KEY
CHEMISTRY 121 - GENERAL CHEMISTRY II
Exam #2 - 100 points
March 27, 2001
Multiple Choice (3 points each, circle the letter of the correct answer)
The decomposition of ozone may occur through the two-step mechanism shown:
Step #1 O3 ® O2 + O
Step #2 O3 + O ® 2O2
The oxygen atom is considered to be a(n)
a.) reactant
b.) product
c.) catalyst
d.) reaction intermediate
e.) activated complex
2.) The following data were obtained for the reaction of NO with O2. Concentrations are in molecules/cm3 and rates are in molecules/cm3·s.
|
[NO]0 (molecules/cm3) |
[O2]0 (molecules/cm3) |
Initial Rate (molecules/cm3·s) |
|
1 x 1018 |
1 x 1018 |
2.0 x 1016 |
|
2 x 1018 |
1 x 1018 |
8.0 x 1016 |
|
3 x 1018 |
1 x 1018 |
18.0 x 1016 |
|
1 x 1018 |
2 x 1018 |
4.0 x 1016 |
|
1 x 1018 |
3 x 1018 |
6.0 x 1016 |
Which of the following is the correct rate law?
a.) Rate = k[NO][O2]
b.) Rate = k[NO][O2]2
c.) Rate = k[NO]2[O2]
d.) Rate = k[NO]2
e.) Rate = k[NO]2[O2]2
3.) For the following reaction...
H2O2 + 3I-1 + 2H+1 ® I3-1 + 2H2O
...the experimentally determined rate law is
Rate = k[H2O2][I-1]
Two mechanisms are proposed for this reaction (both mechanisms do add up to the overall reaction).
I. H2O2 + I-1 ®
H2O + OI-1
II. H2O2 + I-1 + H+1 ®
H2O + HOI
Which of the following describes a potentially correct mechanism?
a.) Mechanism I, if the first step is the rate-determining step.
b.) Mechanism I, if the second step is the rate-determining step.
c.) Mechanism II, if the first step is the rate-determining step.
d.) Mechanism II, if the second step is the rate-determining step.
e.) None of the above could be correct.
4.) The reaction quotient, Q, for a system is 7.2 x 102. If the equilibrium constant for the system is 36, what will happen as equilibrium is approached?
a.) There will be a net gain in product.
b.) There will be a net gain in reactant.
c.) There will be a net gain in both product and reactant.
d.) There will be no net gain in either product or reactant.
e.) The equilibrium constant will decrease until it equals the reaction
quotient.
5.) Equilibrium is reached in chemical reactions when:
a.) the rates of the forward and reverse reactions become equal.
b.) the concentrations of reactants and products become equal.
c.) the temperature shows a sharp rise.
d.) all chemical reactions stop.
e.) the forward reaction stops.
6.) The bicarbonate ion, HCO3-1, is amphoteric...the conjugate acid and conjugate base of HCO3-1 are, respectively:
a.) H3O+1 and OH-1
b.) H3O+1 and CO3-2
c.) H2CO3 and OH-1
d.) H2CO3 and CO3-2
e.) CO3-2 and OH-1
7.) In deciding which of two acids is stronger, one must know:
a.) the concentration of each acid solution.
b.) the pH of each acid solution.
c.) the Ka of each acid.
d.) all of the above
e.) both a and c must be known
Short Answer
8.) (6 points) Calculate the pH and pOH of a 8.2 x 10-2 M HClO4(aq) (perchloric acid) solution.
HClO4 is a strong acid; therefore, we can assume that the H+1 concentration is equal to the concentration given for the acid (8.2 x 10-2 M). So...
pH = -log[H+1] = -log(8.2 x 10-2 M) = 1.09
pOH = 14 - 1.09 = 12.91
9.) (10 points) For the reaction...
A <===>B
...the following reaction coordinate diagram is drawn.

a.) Indicate which number on the chart corresponds to which quantity.
Ea of forward reaction 3
Ea of reverse reaction 2
D
E of reaction 1b.) Which reaction (forward or reverse) will have the higher rate constant? (Circle one.)
c.) Is the forward reaction (A ® B) exothermic or endothermic? (Circle one.)
10.) (10 points) Consider the following system at equilibrium:
N2(g) + 3H2(g) <==> 2NH3(g)
DH = -92.94 kJ/mol N2For each of the following pairs of changes, circle the change that will shift the equilibrium to the right.
|
heating the reaction |
OR |
cooling the reaction
|
|
increasing the flask volume |
OR |
decreasing the flask volume
|
|
removing some NH3 |
OR |
adding some NH3
|
|
removing some N2 |
OR |
adding some N2
|
|
removing some H2 |
OR |
adding some H2
|
Problems
11.) (10 points) When ethyl chloride, CH3CH2Cl, is dissolved in 1.0 M NaOH, it is converted into ethanol, CH3CH2OH, by the reaction
CH3CH2Cl + OH-1 ® CH3CH2OH + Cl-1
At 25°C the (first order) rate constant is 1.0 x 10-3 s-1. If the activation energy is 100.0 kJ/mol, what will the rate constant be at 40°C?
