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Quiz 3
The equilibrium concentrations at 35˚C for the reaction;
2NO2(g) 2NO(g) + O2(g), are [NO] = 0.52M; [O2] = 0.24M; [NO2] =0.18M.
What is the value of Kp at this temperature?
Kp=Kc(RT)Δn
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CHEM 1412 Exam #1(SAMPLE)
Name:___________________________
(chapters 12,13, and 14)
Score:
PART I - ( 3 points each) - Please write your correct answer next to each question number. DO NOT CIRCLE
____1. In which colligative property(ies) does the value decrease as more solute is added?
a. boiling point
c. vapor pressure
b. freezing point
d. freezing point and vapor pressure
____2. What is the molarity of a solution prepared by dissolving 25.2 g CaCO3 in 600. mL
of a water solution?
a. 0.420 M
b. 0.567 M
c. 0.042 M
d. 0.325 M
____3. The solubility of nitrogen gas in water at a nitrogen pressure of 1.0 atm is
6.9 x 10-4 M. What is the solubility of nitrogen in water at a nitrogen pressure of
0.80 atm?
a. 5.5 x 10-4
b. 8.6 x 10-4
c. 3.7 x 10-3
d. 1.2 x 103
____4. What is the freezing point of an aqueous glucose solution that has 25.0 g of glucose,
C6H12O6, per 100.0 g H2O (Kf = 1.86°C /m)?
a. 0.258
b. -0.258
c. 2.58
d. -2.58
____5. What is the osmotic pressure in atm produced by a 1.20 M glucose (C6H12O6)
solution at 25°C?
a. 29.3
b. 4.89
c. 25.1
d. 36.0
____6. The vapor pressure of pure ethanol at 60°C is 349 mm Hg. Calculate vapor pressure
in mm Hg at 60°C for a solution prepared by dissolving 10.0 mol naphthalene
(nonvolatile) in 90.0 mol ethanol?
a.
600
b. 314
c. 34.9
d. 69.8
____7. Which statement is not correct regarding the function of a catalyst?
a. it lowers the activation energy
c. it changes the rate constant of a reaction
b. it affects the rate of a chemical reaction
d. all of these are correct
____8. For first-order reactions the rate constant, k, has the units
a. M s-1
b. M-1 s-1
c. M-2 s-1
d. s-1
____ 9. For second-order reactions the slope of a plot of 1/[A] versus time is
2
3
a. k
b. k/[A]0
c. kt
d. -k
____10. If the reaction 2A + 3D  products is first-order in A and second- order in D,
then the rate law will have the form rate =
a. k[A]2[D]3
b. k[A][D]
c. k[A]2[D]2
d. k[A][D]2
____11. In the first-order reaction A  products, the initial concentration of A is 1.56 M and the concentration
is 0.869 M after 48.0 min. What is the value of the rate constant, k, in min-1?
a. 3.84 x 10-2
b. 2.92 x 10-2
c. 5.68 x 10-2
d. 1.22 x 10-2
____12. The following time and concentration data was obtained for the reaction; 2A  products
Time(min)
[A] M
0
1.20
1.1
1.00
2.3
0.80
4.0
0.60
Refer to the table above. If the reaction is known to be first-order, determine the rate constant for the
reaction.
a. 0.17
b. 0.37
c. 0.49
d. 0.60
____13. Consider the reaction 2HI(g)  H2(g) + I2(g). What is the value of the equilibrium constant, Kc, if at
equilibrium , [H2] = 6.50 x 10-7 M, [I2] = 1.06 x 10-5 M, and [HI] = 1.87 x 10-5 M?
a. 3.68 x 10-7
b. 1.97 x 10-2
c. 1.29 x 10-16
d. 50.8
____ 14. For the elementary reaction NO3 + CO  NO2 + CO2
a.
b.
c.
d.
the molecularity is 2 and rate = k[NO3][CO]/[NO2][CO2]
the molecularity is 4 and rate = k[NO3][CO]/[NO3][CO]
the molecularity is 4 and rate = k[NO3][CO][NO2][CO2]
the molecularity is 2 and rate = k[NO3][CO]
____15. Given the following mechanism, determine which of the species below is a catalyst?
I) C + ClO2  ClO + CO
II) CO + ClO2  CO2 + ClO
III) ClO + O2  ClO2 + O
IV) ClO + O  ClO2
a.
