Name:
Grade:
1:
/33
2:
/33
3:
/33
+1 free point
/100
Midterm Exam II
CHEM 181: Introduction to Chemical Principles
October 9, 2012
Answer Key
Directions:
Do all three problems.
Show all of your work neatly and clearly. Do not skip steps. Partial credit will be
awarded for all problems. Correct answers will not receive credit if your work is not
shown.
If you are not sure exactly what a question means, ask!
Not all problems are of equal difficulty, but all are worth the same fraction of the
overall grade.
1
1.
C-H stretch
C-C bend
C=O stretch
C-H bend
C-C stretch
C-H stretch
C-D bend
C-C bend
C=O stretch
C-C stretch
Shown above are IR spectra for acetone and deuterated acetone.
(a) Identify each of the starred features in the top spectrum as
•
•
•
•
•
C–H stretches
C=O stretch
C–C stretches
C–H bends/scissors/wags
C–C bends
(b) For each transition marked in the top spectrum, mark the analogous transition in the bottom spectrum.
2
Analysis:
• IR peaks table lists values for C–H stretches, C=O stretch, and C–H bends.
√
• C–D stretches and C–D bends should appear at frequencies ∼ 2 times
lower than analogous C–H vibrations.
• C=O stretch, C–C stretches, and C–C bends should be similar in both
compounds.
• Bends are at lower frequency than stretches.
3
2. The 1 H NMR spectra on this page and the next are all of compounds with the
molecular formula C4 H8 O2 :
4
5
3. Complete the following equilibrium problems:
(a) The equilibrium constants for the following reactions are given below:
N2 (g) + 3H2 (g)
1
N2 (g) + O2 (g)
2
4.34 × 10−3 atm−2
3.6 × 10−8 atm−1/2
2 NH3 (g)
NO2 (g)
Calculate the equilibrium constant for the reaction
3
NH3 (g) + O2 (g) NO2 (g) + H2 (g)
2
Set up:
2
PNH
3
PN2 PH3 2
PNO2
K1 =
K2 =
1/2
PN2 PO2
3/2
PNO2 PH2
PNH3 PO2
K3 =
Now express K3 in terms of K1 and K2 . Since the PNO2 in K3 can only
come from K2 , we get
K3 = K2 · X
3/2
PNO2 PH2
PNH3 PO2
=
PNO2
1/2
·X
P N 2 PO 2
1/2
3/2
P N 2 PH 2
X =
PNH3
Looking back, we see that
1
X=√
K1
so
K2
K3 = √
K1
3.6 × 10−8 atm−1/2
= √
4.34 × 10−3 atm−2
= 5.5 × 10−7 atm1/2
6
(b) The equilibrium constant for the equation
2ICl(g) I2 (g) + Cl2 (g)
is Kc = 0.11. Calculate the equilibrium concentrations of ICl(g), I2 (g),
and Cl2 (g) when 0.65 moles of I2 (g) and 0.33 moles of Cl2 (g) are mixed in
a 1.5 L reaction vessel.
Concentrations in moles per liter are 0.43 M and 0.22 M for I2 and Cl2 .
Using these, we set up the problem:
[ICl]
Initial:
Change:
Equilibrium:
0
+2x
2x
[I2 ]
[Cl2 ]
0.43
0.22
−x
−x
0.43 − x 0.22 − x
and so
Kc =
0.11 =
0.44x2
0.44x2
0.56x2 − 0.65x + 0.0953
x
=
=
=
=
[I2 ][Cl2 ]
[ICl]2
(0.43 − x)(0.22 − x)
(2x)2
(0.43 − x)(0.22 − x)
x2 − 0.65x + 0.0953
0
0.17
This gives:
[ICl] = 0.34 M
[I2 ] = 0.26 M
and double-check:
0.26 · 0.05
0.342
0.11 = 0.11
?
0.11 =
7
[Cl2 ] = 0.05 M
(c) The equilibrium constant for the reaction
NH4 HS(s) NH3 (g) + H2 S(g)
is Kc = 1.81 × 10−4 M2 at 25 ◦ C. Follow these steps
• 10 g of NH4 HS(s) is placed in an evacuated 1 L reaction vessel,
• the above reaction is allowed to run until equilibrium,
• an additional 0.01 moles of NH3 (g) is added to the container, and
• equilibrium is once again established.
What will the final concentrations of NH3 (g) and H2 S(g) be?
First off check how much NH4 HS(s) we have. Molar mass is 51 g, so we
have ∼0.2 moles. This tells us that NH4 HS(s) is in excess, and can be
ignored.
The straightforward way is to calculate the first equilibrium:
[NH3 ]
[H2 S]
0
+x
x
0
+x
x
Initial:
Change:
Equilibrium:
Kc = [NH3 ][H2 S]
1.81 × 10−4 = x2
x = 1.35 × 10−2 M
Then we add 0.01 M NH3 so that our new concentrations are
[NH3 ] = 2.35 × 10−2 M
and
[H2 S] = 1.35 × 10−2 M
Calculate equilibrium again:
Initial:
Change:
Equilibrium:
[NH3 ]
[H2 S]
2.35 × 10−2
−x
2.35 × 10−2 − x
1.35 × 10−2
−x
1.35 × 10−2 − x
Kc
1.81 × 10−4
1.81 × 10−4
x2 − 3.7 × 10−2 x + 1.36 × 10−4
x
8
=
=
=
=
=
[NH3 ][H2 S]
(2.35 × 10−2 − x)(1.35 × 10−2 − x)
x2 − 3.7 × 10−2 x + 3.17 × 10−4
0
4.14 × 10−3 M
which gives us
[NH3 ] = 1.94 × 10−2 M
and
[H2 S] = 0.94 × 10−2 M
Double-check:
?
1.81 × 10−4 = (1.94 × 10−2 )(0.94 × 10−2 )
1.81 × 10−4 ≈ 1.82 × 10−4
9