2012FRQANSDRAFT

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Adrian Dingles Chemistry Pages 2004, 2005, 2006, 2007, 2008, 2009, 2010, 2011, 2012. All rights reserved.These materials may NOT be copied or redistributed in any way, except for individual class instruction.

Revised May 2012

Page 1 of65/10/1210:48AM

2012 AP Chemistry FRQ Answers DRAFT

1. Guesstimate Points: 2, 1, 1, 2, 2, 2

(a) The pH at the equivalence point is above 7, suggesting that the salt formed in the

neutralization reaction with a strong base (NaOH) is a basic one. This means that the salt will

undergo a hydrolysis reaction with water that yields hydroxide ions (since it is the conjugate base

of a weak acid), making the pH at the equivalence point basic and according to the equation;

A-+ H2O HA + OH

-.

(b) HA + OH- H2O + A

-

(c) At equivalence point, moles of NaOH =40.0 mL 1.00 L 0.250 mol

1000 mL 1.00 L

= 0.0100 mols

Ratio (from (b)) is 1:1, so moles of HA also = 0.0100 moles

(d) Molar mass =mass in g 1.22 g

mols 0.0100 mols= = 122 g mol

-1

(e)[ ] [ ]

2+ - +

3 3-5 H O A H OKa = 6.3 x 10 = =HA 0.200

(assuming dissociation of HA to be negligible)

[H3O+] = 3.55 x 10

-3,pH = - log (3.55 x 10

-3) = 2.45

(f) Moles of NaOH added = (0.030 L)(0.250 mol/L) = .00750 mols (which is less than 0.0100

moles of acid originally present, so the acid is in excess, the NaOH is limiting, and therefore a

buffer has been created in the flask. Since NaOH is limiting, and the ratio is 1:1, the moles of salt

produced = moles of NaOH = 0.00750 mols

For an acidic buffer, pH = pKa + log ([salt]/[acid])

pH = -log (6.3 x 10-5

) + log ((0.00750/0.08))/((0.0025/0.08)) = 4.68

pH = 4.68 = - log [H3O+]

[H3O+] = 10

-4.68= 2.1 x 10

-5


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(a) P V = n R T

(0.200 atm) (1.00 L) = n (0.0821 L atm mol-1

K-1

) (273 + 127 K)

n = 0.00609 moles

(b) (1.40 atm) (1.00 L) = nTOTAL (0.0821 L atm mol-1

K-1

) (273 + 127 K)

nTOTAL = 0.0426 moles

Moles of O2 = 0.0426 0.00609 = 0.0365

(P) (1.00 L) = (0.0365 mol) (0.0821 L atm mol-1

K-1

) (273 + 127 K)

P = 1.20 atm

(c) Ratio of CO2 to H2O is 0.600:0.800 or 3:4, meaning 3 C atoms to every 8 H atoms,

suggesting; C3H8 + 5O2 3CO2 + 4H2O

0.00609 moles of hydrocarbon would need 5 times as much O 2, i.e., 0.0305 moles of O2.

(P) (1.00 L) = (0.0305 mol) (0.0821 L atm mol-1

K-1

) (273 + 127 K)

P = 1. 00 atm

(d) C3H8 + 5O2 3CO2 + 4H2O

(e) (0.00609 mols) (44.0962 g mol-1

) = 0.269 g

(f) Less than 7 as CO2 will dissolve in water to produce carbonic acid according to;

H2O + CO2 H2CO3


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(a) S = Sproducts Sreactants = (219.3 + 192.8) (284.9) = 127.2 J/(molrxn K)

(b) Net bonds broken (endothermic)

1 C-H = 1(413) = 413

1 C-C = 1 (348) = 348

1 C-N = 1(293) = 293

Total = +1054 kJ

Net bonds made (exothermic)

1 C=C = 1(614) = -614

1 N-H = 1(391) = -391

Total = -1005 kJ

H = sum of bonds broken (endo) and bonds made (exo) = +49 kJ/mol rxn

(c) Endothermic reaction so temperature of surroundings goes down.

(d) ln (3.60 x 10-4

) - ln (4.70 x 10-3

/ 2.00 L) = -k (20.0 min)

k = 0.0938 min-1

(e) Rate = k [CH3CH2NH2]

Rate = 0.0938 min-1

(4.70 x 10-3

mol / 2.00 L) = 2.2 x 10-4

mol L-1

min-1

(f) Curve. A straight line would result if the reciprocal of concentration is plotted against time AND

if the reaction is second order. This is a first order reaction, so no straight line.


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4. Guesstimate Points: 5, 5, 5

(a)(i) SrCO3 + 2H+ Sr

2++ CO2 + H2O

(ii) Bubbles of CO2 being liberated orthe solid carbonate dissolving

(b)(i) 2Mg + O2 2MgO

(ii) 0, +2

(c)(i) Ni2+

+ 2OH- Ni(OH)2

(ii) Cl-


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5. Guesstimate Points: 2, 2, 2, 2, 2

(a) More ordered, stronger, intermolecular London dispersion forces in solid iodine (when

compared to liquid bromine), require a greater amount of energy in order to be broken (and

separate the I2 molecules), and as a result there is a greater endothermic energy for the

formation of I2(g) when compared to Br2(g) from the respective elements.

(b) The formation of I2(g) from the solid would have the greater increase in entropy. Both

substances end up very disordered gases, but the change from a (relatively) very ordered solid to

gas, as opposed to from a liquid to a gas, requires a larger increase in entropy.

(c) IBr. IBr is both a molecule with more electrons than Br2 (IBr is more polarizable with greater

London dispersion forces), and a polar molecule (because of differences in electronegativity

between Br and I that is not present between Br and Br in Br2), so it will have larger

intermolecular forces and hence require more energy to become a vapor.

(d) I2 is a non-polar molecule and therefore will not dissolve in polar water. It will however be

soluble in a non-polar solvent, hexane. The purple color observed in the hexane layer is dissolved

I2. Additionally, the hydrogen bonding between water molecules is much stronger than the

(potential) IMF between I2 and H2O so iodine will tend to not dissolve in water, but the (potential)

IMF between I2 and hexane IS comparable to the LDFs between molecules of hexane so the I 2

will dissolve in hexane.

(e)

(i) I3-is linear with 3LP and 2BP based upon an electron geometry of trigonal bipyramid but with

an atom geometry that is linear.

(N.B. I would add 3 lone pairs to each of the terminal I atoms to complete the Lewis structure)

(ii) Higher concentration of I3-in the water layer. I3

-will dissolve in the water layer, since it is

charged (ionic) and can form ion to dipole bonds with polar H2O that it will not do with non-polar

hexane.


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6. Guesstimate Points: 1, 1, 2, 2, 3

(a) Least: Q, X, Pb : Most

(b) Pb Pb2+

+ 2e-

(c) (i) 0.47 V = X (- 0.13 V)

X = + 0.34 V

(ii) Cu

(d) The lead electrode loses mass as the metal dissolves and turns into soluble ions. The copper

electrode gains mass as ions are reduced and copper metal is deposited.

(e)(i) 0 V. The circuit is broken, no ions can flow through the salt bridge.

(ii) Greater than 0.47 V. The concentration of Pb2+

is reduced by the formation of PbSO4,

according to;

Pb2+

(aq) + SO42-

(aq) PbSO4(s)

According to the Nernst equation,

Q =

2+

2+

Pb

Cu

and as a result Q becomes less than 1. The log of a number less than 1 is negative.

Ecell = 0.47 -

2

0.0592(negative number) = number greater than 0.47 V

(iii) Less than 0.47 V, since the cell will have run down overnight and the potential difference

between the two half-cells will be less.

(NORMALLY, opening the switch would break the circuit and the voltage would become zero,

but because of the way the circuit is set up it is possible to have the switch open and still have the

cell running VERY, very odd).

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2012FRQANSDRAFT