At $25^{0} C$ and 1 atmospheric pressure, the partial pressures in equilibrium mixture of gaseous $N_{2} O_{4}$ and $N O_{2}$ are 0.7 and 0.3 atm respectively. Calculate the partial pressure of the $N_{2} O_{4}$ gas when it is in equilibrium $25^{o} C$ and the total pressure of 10 atm

9.0 a t m

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

1.0 a t m

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

0 .9 a t m

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

0.1 a t m

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

Open in App

Solution

The correct option is A $9.0 a t m$
$\begin{matrix} N_{2} O_{4} ⇌ 2 N O_{2} P r e s s u r e a t e q u i l i b r i u m & 0.7 & 0.3 \end{matrix}$
$∴ K_{P} = \frac{(P_{N O_{2}})^{2}}{P_{N_{2} O_{4}}} = \frac{0.3 \times 0.3}{0.7} = 0.1286 a t m$
Now assume decompostion at 10 atm pressure

$\begin{matrix} N_{2} O_{4} ⇌ 2 N O_{2} I n i t i a l m o l e & 1 & 0 M o l e a t e q u i l i b r i u m & (1 - x) & 2 x \end{matrix}$

$P_{N_{2} O_{4}} = \frac{1 - x}{1 + x} \times 10$ Eq. 1 $P_{N O_{2}} = \frac{2 x}{1 + x} \times 10$ Eq. 2

$K_{P} = \frac{(P_{N O_{2}})^{2}}{P_{N_{2} O_{4}}}$
$\Rightarrow 0.1286 = \frac{(\frac{2 x}{1 + x} \times 10)^{2}}{\frac{1 - x}{1 + x} \times 10}$
putting values, and solving we get;
x=0.0566

Putting values of x in equation 1 and 2 above,

$P_{N O_{2}} = 1.07 a t m \approx 1$
$P_{N_{2} O_{4}} = 8.93 a t m \approx 9 a t m$

Suggest Corrections

Similar questions

Q. At

25^{0} C

and 1 atmospheric pressure, the partial pressures in equilibrium mixture of gaseous

N_{2} O_{4}

and

N O_{2}

are 0.7 and 0.3 atm respectively. Calculate the partial pressure of the

N_{2} O_{4}

gas when it is in equilibrium

25^{o} C

and the total pressure of 10 atm

At 298 K and 1 atm pressure, the partial pressures in an equilibrium mixture of $N_{2} O_{4}$ and $N O_{2}$ are 0.7 and 0.3 atmosphere respectively. What will be the partial pressure of $N O_{2}$ when the gases are in equilibrium at 298 K and at a total (equilibrium) pressure of 10 atmospheres?