K_p Approach to Equilibrium Example
The equilibrium constant (K
p
) for the reaction of NO(g) with O
2
(g) to form NO
2
(g) to be determined.
In an experiment, the initial pressure of NO(g) was 10.65 atm, while the initial pressure of O
2
(g) was 13.78 atm, and the initial pressure of NO
2
(g) was 11.22 atm. Please fill in the following tableau:
ICE table for Equilibrium Computations
P
NO
P
O
2
P
NO
2
Initial
?
10.65
13.78
11.22
?
10.65
13.78
11.22
?
10.65
13.78
11.22
Change
y
?
+y
-y
+2y
-2y
+y/2
-y/2
?
+y
+2y
-2y
-y
+y/2
-y/2
Equilibium
?
10.65+y
10.65-y
10.65+y/2
10.65-y/2
13.78+y
13.78-y
13.78+y/2
13.78-y/2
11.22+y
11.22-y
11.22+y/2
11.22-y/2
?
10.65+y
10.65-y
10.65+y/2
10.65-y/2
13.78+y
13.78-y
13.78+y/2
13.78-y/2
11.22+y
11.22-y
11.22+y/2
11.22-y/2
?
10.65+y
10.65-y
10.65+y/2
10.65-y/2
13.78+y
13.78-y
13.78+y/2
13.78-y/2
11.22+y
11.22-y
11.22+y/2
11.22-y/2
Initial p
NO
appears to be wrong.
Initial p
O
2
appears to be wrong
Initial p
NO
2
appears to be wrong.
Change in p
NO
2
appears to be wrong.
Change in p
O
2
appears to be wrong.
Equilibrium P
NO
appears to be wrong.
Equilibrium P
O
2
= appears to be wrong.
Equilibrium p
NO
2
appears to be wrong.
You can send comments about this question to
Carl.David@uconn.edu
This calculator is quite idiosyncratic, and may not be worth using, but it is included for those without handy nearby calculators.
CHANGE ME
Use the above calculator like a slide rule, if you wish. Since it does not support powers of ten, you need to write everything in scientific notation, and do the power of ten computations separately (just like the old slide rules).
Everything is pretty straightforward, except that asi/aco/ata are arc sine, cosine and tangent respectively.
Further, remember to close brackets on arguments of functions.
Finally, Sto stores your answer, and Rcl retrieves it.
