The equilibrium constant is 0.0900 at 25°C for the reaction H 2 O ( g ) + Cl 2 O ( g ) ⇌ 2 HOCl ( g ) For which of the following sets of conditions is the system at equilibrium? For those that are not at equilibrium, in which direction will the system shift? a .· P H 2 O = 1.00 atm, P CI 2 O = 1.00 atm, P HOCI = 1.00 atm b . P H 2 O = 200. torr, P CI 2 O = 49.8 torr, P HOCI = 21.0 torr c . P H 2 O = 296 torr, P CI 2 O = 15.0 torr , P HOCI = 20.0 torr
The equilibrium constant is 0.0900 at 25°C for the reaction H 2 O ( g ) + Cl 2 O ( g ) ⇌ 2 HOCl ( g ) For which of the following sets of conditions is the system at equilibrium? For those that are not at equilibrium, in which direction will the system shift? a .· P H 2 O = 1.00 atm, P CI 2 O = 1.00 atm, P HOCI = 1.00 atm b . P H 2 O = 200. torr, P CI 2 O = 49.8 torr, P HOCI = 21.0 torr c . P H 2 O = 296 torr, P CI 2 O = 15.0 torr , P HOCI = 20.0 torr
Solution Summary: The author explains the equilibrium constant value and the direction of shift for the reactions that are not at equilibrium.
The equilibrium constant is 0.0900 at 25°C for the reaction
H
2
O
(
g
)
+
Cl
2
O
(
g
)
⇌
2
HOCl
(
g
)
For which of the following sets of conditions is the system at equilibrium? For those that are not at equilibrium, in which direction will the system shift?
a.·
P
H
2
O
=
1.00
atm,
P
CI
2
O
=
1.00
atm,
P
HOCI
=
1.00
atm
b.
P
H
2
O
=
200.
torr,
P
CI
2
O
=
49.8
torr,
P
HOCI
=
21.0
torr
c.
P
H
2
O
=
296
torr,
P
CI
2
O
=
15.0
torr
,
P
HOCI
=
20.0
torr
A mixture of C7H12O2, C9H9OCl, biphenyl and acetone was put together in a gas chromatography tube. Please decide from the GC resutls which correspond to the peak for C7,C9 and biphenyl and explain the reasoning based on GC results. Eliminate unnecessary peaks from Gas Chromatography results.
Is the molecule chiral, meso, or achiral?
CI
.CH3
H₂C
CI
PLEASE HELP ! URGENT!
Chapter 12 Solutions
Bundle: Chemistry: An Atoms First Approach, 2nd, Loose-Leaf + OWLv2, 4 terms (24 months) Printed Access Card
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell