Calculate the rate constant for the reaction

Chem 112, Fall 05
Exam 2a

YOU MUST:

Please keep your eyes on your own paper and your answers covered.

Use the exam as scratch paper. We will not grade anything on the exam itself.

1.	Column A	vs.
1.		vs.	0.0821 (L atm)/(mole K)
2.	1	vs.

3.		vs.
4.	at T= 200K	vs.	at T = 400K
4.	Kc for an exothermic reaction	vs.
5.	at T= 200K	vs.
5.		vs.
6.	Q for a reaction	vs.
6.	Q for a reaction	vs.	Kc for the same reaction
7.	Q for the reaction X (g) ⇄ Y (g)	vs.

8.	Kc for a product-favored reaction	vs.
9.	Kc for a product-favored reaction	vs.
9.	Kc at 298K for a gaseous reaction	vs.
10.	with Δn = -2	vs.	2
10.		vs.	2

A. 0.0162M ( = 0.097% by volume)
B. 0.0173M ( = 0.103% by volume)
C. 0.0254M ( = 0.152% by volume)
D. 0.0434M ( = 0.260% by volume)
E. 0.0507M ( = 0.303% by volume)

12. Which is the correct form of the balanced chemical equation that corresponds to the following equilibrium constant expression?

Experiment [NO]		[H2]
1	0.420 M	0.122 M
2	0.140 M	0.122 M	1.51 x 10-5 M/sec
3	0.140 M	0.244 M	3.02 x 10-5 M/sec

15. What is the activation energy of a reaction if the rate constant k drops from 5.42 x 10-5 sec-1 to 1.15 x 10-5 sec-1 when the temperature is reduced from 500K to 400K?

A. 0.254 kJ/mole
B. 1.29 kJ/mole
C. 11.2 kJ/mole
D. 25.8 kJ/mole
E. 78.3 kJ/mole

A. Reaction 1
B. Reaction 2
C. Both Reactions 1 and 2
D. Neither Reaction 1 nor 2

18. From the data above, determine the equilibrium constant for the following reaction Ca2+(aq) + 2 H2O(liq) ⇄ Ca(OH)2(s) + 2 H+(aq)
A. Kc = 6.5 × 10-34
B. Kc = 1.5 × 10-23
C. Kc = 6.5 × 10-20
D. Kc = 1.5 × 10-9
E. Kc = 3.1 × 10-9

The next three questions refer to the following information:
The splitting of the sugar lactose into glucose and galactose occurs via a first order process with a first order rate constant of 0.0367 sec-1. We initially have 0.200 moles of lactose in a 1 L container and we want to reduce that amount to 0.0200 moles.

Lactose → glucose + galactose

Chem 112, Fall 05
Exam 2a

D. [SO3] will increase because Q > K.

E. [SO3] will remain the same because Q = K.

A. 0.0030 min
B. 0.077 min
C. 0.33 min
D. 3.0 min
E. 2700 min

Please sign the following statement at the completion of the exam: I did not cheat on this exam. _______________________________(name) _______________________________(signature)

	2A	3B	4B	5B	6B	7B	8B	8B	8B	1B	2B	3A	4A	5A	6A	7A	8A

1 H
6.939	4 Be									12.01	7 N		19.00	10 Ne
	24.31								13 Al		30.97	16 S		39.95
19 K		44.96	22 Ti	52.00	25 Mn	58.93	28 Ni	65.39	31 Ga		74.92	34 Se		83.80
37 Rb		88.91	40 Zr	95.94	43 Tc	102.9	46 Pd	112.4	49 In		121.8	52 Te		131.3
55 Cs		138.9	72 Hf	183.8	75 Re	192.2	78 Pt	200.6	81 Tl		209.0	84 Po		(222)
87 Fr		227.0	104 Unq	(263)	107 Uns	(266)

USEFUL INFORMATION:

ln(k2/k1) = (Ea/R)(1/T1-1/T2) ΔG° = -RT ln K t1/2 = 0.693/k Kp = Kc(RT)Δn