 2(g) + O 2(g) 2SO 3(g) 3.5A (aq) + 2B (s) 3C (g) + 4D (l) Remember no solids or liquids K

Solving beginner equilibrium problems 1.At 400 o C an equilibrium mixture of nitrogen, oxygen, and ammonia have the following concentrations; N 2 = [.220M]H 2 = [0.3M]NH 3 = [0.113M] What is the equilibrium constant? N 2(g) + 3H 2(g) 2NH 3(g) K is less than one …means at equilibrium the reaction system will contain mostly products

2.A mixture at 325 o C is calculated to be [1.83 x M] H 2, [3.13 x M] I 2, [1.77 x M] HI Calculate the K constant. H 2(g) + I 2(g) 2HI (g)

3.In the reaction CO (g) + NO 2(g) CO 2(g) + NO (g) [CO]=3.0M, [NO 2 ]=1.44x10 -1 M, [CO 2 }=7.0M [NO]=8.0M

4. The equilibrium reaction 2AB(aq)2A(aq) + B2(g) Concentration of AB and A are 5 moles and 4 moles respectively. Concentration of B 2 is 2.0M. What is the K constant?

5.Consider the reaction: NaBr (aq) + H 2 SO 4(aq) NaHSO 4(aq) + HBr (aq) If molarites are 6.0M, 9.0M, 3.0M, 1.5M respectively, what is the equilibrium constant?

6.What is K for this reaction if the concentration of H+ and HC 2 O 3 are both 5.4 x M and H 2 C 2 O 3 has a molarity of 6.3 x 10 1 ? H 2 C 2 O 3(aq) H + (aq) + HC 2 O 3 - (aq)

7. 2SO 2(g) + O 2(g) 2SO 3(g) [1.50M][1.25M][3.50M] Calculate equilibrium constant.

8.CaCO 3(s) CaO (s) + CO 2(g) [2.35M][1.85][3.2M] Express the equilibrium constant

9. 2NOCl (g) 2NO (g) + Cl 2(g) [0.33M][0.0015M] [0.8M] Calculate K

10. 4Fe (s) + 3O 2(g) 2Fe 2 O 3(s) [0.5M] + [5 x ) [1M] Calculate the equilibrium constant

The student will: 1. compare and contrast forward reactions with reversible reactions. 2. express the equilibrium constant equation. 3. calculate the equilibrium constant