Chem 162-2011 exam III review session

Chem 162-2011 exam III review session - Chemistry 162-2011...

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Unformatted text preview: Chemistry 162-2011 Exam III Review Session from Exam III (4/14/10) Working Copy 1 SOLVING EQUILIBRIA PROBLEMS Step 1: Write balanced equilibrium equation Step 2: Fill in and solve ICE table Step 3: Write equilibrium expression Step 4: Solve equilibrium expression 2 SOLVING EQUILIBRIA PROBLEMS Step 1: Write balanced equilibrium equation Step 2: Fill in and solve ICE table Step 3: Write equilibrium expression Step 4: Solve equilibrium expression e.g., Calculate the equilibrium pressure of NOCl(g) starting with initial pressures of 0.80 atm of NO and 0.30 atm of Cl 2 . 2NO(g) + Cl 2 (g) 2NOCl(g) K p = 1.40 x 10 5 2NO(g) + Cl 2 (g) 2NOCl(g) Initial 0.80 0.30 Change Equilibrium 2NO(g) + Cl 2 (g) 2NOCl(g) Initial 0.80 0.30 Change-2X-X +2X Equilibrium 0.80-2X 0.30-X +2X [NOCl] 2 /[NO] 2 [Cl 2 ] = 1.40 x 10 5 [2X] 2 /[0.80-2X] 2 [0.30-X] = 1.40 x 10 5 [NOCl] = +0.60 atm This format may be used for : Chemical equilibria, 2NO(g) + Cl 2 (g) 2NOCl(g) K c , K p Acid-base equilibria, e.g., HF + H 2 O H 3 O + + F- K a , K cb , K b , K ca Solubility product equilibria AgCl(s) Ag + (aq) + Cl- (aq) K sp = 1.8 x 10-10 Complex ion equilibria Ag + + 2NH 3 Ag(NH 3 ) 2 + K f = 1.6 x 10 7 Combination of solubility product and complex ion equilibria AgCl(s) Ag + (aq) + Cl- (aq) K sp = 1.8 x 10-10 Ag + + 2NH 3 Ag(NH 3 ) 2 + K f = 1.6 x 10 7 AgCl(s) + 2NH 3 Ag(NH 3 ) 2 + + Cl- K = K sp K f = 2.9 x 10-3 3 MATHEMATICAL TRICKS FOR AVOIDING QUADRATIC EQUATIONS (1) Perfect square rule. If possible, take the square root of both sides of the equation. (2) Small K rule. If the equilibrium constant is small, such as 10-3 K, then drop the X in the ICE table when the X is in a format such as Y-X or Y+X. The error caused by this approximation must be < 5% of Y. (3) Large K rule. If the equilibrium constant is large, such as 10 3 , then find the limiting reactant and bring the reaction to completion. (4) Right-to-left rule (sometimes referred to as the 99% rule or back to equilibrium rule). If after using the large K rule, the value of a needed reactant is 0 (requiring a new ICE table going from right to left), or if for any reason we need to go from right to left in the ICE table, then, for convenience, reverse the equation, invert K, and redo the ICE table with a New Initial. An alternate to that, for buffers, might be to use the Henderson- Hasselbalch equation. 4 SIX STRONG ACIDS ET: Discuss strong acids and strong bases and the strengths of their conjugate bases and acids; practice both. Very weak base (i.e., not a base; K a Acid spectator ion)** ~10 6 HCl + H 2 O H 3 O + + Cl- ~10 8 HBr + H 2 O H 3 O + + Br- ~10 9 HI + H 2 O H 3 O + + I- H 2 SO 4 + H 2 O H 3 O + + HSO 4- HNO 3 + H 2 O H 3 O + + NO 3- HClO 4 + H 2 O H 3 O + + ClO 4- (HClO 3 + H 2 O H 3 O + + ClO 3- )* *Strong, but not common acid. Generally not considered as a strong acid....
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This document was uploaded on 11/02/2011 for the course GEN CHEM 162 at Rutgers.

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Chem 162-2011 exam III review session - Chemistry 162-2011...

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