CH131_15B

CH131_15B - Exam 3 CH 131 Tuesday Nov 27 A ­ D last name...

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
    Exam 3  CH 131 Tuesday Nov. 27 A - D  last name CAS 224 E - Z  last name SCI 107 Useful Information  will be posted on  course website/course documents
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
    Exam 3 Topics All material covered since the last exam:  10/25 to 11/20 (today). Chap. 12 (last half) First Law of Thermo •Enthalpy of reactions (exothermic, endothermic), Hess’s law  Heats of formation •Isothermal, reversible and adiabatic processes in gases Chapter 13.  Entropy •Entropy, entropy calculations in simple systems, spontaneity •Statistical interpretation of S •Third Law of Thermodynamics •Gibbs Free Energy 2200∆ G and direction of spontaneous change 2200 G f o  standard molar free energy of formation and chemical changes
Background image of page 2
    Chapter 14.  Chemical Equilibrium •Equilibrium constant (thermodynamic and empirical) •Concentrations (gas phase- partial pressures) at equilibrium from K •Reaction Quotient and predictions of direction of spontaneous  change •Le Chatlier’s Principle => predicting effects of T, P, V, n on systems  at equlibrium Chapter 15.  Acid/Base Equilibium •BL definition of acid/base reactivity, conjugate acid/bases K w , pH, pOH •Strong acid & strong base characteristics •Acid/base equilibria for weak acids and bases, pH of weak acid/base  solutions •Hydrolysis of ionizable salts (qualitatively and quantitatively) •Titration of strong acids and bases
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
    So far K a  considerations have been just qualitative, but K a  can be  used quantitatively to determine equilibrium concentrations e.g. Calculate the pH (i.e. [H 3 O + ]) of a 1.0 M solution of HC 2 H 3 O 2   (aqueous solution of acetic acid)  K a  = 1.8 x 10 -5  (acetic acid) HC 2 H 3 O 2    +  H 2 O      H 3 O +   +  C 2 H 3 O 2    K a = H 3 O + [ ] C 2 H 3 O 2 - [ ] HC 2 H 3 O 2 [ ] = 1.8 ´ 10 - 5 In principle, 2 sources of H 3 O + : (1) ionization of HC 2 H 3 O 2 , (2)  autoionization of H 2 O Neglect contribution from H 2 O (2) first and check at end of  calculation Solve using the same methodology used previously for gas phase  equilibria problems =>  ICE  table (initial conc., change due to  chemical activity, substitute in K expression and solve algebraically) 
Background image of page 4
    x x 1.0 –x E quil. +x +x –x C hange 0 ~ 0 1.0 M I nitial H 3 O + HC 2 H 3 O 2 HC 2 H 3 O 2    +  H 2 O      H 3 O + Let x = concentration  of HC 2 H 3 O 2  that  ionizes  Neglect H 3 O +  from auto- ionization of H 2 O K a = H 3 O + [ ] C 2 H 3 O 2 - [ ] HC 2 H 3 O 2 [ ] = x 2 1.0 - x = 1.8 ´ 10 - 5 Can be solved in 2 ways: 2. Brute force => use quadratic formula:  x 2 + (1.8 ´ 10 - 5 ) x - ´ 10 - 5 ) = 0 So, [HC 2 H 3 O 2 ] = 1.0 - x   1.0 (neglect x with respect to 1.0) x 2  = 1.8 x 10 –5 ;    x = 4.2 x 10 -3  = [H 3 O + ]   pH = -log(4.2 x 10 -3 ) = 2.4 Yes, x small w.r.t. 1.0 and large w.r.t. 10
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 6
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 21

CH131_15B - Exam 3 CH 131 Tuesday Nov 27 A ­ D last name...

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online