18_emmo

=**__Chapter 18: Chemical Equilibrium__**=


 * Some reactions are reversible.
 * When these reactions have the same reaction forwards and reverse, the reaction is at equilibrium.
 * *This does not mean that there are 50/50 mix of products and reactants.

**__Mass Action Expression __**
pA+qB <=> rC+sD Keq= __[products]__ = __[C]__ __ r __ __[D]__ s  [reactants] [B] q  [A] p  

*If Keq = 1, there is an __equal concentration__ of products and reactants.
 * If Keq>1, there are __more products__ than reactants. The equilibrium "favors" the products.
 * If Keq<1, there are __more reactants__ than products. The equilibrium "favors" the reactants.
 * Keq equations only include <span style="color: rgb(232,2,116);">GAS and AQUEOUS compounds. If a compound is solid, do not include it in the equation.

__**Practice problems: (Answers will be at the end)**__
<span style="font-size: 120%; color: rgb(0,153,255);">~2NO2 (g) <=> N2O4(g) Give the Keq. <span style="font-size: 110%; color: rgb(0,153,255);">~N2 +3H2 <=> 2NH3 Give the Keq.

**__<span style="font-size: 110%; font-family: Arial,Helvetica,sans-serif;">LeChatelier's Principle (LCP): __**
<span style="font-size: 110%; color: rgb(2,232,132); font-family: Arial,Helvetica,sans-serif;">** If an equilibrium is stressed, the __reaction will shift__ in such a way as to minimize the stress. ** *If you add more <span style="font-size: 110%; color: rgb(206,13,83); font-family: Arial,Helvetica,sans-serif;">__**reactants**__ <span style="font-size: 110%; color: rgb(2,232,132); font-family: Arial,Helvetica,sans-serif;"> to the equation, the equation will lean toward the left (reactant) side. To keep it balanced, the equation will shift to the **__<span style="font-size: 121%; color: rgb(206,13,83); font-family: Arial,Helvetica,sans-serif;">right. __** <span style="font-size: 110%; color: rgb(2,232,132); font-family: Arial,Helvetica,sans-serif;"> *If you add more **__<span style="font-size: 121%; color: rgb(206,13,83); font-family: Arial,Helvetica,sans-serif;">products __**<span style="font-size: 110%; color: rgb(2,232,132); font-family: Arial,Helvetica,sans-serif;"> to the equation, the equation iwill tilt toward the right (product) side. To keep it balanced, the equation will shift to the <span style="font-size: 121%; color: rgb(206,13,83); font-family: Arial,Helvetica,sans-serif;">__**left.**__

**__<span style="font-size: 110%; font-family: Arial,Helvetica,sans-serif;">Pressure Change: __**
<span style="font-size: 121%; color: rgb(2,232,132); font-family: Arial,Helvetica,sans-serif;">*If pressure is __**increased**__, then the reaction moves to the **<span style="font-size: 121%; color: rgb(206,13,83); font-family: Arial,Helvetica,sans-serif;">smaller side **<span style="font-size: 121%; color: rgb(2,232,132); font-family: Arial,Helvetica,sans-serif;"> (the side with fewer moles).
 * If the pressure is **__decreased__**, then the reaction moves to the side with <span style="font-size: 145.2%; color: rgb(206,13,83); font-family: Arial,Helvetica,sans-serif;">more moles <span style="font-size: 121%; color: rgb(2,232,132); font-family: Arial,Helvetica,sans-serif;">. <span style="font-size: 110%; font-family: Arial,Helvetica,sans-serif;">

<span style="font-size: 121%; color: rgb(0,153,255); font-family: Arial,Helvetica,sans-serif;">
<span style="font-size: 121%; color: rgb(0,153,255); font-family: Arial,Helvetica,sans-serif;">~CH3OH <=> 2H2 +CO. Increase the pressure. Which direction does the equation move?

~3O2 <=> 2O3 What happens if you decrease the pressure?

**__<span style="font-size: 110%; font-family: Arial,Helvetica,sans-serif;">Solubility Product: __**
<span style="font-size: 121%; color: rgb(2,232,132); font-family: Arial,Helvetica,sans-serif;">NaCl(s) <=> Na+(aq) + Cl- (aq) Ksp=[Na+][Cl-] <span style="font-size: 110%; font-family: Arial,Helvetica,sans-serif;"> <span style="font-size: 121%; color: rgb(2,232,132); font-family: Arial,Helvetica,sans-serif;">What will happen if CaCl2 is added? CaCl2 <=> Ca2+ +2Cl- Cl is the <span style="font-size: 133.1%; color: rgb(206,13,83); font-family: Arial,Helvetica,sans-serif;">common ion <span style="font-size: 121%; color: rgb(2,232,132); font-family: Arial,Helvetica,sans-serif;">, because it is the ion common between the two equations.

__**Practice Problem:**__
~What is the common ion in the following situation: 50.0mL of 1.0M NaCH3COO is added to 1.0M CH3COOH. ~What is the common ion if 5.0g of NaCl is added to a 2.0M solution of HCl?

**__<span style="font-size: 110%; font-family: Arial,Helvetica,sans-serif;">Answers: __**
<span style="font-size: 120%; color: rgb(0,153,255);"> <span style="font-size: 130%; color: #c700ff; font-family: Verdana, Geneva, sans-serif;">**<span style="font-size: 70%; color: #0099ff; font-family: Arial, Helvetica, sans-serif;">~Keq = __[N2O4]__ [NO2] This is because NO2 is the reactant, so it goes on the bottom. N2O4 is the product, so it goes on the top. ~Keq= __[NH3]2__ [N2][H2]3 N2 and H2 are the reactants. There are three hydrogens needed for the equation to work, so it is to the third power. NH3 is the product, and there are two needed in the equation. ~The equation moves to the left. (Pressure change rules.) ~The equation will move to the right. (Pressure change rules) ~The common ion is CH3COO. It is the only ion that is found in both compounds. ~Cl is the common ion. It is the only ion common between the two compounds. **

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