1. Chem. 1B – 9/3 Lecture

2. Announcements I Diagnostic Quiz –Should be receiving back in lab –Learn what material to review SacCT – will be set up once adding is done Mastering – class code may be case sensitive Bring a periodic table to class (I will provide on on exam days) Today’s Lecture – note: not covered in same order as in text –Basic Equilibrium (and Questions)

3. Announcements II Today’s Lecture – cont. –Manipulating Reactions –K P vs. K C –Equilibrium Problems: STARTING AT EQUILIBRIUM –Equilibrium Problems: STARTING AT INITIAL CONDITIONS (if time) Chem 1A Review Problem Page 2 (Powerpoint slide #17) Answers: [CaCl 2 ] = 0.00728 M, CaCl 2 = ionic compound, [Cl - ] = 0.0146 M, no rxn, AgCl precipitates

4. Chem 1B - Equilibrium Introduction Graphic example of reaction: A + B ↔ C (assume [A] = [B] at start) In the beginning, forward rate is fast (as A and B are high). At the points the lines cross, [A] = [B] = [C], but the forward rate is still faster than the backwards rate. We can see that [C] is favored over [A] and [B] because its final conc. is higher. Conc. (A and B) Conc. (C) Time

5. Chem 1B - Equilibrium Introduction Some examples: 1. Carbon dioxide (CO 2 ) is not very soluble (you are supposed to know it forms from acid + carbonates) –However, the acidity of rain in “pristine” conditions is supposed to be based on dissolution of CO 2 : –CO 2 (g) + H 2 O (l) ↔ H 2 CO 3 (aq) ↔ H + + HCO 3 - 2.Nitrogen gas (N 2 ) is difficult to react because of the strength of its triple bonds, however under certain conditions, beneficial and problematic reactions involve N 2 where both reactants and products exist. a)N 2 (g) + 3H 2 (g) ↔ 2NH 3 (g) (used in fertilizer) b)N 2 (g) + O 2 (g) ↔ 2NO (g) (source of smog from combustion) Conditions are critical to make a) go but to limit b)

6. Chem 1B - Equilibrium Equilibrium Constant and Equation Whether an equation favors reactants or products is determined by an EQUILIBRIUM CONSTANT (K) – as well as reaction stoichiometry Generic reaction: aA + bB ↔ cC + dD K = equilibrium constant and for above reaction, A larger K value means products are more favored

7. Chem 1B - Equilibrium Equilibrium Equation – Further Details Technically, instead of concentration we have a concentration ratio to a “standard state” (e.g. 1 mol/L) – we can ignore this, but this is why K is unitless Only species in gas phase or in solution will have concentrations. Solids and pure liquids are not included. When gases are involved, there are two Ks: K C for concentration units (including gases) and K P in which pressure (atm) replaces concentration (assume K = K C unless specified) Even further corrections (activity and fugacity) are needed under certain conditions but are beyond this class

8. Chem 1B - Equilibrium Equilibrium Equation – Example N 2 (g) + O 2 (g) ↔ 2NO (g) or At T = 25°C K C = K P (in this case) = 4 x 10 -31 Almost no NO will form in air: P N2 = 0.8 atm, P O2 = 0.2 atm; rearranging the above equation, P NO = [K P P N2 P O2 ] 0.5 = 2 x 10 -16 atm Equilibrium at room temp. is not realistic because kinetics is too slow (insufficient collision energy to break triple bond), so even less NO expected At car engine temp., reactions are faster and K is much larger (even if <1). This leads to significant NO formation. Fuel rich conditions (low P O2 ) can limit NO formation (reducing P O2 reduces P NO ).

9. Chem 1B - Equilibrium Equilibrium Equation – Questions I If the following reaction is at equilibrium, which of the following statements is true: 2NO 2 (g) ↔ N 2 O 4 (g) a) the concentrations of reactants and products are equal b) the rate of formation of N 2 O 4 (M/s) is equal to its loss (in M/s) c) the rate of loss of NO 2 in forming N 2 O 4 is equal to the rate of loss of N 2 O 4 in forming NO 2 d) neither molecule is reacting at all as their concentrations are constant

10. Chem 1B - Equilibrium Equilibrium Equation – Questions II For the following reactions, give an equilibrium equation: CH 4 (g) + H 2 O(g) ↔ CO(g) + 3H 2 (g) H 2 (g) + I 2 (s) ↔ 2HI(g) S(s) + 3F 2 (g) ↔ SF 6 (g) AgCl(s) + 2NH 3 (aq) ↔ Ag(NH 3 ) 2 + (aq) + Cl - (aq) What are the units for the K value for the first reaction? Given the reactions and K values, for which reaction is formation or products most likely? 4Cu(s) + O 2 (g) ↔ 2Cu 2 O(s)K = 3.9 x 10 25 2Cu(s) + O 2 (g) ↔ 2CuO(s)K = 1.8 x 10 22 2Cu(s) + H 2 O(l) ↔ Cu 2 O(s) + H 2 (g) K = 1.0 x 10 -16

