Answers to 1St Sem

Answers to 1st Sem. AP Chemistry Review 16

1. The structures of a water molecule and a crystal of LiCl(s) are represented on the right. A student prepares a solution by dissolving 4.2 g of LiCl (s) in enough water to make 100 mL.

a) In the space provided below, show the interactions of the components of LiCl(aq) by making a drawing that represents the different particles present in the solution. Base the particles in your drawing on the particles shown in the representations above. Include only one formula unit of LiCl and no more than ten molecules of water. Your drawing must include the following details.

· Identity of ions (symbol and charge)

· The arrangement and proper orientation of the particles in the solution

b) The student adds 100 mL of a 0.10 M Pb(NO3)2 to the beaker.

(i) What would you observe? A precipitate of PbCl2 will be produced.

(ii) What concentration would be present for each ion in solution after the reaction has taken place? Note: Don’t forget that the volume is now 0.20 L. 0.50 M Li1+, 0.10 M NO31-, 0.0 M Pb2+, 0.40 M Cl-,

(iv) If any precipitate exists, what is its identity and how many grams would be present? 2.8 grams of PbCl2

Ch. 4 - Reactions/Stoichiometry

1. (a) x 100 = x 100 = 15.9%

(b) 3.21 g x x x = 2.06 g

(ii) 17.80 mL x = 2.67x10–4 mol MnO4-

2.67x10–4 mol MnO4- x = 6.68x10–4 mol C2O42-

(iii) 100. mL x = 3.34x10–3 mol C2O42-

(iv) 3.34x10–3 mol C2O42- x x = 0.324 g BeC2O4(s)

x 100 = 93.9%

Kinetics:

3. (a) A = abc; 0.600 = (5000 cm–1M–1)(1.00 cm)(c) c = 1.20x10–4 M

(b) ln[X]t – ln[X]0 = –kt

ln(4.00 x 10–5) – ln(1.20 x 10–4) = –k(35 min) k = 0.0314 min–1

ln[1.50 x 10–5] – ln[1.20 x 10–4] = –0.0314 min–1t t = 66.2 min.

(d) t1/2 = = = 22.1 min

(e) (i)

(ii) = slope of the line, multiply the slope by –R to obtain Ea

4. (a) (i) 1st order with respect to Br–; in experiments 1 and 2, a doubling of the [Br–] results in the doubling of the initial rate, and indication of 1st order

(ii) 1st order with respect to BrO3– using expt. 1 & 3

rate1 = k[Br–]1[BrO3–]m[H+]n

= k[Br–]1[H+]n

rate3 = k[Br–]1[BrO3–]m[H+]n

= k[Br–]1[H+]n

m = 1

(iii) 2nd order with respect to H+

using expt. 3 & 4

rate3 = k[Br–]1[BrO3–]1[H+]n

= k[Br–]1

rate4 = k[Br–]1[BrO3–]1[H+]n

= k[Br–]1

n = 2

(b) rate = k[Br–]1 [BrO3–]1 [H+]2

k = 5000 mol–3L3s-1

Equilibrium

5. (a) PNH3 does not change. Since NH4Cl(s) has constant concentration (a = 1), equilibrium does not shift.

(b) PNH3 increases. Since the reaction is endothermic, increasing the temperature shifts the equilibrium to the right and more NH3 is present.

(c) PNH3 does not change. As V increases, some solid NH4Cl decomposes to produce more NH3. But as the volume increases, PNH3 remains constant due to the additional decomposition.

(d) PNH3 decreases. Some NH3 reacts with the added HCl to relieve the stress from the HCl addition.

(e) PNH3 increases. Some of the added NH3 reacts with HCl to relieve the stress, but only a part of the added NH3 reacts, so PNH3 increases.

6. (a) CO = f(0.55 mol/ 1.6 mol) = 0.34

(b) Kc = ([H2O][CO])/([H2][CO2]) = (0.55´0.55)/(0.20´0.30) = 5.04

(d) [CO] = 0.55 - 30.0% = 0.55 - 0.165 = 0.385 M

[H2O] = 0.55 - 0.165 = 0.385 M

[H2] = 0.20 + 0.165 = 0.365 M

[CO2] = 0.30 + 0.165 = 0.465 M

K = (0.385)2/(0.365´0.465) = 0.87

(e) let X = D[H2] to reach equilibrium

[H2] = 0.50 mol/3.0L - X = 0.167 - X

[CO2] = 0.50 mol/3.0L - X = 0.167 - X

[CO] = +X ; [H2O] = +X

K = X2/(0.167 - X)2 = 5.04 ; X = [CO] = 0.12 M

Thermodynamics

7. -1323kJ per 2 moles of water vapor are released. If those two moles of water vapor are changed to liquid water, -88 kJ would also be released (2 x -44 kJ/mol). A total of -1411 kJ would be released.

8. (a) x would be endothermic; Y would be exothermic; Z would be exothermic.

(b) delta H is positive for X and negative for Y and Z

(d) X has the highest activation energy, therefore it would benefit the most from a catalyst. A catalyst can change the activation energy, but not the delta H of the reaction.

(e) TS = transition state which is the molecule produced when all of the reactants are combined into an unstable structure. Eact is the activation energy. This is the energy needed to activate the reaction.

9. (a) ∆H°f (C7H16) = -191 kJ/mol

(b) qreleased = 0.0108 mol C7H16(-4850 kJ/mol C7H16) = 52.4 kJ of heat released.