7.17. Kp =0.04 atm at 898 K for the equilibrium shown below. What is the equilibrium concentration ok C2H6 when it is placed in a flask at 4 atm pressure,and allowed to come to equilibrium.
C2H6(g) ⇌ C2H4(g) + H2(g).
7.17. Kp =0.04 atm at 898 K for the equilibrium shown below. What is the equilibrium concentration ok C2H6 when it is placed in a flask at 4 atm pressure,and allowed to come to equilibrium.
C2H6(g) ⇌ C2H4(g) + H2(g).
The equilibrium reaction is
C2?H6?(g) ? C2?H4?(g)+H2?(g).
Initial | 4 | 0 | 0 |
Change | −x | x | x |
Equilibrium | 4−x | x | x |
The expression for the equilibrium constant is Kp?= (??PC2?H4??) (PH2) / PC2?H6???.
Substituting the values in the above equation, we get
0.04=x2 / (4−x)
? or x=0.38
Thus, the pressure of ethane is, P
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 &nbs
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.
On Shiksha, get access to
Learn more about...

Chemistry Ncert Solutions Class 11th 2023
View Exam DetailsMost viewed information
SummaryDidn't find the answer you were looking for?
Search from Shiksha's 1 lakh+ Topics
Ask Current Students, Alumni & our Experts
Have a question related to your career & education?
See what others like you are asking & answering

