Question

For the reaction: n2o3→no2 +no; total pressure p, degree of dissociation=50%. Then kp wouldbe

Answer 1

Answer: The value of Kp is, 0.33 P

Solution :

Degree of dissociation = 50 % = 0.5

Total pressure = P

The given equilibrium reaction is,

                       $N_2O_3\rightleftharpoons NO_2+NO$

Initially                  1         0       0

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

$\text{ Total number of moles}=1-\alpha+\alpha+\alpha=1+\alpha=1+0.50=1.5$

Moles of $N_2O_3$ = $(1-\alpha)=1-0.50=0.50$

Moles of $NO_2$ = $(\alpha)=0.50$

Moles of $NO$ = $(\alpha)=0.50$

Now we have to calculate the partial pressure of $N_2O_4$, $NO_2$  and $NO$

$\text{ Partial pressure of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Moles of }N_2O_4+\text{Moles of }NO_2+\text{Moles of }NO}\times P=\frac{0.50}{1.5}\times P=0.33P$

$\text{ Partial pressure of }NO_2=\frac{\text{Moles of }NO_2}{\text{Moles of }N_2O_4+\text{Moles of }NO_2+\text{Moles of }NO}\times P=\frac{0.5}{1.5}\times P=0.33P$

$\text{ Partial pressure of }NO=\frac{\text{Moles of }NO}{\text{Moles of }N_2O_4+\text{Moles of }NO_2+\text{Moles of }NO}\times P=\frac{0.5}{1.5}\times P=0.33P$

The relation between the $K_p$ and the total pressure is,

$K_p=\frac{(p_{NO_2})\times (p_{NO})}{p_{N_2O_4}}$

Now put all the values of partial pressure, we get

$K_p=\frac{(0.33P)\times (0.33P)}{0.33P}$

$K_p=0.33P$

Therefore, the value of Kp is, 0.33 P

Answer 2

Answer:

P/3

Explanation: