GENERAL CHEMISTRY 2
THIRD HOUR EXAM
JUNE 29, 2010
Name ______
Panthersoft ID ______
Signature ______
Part 1 ______(28 points)
Part 2 ______(20 points)
Part 3 ______(52 points)
TOTAL ______(100 points)
NA = 6.022 x 1023 °C = (5/9) (°F - 32) °F = (9/5)(°C) + 32
1 amu = 1.661 x 10-27 kg °C = K - 273.15 K = °C + 273.15
1 atm = 760 torr = 760 mm Hg 1 atm = 1.013 bar pV = nRT
R = 0.08206 L.atm/mol.K 1 L.atm = 101.3 J
R = 8.314 J/mol.K 1 J= 1 kg.m2/s2
If ax2 + bx + c = 0, then x = ( - b ± [b2 - 4ac]1/2 )/2a KP = Kc (RT)Dn
Ka.Kb = Kw = 1.0 x 10-14 (at T = 25°C) pH = pKa + log10{[base]/[acid]}
H = E + pV G = H - TS
DGrxn = DG°rxn + RT ln Q ln K = - DG°rxn/RT
Do all of the following problems. Show your work.
Part 1. Multiple choice. Circle the letter corresponding to the correct answer. There is one and only one correct answer per problem. [4 points each]
1) Which of the following neutralization reactions is expected to go essentially to completion?
a) Reaction of a strong acid with a strong base
b) Reaction of a strong acid with a weak base
E c) Reaction of a weak acid with a strong base
d) Both b and c
e) All of the above
2) Consider the titration of a weak base with a strong acid. At the equivalence point the pH
a) Is expected to be smaller than 7.0
b) Is expected to be equal to 7.0
A c) Is expected to be larger than 7.0
d) Can be smaller than, equal to, or larger than 7.0
e) Is not defined, since complete reaction has occurred
3) Which of the following solutions will be a buffer solution?
a) A solution where 0.100 mol of acetic acid (CH3COOH) and 0.100 mol of sodium acetate (NaCH3COO) are added to 1.000 L of water.
b) A solution where 0.100 mol of acetic acid (CH3COOH) and 0.050 mol of sodium hydroxide (NaOH) are added to 1.000 L of water.
D c) A solution where 0.100 mol of hydrochloric acid (HCl) and 0.050 mol of sodium hydroxide (NaOH) are added to 1.000 L of water.
d) Both a and b
c) Both b and c
4) Which of the following substances is expected to have the largest value for entropy for standard conditions?
a) CH3OH(g)
b) C2H5OH(g)
B c) CH3OH()
d) C2H5OH()
e) All of the above will have the same value for entropy for standard conditions.
5) For which of the following substances is DG°f = 0.00 kJ/mol at T = 25.0 °C?
a) Cu(s)
b) N2(g)
D c) NO2(g)
d) Both a and b
e) None of the above
6) For which of the following substances is S° = 0.00 J/mol.K at T = 25.0 °C?
a) Cu(s)
b) N2(g)
E c) NO2(g)
d) Both a and b
e) None of the above
7) For a particular chemical reaction it is found that DH°rxn > 0 and DS°rxn < 0. For cases where the reactants and products are at standard concentration
a) The reaction is never spontaneous
b) The reaction is always spontaneous
A c) The reaction is spontaneous at low temperatures but not at high temperatures
d) The reaction is spontaneous at high temperatures but not at low temperatures
e) The system is always at equilibrium
Part 2. Short answer.
1) What is the difference, if any, between the end point and the equivalence point for a titration? [6 points]
The equivalence point in an acid/base titration is the point where sufficient titrant (the solution being added) is present to otain complete reaction between the acid and base, with no excess of either present. The end point of the titration is the point where the color change in the indicator is observed. One chooses the indicator to try and make the end point as close as possible to the equivalence point.
2) Is it possible to carry out a process where DSsyst, the entropy change for the system, is negative (yes/no, and a brief justification of your answer)? [6 points]
Yes. Since the second law states DSuniv = DSsyst + DSsurr ³ 0, it follows that DSsyst < 0 is possible, as long as DSsurr > 0 and is large enough to make sure that the sum, DSuniv, is greater than or equal to zero.
3) A saturated solution is formed by adding excess solid manganese II hydoxide (Mn(OH)2) to water, at T = 25.0 °C. The pH of the saturated solution is pH = 9.53. Based on this, find the value for Ksp for manganese II hydroxide at this temperature. [8 points]
Mn(OH)2(s) D Mn2+(aq) + 2 OH-(aq) Ksp = [Mn2+] [OH-]2
We may find [OH-] from the pH. pOH = 14.00 – pH = 14.00 – 9.53 = 4.47
So [OH-] = 10-pH = 3.4 x 10-5 M
Based on the stoichiometry of the reaction, [Mn2+] = ½ [OH-] = 1.7 x 10-5 M
So Ksp = (1.7 x 10-5) (3.4 x 10-5)2 = 1.9 x 10-14
Part 3. Problems. Note: You may assume T = 25.0 °C in all of the problems below.
