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Chem 112 - Week 4 Notes (Equilibrium)

by: Christopher Cooke

Chem 112 - Week 4 Notes (Equilibrium) CHEM 112

Marketplace > Pennsylvania State University > Chemistry > CHEM 112 > Chem 112 Week 4 Notes Equilibrium
Christopher Cooke
Penn State
GPA 3.5

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Notes from week 4 of Chem 112. Beginning concepts of chemical equilibrium.
Chemical Principles II
Dr. Raymond Shaak
Class Notes
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This 3 page Class Notes was uploaded by Christopher Cooke on Monday September 19, 2016. The Class Notes belongs to CHEM 112 at Pennsylvania State University taught by Dr. Raymond Shaak in Fall 2016. Since its upload, it has received 11 views. For similar materials see Chemical Principles II in Chemistry at Pennsylvania State University.

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Date Created: 09/19/16
Chem 112 9-14-16 Equilibrium  A state in which opposing forces or influence are balanced. (Google) o Chemistry-specific: A state in which a process and its reverse are occurring at equal rates so that no overall change is taking place. (Google)  Equilibrium process example: o 2NO + Br 2 2NOBr  Chemical Equilibrium o Preferred state of 2 O at:  120 deg C: gas  Equilibrium reaction: 2 O (l)  H 2(g)  37 deg C: liquid  Equilibrium reaction: 2 O (l)  H 2 (g)  -10 deg C: solid  Equilibrium reaction: 2 O (s)  H 2 (l)  100 deg C: gas  Equilibrium reaction: 2 O (l)  H 2 (g)  0 deg C: solid  Equilibrium reaction: H O (s) H O (l) 2 2 o Note that, for different temperatures, the equilibrium reactions may be the same  They are further specified by K, the equilibrium constant o What state(s) of H2O are present at:  37 deg C: liquid  0 deg C: solid, liquid  100 deg C: liquid, gas  Key points about chemical equilibrium o At equilibrium, the relative concentration of reactants and products does not change over time. o The ratio of the concentrations of products to reactants at equilibrium is the same regardless of starting conditions.  Macroscopic vs. microscopic o Relative concentrations remain constant at equilibrium o The reaction still occurs  The forwards and backwards expression of a same reaction will have equal rates, unequal to zero  Equilibrium constant o Equilibrium constant (Keq defines the ratio of the amounts of products and reactants in equilibrium reactions o Keqs a constant; does not change when concentrations change  Concentrations change in order to match up with K eqlue  Keqill change with temperature o Keqxpression for a reaction αA + ßB = γC + δD: γ δ α ß  Keq [C] [D] /[A] [B] o K expression for CH COOH (aq)  CH COO(aq) + H (aq)- + eq 3 3  Keq [CH CO3][H ]/[CH COOH]3 o We can express amounts of chemical species as partial pressures or concentrations  Kp K (eqrtial pressures)  K  K (concentrations) c eq  Kp= K cRT) Δn  Δn = change in number of moles of gas  Heterogeneous equilibria o Homogeneous equilibria: all species are in the same phase o Heterogeneous equilibria: species are in different phases o How to write K eeqressions for heterogeneous equilibria:  CaCO (3)  CaO (s) + CO (g) 2  K = K = PCO ; K = [CO ] eq p 2 c 2  Pb (aq) + 2Cl (aq)  PbCl (s) 2  Keq 1/[Pb ][Cl ] - 2 + -  CH 3OOH (aq)  H O (g) 3 CH COO (aq)3 - +  Keq [CH CO3][H O ]/3CH COOH] 3 Chem 112 9-16-16 Expressing K eq  Remember: pure liquids and solids will not appear in K equateqns - Only aqueous and gaseous products molecules. Remember that K is exeqessed as /reactants Examples  CaCO (3)  CaO (s) + CO (g) 2 o Autopilot first  Keq [CO ][2aO]/[CaCO ] 3 o Remove solids and liquids  Keq [CO ][2aO]/[CaCO ] 3  K = [CO2] eq  Pb (aq) + 2Cl (aq)  PbCl (s) 2 o Autopilot first 2+ - 2  Keq [PbCl ]2[Pb ][Cl ] o Remove solids and liquids  Keq [PbCl ]2[Pb ][Cl ] - 2 2+ - 2  Keq 1/[Pb ][Cl ] - +  CH 3OOH (aq) + H O (2)  CH COO (a3) + H O (aq) 3 o Autopilot first  Keq [H O3][CH CO3]/[CH COOH][3 O] 2 o Remove solids and liquids  K = [H O ][CH COO]/[CH COOH][H O] eq 3 3 3 2  Keq [H O3][CH COO3/[CH COOH] 3  2PO 43-(aq) + 3Sr (aq)  Sr (PO 3 (s)4 2 o Autopilot first  K = [Sr (PO ) ]/[PO ] [Sr ] 2+ 3 eq 3 4 2 4 o Remove solids and liquids  Keq [Sr 3PO )4 2[PO ] 4Sr ] 2+ 3 3- 2 2+ 3  Keq 1/[PO ] 4Sr ] Given generic formulae, we can determine how inverse reactions or reactions with the same products/reactions are related.  We will assume that there are no pure solids or liquids in any of the following equations. 1. A  C Keq1= [C]/[A] 2. B  D Keq2= [D]/[B] 3. A + B  C + D K = K * K = [C][D]/[A][B] eq3 eq1 eq2 4. C + D  A + B Keq4= 1/K eq3 = [A][B]/[C][D] 2 2 2 2 2 2 5. 2A + 2B  2C + 2D Keq5= K eq1 * Keq2 = [C] [D] /[A] [B]


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