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Sample Examination # 1A
Chemistry 1412
1
CHEM 1412 Exam #1 Sample Exam # 1A
(chapters 12,13, and 14)
Name: ____________________
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 affects the equilibrium constant
b. it affects the rate of a chemical reaction
d. it changes the rate constant of a reaction
____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
2
d. s-1
____ 9. For second-order reactions the slope of a plot of 1/[A] versus time is
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
c. k[A]2[D]2
b. k[A][D]
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)
0
1.1
2.3
4.0
[A], M
1.20
1.00
0.80
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.
3
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]
____17. 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
c. 5.74 x 10-4
b. 1.23
d. 0.658
____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
b. decreasing the temperature
c. increasing the container volume
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)
4
24. Initial rate data were obtained for the following reaction: A(g) + 2B(g)  C(g) + D(g)
Experiment
1
2
3
initial
initial
[A], mol/L [B], mol/L
0.15
0.30
0.15
initial
rate
0.10
0.10
0.20
0.45
1.8
0.9
What are the rate law and k value for the reaction?
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
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
5
Initial rate M/s
1412 EX#1A 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 = 196 g/mol
OR
g.R.T.i
(6.00)(0.0821)(298)(1)
MW = ------------ = ---------------------------- = 196 g/mol
π . 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)
6
 -0.8313 = -Ea (0.000058)  Ea = 14332.716 J/mol (divided by 1000)  Ea = 14 kJ/mol
24. A is second- order and B is first-order  rate = k[A]2[B]
rate
( 0.45 M s-1)
k = ------------ = ------------------------- = 200 M –2 s-1
[A]2[B]
(0.15 M)2 (0.10M)
[NO] = 0.100 mol/1L = 0.100 M , [H2O] = 0.100 mol/1L = 0.100 M , [H2] = 0.050 mol/1L = 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)
rate = k[A]2[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
7
Sample Examination # 1B
Chemistry 1412
8
CHEM 1412 Exam #1 Sample Exam # 1B
(chapters 12,13, and 14)
Name: ____________________
Score:
PART I - ( 3 points each) - Please write your correct answer next to each question
number. DO NOT CIRCLE
_____ 1. The number of moles per of solute per one liter of solvent is called __________ .
a) molarity
b) molality
c) normality
d) none of these
____ 2. How many grams NaOH (40.0 g/mol) are required to make 250 mL of a 0.500 M
solution?
a) 5
b) 5000
c) 0.125
d) 125
_____3. If 200 mL of 1.60 M NaOH are diluted with water to a volume of 350 mL, the new
concentration of the solution is ……..
a) 0.257 M
b) 0.914 M
c) 2.8 M
d) 0.582 M
____ 4. Of the following salts, the one that is LEAST soluble in water is ……….
a) MgCl2
b) FeCl2
c) AgCl
d) CaCl2
____ 5. Which one of the following 0.15 m aqueous solutions lowers freezing point the most?
a) NaCl
b) C6H12O6
c) K2SO4
d) NaNO3
____ 6. What is the normality of a 2.0 M solution of phosphoric acid ?
a) 2
b) 3
c) 6
d) 0.67
____ 7. A catalyst
a) increases the yield of product
b) increases the energy of activation
c) decreases the enthalpy of the reaction
d) provides a new pathway which requires lower activation energy
_____8. The unit for a first order rate constant is ………
a) M s-1
____ 9.
b) M-1 s-1
c) s-1
d) M-2 s-1
If the rate of a reaction is second order with respect to component A, how will the rate
change if the concentration of A tripled?
a) It will double
c)It will be six times as great
b) It will tripled
d) It will be nine times as great
9
_____10. For the chemical reaction A + B  C , a plot of ln[A]t vs time is found to give a straight
line with a negative slope. What is the order of the reaction?
a) zero
b) first
c) second
d) third
_____11. The rate constant for the first order decomposition of C4H8 at 500 oC is
9.2x10-3 s-1. How long will it take for 10.0% of 0.100 M sample of C4H8
to decompose at 500 oC?
a) 12 sec
b) 0.0084 sec
c) 512 sec
d) none of these
_____ 12. For the hypothetical reaction A + 3 B  2C, the rate of appearance of C,  [C]/  t
may also be expressed as
a)  [C]/  t = - [A] /t
c)  [C]/  t = -2/3 [B]/ t
b)  [C]/  t = - 3/2 [B] /t
d)  [C]/  t = -1/2 [A]/ t
_____ 13. Consider the reaction 2HI  H2 + I2
time, sec:
[HI], M :
20
0.531
21
0.529
22
0.527
What is the rate of reaction of HI between the interval 21 sec and 20 sec?
