EXAM USEFUL INFORMATION. Energy states of the hydrogen atom: E = ( x10-18 J)(1/n 2 )

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1 220 Chapter 15 CHEMISTRY 1220 products at equilibrium. EXAM USEFUL INFORMATION This expression is based on the law of mass action. For a general reaction, CONVERSIONS AND CONSTANTS aa + bb dd + ee 1 inch= 2.54 cm, 1 nm = 10 The - 9 equilibrium-constant m, 1 pm = m, 1 expression Å = m, is given 1 cmby: 3 = 1 ml 220 Chapter 15 We can write an expression for the relationship between the concentration of the reactan N = kg- m/s 2, Pa = N/m 2, J = kg- m 2 /s We mmhg=760 can write an expression torr=1 atm= for the relationship bar=101,325 between the Pa= concentration K c [Dd ][E] e [A] kpa= of a [B] the b reactants psi and products at equilibrium. This expression Molar is Volume based on at the STP law = of 22.4 mass L, action. T(K) = T( C) Where K The For speed a general of light reaction, is c = 3.00 x 10 8 c is the equilibrium constant. The subscript m/s, c Planck s indicates constant, that molar h concentrations = x were J s used to evaluate the cons aa + bb dd + ee NA = 6.022x10 23, Note R = that the equilibrium L- atm/mol- K constant = expression J/mol- K has products in the numerator and reac The equilibrium-constant expression is given by: denominator. d = m/v, density of H2O at 25ºC = 1.00 g/cm 3, density of Hg at 20ºC = g/cm 3 Evaluating K c K c [Dd ][E] e FORMULAS The value of K c does [A] not a depend [B] b on initial concentrations of products or reactants. Consider the reaction: Energy states of the hydrogen atom: E = ( x10-18 J)(1/n 2 ) Where K c is the equilibrium constant. N 2 O 4 (g) 2NO 2 (g) The subscript c indicates The that λ equilibrium = molar h/mv, concentrations E constant = hc/λ is given were used by: to evaluate the constant. Note ΔH rxn = Σ that the equilibrium ΔH products - Σ nδh reactants, constant expression ΔH rxn = Σ has bonds products broken in the - numerator Σ bonds and formed reactants in the denominator. K c [NO 2] 2 q = mass x specific heat x ΔT, PE of two interacting charges E [N = k(q1q2)/d 2 O 4 ] Evaluating K c The value of this constant (at 100 C) is 6.50 (regardless of the initial concentrations The value of K F = ma, P = F/A, KE = ½ mv 2 c does not depend or NO on 2 (g). initial concentrations of products or reactants. Consider the P + n2 reaction: a The equilibrium expression depends on stoichiometry. ( V nb) = nrt, and for an ideal gases: PV = nrt V 2 It does not Ndepend 2 O 4 (g) on 2NO the reaction 2 (g) mechanism. The equilibrium constant The is v = 3RT given value by: of K c varies with temperature. We generally omit the where v is rms speed M K c [NO units of the equilibrium constant. 2 Equilibrium Constants ]2 [N in 2 O Terms 4 ] of Pressure, K p The value z 2 = of x 2 this + y constant 2 (diagonal When (at the 100 of reactants right C) is angle 6.50 and products (regardless triangle), are of gases, Vbox the = initial we l w can concentrations h write an equilibrium of N 2 O 4 (g) expression using or NO 2 (g). pressures rather than molar concentrations. Sg = khpg, PA = XAP A, ΔTb = Kfm, ΔTf = Kfm, Π = (n/v)rt The equilibrium expression The depends equilibrium on stoichiometry. constant is K p where p stands for pressure. o o It does not depend P on For the X the solute reaction reaction: P solvent mechanism. P solution X solvent P solvent Chapter 15 The value of K c varies with temperature. aa + bb dd + ee directly proportional to th We generally omit the units of the equilibrium constant. We can write an expression Equilibrium for the Constants relationship in between Terms the of concentration Pressure, Kof p the reactants Kand p (P D )d (P E ) e (P products at equilibrium. A ) a (P B ) b When the reactants and products are gases, we can write an equilibrium expression using partial This expression is based pressures on the rather law of than mass For molar the action. They can be interconverted using the ideal gas equation and our definition of molarity: concentrations. general equation: aa + bb dd + ee For a general reaction, PV = nrt thus P = (n/v)rt The equilibrium constant is K p where p stands for ression pressure. is given by: aa + bb If 1dD we A+ express ee volume in liters the quantity (n/v) 1 B 1 C 1 D Rate Q D is dequivalent E e to molarity. For the reaction: The equilibrium-constant expression is given Thus aby: the partial pressure of a substance, A, is given t b aa t + bb c t dd d + tee A a as: P B b A =(n A /V)RT = [A]RT We can olarities (for subs K c [Dd ][E] e use this to obtain K [A] a [B] b p (P a general D )d (P E ) e expression relating K c and K p: K p = K c (RT) n Where n = (moles (P A of ) a gaseous (P B ) b products) (moles of gaseous reactants). They can be interconverted using The numerical the ideal gas values equation of K c and our K p will definition differ of if n molarity: = 0. Where K c is the equilibrium constant. PV = nrt thus P = (n/v)rt The subscript c indicates If we express that molar volume concentrations in liters the were quantity used (n/v) to evaluate is equivalent the constant. to molarity. Note that the equilibrium Thus the constant partial expression pressure of has a substance, products in A, the is numerator given as: and reactants in the denominator. P A =(n A /V)RT = [A]RT Copyright 2012 Pearson Education, Inc. aluating K We can use this to obtain a general expression relating K c and K p: c K p = K c (RT) n The value of K c does not depend Where on n initial = (moles concentrations of gaseous of products) products or (moles reactants. of gaseous reactants).

