Camphor
Sample Examination # 1A
Chemistry 1412
1
CHEM 1412 Exam #1 Sample Exam # 1A
(chapters 11& 12)
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.
VOID
____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
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 =
2
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)
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
____13. VOID
____ 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]
PART II- ( 8 points each) Please show all your work .
3
d. 0.60
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
1
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
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
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. VOID
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
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
5
11. D
ln[A]t = -kt + ln[A]0  ln[0.869] = -k(48) + ln[1.56]  k = 1.22x10-2 s-1
12. A
13. VOID
14. D
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)
 -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
6
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 11&12)
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.
Osmotic pressure is
A) inversely proportional to mole fraction.
B) inversely proportional to mass fraction.
C) inversely proportional to molality.
D) directly proportional to molarity.
____ 4.
For a 0.001 M solution of NH4HCO3, the van’t Hoff factor (i) would be about
A) 1.
B) 2.
C) 3.
D) 4.
____ 5.
What type of colloid is formed when a solid is dispersed in a liquid?
a. Foam
b. Aerosol
c. Emulsion
d. Sol
e. Gel
____ 6.
The reaction of a mixture of SO2 and O2 at a given temperature is represented by the
equation
2SO2(g) + O2(g)
2SO3(g)
When the equilibrium is established, which of the following ratios is constant regardless
of the initial concentrations of SO2 and O2?
A)
[SO3 ]
[SO 2 ][O 2 ]
9
B) [SO 2 ][O 2 ]
[SO3 ]
C)
[SO3 ]
[SO 2 ][O 2 ]1/2
D)
[2SO3 ]
[2SO 2 ][O 2 ]
____ 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
_____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.
_____ 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?
10
a) 0.531 M/s
b) 0.002 M/s
c) 0.529 M/s
d) 0.527 M/s
_____ 14.
_____ 15
_____ 16.
_____ 17.
_____ 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.
_____ 20.
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.
23.
.
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).
11
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
1412 EX#1B Sample(key)
PART I
1. D
moles solute/liters solvent is not defined
12
4 C?
2. A
3.D
4. B
5. D
6.
7. D
8. C
9. D
10. C
11. A
12.
13. B
14.
15.
16.
17.
18.
19.
( 0.5)(0.250)(40) = 5 g/mol
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
rate = -(0.29 – 0.531) M / (21 –20)sec = 0.002 M/s
20.
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)
4.05x101
4.05x101
=
=
= 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
23.
PCl2 . P2HI
Keq =

P2HCl
3.9x10-33
=
(x) . ( 2x)2
=
2
(0.445 – 2x)
13
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
(0.10 M) (0.10 M)2
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