Chemistry 4511
Problem Set 2
1.
For the reaction 2CO2 Æ 2CO + O2
a.
If there are initially 0.534 moles of CO2 and 0.42 moles of CO are generated at
some time, demonstrate that ξ is the same with respect to each component.
b.
Once the reaction goes to completion, there is a ∆H value measured at T = 1000
K. Calculate the expected value for ∆H two ways and compare the values (use heats of
formation and bond enthalpies and adjust the ∆H value away from standard T using
∆CP,m(T))
a.
ξ = ∆ni/νi
i
CO2
CO
O2
ξCO2 = -0.42/-2 = 0.21
b.
νi
-2
+2
+1
ξCO = 0.42/2 = 0.21
∆ni
-0.42
+0.42
+0.21
ξO2 = 0.21/1 = 0.21
∆Ho = Σ∆fHoprod – Σ∆fHoreact = 2(-110.53)+0-2(-393.51) = 565.96 kJ/mol
∆Ho = Σ∆Hbroken – Σ∆Hformed = 2(531) – 497 = 565 kJ/mol
∆CP,m = ∆d + ∆eT + ∆fT-2
∆d = 2(28.41) + 29.96 – 2(44.22) = -1.44 J K-1 mol-1
∆e(x 103) = 2(4.1) + 4.18 – 2(8.79) = -5.2 Æ -5.2 x 10-3 J K-2 mol-1
∆f(x 104) = 2(-4.6) + (-16.7) – 2(-86.2) = 146.5 Æ 1.465 x 106 J K mol-1
∆CP,m = -1.44 – 5.2 x 10-3T + 1.465 x 106 T-2
T2
∆∆H (T ) = ∫ C P ,m dT = ∆d (T2 − T1 ) +
T1
⎛1 1⎞
∆e 2
2
(T2 − T1 ) − ∆f ⎜⎜ − ⎟⎟
2
⎝ T2 T1 ⎠
− 5.2 x10 −3
1 ⎞
⎛ 1
−
(1000 2 − 298.15 2 ) − 1.465 x10 6 ⎜
⎟
2
⎝ 1000 298.15 ⎠
∆H1000 = ∆Ho + ∆∆H
∆∆H (T ) = −1.44(1000 − 298.15) +
∆∆H (T ) = 69 Jmol −1
so
∆H1000 = 565.96 + 0.069 = 566.029 kJ/mol for formation constants
∆H1000 = 565 + 0.069 = 565.069 kJ/mol for bond enthalpies
We can see that the T dependence is not significant compared to the process, and this is
typically the case. Also, we see that the two ways of deriving ∆H are close to the same.
Classically, ∆fH values are more accurate than bond energies, although methods of
measuring the latter are improving all the time.
2.
An expandable vessel is filled with 1 mol of an ideal gas…
a.
The gas is compressed reversibly from 2L to 1L at a constant pressure and the
temperature changes from 300K to 150K. What is the work done on the system? If ∆U =
-19 kJ/mol, then what is Cp,m for the gas?
wrev = R(T2 – T1) = 8.315(-150) = -1.247 kJ mol-1
∆U = CV,m∆T Æ -19/-150 = 126.7 J K-1 mol-1 = CV,m
also, P = RT/V = 8.315(300)/.002 = 1.24 x 106 Pa
b.
The vessel is then made rigid and the temperature is dropped to 100K. What is ∆U
for this process?
∆U = CV,m∆T = 126.7(-50) = -6.34 kJ mol-1
also, P2 = P1T2/T1 = 1.24 x 106 (100/150) = 8.27 x 105 Pa
c.
From here, T is held constant and the pressure is reversibly changed by +1.4 x 104
Pa. What is the work and heat derived from this process?
wrev = RTln(P2/P1) = 8.315*100*ln(8.41/8.27) = 14.0 J mol-1
qrev = -wrev = -14.0 J mol-1
also, ∆U = 0
d.
Finally, the vessel is adiabatically changed back to 300K, the original pressure
and 2L. What is ∆U for this process? What is ∆U for the entire process? Is this what you
expect? Why or why not?
∆U = CV.m∆T = 126.7 (+200) = 25.34 kJ mol-1
∆UTotal = Σ∆Ui = -19 – 6.34 + 25.34 = 0
also, ∆U = CV,m∆T = 0 holds true, regardless of path because these are state functions
3.
