1. Based on CaCO 3 (s) CaO(s) + CO 2 (g), which of the following can attain equilibrium? Start with... (a)...pure CaCO 3 (b)...CaO and some CO 2 at P > K p (c)...CaCO 3 and some CO 2 at P > K p (d)...CaCO 3 and CaO “YEP.” “YEP.” “YEP.” “NOPE.” (it breaks down, forming products until the amt. of CO 2 is “right”) (too much CO 2 ; it reacts w/available CaO until amt. of CO 2 is “right”) (too much CO 2, but no CaO for it to react with in order to attain the “right” amt. of CO 2 ) (K p = P CO2 )

2. The Magnitude of the Equilibrium Constant If K >> 1... If K << 1... products are favored. Eq. “lies to the right.” reactants are favored. Eq. “lies to the left.” The K for the forward and reverse reactions are NOT the same. --they are reciprocals -- You must write out the equation and specify the temperature when reporting a K.

3. Calculating Equilibrium Constants (1) If the concentrations of all substances at equilibrium are known, plug and chug. = 0.105 Find K c @ 472 o C forN 2 (g) + 3 H 2 (g) 2 NH 3 (g) At equilibruim @ 472 o C…[NH 3 ] = 0.00272 M [N 2 ] = 0.0402 M [H 2 ] = 0.1207 M.

4. (2) If you know the concentrations of only some substances at equilibrium, make a chart and use reaction stoichiometry to figure out the other concentrations at equilibrium. THEN plug and chug. “Mr. B, what kind of chart?” “Ice, ice, baby…” I = “initial” C = “change” E = “equilibrium”

5. At 1000 K, the amounts shown below are known. Find K c @ 1000 K. 2 SO 3 (g) 2 SO 2 (g) + O 2 (g) initial 6.09 x 10 –3 M 0 M 0 M  –3.65 x 10 –3 M +3.65 x 10 –3 M +1.825 x 10 –3 M at eq. 2.44 x 10 –3 M +3.65 x 10 –3 M +1.825 x 10 –3 M = 4.08 x 10 –3 Put “at eq.” #s into K c expression…

6. At 1000 K, the amounts shown below are known. Find K c @ 1000 K. 2 SO 3 (g) 2 SO 2 (g) + O 2 (g) initial 6.09 x 10 –3 M 0 M 0 M  –3.65 x 10 –3 M +3.65 x 10 –3 M +1.825 x 10 –3 M at eq. 2.44 x 10 –3 M +3.65 x 10 –3 M +1.825 x 10 –3 M = 4.08 x 10 –3 Put “at eq.” #s into K c expression…

7. By knowing K, we can... (1) (2) predict the direction of a reaction calc. amts. of R and P once eq. has been reached, if we know init. amts. of everything reaction quotient, Q: what you get when you plug the R and P amts. at any given time into the eq.-constant expression If Q > K... If Q < K... If Q = K... we are at eq. rxn. proceeds (Be sure you know WHY.)

8. For H 2 (g) + I 2 (g) 2 HI(g), K c = 51 at a temp. of 488 o C. If you start with 0.020 mol HI, 0.010 mol H 2, and 0.030 mol I 2 in a 2.0-L container, which way will the reaction proceed? = 1.3< 51 (need more product) rxn. proceeds to eq. From an MSDS for HI(aq)… (i.e., hydroiodic acid) “May cause congenital malformation in the fetus. Corrosive - causes burns. Harmful if swallowed, inhaled, or in contact with skin. Very destructive of mucous membranes. May affect functioning of thyroid. May increase size of skull.” HI!

9. (Did this rxn start with only PCl 5 ?) For PCl 5 (g) PCl 3 (g) + Cl 2 (g), K p = 0.497 at 500 K. At equilibrium, the partial pressures of PCl 5 and PCl 3 are 0.860 atm and 0.350 atm, respectively. Find the partial pressure of Cl 2. 1.22 atm

10. For the reaction in the previous problem, if a gas cylin- der at 500 K is charged with PCl 5 at 1.66 atm, find the partial pressures of all three substances at equilibrium. PCl 5 (g) PCl 3 (g) + Cl 2 (g) initial 1.66 atm 0 atm 0 atm  –x x x at eq. 1.66 – x x x a b c x = 0.693 atm or –1.190 atm

11. For the reaction in the previous problem, if a gas cylin- der at 500 K is charged with PCl 5 at 1.66 atm, find the partial pressures of all three substances at equilibrium. PCl 5 (g) PCl 3 (g) + Cl 2 (g) initial 1.66 atm 0 atm 0 atm  –x x x at eq. 1.66 – x x x a b c x = 0.693 atm or –1.190 atm