GENERAL CHEMISTRY 2
FIRST HOUR EXAM
JULY 9, 2012
Name ______KEY – Version 4______
Panthersoft ID ______
Signature ______
Part 1 ______(20 points)
Part 2 ______(34 points)
Part 3 ______(46 points)
TOTAL ______(100 points)
NA = 6.022 x 1023 °C = (5/9) (°F - 32) °F = (9/5)(°C) + 32
1 amu = 1.661 x 10-27 kg °C = K - 273.15 K = °C + 273.15
1 atm = 760 torr = 760 mm Hg 1 atm = 1.013 bar pV = nRT
R = 0.08206 L.atm/mol.K 1 L.atm = 101.3 J
R = 8.314 J/mol.K 1 J= 1 kg.m2/s2
pA = XA pA° [B] = KB pB DpA = XBpA°
DTb = Kb mB DTf = Kf mB P = [B]RT
[A]t = [A]0 e-kt ln[A]t = ln[A]0 - kt t1/2 = ln2/k
[A]t = [A]0/(1 + kt[A]0) (1/[A]t) = (1/[A]0) + kt t1/2 = 1/(k[A]0)
k = A e-Ea/RT ln k = ln A - (Ea/R)(1/T) ln(k2/k1) = - (Ea/R) [ (1/T2) - (1/T1) ]
Do all of the following problems. Show your work.
Part 1. Multiple choice. Circle the letter corresponding to the correct answer. There is one and only one correct answer per problem. [4 points each]
1) Which pairs of liquids would be expected to be immiscible?
a) A polar liquid and a nonpolar liquid
b) Two polar liquids
A c) Two nonpolar liquids
d) Both b and c
e) Both a and b and c
2) Copper II nitrate (Cu(NO3)2) is soluble in water. The theoretical value for i, the van't Hoff factor, for copper II nitrate in water is
a) i = 1
b) i = 2
C c) i = 3
d) i = 9
e) None of the above
3) A solution is prepared by dissolving 10.7 g of naphthalene (C10H8), a nonvolatile solute, in 500.0 g of benzene (C6H6) a volatile solvent. Compared to pure benzene
a) The vapor pressure of benzene above the solution will be smaller than the vapor pressure of benzene above pure benzene
b) The normal freezing point of the solution will be lower than the normal freezing point of pure benzene
D c) The normal boiling point of the solution will be lower than the normal boiling point of pure benzene
d) Both a and b
e) Both a and c
4) Data plots are given below for an irreversible reaction obeying the rate law Rate = k [A]n.
Based on the data plots
a) We may conclude that the reaction is third order homogeneous
b) We may conclude the reaction is second order homogeneous
B c) We may conclude the reaction is first order homogeneous
d) We may conclude that the reaction is zero order
e) We cannot determine the order of the reaction
5) The value of the parameter A in the Arrhenius equation depends on
a) The frequency of collisions between reactant molecules
b) The fraction of collisions that have a favorable orientation for reaction
D c) The activation energy for the reaction
d) Both a and b
e) Both a and c
Part 2. Short answer.
1) Define the following terms [4 points each]
a) lattice energy – The energy required to convert one mole of a crystalline solid into particles in the gas phase (ions for ionic solids, atoms or molecules for other crystalline solids).
b) unimolecular reaction – n elementary reaction where a single reactant is converted into products. The general form is
A ® “products”
2) Information about liquid ethyl alcohol (C2H5OH) is given below and may be of use in doing the following problem.
Normal boiling point 78.3 °C Boiling point elevation constant 1.22 kg.°C/mol
Normal freezing point - 114.1 °C Freezing point depression constant 1.99 kg.°C/mol
The normal freezing point of a solution of a nonvolatile solute dissolved in ethyl alcohol is - 117.0 °C. What is the normal boiling point of the solution? [8 points]
We will use the freezing point data to find mB (solute molality), and then proceed to find DTb.
DTf = Kf mB, so mB = (DTf/Kf). DTf = - 114.1 °C – (- 117.0 °C) = + 2.9 °C
mB = (2.9 °C/1.99 kg.°C/mol) = 1.46 mol/kg
DTb = Kb mB = (1.46 mol/kg) (1.22 kg.°C/mol) = 1.78 °C
The boiling point of the solution is then T = 78.3 °C + 1.78 °C = 80.1 °C
Version 1 T = 79.8 °C Version 2 T = 79.3 °C Version 3 T = 79.6 °C
3) An irreversible chemical reaction is known to be second order homogeneous, that is, Rate = k [A]2. The value for the half-life for this reaction was determined in an experiment where the initial concentration of A was 0.558 mol/L, and was t1/2 = 48.6 min. What is k, the rate constant, for the reaction (including correct units)? [8 points]
Reaction is second order, and so t1/2 = 1/(k[A]0).
