Name:__________________________
Chem 202
Section: _____________
Quiz 8
10/31/2008
1) The Haber process is of great commercial value because ammonia is an important fertilizer for the
growth of corn and other crops. It can be represented by the following reaction:
N2 (g) + 3 H2 (g)
2 NH3 (g)
Suppose that, at equilibrium, the concentrations are [H2] = 5.0 M, [N2] = 8.0 M, and [NH3] = 4.0
M.
a.) Calculate Kp for the Haber process (include units) at 37˚C.(1.5)
b.) Determine the concentrations of each reactant and product that were reacted initially. (1.5)
Answer: a.) Keq = [NH3]2
[H2]3[N2]
At the concentrations given, Keq will equal 0.016 M-2. However, the question asks for Kp.
Kp = K(RT)!n where !n = 2 mol – 4 mol = -2 mol. T = 310 K. R = 0.0821 L atm/mol K.
Thus, Kp = 2.5 x 10-5 atm-2
b.)
Thus, [NH3] = 0 M, [H2] = 11 M, and [N2] = 10 M at initial conditions.
2) Oxalic acid (H2C2O4) can be used to standardize a solution of permanganate (MnO4-). If 28.97 mL of
permanganate is required to react completely with 0.001175 mol oxalic acid, determine the molarity of the
permanganate.(3)
The unbalanced redox reaction is:
MnO4- (aq) + H2C2O4 (aq)
Mn2+ (aq) +
CO2 (g)
Thus, the molarity of the permanganate is 1.622 x 10-2 M.
3) Bismuth sulfide in aqueous solution is shown in the following reaction:
a) Calculate the Ksp for bismuth sulfide if when the system reaches equilibrium there are 1.026x10-14 mol
Bi3+ and 1.539x10-14 mol S2- in a 3.00 L container. (1)
b) Calculate the solubility of bismuth sulfide in H2O. (1)
c) Calculate the solubility of bismuth sulfide in 0.2 M solution of Na2S. (1)
d) Is the solubility greater or worse when mix in a solution of Na2S? Why? (1)
7.02e-36 M
Name:__________________________
Chem 202
Section: _____________
Quiz 8
10/31/2008
1) Calculate the solubility of the following solid – Ca3(PO4)2, Ksp = 1.3 x 10-32.
a) The solid is added to water (1.5)
b) The solid is added to a .01 M Na3PO4 solution. (1.5)
a)
Ca3(PO4)2 (s) !" 3Ca2+(aq) + 2PO43- (aq)
Ksp = [Ca2+]3[PO43-]2
Ca2+
PO43I
0
0
C
+3x
+2x
E
3x
2x
-32
3
2
5
1.3 x 10 = [3x] [2x] = 108x
x = 1.65 * 10-7
1.65 * 10-7 mol of Ca3(PO4)2 are dissolve in every liter
b)
Ca2+
PO43I
0
.01
C
+3x
+2x
E
3x
.01 + 2x
1.3 x 10-32 = [3x]3[.01+2x]2
.01 >>> 1.65 * 10-7 therefore you can assume [.01+2x] = [.01]
1.3 x 10-32 = [3x]3[.01]2
x = 1.689 * 10-10
1.689 * 10-10 mol of Ca3(PO4)2 are dissolved in every liter of .01 M Na3PO4 solution
2) Alkaline batteries (like those in your calculator) function by using electrochemistry
and generate a voltage. Alkaline batteries reactions occur in basic environments (i.e. free
OH-) The following is the formula for a typical reaction in an alkaline battery:
Zn(s) + MnO2(s) " Mn2O3(s) + ZnO(s) (unbalanced)
a) Write the half reactions for the reaction (3)
b) Label which half reaction is for the oxidizing agent and which is the reducing
agent (1)
Answer:
a) Zn (s) + 2OH! (aq) " ZnO (s) + H2O (l) + 2e!
2MnO2 (s) + H2O (l) + 2e! "Mn2O3 (s) + 2OH! (aq)
b) Zn is the reducing agent and MnO2 is the oxidizing agent
3) Identify the formal charge of all atoms in the following (0.25 pt each):
a. XeOF4
Xe:
O:
F:
b. NaBiO3
Na:
Bi:
O:
c. Mg2P2O7
Mg:
P:
O:
d. Na2S2O3
Na:
S
O:
Answer:
a.
b.
c.
d.
+6; -2; -1
+1; +5; -2
+2; +5; -2
+1; +2; -2
Name:__________________________
Chem 202
Section: _____________
Quiz 8
10/31/2008
1) You and your friends go to the Green St. bars one night (never telling your Chem TA,
of course!) and you get pulled aside by a cop. He asks one of your friends to undergo a
breathalyzer test. Since your friend had one too many (because you would never do that),
the orange testing solution turns lime green indicating a positive test. This piques your
chemical curiosity.
Since your friend tested positive, the cop gives you a choice. If you can tell him
the sum of the coefficients of the balanced redox reaction taking place in the breathalyzer
machine (the unbalanced reaction is given below), your friend will go free. Otherwise,
it’s a drinking ticket for you and your friend. What is the sum of the coefficients for the
breathalyzer reaction?(4)
H+ (aq) + Cr2O72- (aq) + C2H5OH (g) CO2 (g) + Cr3+ (aq) + H2O (l)
1) Half Rxn 1: 2[6e- + 14H+ (aq) + Cr2O72- (aq) ! 2Cr3+ (aq) + 7H2O (l)]
Half Rxn 2: 3H2O (l) + C2H5OH (g) ! 2CO2 (g) + 12H+ (aq) + 12eCombined Rxn: 12e- + 28H+ (aq) + 2Cr2O72- (aq) + 3H2O (l) + C2H5OH (g)
! 2CO2 (g) + 12H+ (aq) + 12e- + 4Cr3+ (aq) + 14H2O (l)
Final Rxn: 16H+ (aq) + 2Cr2O72- (aq) + C2H5OH (aq) ! 2CO2 (g) + 4Cr3+
(aq) + 11H2O (l)
Answer: 16 + 2 + 1 + 2 + 4 + 11 = 36.
2) Given a sample of silver hydroxide (AgOH), with a Ksp of 2.0 x 10-8 calculate its
solubility in:
a) Pure Water(1)
b) 0.1 M NaOH(1)
c) 0.1 M NH4Cl (NH4+ + OH- NH3 + H2O; Ksp = 5.6 x 104)(1)
Answer: on page 7
3) SrO (s) ! Sr(l) + ! O2 (g) K1 = 3.0 x 10 -2
Sr2O2 (s) ! 2Sr(l) + O2 (g) K2 = 1.2 x 10 -13
Sr2O(g) ! 2Sr(l) + ! O2 (g) K3= 5.2 x 10 -41
SrO2(l) ! 2Sr(l) + O2 (g)
K4= 2.2 x 10 -22
For the following reactions, determine the equilibrium constant:
a. SrO2(l) !SrO(s)+ ! O2(g) + Sr(l)(1.5)
Answer: K4(1/K1) = 7.3 x 10-21
b. 4Sr2O2(s) ! Sr2O(g) + 3SrO2(l) + ! O2(g)(1.5)
Answer: (K2)4 (1/K3) (1/K4)3 = 3.8 x 1053