You visited us 260 times! Enjoying our articles? Unlock Full Access!

Characteristics of Equilibrium Constant

For the react...

Question

For the reaction, $N_{2} O_{4} ⇌ 2 {N O}_{2}$ , if degree of dissociation of $N_{2} O_{4}$ are $25$ %, $50$ %, $75$ % and $100$ %, the gradation of observed vapour densities is:

d_{1} > d_{2} > d_{3} > d_{4}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

d_{4} > d_{3} > d_{2} > d_{1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

d_{1} = d_{2} = d_{3} = d_{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of the above

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $d_{1} > d_{2} > d_{3} > d_{4}$
$E x p l a n a t i o n$

$N_{2} O_{4} ⇌ 2 N O_{2}$
$1$ $0$ At time $t = 0$
$1 - α$ $2 α$ At equilibrium

There is a relationship between the degree of dissociation and vapour density:

$α = \frac{(D - d)}{(n - 1) d} \to (1)$

$α =$ Degree of dissociation
$d =$ Vapor density
$n =$ No. of moles of gaseous product for $1$ mole of reactant

From the given equation $n = 2$ substitute in equation $1$
then we have:

$α = \frac{D - d}{d} \Rightarrow \frac{D}{d} - 1$

$\Rightarrow 1 + α = \frac{D}{d}$

$\Rightarrow d = \frac{D}{1 + α}$

If $D \sim 1$ then $d = \frac{1}{1 + α}$

$∴$ The higher the $α$ lower the $d$ value.

$∵ α_{1} < α_{2} < α_{3} < α_{4}$

$∴ d_{1} > d_{2} > d_{3} > d_{4}$

Hence the correct answer is option $A$ .

Suggest Corrections

Similar questions

1, d_{1}, d_{2}, d_{3}, d_{4}

x^{5} = 1

E = \frac{ω - d_{1}}{ω^{2} - d_{1}} \cdot \frac{ω - d_{2}}{ω^{2} - d_{2}} \cdot \frac{ω - d_{3}}{ω^{2} - d_{3}} \cdot \frac{ω - d_{4}}{ω^{2} - d_{4}}

ω

D_{1}, D_{2}, D_{3}

D_{4}

1, 2, 3, 4, 5

6,

D_{4}

D_{1}, D_{2}

D_{3}

D_{p} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} p & 15 & 8 p^{2} & 35 & 9 p^{3} & 25 & 10 \end{matrix} ∣ ∣ ∣ ∣ ∣

D_{1} + D_{2} + D_{3} + D_{4} + D_{5}

D_{p}

∣ ∣ ∣ ∣ ∣ \begin{matrix} p & 15 & 8 p^{2} & 55 & 9 p^{3} & 225 & 10 \end{matrix} ∣ ∣ ∣ ∣ ∣

D_{1} + D_{2} + D_{3} + D_{4} + D_{5}

D_{P} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} P & 15 & 8 P^{2} & 35 & 9 P^{3} & 25 & 10 \end{matrix} ∣ ∣ ∣ ∣ ∣,

D_{1} + D_{2} + D_{3} + D_{4} + D_{5}

Join BYJU'S Learning Program