2 moles of solute moles of solute Molarity, M Molality, m liters of solution kilograms of solvent ty will change with a change in tempe vary with essure temperature. is: n A = εbc RT MRT V e nic exclude if they have it from the same the expression os at 25 o C, Kw = 1.0x10-14 K c = [H 3 O + ][OH ] = K w. nstant. K a H A HA % ionization ph. ph = log[h + ] = log[h 3 O + ]. c scale. H equilibrium 100% HA initial uals the ion-produc be the [OH ]. K a K b = K w as: poh = log[oh ] S = kb ln W, G = H T S, G = G + RT ln Q, G = RT ln K V E cell = E red (cathode) E red (anode), G = nfecell, Ecell = E cell logq n 1 e = 1.60 x C, 1 V = 1 J / 1 C, 1 W = 1 J / 1 s, 1 kw h=3.6x10 6 J F = 96,485 J/V mol = 96,486 C/mol e ln K = ΔH /R(1/T) + C

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5 Equilibrium constants for of complex- ion formation reactions. Cations that form complex ions with OH and NH3 are given for: Ag, Al, Bi, Co, Cr, Cu, Fe, Ni, Zn. K f Formation reaction Ag Cl AgCl 2 1.8x10 5 Ag NH 3 Ag(NH 3 ) x10 7 Pb + 3 Cl PbCl 3 2.4x10 1 Co + 6 NH 3 Co(NH 3 ) 6 5.0x10 4 Co NH 3 Co(NH 3 ) x10 33 Cr NH 3 Cr(NH 3 ) x10 8 Cu + 4 NH 3 Cu(NH 3 ) 4 1.1x10 13 Ni + 6 NH 3 Ni(NH 3 ) 6 2.0x10 8 Zn + 4 NH 3 Zn(NH 3 ) 4 7.8x10 8 Cu + 4 OH Cu(OH) x10 16 Zn + 4 OH Zn(OH) x10 17 Pb + 3 OH Pb(OH) x10 14 Al OH Al(OH) 4 7.7x10 33 Cr OH Cr(OH) 4 8x10 29 Spectrochemical series: Cl < F < H2O < NH3 < en < NO2 (N- bonded) < CN Violet Blue Green Yellow Orange Red nm nm nm nm nm nm

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