Question

The value of Kp for the equilibrium reaction N2O4;;;;;;;;;;;;; 2NO2(g) is 2. The percantage dissociation of N2O4(g) at a pressure of 0.5 atm is: .....................

Answer 1

Answer: The percentage dissociation is, 61.8%

Solution :

Total pressure = 0.5 atm

The given equilibrium reaction is,

                        $N_2O_4\rightleftharpoons 2NO_2$

Initially                   1         0

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

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

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

Moles of $NO_2$ = $(2\alpha)$

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

$\text{ Partial pressure of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Total number of moles}}\times P_T=\frac{1-\alpha}{1+\alpha}\times P_T$

$\text{ Partial pressure of }NO_2=\frac{\text{Moles of }NO_2}{\text{Total number of moles}}\times P_T=\frac{2\alpha}{1+\alpha}\times P_T$

The expression for $K_p$ is,

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

Now put all the given values, we get

$K_p=\frac{4\times (\alpha)^2}{1-(\alpha)^2}\times P_T$

$2=\frac{4\times (\alpha)^2}{1-(\alpha)^2}\times 0.5$

$\alpha=0.618$

Percentage dissocistion = $\frac{\alpha}\times 100=0.618\times 100=61.8\%$

Therefore, the percentage dissociation is, 61.8%

Answer 2

Answer:

Degree of Dissociation is 71%