a) 0.531 M/s
b) 0.002 M/s
_____ 14. For the reaction CaCO3(s)
c) 0.529 M/s
d) 0.527 M/s
CaO(s) + CO2 (g) , increasing the pressure on the
system at equilibrium causes
a) increased amount of CaCO3 and CaO
b) decreased the amount of CaO and CO2
c) increased the amount of CO2 and CaCO3
d) increased the amount of CaCO3 and CO2
_____ 15. Equilibrium is reached in all reversible reaction when the
a)
b)
c)
d)
forward reaction stops
reversed reaction stops
concentrations of reactants and the products become equal
rates of the opposing reactions become equal
_____ 16. The value of Kc for the following reaction is 1.60. C(s) + CO2(g)  2 CO(g)
What is the equilibrium concentration of CO if the equilibrium concentration of
CO2 is 0.50 M?
a) 0.79
b) 0.40
c) 0.894
10
d) 2.24
_____ 17. Phosgene, COCl2 , a poisonous gas decomposes according to the following equation;
COCl2 (g)  CO(g) + Cl2 (g)
If Kc = 0.083 at 900
what is the value of Kp?
oC,
a) 0.125
b) 8.0
c) 6.1
d) 0.16
_____ 18. Consider the two gaseous equilibria;
SO2(g) + ½ O2 (g) SO3 (g)
2SO3 (g)  2 SO2 (g) + O2 (g)
, K1
, K2
The value of the equilibrium constant s are related by
a) K2 = K1
b) K2 = (K1) -1
c) K2 = (K1) -2
d) K2 = (K1)2
_____ 19. Consider the reaction N2g) + O2(g)
2 NO(g) , Kc = 0.10 at 200 oC.
Starting with initial concentration of 0.04 mol/L of N2 and 0.040 mol/L of O2, calculate
the equilibrium concentration on NO in mol/L.
a) 5.4x10-3
b) 0.0096
c) 0.013
d) 1.6x10-4
_____ 20. At 700 K the reaction, 2 SO2 (g) + O2 (g)  2SO3 (g) , Kc = 4.3x106
has an equilibrium concentration of [SO2]o = 0.10M , [SO3]o = 1.0 M , [O2]o = 0.10 M.
a) The reaction mixture is at equilibrium
c) The reaction direction is reversed
b) The reaction direction is forward
d) none of the above is correct
PART II- ( 8 points each) Please show all your work .
21. At 35 oC and 75 oC the second order rate constants of a reaction are 2.50 x 10-5 and
3.26 x 10- 3 M-1 s-1 respectively. What is the enthalpy of activation (kJ/mole)?
(R = 8.314 J/mole K).
22. The equilibrium constant for the following reaction is 10.5 at 500 K. A system at equilibrium
has [CO] = 0.250 M and [H2] = 0.120 M. What is the concentration of [CH3OH]?
CO (g) + 2 H2 (g)  CH3OH (g)
11
23. HCl(g) initially at a partial pressure of 0.445 atm; is reacting with I2(s) ;
2 HCl(g) + I2 (s)  2 HI (g) + Cl2 (g) , Keq = 3.9x10 -33 , at 25 0C .
Calculate the final partial pressures at equilibrium.
24. A 1.450 g sample of an unknown organic compound , X, is dissolved in 15.0 g of toluene
( C7H8 = 92 g/mol) and the freezing point is lowered by 1.33 oC. What is the molecular weight
of the organic compound? (Kf = 5.12 oC/m).
25. Calculate the osmotic pressure in torr of a solution of 0.050 g of an unknown organic
compound in 10.0 mL of water at 25 oC. Molecular weight of the unknown organic compound
is 195 g/mol ? ( R = 0.0821 L.atm/ mol.K) ( 1 atm = 760 torr)
BONUS QUESTION - (10 points)
The following rate data were obtained at 25oC for the indicated reaction. 2 A + B
Exp.
