$K_{P} = 0.04 a t m$ 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 $4.0 a t m$ pressure and allowed to come to equilibrium?
$C_{2} H_{6} (g) ⇌ C_{2} H_{6} (g) + H_{2} (g)$ :

7.24 a t m

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3.62 a t m

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1 a t m

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1.5 a t m

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Solution

The correct option is B $3.62 a t m$
The equilibrium reaction is
$C_{2} H_{6} (g) ⇌ C_{2} H_{4} (g) + H_{2} (g)$ .

Initial
$4$
$0$
$0$
Change
$- x$
$x$
$x$
Equilibrium $4 - x$
$x$
$x$
The expression for the equilibrium constant is $K_{p} = \frac{P_{C_{2} H_{4}} P_{H_{2}}}{P_{C_{2} H_{6}}}$ .
Substitute values in the above expression.
$0.04 = \frac{x^{2}}{4 - x}$ or $x = 0.38$

Thus the pressure of ethane is, $P_{C_{2} H_{6}} = 3.62 a t m$ .
Option B is correct.

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Similar questions

K_p = 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of C₂H₆ when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium?

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) ⇌ C_{2} H_{4} (g) + H_{2} (g)

K_p = 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of C₂H₆ when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium?