Question

At 27 c and 1 atmosphere pressure N2O4 is 20% dissociated into NO2 find kp

Answer 1

Answer : The value of $K_p$  for this reaction is, 0.2

Solution :  Given,

Initial pressure of $N_2$ = 1.42 bar

Initial pressure of $H_2$ = 2.87 bar

$K_p$ = 0.036

The given equilibrium reaction is,

                            $N_2O_4(g)\rightleftharpoons 2NO_2(g)$

Initially                     1                 0

At equilibrium    $(1-\alpha)$           $2\alpha$

The expression of $K_p$ will be,

$K_p=\frac{(p_{NO_2})^2}{(p_{N_2O_4})}$

$K_p=\frac{(2\alpha)^2}{(1-\alpha)}$

where,

$\alpha$ = degree of dissociation = 20 % = 0.2

Now put all the given values in the above expression, we get:

$K_p=\frac{(2\alpha)^2}{(1-\alpha)}$

$K_p=\frac{(2\times 0.2)^2}{(1-0.2)}$

$K_p=0.2$

Therefore, the value of $K_p$  for this reaction is, 0.2

Answer 2

Explanation:

Answer : The value of K_pK

p

for this reaction is, 0.2

Solution : Given,

Initial pressure of N_2N

2

= 1.42 bar

Initial pressure of H_2H

2

= 2.87 bar

K_pK

p

= 0.036

The given equilibrium reaction is,

N_2O_4(g)\rightleftharpoons 2NO_2(g)N

2

O

4

(g)⇌2NO

2

(g)

Initially 1 0

At equilibrium (1-\alpha)(1−α) 2\alpha2α

The expression of K_pK

p

will be,

K_p=\frac{(p_{NO_2})^2}{(p_{N_2O_4})}K

p

=

(p

N

2

O

4

)

(p

NO

2

)

2

K_p=\frac{(2\alpha)^2}{(1-\alpha)}K

p

=

(1−α)

(2α)

2

where,

\alphaα = degree of dissociation = 20 % = 0.2

Now put all the given values in the above expression, we get:

K_p=\frac{(2\alpha)^2}{(1-\alpha)}K

p

=

(1−α)

(2α)

2

K_p=\frac{(2\times 0.2)^2}{(1-0.2)}K

p

=

(1−0.2)

(2×0.2)

2

K_p=0.2K

p

=0.2

Therefore, the value of K_pK

p

for this reaction is, 0.2