ClO2
b.
CO2
c.
O
d. CO
____16. For the system CaO(s) + CO2(g)  CaCO3(s) the equilibrium constant expression is
a. [CO2]
b. 1 / [CO2]
c. [CaO] [CO2] / [CaCO3]
d. [CaCO3] / [CaO] [CO2]
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The value of Kp for the reaction 2NO2(g)  N2O4(g) is 1.52 at 319 K. What is the
value of Kp at this temperature for the reaction
N2O4(g)  2NO2(g)?
a. -1.52
b. 1.23
c. 5.74 x 10-4
d. 0.658
__17.
____18. The value of Kc for the reaction C(s) + CO2(g)  2CO(g) is 1.6. What is the equilibrium concentration
of CO if the equilibrium concentration of CO2 is 0.50 M?
a. 0.31
b. 0.80
c. 0.89
d.
0.75
____19. Consider the reaction below:
2SO3(g)  2SO2(g) + O2(g) , ∆H° = +198 kJ
All of the following changes would shift the equilibrium to the left except one. Which one would not
cause the equilibrium to shift to the left?
a removing some SO3
c. increasing the container volume
b. decreasing the temperature
d. adding some SO2
____ 20. For which of the following reactions is Kc equal to Kp?
a.
c.
N2O4(g)  2NO2(g)
H2(g) + Cl2(g)  2HCl(g)
b. 2SO3(g)  2SO2(g) + O2(g)
d. C(s) + CO2(g)  2CO(g)
PART II- ( 8 points each) Please show all your work .
21. What is the boiling point (in °C) of a solution prepared by dissolving 11.5 g of Ca(NO3)2
(formula weight = 164 g/mol) in 150 g of water? (Kb for water is 0.52°C/m)
22. A solution is prepared by dissolving 6.00 g of an unknown nonelectrolyte in enough water to make 1.00 L of
solution. The osmotic pressure of this solution is 0.750 atm at 25.0°C. What is the molecular weight of the
unknown solute (R = 0.0821 L·atm/K·mol)?
23. The rate constant for a particular reaction is 2.7 x 10-2 s-1 at 25°C and 6.2 x 10-2 s-1 at 75°C. What is the
activation energy for the reaction in kJ/mol? ( R = 8.314 J/mol.K)
24. Initial rate data were obtained for the following reaction: A(g) + 2B(g)  C(g) + D(g)
Experiment
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2
3
initial
initial
[A], mol/L [B], mol/L
0.15
0.30
0.15
What are the rate law and k value for the reaction?
4
0.10
0.10
0.20
initial
rate
0.45
1.8
0.9
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25. A mixture of 0.100 mol of NO, 0.0500 mol of H2, and 0.100 mol of H2O is placed in a 1.00-L vessel. The
following equilibrium is established:
2NO(g) + 2H2 (g)  N2(g) + 2H2O(g)
At equilibrium [NO] = 0.0620 M. Calculate the equilibrium concentrations of H2, N2, and H2O.
