Chemistry 111
KEY
Page
1
out of 8
Final Exam D / Fall 2004
1.
Consider the following reaction…
2 A (g)
+
3 B (g)
4 C (g)
+
D (g)
Which of the following is the mathematical definition of Kc for this reaction?
a)
[A]
2
[C]
4
[D]
b)
[A]
2
[B]
3
[C]
4
[D]
c)
[C]
4
[D]
[A]
2
d)
[C]
4
[D]
[A]
2
[B]
3
e)
[A] [B]
[C] [D]
2.
Consider the reaction below with a Kc = 9.00.
In a given 1L flask, we place 0.150 mol of A,
0.150 mol of B, and 0.600 mol of C.
What will be the concentration of C at equilibrium?
A (g)
+
B (g)
2 C (g)
a)
0.63 M
b) 0.57 M
c)
0.51 M
d) 0.54 M
e)
None of these.
First set up the initial conditions and calculate Q in order to decide the direction of the shift.
A
+
B
2 C
init:
0.15 M
0.15 M
0.60 M
Calculation of Q is given by:
Q=
[A][B]
[C]
2
=
(0.15)
2
(0.60)
2
=1
6
Since Q > Kc, then the equilibrium will shift to the left until Q = 9.00.
Therefore,
A
+
B
2 C
init:
0.15 M
0.15 M
0.60 M
@ eq:
0.15 + x
0.15 + x
0.60  2x
Substituting into the mathematical expression for Kc, we get:
Kc =
[A][B]
[C]
2
=
(0.15 + x)
2
(0.60  2x)
2
=9
.
0
0
Taking square roots on both sides, we get:
0.15 + x
0.60  2x
=3
∴
x
=
0.03 M
However, the answer must be 0.60 – 2(0.03)
=
0.54 M
.
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View Full DocumentChemistry 111
KEY
Page
2
out of 8
Final Exam D / Fall 2004
3.
Consider the following reaction for which Kp = 2.50
×
10
5
.
In a given flask we place gases
A, B, and C at initial pressures of 0.100, 0.200, and 0.300 atm, respectively.
What will be
the pressure of gas D at equilibrium?
2 A (g)
+
3 B (g)
4 C (g)
+
3 D (g)
a) 1.59 torr
b) 4.77 torr
c) 9.01 torr
d) 14.4 torr
e)
None of these.
Once more, we'll start by examining the initial conditions and calculating Q to decide the
direction of the shift.
init:
0.100 atm
0.200 atm
0.300 atm
0
2 A
+
3 B
4 C
+
3 D
Since the initial pressure of D is zero, then Q must be equal to zero.
This means that we are
very close to Kc from the beginning.
So, we will shift to the right.
How much?
Just a tiny
bit.
This will make x negligible…
init:
0.100 atm
0.200 atm
0.300 atm
0
2 A
+
3 B
4 C
+
3 D
@ eq:
0.100  2x
0.200  3x
0.300 + 4x
3x
Substitution into Kc yields:
43
4
3
C
23
2
3
[C] [D]
(0.3) (3x)
K=
=
=
[A] [B]
(0.1) (0.2)
2.50 x 10
3
∴
3x
=
[D]
=
6.28
×
10
3
atm.
That is a small number so we will convert it to torr (also because those are the units given for
the answer).
3
760torr
6.28x10 atm x
1atm
=
4.77 torr
.
31. Consider the reaction below with a Kc of 2.25
×
10
+12
.
In a given 1L flask, we place 2.00
mol of A, 8.00 mol of B, 2.00 mol of C and no D.
What will be the concentration of gas A at
equilibrium?
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