Question

Question 7.17 Kp = 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?

C2H6(g) <---> C2H4(g) + H2(g)

Class XI Equilibrium Page 226

Answer 1

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: C_2H_6(g) < = > C_2H_4(g) + H_2(g) \\ at \: t = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 4atm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 0 \\ at \: eqlm \: \: \: \: \: \: \: \: \: \: \: (4 - p)atm \: \: \: \: \: \: \: \: \: \: p \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: p$
we know,
$K_p=\frac{P_{C_2H_4}.P_{H_2}}{P_{C_2H_6}}$
Kp = P. P/(4 - P)
0.04 = P²/(4 - P)
1/25 = P²/(4 - P)
4 - P = 25P²
25P² + P - 4 = 0
P = {-1 ± √(1 + 400)}/50
P = { -0.02 ± 0.4}
P = 0.38

Hence, pressure of C2H6 = 4 - p = 4 - 0.38 = 3.62 atm

Answer 2

Answer:

refer to the image given below :-

hope it helps