7.33. The value of Kc for the reaction 3O2(g) ⇌2O3(g) is 2.0 x 10-50 at 25°C. If equilibrium concentration of O2 in air at 25°C is 1.6 x 10-2, what is the concentration of O3?
7.33. The value of Kc for the reaction 3O2(g) ⇌2O3(g) is 2.0 x 10-50 at 25°C. If equilibrium concentration of O2 in air at 25°C is 1.6 x 10-2, what is the concentration of O3?
-
1 Answer
-
The equilibrium constant expression is
Kc = [O3]2 / [O2]3
2.0×1050 = [O3]2 / (1.6×102)3
[O3]2=2.0×1050× (1.6×102)3
=8.192×1056
Hence, the equilibrium concentration of ozone is
[O3]= 2.86×1028M
Similar Questions for you
0.01 M NaOH,
M = 1 * 10-2
pOH = 2
pH = 2
Kp = Kc (RT)Dng
36 * 10–2 = Kc (0.0821 * 300)–1
Kc = 0.36 * 0.0821 * 300 = 8.86 » 9
A(g) ->B(g) + (g)
Initial moles n 0 0
Eqb. moles n(1 – a) na
total moles =
Eqb. pressure
On increasing pressure, equilibrium moves in that direction where number of gaseous moles decreases.
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else
Sign Up on ShikshaOn Shiksha, get access to
- 65k Colleges
- 1.2k Exams
- 687k Reviews
- 1800k Answers