11. Chem 1B - Equilibrium Equilibrium Equation – Manipulating Equations This is similar to Hess’s Law (used to calculate  H for a reaction which can be made from combinations of other reactions) Rules: –Flipping Directions If for A ↔ B, K = K 1, then for B ↔ A, K = 1/K 1 –Multiplication ½N 2 (g) + ½O 2 (g) ↔ NO (g) K = K 1 N 2 (g) + O 2 (g) ↔ 2NO (g) = 2·RXN1, then K = K 1 2

12. Chem 1B - Equilibrium Equilibrium Equation – Manipulating Equations Rules: - cont. –Adding Reactions: N 2 (g) + O 2 (g) ↔ 2NO (g) K = K 1 2NO (g) + O 2 (g) ↔ 2NO 2 (g) K = K 2 N 2 (g) + 2O 2 (g) ↔ 2NO 2 (g) K = K 3 Rxn 3 = rxn 1 + rxn 2 K 3 = K 1· K 2 –Why is math different than Hess’s Law? Covered in detail in Ch. 17, but short reason:  G (somewhat like  H) = -RTlnK And from math we know ln(K 1· K 2 ) = lnK 1 + lnK 2 Thus addition in reaction becomes multiplication in K

13. Chem 1B - Equilibrium Equilibrium Equation – Manipulating Equations Example Problem: If the following reactions have the given equilibrium constants: 1)Ag + + 2NH 3 (aq) ↔ Ag(NH 3 ) 2 + K = 1.70 x 10 7 2)NH 3 (aq) + H 2 O(l) ↔ NH 4 + + OH - K = 1.76 x 10 -5 3)H 2 O(l) ↔ H + + OH - K = 1.0 x 10 -14 Determine the equilibrium constant for the following reaction: Ag(NH 3 ) 2 + + 2H + ↔ Ag + + 2NH 4 +

14. Chem 1B - Equilibrium Equilibrium Constants – K P vs. K C Are K P and K P the same? If not how are they related? Q1 – No (except for some reactions) Q2 – How is P related to C? Ideal gas law: PV = nRT And we know [] is moles gas/volume = n/V We can rearrange the ideal gas law to: P = (n/V)RT = []RT

15. Chem 1B - Equilibrium Equilibrium Constants – K P vs. K C Example: 2NO 2 (g) ↔ N 2 O 4 (g) K P = P N2O4 /P NO2 2 Or K P = ([N 2 O 4 ]RT)/([NO 2 ]RT) 2 = K C (RT) -1 General Rule: K P = K C (RT)  n where  n = change in number of moles (moles gas product – moles gas reactants) = 1 – 2 = -1 in above example

16. Chem 1B - Equilibrium Equilibrium Problems – AT EQUILIBRIUM In this case the equilibrium equation is used with concentrations (or pressures) given AT EQUILIBRIUM These types of problems are very important for environmental chemistry, but underemphasized in text For example, an atmospheric chemist measured high NO in air near fresh lava. He wondered if it came from the N 2 (g) + O 2 (g) ↔ 2NO(g) reaction. If K P (T = 1000 K) = 7 x 10 -9, calculate P NO in equilibrium with N 2 and O 2 in air.

17. Chem 1B - Equilibrium Equilibrium Problems – AT EQUILIBRIUM 2 nd Example Problem: A rich chemist wants to measure K C for the reaction: N 2 O 4 (g) ↔ 2NO 2 (g) He puts N 2 O 4 in a container at the temperature he wants to measure K C. He measures [NO 2 ] and [N 2 O 4 ] (using an expensive mass spectrometer) until the concentrations stop changing. He finds [NO 2 ] = 0.0311 M and [N 2 O 4 ] = 0.000170 M. What is K C ?

18. Chem 1B - Equilibrium Equilibrium Problems – Starting from initial conditions In this case an initial concentration or pressure is given (typically of reactants) The reaction then proceeds to equilibrium The student calculates K or the concentration of a reactant or product An important part of working out this problem is to make an ICE table ICE stands for initial change equilibrium

19. Chem 1B - Equilibrium Equilibrium Problems – Starting from initial conditions Example Problem: The rich chemist lost his research grant and had his mass spectrometer repossessed. He still has a UV-Visible spectrometer to measure [NO 2 ] (it’s a brown gas – while N 2 O 4 is invisible). –Can he still calculate K? –Yes, but we need to define the experiment more carefully –Initially, the chemist puts 0.0100 mol N 2 O 4 into a 5.0 L container and sets T. He measures [NO 2 ]. When the concentration stops increasing, he finds [NO 2 ] = 0.0028 M. What is K?