Let's use the modified form of the Arrhenius equation...
ln(k1/k2) = Ea/R (1/T2 - 1/T1)
T1 = 25°C = 298.15 K (and you must switch temperatures to
Kelvin!!!)
T2 = 40°C = 313.15 K
k1 = 1.0 x 10-3 s-1
k2 = ????
Ea = 100 kJ/mol = 1.x 105 J/mol (and you must
switch the activation energy to Joules!!!)
R = 8.3145 J/mol·K
ln(1.0 x 10-3/k2) = 1 x 105/8.3145 (1/313.15 - 1/298.15)
ln(1.0 x 10-3/k2) = 12,027 (0.0031934 - 0.0033540)
ln(1.0 x 10-3/k2) = -1.93154
Take the inverse natural log of both sides...
1.0 x 10-3/k2 = 0.14492
k2 = 1.0 x 10-3/0.14492 = 6.9 x 10-3 s-1
12.) (13 points) The decomposition of NO2 is a very slow process.
2NO2 ® 2NO + O2
The rate constant for this reaction is 2.80 x 10-10 L/mol·s at 25°C. If 3.00 moles of NO2 is placed in a 2.00 L flask at 25°C, in 75.5 years there will be 1.50 moles of NO2 left in the flask. What is the order of the reaction with respect to NO2? (3 points will buy you a hint...)
If we start out with 3.00 moles of NO2 in the flask and end up with 1.5 moles after 75.5 years, 75.5 years must be the half-life of the reaction (this was the 3-point clue, in case you're curious). So, if we plug the rate constant and the initial concentration of NO2 (if needed) into the expressions for half-life given on the formula page, one of these expressions will give us an answer equivalent to 75.5 years...that expression will be for the correct order of the reaction.
k = 2.8 x 10-10 L/mol·s
75.5 years x 365 days/yr x 24 hrs/day x 60 min/hr x 60 sec/min = 2.38 x 109 seconds
[NO2]0 = 3.00 moles/2 liters = 1.5 M
For a first order reaction...
t1/2 = 0.693/k = 2.48 x 109 seconds ≠ 75.5 years...must not be a first order reaction
For a second order reaction...
t1/2 = 1/[A]0k = 2.38 x 109 seconds...bingo! Second order!
Check zeroth order reaction...
t1/2 = [A]0/2k = 2.68 x 109 seconds...not zeroth order.
An alternative method of solving this problem lies in evaluating the units of the rate constant and determining whether they correspond to a first, second, or zeroth order reaction. For a first order reaction, the units on the rate constant must be s-1 (or some inverse time unit) in order for the units to cancel in the rate law...
Rate = k[A]
mol/L·s = k x mol/L
k = s-1
For a second order reaction, the units of k can be found similarly...
k = rate/[A]2
Units for k = (mol/L·s)/(mol/L)2 = L/mol·s
For a zeroth order reaction, Rate = k, so the units on rate (mol/L·s) must also be the units for k.
Since the units on the rate constant are L/mol·s, we know the reaction must be second order.
13.) (15 points) Bromine chloride (BrCl), a reddish gas with properties similar to those of chlorine gas, may eventually replace chlorine as a water disinfectant. It is formed by the reaction of chlorine and bromine gases. 100.0 g of Cl2 and 100.0 g of Br2 are placed in a 5.00 L reaction flask and allowed to reach equilibrium at a certain temperature.
Cl2(g) + Br2(g) <===> 2BrCl(g)
The equilibrium constant for this reaction is 4.7 x 10-2. What are the concentrations of Cl2, Br2, and BrCl at equilibrium?
Let's first calculate the initial concentrations of Cl2 and Br2.
100.0 g Cl2/70.9 g/mol = 1.41 moles Cl2
1.41 moles Cl2/5 L = 0.282 M
100.0 g Br2/159.8 g/mol = 0.626 moles Br2
0.626 moles Br2/5 L = 0.125 M
Since we don't start out with any BrCl, we know that this reaction must shift to the right to reach equilibrium (produce products and consume reactants). Let's set up our equilibrium table...
|
[Cl2] (mol/L) |
[Br2] (mol/L) |
[BrCl] (mol/L) |
|
|
Initial |
0.282 M |
0.125 M |
0 |
|
D |
-x |
-x |
+2x |
|
Equilibrium |
0.282 - x |
0.125 - x |
2x |
K = 4.7 x 10-2 = [BrCl]2/[Cl2][Br2] = (2x)2/(0.282 - x)(0.125 - x)
4.7 x 10-2 = 4x2/(0.03525 - 0.157x - x2)
4x2 = 1.66 x 10-3 - 0.407x - 4.7 x 10-2x2
3.953x2 + 0.407x - 1.66 x 10-3 = 0
Using the quadratic formula...
x = (-0.407 ± √[0.407)2 - 4(3.953)(-1.66 x 10-3)])/2(3.953) = 0.0182 (the other value for x is negative and, therefore, impossible).
