Monday, March 17 th : “A” Day Tuesday, March 18 th : “B” Day Agenda  Ch. 13 Tests  Begin chapter 14: “Chemical Equilibrium” Sec. 14.1: “Reversible Reactions and Equilibrium”  In-class: Section 14.1 review, pg. 501: #1-5  Homework: Concept Review: “Reversible Reactions and Equilibrium” Be ready for a quiz covering this section next time!

Ch. 13 Tests “Solutions” ClassAverage Grade (out of 65) Average Percentage 3A % 4B %

Completion Reactions  If enough oxygen gas is provided for the following reaction, almost all of the sulfur will react: S 8 (s) + 8 O 2 (g) → 8 SO 2 (g)  Reactions such in which almost all of the reactants react to form products are called completion reactions.

Reversible Reactions  In other reactions, called reversible reactions, the products can re-form reactants.  Reversible reaction: a chemical reaction in which the products re-form the original reactants.  Another way to think about reversible reactions is that they form both products and reactants.

Reversible Reactions Reach Equilibrium  A reversible reaction occurs when you mix solutions of calcium chloride and sodium sulfate: CaCl 2 (aq) + Na 2 SO 4 (aq) → CaSO 4 (s) + 2 NaCl (aq)  Because Na + and Cl - are spectator ions, the net ionic equation best describes what happens. Ca 2+ (aq) + SO 4 2- (aq) CaSO 4 (s)

Reversible Reactions Reach Equilibrium  Solid calcium sulfate, the product, can break down to make calcium ions and sulfate ions in a reaction that is the reverse of the previous one. CaSO 4 (s) Ca 2+ (aq) + SO 4 2- (aq)  Use arrows that point in opposite directions when writing a chemical equation for a reversible reaction. Ca 2+ (aq) + SO 4 2- (aq) CaSO 4 (s)

Chemical Equilibrium  The reactions occur at the same rate after the initial mixing of CaCl 2 and Na 2 SO 4.  The amounts of the products and reactants do not change.  Reactions in which the forward and reverse reaction rates are equal are at chemical equilibrium.  Chemical Equilibrium: a state of balance in which the rate of a forward reaction equals the rate of the reverse reaction and the concentrations of products and reactants remains unchanged.

Opposing Reaction Rates are Equal at Equilibrium  The reaction of hydrogen, H 2, and iodine, I 2, to form hydrogen iodide, HI, reaches chemical equilibrium. H 2 (g) + I 2 (g) 2 HI (g)  At first, only a very small fraction of the collisions between H 2 and I 2 result in the formation of HI. H 2 (g) + I 2 (g) → 2 HI(g)

Opposing Reaction Rates are Equal at Equilibrium  After some time, the concentration of HI goes up.  As a result, fewer collisions occur between H 2 and I 2 molecules, and the rate of the forward reaction drops.  Similarly, in the beginning, few HI molecules exist in the system, so they rarely collide with each other.

Opposing Reaction Rates are Equal at Equilibrium  As more HI molecules are made, they collide more often and form H 2 and I 2 by the reverse reaction. 2 HI (g) → H 2 (g) + I 2 (g)  The greater the number of HI molecules that form, the more often the reverse reaction occurs.

Rate Comparison for H 2 (g) + I 2 (g) 2 HI (g)

Opposing Reaction Rates are Equal at Equilibrium  When the forward rate and the reverse rate are equal, the system is at chemical equilibrium.  If you repeated this experiment at the same temperature, starting with a similar amount of pure HI instead of the H 2 and I 2, the reaction would reach chemical equilibrium again and produce the same concentrations of each substance.

Chemical Equilibria are Dynamic  If you drop a ball into a bowl, it will bounce.  When the ball comes to a stop it has reached static equilibrium, a state in which nothing changes.  Chemical equilibrium is different from static equilibrium because it is dynamic.  In a dynamic equilibrium, there is no net change in the system.  Two opposite changes occur at the same time.

Chemical Equilibria are Dynamic  In equilibrium, an atom may change from being part of the products to part of the reactants many times.  But the overall concentrations of products and reactants stay the same.  For chemical equilibrium to be maintained, the rates of the forward and reverse reactions must be equal.  Arrows of equal length are used to show equilibrium. Reactants Products

Chemical Equilibria are Dynamic  In some cases, the equilibrium has a higher concentration of products than reactants.  This type of equilibrium is also shown by using two arrows. Reactants Products  The forward reaction has a longer arrow to show that the products are favored.

Another Example of Equilibria  Even when systems are not in equilibrium, they are continuously changing to try to reach equilibrium.  For example, combustion produces carbon dioxide, CO 2, and poisonous carbon monoxide, CO. As CO and CO 2 cool after combustion, a reversible reaction produces soot, solid carbon. 2 CO (g) C (s) + CO 2 (g)  This reaction of gases and a solid will reach chemical equilibrium.  Equilibria can involve any state of matter, including aqueous solutions.

Equilibria Involving Complex Ions  Complex ion, or coordination compound: any metal atom or ion that is bonded to more than one atom or molecule.  Ligands: a molecule or anion that readily bonds to a metal ion. (Ex: NH 3, CN - )  Complex ions may be positively charged cations or negatively charged anions.  (Remember, in order to be an ION, an atom or group of atoms has to have a CHARGE.)

Equilibria Involving Complex Ions  In this complex ion, [Cu(NH 3 ) 4 ] 2+, ammonia molecules bond to the central copper(II) ion.  What is the ligand in this complex ion? NH 3

Equilibria Involving Complex Ions  Complex ions formed from transition metals are often deeply colored.

Equilibria Involving Complex Ions  The charge on a complex ion is a sum of the charges on the species from which the complex ion forms. For example, when the cobalt ion, Co 2+, bonds with four Cl − ligands, the total charge is (+2) + 4(−1) = −2  Metal ions and ligands can form complexes that have no charge. These are not complex ions. Why not?  Complex ions often form in systems that reach equilibrium.

Equilibria Involving Complex Ions  Consider zinc nitrate dissolving in water: Zn(NO 3 ) 2 (s) Zn 2+ (aq) + 2 NO 3 - (aq)  In the absence of other ligands, water molecules bond with zinc ions. So, this reaction can be written: Zn(NO 3 ) 2 (s) + 4 H 2 O [Zn(H 2 O) 4 ] 2+ (aq) + 2 NO 3 - (aq) (complex ion)

Equilibria Involving Complex Ions  If another ligand, such as CN −, is added, the new system will again reach chemical equilibrium.  Both water molecules and cyanide ions “compete” to bond with zinc ions, as shown in the equation below. [Zn(H 2 O) 4 ] 2+ (aq) + 4CN - (aq) [Zn(CN) 4 ] 2- (aq) + 4H 2 O(l)  All of these ions are colorless, so you can’t see which complex ion has the greater concentration.

Equilibria Involving Complex Ions  In the chemical equilibrium of nickel ions, ammonia, and water, the complex ions have different colors.  You can tell which ion has the greater concentration based on color: [Ni(H 2 O) 6 ] 2+ (aq) + 6 NH 3 (aq) [Ni(NH 3 ) 6 ] 2+ (aq) + 6 H 2 0(l) Green Blue-violet  The starting concentration of NH 3 will determine which one will have the greater concentration.

Demo Fantastic Four-Color Oscillator

In-Class Assignment/Homework  In-Class: Section 14.1 review, pg. 501: #1-5  Homework: Concept Review: “Reversible Reactions and Equilibrium”  Word to the wise: don’t leave the concept review to the end… Quiz over this section next time…