1
Chem 6 Sample exam 3 (final) brief answers
1.
Because E = hc/
λ
, shortest wavelength means biggest energy.
This must
correspond to a transition
from n = infinity to the unknown quantum number n
.
The
energy of this transition is given by
∆
E = –2.18x10
–18
J Z
2
(1/n
f
2
– 1/n
i
2
)
Plug in Z = 1 for hydrogen, n
f
= n, n
i
= infinity and solve for n
∆
E = hc/
λ =
I(–2.18x10
–18
J (1/n
f
2
))I
So [(6.626x10
–34
J)(3x10
8
m/s)/(3280 nm x 1m/10
9
nm)]/ 2.18x10
–18
J = 1/n
f
2
And
n
f
= 6
.
Now we can find the wavelength for line B.
The farright line (largest wavelength)
must be the smallest energy transition, from n = 7 to n = 6.
So, line B must be due
to the transition from n = 8 to n = 6.
Plugnchug:
∆
E = –2.18x10
–18
J (1/6
2
– 1/8
2
) = –2.65x10
–20
J
Convert to wavelength
I
∆
EI = hc/
λ
or
λ
= hc/ I
∆
EI = [(6.626x10
–34
J)(3x10
8
m/s)]/ 2.65x10
–20
J
= 7.5x10
–6
m x 10
9
nm/m =
7500 nm
2
. Rate = k
2
[A*]
Steadystate: d/dt[A*] = k
1
[A]
2
– k
–1
[A*][A] – k
2
[A*] = 0
Solve for [A*]
k
–1
[A*][A] + k
2
[A*] = k
1
[A]
2
(k
–1
[A] + k
2
)[A*] = k
1
[A]
2
[A*] = k
1
[A]
2
/(k
–1
[A] + k
2
)
Plug in to the rate expression
Rate = k
2
k
1
[A]
2
/(k
–1
[A] + k
2
)
There are two possible extremes with the magnitude of the terms in the
denominator.
If k
–1
[A] >> k
2
, then the denominator ~ k
–1
[A].
Then the rate =
k
2
k
1
[A]
2
/(k
–1
[A]) = k
2
k
1
[A]/(k
–1
) or k
2
K
eq
[A] which is firstorder in [A].
Physically, this means
that the rate of the second step is very slow, in comparison to the faster first step.
Note
that if this 2
nd
step is slow (ratedetermining), then
overall rate
= rate of 2
nd
step = k
2
[A*]
but [A*] from the fast 1
st
equilibrium step is given by [A*][A]/[A]
2
= K
eq
or K
eq
[A] = [A*].
Plug this in to the rate expression to get
rate = k
2
K
eq
[A], same as the result above from the steadystate approach.
For the other extreme, at low pressure (i.e. low [A]), then k
–1
[A] << k
2
, so that the
denominator ~ k
2
.
Then the rate = k
2
k
1
[A]
2
/(k
2
) = k
1
[A]
2
which is secondorder in [A].
Physical significance: the slow, ratedetermining, step is now the first step, bimolecular
formation of A*, and its decomposition (step 2) is fast in comparison.
Again writing the overall rate as equal to the rate of the first, slow step:
Rate = k
1
[A]
2
which, as it should, matches the steadystate result.
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2
3a. Here's the MO energy level scheme for HF.
As described in class, the F 2s is too
different in energy from the H 1s for these orbitals to form MOs; the F 2p
x
and 2p
y
AOs
have the wrong symmetry to combine with the H 1s.
Therefore these orbitals are
nonbonding (same as they started, unchanged F AOs).
The F 2p
z
and the H 1s AOs
combine to form sigmabonding and sigmaantibonding MOs.
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 Spring '08
 Gribble
 Electron, Atomic orbital, Lattice Energy, MO energy level

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