So, if x = 0.0182, then the concentrations of Cl2, Br2, and BrCl at equilibrium are as follows...
[Cl2] = 0.282 - x = 0.264 M
[Br2] = 0.125 - x = 0.107 M
[BrCl] = 2x = 0.0364 M
14.) (15 points) What is the pH of a solution made by mixing 100 mL of 1.0 M hydrofluoric acid (HF) and 100 mL of 1.0 M hydrocyanic acid (HCN)?
Ka for HF = 7.2 x 10-4, Ka for HCN = 6.2 x 10-10.
What are the species present in solution?
HF, HCN, and H2O
All three of these will produce H+1 ions, but HF will be the primary H+1 producer since it has the highest Ka by far. So, we can safely ignore HCN and H2O, and concentrate on the amount of H+1 that will be produced by HF. This becomes an equilibrium problem. What is our initial concentration of HF?
0.1 L HF solution x 1.0 M = 0.1 moles of HF present in sample
0.1 moles HF/0.2 L of solution (total...remember that we're mixing the two acids together and, thus, diluting the initial HF solution by 2) = 0.5 M HF
|
[HF] (mol/L) |
[H+1] (mol/L) |
[F-1] (mol/L) |
|
|
Initial |
0.5 |
0 |
0 |
|
D |
-x |
+x |
+x |
|
Equilibrium |
0.5 - x |
x |
x |
Ka = 7.2 x 10-4 = [H+1][F-1]/[HF] = x2/0.5 - x ≈ x2/0.5
x = 0.019
Notice that we've assumed that x is so small compared to the initial concentration of HF that we can ignore it in the denominator of our equilibrium constant. Let's test that assumption...
0.019/0.5 x 100% = 3.8% (< 5%...OK)
If x = 0.019, [H+1] = 0.019 M, so pH = 1.72
EXTRA CREDIT (5 points)
15.) Consider the reaction
A(g) + B(g) C(g)
for which K = 1.30 x 102. Assume that 0.406 mol C(g) is placed in the cylinder represented below. The temperature is 300.0 K, and the barometric pressure on the piston (which is assumed to be massless and frictionless) is constant at 1.00 atm. The original volume (before the 0.406 mol C(g) begins to decompose) is 10.00 L. What is the volume in the cylinder at equilibrium? (aka Marathon Problem from Chapter 13.)
Here's my solution to this problem...there's probably an easier way, but this is all I could come up with.
|
|
# moles A |
# moles B |
# moles C |
|
Initial |
0 |
0 |
0.406 |
|
D |
+x |
+x |
-x |
|
Final |
x |
x |
0.406 - x |
Notice that we've
used the number of moles of the reaction components here, not the
concentrations. Let's use these
expressions in the equations for mole fraction...
# moles B = 0.162
# moles C = 0.244
Total # moles = 0.568
16.) Most reactions occur by a series of steps. The energy profile for a certain reaction that proceeds by a two-step mechanism is...

On the energy profile, indicate (i.e. label with the corresponding letter)
a.) the positions of reactants and products
b.) the activation energy for the overall reaction
c.)
DE for the reactiond.) Which point on the plot represents the energy of the intermediate in the two-step reaction?
Which step in the mechanism for this reaction is rate determining, the first or second step? (Circle one.)
See labels on diagram above. The second step is slower because it has a higher activation energy than the first step...it is therefore the rate-determining step.
The overall activation energy of the reaction is the difference between the energy of the highest "hill" the reaction has to "climb" minus the energy of the reactants.
FORMULAS/CONSTANTS
Integrated Rate Laws:
ln[A] = -kt + ln[A]0
1/[A] = kt + 1/[A]0
[A] = -kt + [A]0
Half-life expressions:
t1/2 = [A]0/2k (zeroth order)
t1/2 = 0.693/k (first order)
t1/2 = 1/k[A]0 (second order)
Arrhenius Equation:
k = Ae-Ea/RT
ln(k) = -Ea/RT + ln(A)
ln(k1/k2) = Ea/R x (1/T2 - 1/T1)
R = 0.08206 L·atm/mol·K
= 8.3145 J/K·mol
Kw = 1.0 x 10-14
Kp = K(RT)