Chem 206 Exam 2 Answers

Answers
Exam 2
Chem 206 Section B
Williamsen
23 February 2000
Potentially Useful Information:
J
R = 8.31451
mol ⋅ K
= 0.0820578
L ⋅ atm
mol ⋅ K
x =
− b ±
b
2a
2 −
4ac
As completely, yet as concisely as possible, answer the following questions. Read carefully and only provide
the requested information. When answering problems that require calculations, provide the correct number of
significant figures, the correct units, and show your work.
Good Luck!
1) The following species can be classified as acids or bases under some combination of the Arrhenius,
Brønsted-Lowry, or Lewis definitions.
a) For the following examples, classify each species as an acid or base. In addition state under what
definition(s) this species is classified as an acid or base (state all definitions that apply).
i) HCl (aq) acid Arrhenius, Brønsted-Lowry, Lewis
ii)
iii)
iv)
NH 3 (g) base Brønsted-Lowry, Lewis
KOH (aq) base Arrhenius, Brønsted-Lowry, Lewis
Cu 2+ (aq) acid Lewis
b) Briefly explain the answer you gave for your classification of KOH (aq). In your answer use the
definitions of Arrhenius, Brønsted-Lowry, or Lewis acids or bases, if they apply.
KOH (aq) is an Arrhenius base because it produces OH – in water, a Brønsted-Lowry base because the
OH – can accept a H+, and a Lewis base because OH – has a lone pair available to be donated to another
species that can accept a lone pair.
2) You are investigating the decomposition of tert-butyl chloride < ( CH
3
)
3CCl
> in the gas phase into
isobutylene in the gas phase and HCl in the gas phase. At 500 K, the equilibrium
constant is 0.840.
a) Write the chemical equation for this reaction. In writing the chemical equation, careful write the
arrow(s) in the correct proportion.
( CH 3
) 3
CCl(g)
( CH
3
) C = CH
2
(g) + HCl(g)
Arrow needs to be larger in the reverse direction because the equilibrium constant is less than 1.
b) Into a 0.500-L container at 500 K, you introduce 2.00-mol isobutylene, 2.00-mol HCl, and 4.00-mol
tert-butyl chloride. What are the equilibrium concentrations of all three species?
First, you need to check in which direction the reaction will occur.
[
Q = HCl]
[( CH 3
) 2
C = CH 2 ] 4.00 × 4.00
= = 2.00
[( CH 3 ) 3
CCl]
8.00
Because Q>K, the reaction will go in the reverse direction.
( CH
3
)
3CCl
( CH
3) C = CH
2
2
HCl
Initial 8.00 4.00 4.00
Change +x -x -x
Equilibrium 8.00+x 4.00-x 4.00-x
2
1

( 4.00 − x)( 4.00 − x)
( 8.00 + x)
K =
= 0.840
16.0 − 8.00x + x
0.840 =
8.00 + x
x 2 − 8.840x + 9.2 = 0
( )
x = 8.840 ± −8.8402 − 4.00 ×1.00 × 9.2
2.00 ×1.00
2
= 1.2or 7.6
Can't use 7.6, as this will give negative concentrations. Therefore, the equilibrium concentrations are
( CH
3
)
3CCl
( CH
3) C = CH
HCl
2
2
Concentrations 9.2 2.8 2.8
2.8 × 2.8
Check, K = = 0.85
9.2
c) You study the kinetics of both the forward and reverse reactions and determine that both the forward
and reverse reactions occur in a single elementary step. At 500 K you determine that the rate constant
for the reverse reaction is 3.45 M -1 s -1 . At equilibrium what is the rate (not rate constant) for the forward
reaction?
Rate law for the forward reaction is Rate = k
f
[( CH
3)
CCl]
3
k
f
K = so k
k
f
= 0.840 × 3.45 = 2.90 s −1
r
Therefore, the rate is 2.90 s −1 × 9.2 = 27 M ⋅ s −1 . Note: You must use the equilibrium concentration.
Or: Because at equilibrium k f =k r , 3.45 M −1 ⋅ s −1 × 2.8 × 2.8 = 27 M ⋅ s −1
d) After equilibrium is obtained, you add a catalyst and 3.00 additional moles of HCl. What will happen?
The addition of a catalyst will not change the equilibrium but will only increase the rate at which
equilibrium will be obtained. The addition of HCl will cause the reaction to go to the left (reverse
direction) to relieve the stress (Le Châtlier's Principle)
3) Acid rain is a topic that is often in the news. Even without pollution, however, rain is naturally acidic.
a) What is the reason for the natural acidity of rain water? Briefly explain. Include in your answer the
chemical equation for the species that are in equilibrium to cause the naturally acidic rain.
One of the main constituents of the atmosphere is CO 2 . As rainwater falls to the earth, CO 2 diffuses into
the raindrop to form carbonic acid. Therefore, rainwater has a natural pH of approximately 5.6.
CO
2
(g) + H
2O(l)
H
2CO3(aq)
2

) The [H 3 O + ] for naturally acidic rain is 2.51x10 -6 M. What is the pH? Show your work and write the
correct number of significant figures.
pH ≈ −log[ 2.51 ×10 −6
] ≈ 5.600
c) If you experimentally measured the pH, would your measured value equal the value that you calculated
in part b)? Why or why not?
No. pH is not really defined as the negative log of the concentration of hydrogen cation but as the
negative log of the activity of the hydrogen cation. The activity is a H
+ = [ H +
]f .
3) For each pair of acids, state which acid is strongest. For each comparison briefly explain your answer. For
polyprotic acids (more than one acidic proton), just consider the dissociation of the first proton.
a) HCl (aq) or H 2 S(aq)
HCl. HCl is one of the few strong acids and H 2 S is not. Also, Cl – is more electronegative than S 2– so
the bond is more polar in HCl, and Cl – will be the more stable conjugate base.
b) H 2 SO 3 (aq) or HClO 3 (aq)
HClO 3 . Cl is more electronegative than S. Because there are the same number of O's, the more
electronegative central atom in the anion will make the conjugate base more stable as it can handle the
negative charge better.
c) HF (aq) or HBr (aq)
HBr. One of the strong acids and HF is not. F and Br are both halogens. Because Br is larger than F,
the bond is weaker.
4) You are monitoring the following reaction: HCl (aq) + NaCN(aq)
HCN (aq) + NaCl(aq)
.
a) You add equal concentrations of the species in the reaction pot. In which direction will the reaction
occur. Why?
The reaction will occur in the forward direction. HCl is a stronger acid than HCN, and NaCN is a
stronger base than NaCl. The reaction goes in the direction so that the stronger acid and base react.
b) What are conjugate acid-base pairs in the above reaction?
HCl (acid) and NaCl (conjugate base)
NaCN (base) and HCN (conjugate acid)
6) Briefly explain what the difference is between Q and K.
The method of calculation for both Q and K is the same for both, but a value for Q can be computed at any
time during the reaction, but the value for K only holds at equilibrium.
Last Update: 26 February 2000 EJW
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