As discussed in class, if we want to determine the functionality of a certain
variable by holding the other pertinent variables constant. Let’s do this analysis for the
following data:
w
2
2
2
2
2
2
1
3
4
5
6
2
2
2
2
2
x
2
2
2
2
2
2
2
2
2
2
2
1
3
4
5
6
y
1
2
3
4
5
6
2
2
2
2
2
2
2
2
2
2
z
0.693147
5.545177
18.71497
44.36142
86.6434
149.7198
11.09035
3.696785
2.772589
2.218071
1.848392
0
8.788898
11.09035
12.8755
14.33408
Determine the functionality of each variable
with respect to z. From this, derive z(w,x,y)
and prove this equation is accurate using a
few data points from the table. Then predict
the value of z at w = 9.2, x = 3.4, and y = 2.3.
Remember, when you are pulling the
constants off the graphs, they are truly a
function of the other variables. The final
equation should have one constant that is
independent of w, x, and y.
First, we organize the data in the following fashion:
Here we understand that each value for Z we observe includes
whatever information the other two variables are adding to it. We
then begin by plotting F(W) vs. Z with simply F(W) = W.
W
Basic W vs. Z
12
X
10
6
4
2
F(X)
Y
0
0
1
2
3
4
W
5
6
1
2
3
4
5
6
1
2
3
4
5
6
y = -1.6319x + 10.24
R2 = 0.7717
8
Z
F(W)
1
2
3
4
5
6
7
Z
1
2
3
4
5
6
0
5.545177
8.788898
11.09035
12.8755
14.33408
1
2
3
4
5
6
Z
0.693147
5.545177
18.71497
44.36142
86.6434
149.7198
F(Y)
1
2
3
4
5
6
We can see that the straight line does not fit well currently (R2 =
1 is the best fit possible). Now we alter F(W). The plot looks like
the PV plot, and we know that P α 1/V so let’s plot 1/W. This gives a much better line,
as shown on the next page.
Z
11.09035
5.545177
3.696785
2.772589
2.218071
1.848392
Now, the straight line gives a
slope of 11.09 for (dz/dw)x,y. Now
we repeat this process for x and y,
and get the following graphs and
slopes as results.
12
10
8
Z
From here, we can understand
that
1/W vs. Z
6
y = 11.09x
R2 = 1
4
2
Z(W) = Cw(X,Y)1/W
0
0
Z(X) = Cx(W,Y)lnX
0.2
0.4
0.6
0.8
1
1.2
1/W
Z(Y) = CY(W,X)Y3
ln(X) vs. Z
16
14
12
10
Z
From any of these equations, or
from the combined equation, we
can assess the true constant C
such that
8
6
y = 8x
R2 = 1
4
Z(W,X,Y) = C F(W,X,Y) where
F(W,X,Y) = Y3lnX/W
2
0
0
Using (W,X,Y) = (2,2,2)
0.5
1
1.5
2
ln(X)
5.545177 = C 23ln2/2 so
Y^3 vs. Z
C=2
160
The final answer is, therefore,
140
120
2Y ln X
W
Z at the specified variables then is
100
Z
Z=
3
80
y = 0.6931x
R2 = 1
60
40
20
Z=
2(2.3) 3 ln(3.4)
= 3.236
9.2
0
0
50
100
150
Y^3
200
250
4.
The following apparatus is set up:
The setup is designed so that the
gases are stoiciometrically ready to
completely undergo:
barrier
H2 (g) + ½ O2 (g) Æ H2O (g)
H2
O2
V1
V2
The barrier is removed and the gases
mix and react, forming 1 mol of H2O
(g). P and T are held at the standard
state (1 bar and 298 K). What is ∆G
for this process, calculated explicitly
using ∆H and ∆S values? Does this
differ from the ∆G value derived from tables?
∆S = ∆Smix + ∆Srxn
∆Smix = -R(XO2lnXO2 + XH2lnXH2) = -8.315(1/3 ln (1/3) + 2/3 ln (2/3)) = 5.293 J K-1 mol1
Now we turn to the reaction
∆Ho = Σ∆fHoprod – Σ∆fHoreact = -241.82 – (0 + ½ (0)) = -241.82 kJ mol-1
∆So = ΣSprod – ΣSreact = 188.83 – (130.684 + ½ (205.138)) = -44.423 J K-1 mol-1
∆Go = ∆Ho – T(∆So + ∆Smix) = -241.82 – 298 (-0.044423 + 0.005293) = -230.16 kJ mol-1
∆Go = Σ∆fGoprod – Σ∆fGoreact = -228.57 kJ mol-1
and ∆Gmix = -T∆S = -298(0.005293) = -1.58 kJ mol-1
so ∆Gprocess = Σ∆G = -228.57 – 1.58 = -230.15 kJ mol-1 – hey, nope… it works like a
charm.