Question

Kp for N204 2 NO2 at equilibrium pressure of 3P, is 0.5 P. On halving the volume of container
equilibrium is distributed. The new equilibrium pressure is-?​

Answer 1

Answer:

N2O4<−−>2NO2

pressure initial: P

At equilibrium: P−X+2X

0.66=4X2/P−X-1

0.5=P+X-2

Solving 1 and 2 we get

x=0.16

y=0.34

2x=0.32=P(NO2)

Answer 2

The $K_p$ of the recation is 0.0833.

The new equilibrium pressure is 7P.

Explanation:

$N_2O_4\rightleftharpoons 2NO_2$

Equilibrium pressure of the $N_2O_4=p_1=3P$

Equilibrium pressure of the $NO_2=p_2=0.5P$

The expression of $K_p$ is given by :

$K_p=\frac{(p_2)^2}{p_1}=\frac{(0.5 P)^2}{3P}=0.0833 P$

Total pressure at equilibrium,p = $p_1+p_2=3P+0.5P = 3.5 P$

Total pressure at the equilibrium when volume is reduced to half = P'

V' = 0.5 V

$pV=P'V'$ (Boyle's law)

$P'=\frac{pV}{V'}=\frac{3.5P\times V}{0.5V}=7P$

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