~~~~~~~~~~~~~~~~~~~~~~~~~~~~CHEM202~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Quiz 9.1
20min
11/02/2007
1. Na2O (s)
NaO (g)
Na2O2 (s)
NaO2 (s)
2Na (l) + 1/2O2 (g) K1 = 2 × 10-25
Na (l) + 1/2O2 (g)
K2 = 2 × 10-5
2Na (l) + O2 (g)
K3 = 5 × 10-29
Na (l) + O2 (g)
K4 = 3 × 10-14
Determine the equilibrium constant for the following reactions using the data above.
You may not need some equations.
a. NaO (g) + Na2O (s)
Na2O2 (s) + Na (l)
NaO (g)
Na (l) + 1/2O2 (g)
K2 = 2 × 10-5
Na2O (s)
2Na (l) + 1/2O2 (g) K1 = 2 × 10-25
2Na (l) + O2 (g)
Na2O2 (s)
1/K3 = 1/5 × 10-29
K=K2K1(1/K3) = 8 × 10-2 L/mol
b) 2NaO(g)
Na2O2 (s)
2(NaO (g)
Na (l) + 1/2O2 (g))
K22 = (2 × 10-5)2
2Na (l) + O2 (g)
Na2O2 (s)
1/K3 = 1/5 × 10-29
2
18
K=K2 (1/K3)= 8 × 10 L/mol
2.. H2C2O4 (aq) + CrO42- (aq)  CO2 (g) + Cr3+ (aq)
Carbon is +3 Carbons is +4
2CO2 + 2e- + 2H+
H2C2O4 
CrO42-

Cr is +6
CrO42- + 3e- + 8H+ 
Cr3+
Cr is +3
Cr3+ + 4H2O
Overall Equation:
3H2C2O4 (aq) + 2CrO42-(aq) + 10H+ (aq)  6CO2 (g) + 2Cr3+ (aq) + 8H2O
Since in the balanced equation the ratio of oxalic acid to CrO42- is 3:2, if the solution used
up 50.0 mL of 0.0350 M it used up:
50
L × 0.0350 = .00175moles of CrO421000
Which means there was
3
× 0.00175 = .002625moles of oxalic acid in 10 mL
2
Which makes for 0.2625 M Oxalic Acid.
3
Indicate the direction in which the equilibrium shifts for the following reactions if
the reactions were in equilibrium when the change was affected.
PCl3 + Cl2 <===> PCl5 + heat
a. increase Cl2 concentration (Right)
b. decrease pressure ( Left)
N2O4 + heat <===> 2 NO2
c.. decrease pressure (Right)
d. remove N2O4 (Left)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~CHEM202~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Quiz 9.2
20min
11/02/2007
1. NO and Br2 react by the following reaction:
2NO (g) + Br2 (g)
2NOBr (g)
A container contains NO with partial pressure of 98.4 atm and Br2 with partial
pressure of 41.3 atm. At equilibrium the total pressure of the gasses is 110.5 atm.
Calculate the value of Kp.
2NO
98.4
-2x
98.4-2x
I
C
E
Br2
41.3
-x
41.3-x
Ptotal = (98.4-2x)+(41.3+x)+2x = 139.7-x = 110.5 atm
X = 29.2 atm
PNO = 40 atm PBr2 = 12.1 atm PNOBr = 58.4 atm
2
PNOBr
(58.4) 2
Kp = 2
=
= 0.176
PNO PBr 2 (40) 2 (12.1)
2. You have the following reaction in equilibrium:
A4B2
↔
A2B2 +
A2
2NOBr
0
+2x
2x
You know that initially you have A4B2 at the partial pressure of 1.500 atm. After it goes
to equilibrium you remove all the A2 from the reaction flask. You measure that you had
the partial pressure of A2 equal to 0.21 atm. After letting the reaction to equilibrium
again, what is the concentration of A2B2 and A2? (Hint: Calculate the Kp and then
proceed)
Solution:
A4B2
1.500 atm
-x
1.5 –x
↔
A2B2 +
0
+x
x
A2
0
+x
0.21
So x = 0.21, [A4B2] =1.29
Kp = x2/(1.5-x) = (0.21)2/1.29 = 0.0341
Thus we resolve an ICE problem and say
1.29
-x
1.29-x
0.21
+x
0.21+x
0
+x
x
Calculate x from Kp = (0.21 +x)*x/(1.29-x) = 0.0341.
Calculating the roots of x, neglecting the unreal solution of x, we get x =0.121
The concentration of A2B2 = 0.21 +0.121 = 0.331 [A2] = 0.121
3. Indicate the direction in which the equilibrium shifts for the following reactions if the
reactions were in equilibrium when the change was affected. (2 pts)
N2 + 3 H2 <===> 2 NH3 + heat
a. remove NH3 gas (right)
b. decrease pressure (left)
c. add N2 gas (right)
d. increase temperature (left)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~CHEM202~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Quiz 9.3
20min
11/02/2007
1. You are running the following reaction at STP:
Cr2O72- (aq) + Cl- (aq)  Cr3+ (aq) + Cl2 (g)
Solution:
Cr2O72- (aq)
 Cr3+ (aq)
Cr is +6
Cr is +3
Thus:
Cr2O72- (aq) + 6e- + 14H+
Cl- (aq)
Cl is –1

 2Cr3+ (aq) + 7H2O
Cl2 (g) + 2eCl is 0
Overall:
Cr2O72- (aq) + 6Cl- (aq) 14H+  2Cr3+ (aq) + 3Cl2 (g) + 7H2O
By PV=nRT at STP we know that
n=
PV
(1)(1)
=
= .0409moles
RT (0.08206)(298)
Thus since Cl2 gas is in a 3:1 ratio with Cr2O72- : we had .01367 moles of Cr2O72Converting this to molarity gives .01367 M.
2. For the gas phase reaction
H2(g) + I2(g) <===>2 HI(g)
Kc = 50.3 at 731 K. Equal amounts (0.100 M each) is introduced to a container, and
then the temperature is raised to 731 K.
Calculate the concentration of each when the system is at equilibrium.
Solution
We assume x M of H2 or I2 to react to give 2x M of HI.
H2(g) + I2(g) = 2 HI(g)
0.100
0.100
0.100
initial concentration
-x
-x
2x
amounts change
0.1-x
0.1-x
0.1+2x equilibrium concentration
By definition of equilibrium constant, we have
Kc
[HI]2
(0.1+2x)2
= ---------- = -------------- = 50.3
[H2] [I2]
(0.1-x)(0.1-x)
Expanding the above equation gives,
46.3 x2 - 10.43x + 0.493 = 0
Solution of the quadratic equation gives x = 0.067 and x' = 0.158. When x' = 0.158, 0.1 0.158 = -0.058. Since concentration should not be a negative value, this solution does not
agree with reality. The only acceptable solution is x = 0.067, and thus we have:
[H2] = [I2] = 0.100 - 0.067 = 0.033 M
[HI] = 2x = 0.100 + 0.134 = 0.234 M
3. Indicate the direction in which the equilibrium shifts for the following reactions if the
reactions were in equilibrium when the change was affected.
H2 + Cl2 <===> 2 HCl + heat
a.. increase H2 concentration (Right)
b.. increase pressure (No Change )
N2 + O2 + heat <===> 2 NO
c.. decrease O2 concentration ( Left)
d.. add catalyst (No change )