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K(p) = 0.04 atm at 899 K for the equilib...

`K_(p) = 0.04` atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of `C_(2)H_(6)` when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium?
`C_(2) H_(6)(g) to C_(2)H_(4) (g) +H_(2)(g)`

A

7.24 atm

B

3.62 atm

C

1 atm

D

1.5 atm

Text Solution

Verified by Experts

The correct Answer is:
B
`C_(2)H_(6) (g) to C_(2) H_(4) (g) +H_(2) (g)`

`0.04 = (x^(2))/( 4-x)`
On solving, `x =0.38 , P_(c_(2)H_(6))=3.62` atm
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