BONUS QUESTION - (10 points)
Using the following experimental data, determine; a) the rate law expression
b) the rate constant
c) the initial rate of this reaction when [A] = 0.60 M, [B] = 0.30 M, and [C] = 0.10 M
2 A + B2 + C  A2B + BC
Trial
Initial [A],M
Initial[B2],M
Initial[C], M
Initial rate M/s
1
0.20
0.20
0.20
2.4x10-6
2
0.40
0.20
0.20
9.6x10-6
3
0.20
0.30
0.20
2.4x10-6
4
0.20
0.20
0.40
4.8x10-6
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1412 EX#1 Sample(key)
PART I
1. D
25.2/100 mol
2. A M = n/VL =
= 0.420 mol/L
600/1000 L
3. A S1/S2 = P1/P2  S2 = (0.6)(6.9x10-4) /1 = 5.52x10-4 M
4. D
Δ Tf = Kf. M.i = (1.86)[( 25.0/18) /0.100](1) = 2.58 0C
Tf = nTf - Δ Tf = 0.00 –2.58 = -2.58 0C
5. A
6. B
7. C
8. D
9. A
10. D
11. D
12. A
π = M.R.T.i = (1.2)(0.0821)(298)(1) = 29.35 atm
PA = XA . P0A = [ 90/(10+90)](349) = 314 mmHg
K = M (1-n) .s-1 = M(1-1) s -1 = s -1
ln[A]t = -kt + ln[A]0  ln[0.869] = -k(48) + ln[1.56]  k = 1.22x10-2 s-1
[H2] [I2]
(6.50x10-7)(1.06x10-5)
Kc = ------------- = ------------------------------- = 1.97x10-2
[HI]
(1.87x10-5)2
13 B
14. D
18. C
19. C
15. A
16. B
17. D
for reverse reaction n = - 1  K′ = (Kc)n = (Kc)-1 = ( 1.52)-1 = (1/1.52) = 0.658
20. C
PART II
21. Δ Tb = Kb. M.i = (0.52)[(11.5/164)/(0.150)] (3) = 0.73 0C,
22.
Tb = nTb + ΔTb = 100.00 + 0.73 = 100.73 0C
π = M.R.T.i  M = (0.750) / (0.0821)(298)(1)  M = 0.031 mol/L
M = n/VL  n = M.VL = ( 0.031 mol/L)(1.00L) = 0.031 mol
MW = grams/moles = 6.00 g/0.031 mol = 194 g/mol
OR
g.R.T.i
(6.00)(0.0821)(298)(1)
MW = ------------ = ---------------------------- = 196 g/mol
M. VL
(0.750)(1.00)
23. ln(k1/k2) = -(Ea/R)(1/T1 –1/T2)  ln 2.7x10-2/6.2x10-2) = -(Ea/8.314)(1/298 – 1/348)
 -0.8313 = -Ea (0.000058)  Ea = 14332.716 J/mol  Ea = 14 kJ/mol
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24. A is second- order and B is first-order 
rate = k[A]2[B]
-1
rate
( 0.45 M s )
k = ------------ = ------------------------- = 200 M –2 s-1
[A]2[B]
(0.15 M)2 (0.10M)
[NO] = 0.100 mol/1 L = 0.100 M , [H2O] = 0.100 mol/1 L = 0.100 M , [H2] = 0.050 mol/ 1 L = 0.050 M
2 NO + 2 H2
 N2
(0.100-2x) (0.050-2x)
+x
+
2 H2O
(0.100+2x)
0.100 –2x = 0.0620  2x = -0.062 – 0.100 = 0.038  x = 0.019 M
[H2] = 0.050 – 2x = 0.050 – 2(0.019) = 0.012 M , [H2O] = 0.100 + 2x = 0.100 + 2(0.0190 = 0.138 M
[N2] = x = 0.019 M
Bonus
rate = k[A]x [B]y [C]z
[B]y1
R1/R3 = ---------  (2.4x10-6/ 2.4x10-6) = 1 = (0.20/0.30)y  y = 0 , B is zero-order
[B]y3
[A]1x
R1/R2 = ---------  (2.4x10-6/9.6x10-6) = ( 0.20/0.40)x  (0.25) = (0.5)x  x =2 , A is second-order
[A]2x
[C]1z
R1/R4 = ---------  (2.4x10-6/4.8x10-6) = (0.20/0.40)z  (0.50) = (0.50)z  z =1 , C is first-order
[C]4z
(2.4x10-6 M s-1)
2
rate = k[A] [C] , k = -------------------------  k = 3.0x10-4 M-2 s-1
(0.20 M)2 (0.20M )
rate = k[A]2[C] = (3.0x10-4 M-2 s-1) (0.60 M)2 (0.10 M)  rate = 1.08x10-5 M s-1
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