k = 1/(t1/2[A]0]) = 1 = 0.0369 L/mol.min
(48.6 min)(0.558 mol/L)
Version 1 k = 0.0534 L/mol.min Version 2 k = 0.0424 L/mol.min Version 3 k = 0.0317 L/mol.min
4) The following reaction mechanism has been proposed for the gas phase decomposition of nitrous oxide
stoichiometric reaction 2 N2O(g) ® 2 N2(g) + O2(g)
mechanism step 1 N2O(g) ® N2(g) + O(g) slow
step 2 N2O(g) + O(g) ® N2(g) + O2(g) fast
a) List all of the substances in the above mechanism that are reaction intermediates [4 points]
______O(g)______
b) Find the rate law corresponding to the proposed mechanism, and explain how your rate law was found.
[6 points]
Overall rate = rate of slow step (since there is a single slow step) and so
Rate = k1 [N2O]
Since there are no reaction intermediates in the above expression for rate, this is the final answer.
Part 3. Problems.
1) The molality of benzene in a solution of benzene (C6H6, MW = 78.12 g/mol) in carbon tetrachloride (CCl4, MW = 153.82 g/mol) is m = 0.3104 mol/kg. The density of the solution is r = 1.557 g/mL. Find MB, the molarity of benzene in the solution , and XB, the mole fraction of benzene in the solution [16 points]
Assume 1.000 kg CCl4 (convenient, since molality is given and has units of mol/kg).
mol benzene = nB = 0.3104 mol mol CCl4 = mC = 1000.0 g 1 mol = 6.501 mol
153.8 g
g soln = g CCl4 + g benzene g benzene = 0.3104 mol 78.12 g = 24.25 g
1 mol
g soln = 1000.0 g + 24.25 g = 1024.2 g V = volume soln = 1024.2 g 1.0 mL = 657.8 mL = 0.6578 L
1.557 g
mol fraction benzene = XB = nB = 0.3104 mol = 0.0456
(nC + nB) (6.501 + 0.3104) mol
molarity benzene = MB = nB = 0.3104 mol = 0.4719 mol/L
(L soln) 0.6578 L
Version 1 XB = 0.888 MB = 0.9225 mol/L Version 2 XB = 0.720 MB = 0.7468 mol/L
Version 3 XB = 0.057 MB = 0.5922 mol/L
2) Consider the following irreversible chemical reaction
A + B ® “product”
The reaction obeys the rate law
Rate = - D[A] = k [A]2
Dt
a) Which of the following is a possible correct set of units for k, the rate constant, for the above reaction (circle the correct answer)? [4 points]
mol L mol2 1
L.min mol.min L2.min min
b) In a particular experiment, the initial concentration of A is 0.5862 mol/L. After 20.0 minutes, the concentration of A is 0.4624 mol/L. What will be the value for the concentration of A after 200.0 minutes?
[12 points]
First find k, then find [A]200.
(1/[A]t) = (1/[A]0) + kt , and so k = (1/t) { (1/[A]t) - (1/[A]0) }
k = (1/20.0 min) { (1/0.4624 mol/L) - (1/0.5862 mol/L) } = 0.02284 L/mol.min
Then [A]t = [A]0 = (0.5862 mol/L) = 0.1594 mol/L
1 + kt[A]0 { 1 + (0.02284 L/mol.min)(200.0 min)(0.5862 mol/L) }
Version 1 [A]t = 0.1522 mol/L Version 2 [A]t = 0.1503 mol/L Version 3 [A]t = 0.1501 mol/L
3) A particular chemical reaction with stoichiometry
A + B ® "products"
is found to fit a rate law of the form
Rate = k [A]m [B]n
In a series of experiments carried out at the same temperature the initial rate of the above reaction is measured. The data are given below. Based on these data find the values for m, n, and k (including correct units). Show your work. [14 points]
Experiment [A]initial [B]initial (Rate)initial
(mol/L) (mol/L) mol/L.s
1 0.0400 0.0100 9.2 x 10-6
2 0.0400 0.0300 2.7 x 10-5
3 0.0800 0.0100 3.7 x 10-5
Compare 3 and 1 (0.08/0.04)m = (3.7 x 10-5/9.2 x 10-6) 2m = 4.02 m = 2
Compare 2 and 1 (0.03/0.01)n = (2.7 x 10-5/9.2 x 10-6) 3n = 2.93 n = 1
So rate = k [A]2[B]
k = (rate)
[A]2[B]
If we use the first trial, then k = (9.2 x 10-5 mol/L.s) = 0.575 L2/mol2.s
(0.0400 mol/L)2(0.0100 mol/L)
4