Question

For the dissociation reaction N2O4(g) ⇌2NO2(g), the degree of dissociation (α) in terms of Kp and total equilibrium pressure P is:-

(1) α=√4P+KpKp (2) α=√Kp4P+Kp (3) α=√Kp4P (4) α=√Kp2P

Answer 1

Answer: α = √Kp/4P

Explanation: $K_p=\frac{[p_{NO_2}]^2}{[p_{N_2O_4]}}$

$N_2O_4 \leftrightarrow 2NO_2$

if initial mole is 1 and α is the degree of dissociation.

Initial moles      1                           0

at equilibrium       1-α                          2α    

Total moles at equilibrium = 1-α+2α  = 1+α

According to Dalton's law of partial pressure

$\text{partial pressure}=mole fraction\times Total Pressure$

$/text{partial pressure of N_2O_4}=\frac{no of moles of N_2O_4}{total no of moles}\times {Total Pressure}$

$p_{NO_2}=\frac{2\alpha} {1+\alpha}\times P$

$p_{N_2O_4}=\frac{1-\alpha} {1+\alpha}\times P$

Substituting the values we get:

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

On solving we get

$K_p=\frac{4\alpha^2}{1-\alpha^2}\times P$

as α<<<1

Rearranging we get α =√Kp/4P

Answer 2

Explanation:

Answer: α = √Kp/4P

Explanation: K_p=\frac{[p_{NO_2}]^2}{[p_{N_2O_4]}}K

p

=

[p

N

2

O

4

]

[p

NO

2

]

2

N_2O_4 \leftrightarrow 2NO_2N

2

O

4

↔2NO

2

if initial mole is 1 and α is the degree of dissociation.

Initial moles 1 0

at equilibrium 1-α 2α

Total moles at equilibrium = 1-α+2α = 1+α

According to Dalton's law of partial pressure

\text{partial pressure}=mole fraction\times Total Pressurepartial pressure=molefraction×TotalPressure

/text{partial pressure of N_2O_4}=\frac{no of moles of N_2O_4}{total no of moles}\times {Total Pressure}/textpartialpressureofN

2

O

4

=

totalnoofmoles

noofmolesofN

2

O

4

×TotalPressure

p_{NO_2}=\frac{2\alpha} {1+\alpha}\times Pp

NO

2

=

1+α

2α

×P

p_{N_2O_4}=\frac{1-\alpha} {1+\alpha}\times Pp

N

2

O

4

=

1+α

1−α

×P

Substituting the values we get:

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

p

=

(1+α)

2

(2α)

2

P

2

×

(1−α)P

(1+α)

On solving we get

K_p=\frac{4\alpha^2}{1-\alpha^2}\times PK

p

=

1−α

2

4α

2

×P

as α<<<1

Rearranging we get α =√Kp/4P