Ag2(C2O4)(s) 2 Ag+(aq) + C2O42-(aq)       Ksp = [Ag+(aq)]2 [C2O42-(aq)] And if x moles dissolve per liter, Ksp = (2x)2(x) = 4x3 where 2x is the molar concentration of the Ag+(aq). But we can know that from E = E° - (0.0592/n) log10(Q) = 0.80 V - 0.0592 log( 1 / [Ag+(aq)] ) = 1.017 V where 0.80 V comes from Table A5.5 on page A26 for Ag+ + e- Ag. [Ag+] = 10-(1.017 - 0.80)/0.0592 = 10-3.67 = 2.2×10-4 = 2x     x = 1.1×10-4 M Thus, Ksp = 4(1.1×10-4)3 = 5.0×10-12

 

1. What is the rate expression for the appearance of P?

   From expts. 1 and 2, doubling [A]0 doubles Rate. Hence first order in [A]. From expts. 3 and 4, doubling BOTH [A]0 and [C]0 doesn't change the Rate at all. So it must be INVERSE first order in [C]. That is, rate is proportional to [A] / [C] thus far. Now, from expts. 2 and 3, doubling [B]0 increases the Rate by a factor of 0.34/0.24 = 1.42 ? What if we didn't recognize that as the moral equivalent of the square root of 2? How would we proceed? We'd say (2)b = 1.42 and take the ln of both sides to be able to deal with b by itself, viz., b×ln(2) = ln(1.42), or b = 0.50 which would prompt us to conclude that Rate = k [A] [B]½ / [C]

   
    

2. What is its rate constant? (Don't forget the units.)

   Whatever expression you got in part (a) gets set equal to the rate for any of the experiments, say #1: k [0.1 M] [0.1 M]½ / [0.1 M ] = 0.0024 M/s k = 0.0024 M/s / (0.1 M)½ = 7.6×10-3 mol½ L-½ s-1

2 A + 2 B P

By mechanism:

A + B C	K (fast equilibrium)
A + C D	k2 (slow)
D + B P	k3 (fast)

Notice that K is an equilibrium constant not a rate constant.

Go for the Slow! (Sort of the opposite of "I brake for Animals.") means that the rate of appearance of product, P, is Rate = k2 [A] [C] And the apparent complication of the fast final production of P is irrelevant because D becomes P as fast as D is formed; so it's rate of appearance is P's rate of appearance. However, we would like to get rid of the intermediate [C} in that expression. We do that by checking the fast equilibrium's constant K = [C] / [A] [B]       or       [C] = K [A] [B] substituted above becomes Rate = k2K [A]2 [B]

 

(The SI power unit is the Watt = J s^-1. A metric ton = 1000 kg.)

(Watch out for the typo in Appendix 4; it says "Al₂O₂" when it obviously means Al₂O₃.)

Page 855 shows the relevant reaction to be 2 Al2O3 + 3 C(graphite) 3 CO2(g) + 4 Al(s) eating up its own electrodes. But that section on the Hall Process doesn't tell us the voltages or the currents for the reaction! Not a problem; we only wanted them so we could calculate DG° to know how much work to do to recover a mole of Al. Pish! We can get that from Appendix 4 (page A21) as long as we don't take its "Al2O2" seriously. So this is a closet Chapter 16 problem: DG° = 3 DGf°[CO2(g)] - 2 DGf°[Al2O3(s)]     = 3(-393.5) - 2(-1676) = + 2.171×106 J/mol of reaction or 5.428×105 J/mol Al

World production is 3.57×106 / 0.17 = 2.1×107 metric tonnes (very British) or 2.1×1010 kg. At an atomic weight of 0.02698 kg for Al, that means that world production is 7.8×1011 moles. Annual electrical work = 5.428×105 J/mol × 7.8×1011 mol = 0.42×1018 J Now while it might be really cool to know that's 0.42 EJ (eka joules), the rest of the world goes by kilowatt hours when talking about electrical energy consumption. And a kW = 1000 W = 1000 J/s; so 1 kW hr = 1000 J/s (3600 s/hr) 1 hr = 3.6×106 J or 1 J = 2.8×10-7 kW hr so Annual electrical work for Al is 0.42×1018 × 2.8×10-7 = 1.2×1011 kW hr (And if you want the cool version of that, 1.2×1014 W hr = 0.12 PW hr (peta Watt hours). America's share is 17% or 20 TW hr (tera Watt hours), and if that truly represents only 5% of our energy consumption, then in 1997, we put away 0.4 PW hr as a nation. That's 0.4×1015 W hr; I'm impressed since that's 1.6 MW hr for every man, woman, and child.)

 

Don't fall into the trap of assuming 50% loss in 360 s! That's linear thinking, and this is an exponential decay problem. At the end of 180 s, there's 75% of the stuff left, so 0.75 = e-k(180 s)     or     ln(0.75) = - k(180 s)     or     k = 1.6×10-3 s-1 t½ = ln(2) / k = 434 s

 

CH₄(g) + ·OH H₂O + ·CH₃

at temperatures relevant to atmospheric chemistry.

1. What is the activation energy for this radical reaction?

   Since k=Ae-Ea/RT, we need to plot ln k vs 1/T whose slope will be -Ea/R and intercept ln(A). So we augment the table: T (K) k (106 L mol-1 s-1) 1/T (10-3 K-1) ln k 223 0.494 4.48 13.110 218 0.452 4.59 13.021 213 0.379 4.69 12.845 206 0.295 4.85 12.595 200 0.241 5.00 12.393 195 0.217 5.13 12.288

   And set our trusty spreadsheet linear least squares regression to work on it. The spreadsheet obligingly tells us slope = - 1356 +/- 62 and intercept is 19.2 +/- 0.3 from a graph like: OK, so it's a wiggly straight line. Still Ea = - R (slope) = 11.3 kJ/mol or riduculously small for an activation energy. Remember we estimate a garden variety activation energy in class to be over 50 kJ/mol. From activation energy alone, then, at 298 K, this guy goes 6 million times faster than garden variety reactions! Radical chemistry is fast! (e(50-11)/RT = 6×106)

   
    

2. What is its Arrhenius constant?

   And A = eintercept = e19.2 = 2.2×108 L mol-1 s-1 (not shabby at all).