[A] mol/L
[B] mol/L
1
2
3
0.10
0.10
0.20
0.10
0.20
0.20
Rate of formation of C
(M/min)
2.0 x 10-4
8.0 x 10-4
1.6 x 10-3
Calculate;
a) the rate law expression
b) the rate constant
c) the initial rate of this reaction when [A] = 0.40 M, [B] = 0.30 M
12
4 C?
1412 EX#1B Sample(key)
PART I
1. D
2. A
3.B
4. C
5. C
6. C
7. D
8. C
9. D
10. C
11. A
12. C
13. B
14. B
15. D
16. C
17. B
moles solute/liters solvent is not defined
( 0.5)(0.250)(40) = 5 g/mol
(1.60)(200)/350 = 0.914 M
AgCl is insoluble in water
K2SO4 has three ions 2K+ and SO42- and lowers the freezing point more.
N = nM = ( 3H)(2) = 6 Eq/L
Catalyst lowers the activation energy.
rate will increase by (3)2 = 9 times.
ln(0.09) = -9.2x10-3 t + ln (0.1)  solve for t = 12 sec
 [C]/  t = -1/3 [B]/ t = - ½ [A]/ t   [C]/  t = -2/3 [B]/ t
rate = -(0.29 – 0.531) M / (21 –20)sec = 0.002 M/s
reaction will shift to left, decreases the amounts of CO2 and CaO
18. C
19. C
20. B
Kc = [CO]2 / [CO2]  1.60 = [CO]2 / 0.50  [CO]2 = 0.80  [CO] = 0.894 M
Δ n = 2-1 = 1 , T = 900 + 273 = 1173 K
Kp = Kc (RT) Δ n = (0.083)(0.0821x 1173)1 = 8.0
Reverse and double K2 = ( K1)-2
Kc = [NO]2 / [N2][O2]  0.1 = [NO]2/ (0.04)(0.040)  solve for [NO] = 0.013 mol/L
Q = [SO3]2 / [SO2]2 [O2]  Q = (0.1)2 / (0.1)2(0.10  Q= 1000
Kc = 4.3x106 > Q = 1000 direction ( ) to forward, right- more products formed.
PART II
21.
-( 8.314) ln ( 2.50x10-5 / 3.26x10-3)
ln(k1/k2) = (-Ea/R)(1/T1 – 1/T2)  Ea =
( 1/ 308 – 1/ 348)
1
1
4.05x10
4.05x10
=
=
= 1.08x105 J/mol
= 109 kJ/mol
( 0.0032468-0.0028736)
0.000375
(divided by 1000)
22. .
Keq =
[CH3OH]
[CH3OH]
 10.5 =
[CO][
H2]2
 [CH3OH] = 0.0373 M
(
0.250)(0.120)2
13
23.
PCl2 . P2HI
Keq =
 3.9x10-33 =
P2HCl
(x) . ( 2x)2
=
2
(0.445 – 2x)
4 x3
= 3.9x10-33
(0.445)2
Ignore (Keq is too small)
4 x3 = (3.9x10-33) (0.445) = 7.72x10-34 
x3 = 1.93x10-34  x = 5.78x10-12 atm
PCl2 = x = 5.78x10-12 atm
PHCl = 0.445 atm – 2x = 0.445 atm
PHI = 2 x = 2(5.78x10-12 ) = 1.16x10-11 atm
24
(g) . (Kf) . (i)
MW =
(kg) . Δ T
25.
(1.450)(5.12)(1)
=
= 372 g/mol
(15.0/1000) (1.66)
π = M. R. T. I = [( 0.050/195)/(10.0/1000)] x (0.0821) x (25+273) x (1) = 0.63 atm
0.63 atm x 760 = 477 torr
Bonus
Rate = k[A]m [B]n
R1/R2 = [A]1m / [A]2 m  (2.0x10-4 / 8.0x10-4) = (0.10/0.20)m
(2/8) =(1/4) = (1/2)m  (1/2)2 = (1/2)m  m =2 ( order of B – second oreder)
R2/R3 = [A]2m [B]2n / [A]3m [B]3n  (8.0x10-4 / 1.6x10-3) = (0.10/0.20)n
(1/2) = (1/2) n  (1/2)1 = (1/2)n  n = 1 ( order of A – first oreder)
rate = k[A][B]2
rate
k=
=
[A]
[B] 2
2.0x10-4 M min-1
= 0.20 M-2 min-1
2
(0.10 M) (0.10 M)
rate = k[A][B] 2 = (0.20 M-2 min-1) ( 0.30 M)(0.40 M)2 = 7.2 x10-3 M min-1
14