Question

The degree of dissociation of N2O4 into NO2 at 1.5 atm and 400C is 0.25.Calculate its Kp at 400C.

Answer 1

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Answer 2

Answer: The value of Kp at $400^oC$ is, 0.4 atm

Solution :

Degree of dissociation = 0.25

Total pressure = 1.5 atm

The given equilibrium reaction is,

                        $N_2O_4\rightleftharpoons 2NO_2$

Initially                   1         0

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

$\text{ Total number of moles}=1-\alpha+2\alpha=1+\alpha=1+0.25=1.25$

Moles of $N_2O_4$ = $(1-\alpha)=1-0.25=0.75$

Moles of $NO_2$ = $(2\alpha)=2\times 0.25=0.5$

Now we have to calculate the partial pressure of $N_2O_4$ and $NO_2$

$\text{ Partial pressure of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Moles of }N_2O_4+\text{Moles of }NO_2}\times P_T=\frac{0.75}{1.25}\times 1.5=0.9$

$\text{ Partial pressure of }NO_2=\frac{\text{Moles of }NO_2}{\text{Moles of }N_2O_4+\text{Moles of }NO_2}\times P_T=\frac{0.5}{1.25}\times 1.5=0.6$

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

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

Now put all the values of partial pressure, we get

$K_p=\frac{(p_{NO_2})^2}{p_{N_2O_4}}=\frac{(0.6)^2}{0.9}=0.4atm$

where,

Therefore, the value of Kp at $400^oC$ is, 0.4 atm