1) Consider the following two solutions
Solution A – A 0.1000 mol/L solution of potassium hydroxide (KOH), a strong soluble base.
Solution B – A solution of hypochlorous acid (HOCl, Ka = 3.0 x 10-8) of unknown concentration.
A 20.00 mL sample of the hypochlorous acid solution is titrated with the potassium hydroxide solution. After 32.05 mL of the potassium hydroxide solution has been added the equivalence point of the titration is reached.
a) What is the concentration of hypochlorous acid in solution B? [8 points]
The neutralization reaction is HOCl(aq) + KOH(aq) ® K+(aq) + OCl-(aq) + H2O
At the equivalence point
mol KOH = 0.03205 L 0.1000 mol = 0.003205 mol KOH
1 L
But based on the stoichiometry of the neutralization reaction, at this point mol KOH = mol HOCl.
Therefore [HOCl] = 0.003205 mol = 0.1602 M
0.0200 L
b) What is the pH for the titration at the point where 20.00 mL of the potassium hydroxide solution has been added to the 20.00 mL sample of the hypochlorous acid solution? [8 points]
Since at this point in the titration we have a buffer we can use the Henderson equation
pH = pKa + log10 [base]/[acid]
Initially we have 0.003205 mol HOCl. After the addition of 20.00 mL of KOH solution, the moles of added KOH is
mol KOH added = 0.0200 L 0.100 mol = 0.00200 mol KOH. Note at this point KOH is the limiting reactant
1 L
After the neutralization reaction we have
mol OCl- formed = mol KOH added = 0.00200 mol mol HOCl remaining = 0.003205 – 0.00200 = 0.001205 mol
So pH = - log10(3.0 x 10-8) + log10 (0.00200/0.001205) = 7.52 + 0.22 = 7.74
2) The solubility product for lead II bromide, PbBr2, is Ksp = 2.1 x 10-6.
a) What is the molar solubility of lead II bromide in pure water? [8 points]
The solubility reaction is
Pb(Br)2 D Pb2+(aq) + 2 Br-(aq) Ksp = [Pb2+] [Br-]
Initial Change Equilibrium
Pb2+ 0 x x
Br- 0 2x 2x
So (x) (2x)2 = 4x3 = 2.1 x 10-6 x = [ (2.1 x 10-6)/4 ]1/3 = 8.1 x 10-3 M
Based on the stoichiometryof the reaction this is the molar solubility.
b) What is the molar solubility of lead II bromide in a 0.0400 mol/L solution of lead II nitrate, Pb(NO3)2, a soluble ionic compound. [8 points]
Everything is the same as in part a except that there is now an initial concentration of Pb2+ from the Pb(NO3)2.
Initial Change Equilibrium
Pb2+ 0.0400 x 0.0400 + x
Br- 0 2x 2x
(0.0400 + x) (2x)2 = 2.1 x 10-6 If we assume x < 0.0400, then
(0.0400) (4x2) = 2.1 x 10-6 x = [ (2.1 x 10-6)/(0.0400)(4) ]1/2 = 3.6 x 10-3 M
Our assumption that x < 0.0400 was correct, so the above represents the molar solubility.
3) The gas phase reaction of sulfur dioxide (SO2) with molecular oxygen (O2) can produce sulfur trioxide (SO3). The balanced equation for the reaction is
2 SO2(g) + O2(g) D 2 SO3(g)
Thermodynamic data for the reactants and products are given below
Substance DH°f (kJ/mol) DG°f (kJ/mol) S° (J/mol.K)
O2(g) 0.0 0.0 205.2
SO2(g) - 296.8 - 300.1 248.2
SO3(g) - 395.7 - 371.1 256.8
Based on the above information find the following.
a) DSsyst and DSsurr [8 points]
DSsyst = DSrxn = [ 2 S°(SO3(g))] – [ 2 S°(SO2(g)) + S°(O2(g)) ] = [ 2 (256.8) ] – [ 2 (248.2) + 205.2 ] = - 188.0 J/mol.K
DSsurr = - DH°rxn/T
DH°rxn/= [ 2 DH°f(SO3(g))] – [ 2 DH°f(SO2(g))] = [ 2 ( - 395.7) ] – [ 2 ( - 296.8) ] = - 197.8 kJ/mol
DSsurr = (197800 J/mol) = 663. J/mol.K
(298.2 K)
b) DG°rxn [4 points]
DG°rxn/= [ 2 DG°f(SO3(g))] – [ 2 DG°f(SO2(g))] = [ 2 ( - 371.1) ] – [ 2 ( - 300.1) ] = - 142.0 kJ/mol
c) K (the equilibrium constant for the reaction) [6 points]
ln K = - DG°rxnRT = 142000. J/mol = 57.27
(8.314 J/mol.K) (298.2 K)
K = e57.27 = 7.5 x 1024
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